Pure Mathematics (Sixth Form) Vol.2

7/21/2019 Pure Mathematics (Sixth Form) Vol.2

http://slidepdf.com/reader/full/pure-mathematics-sixth-form-vol2 1/454

CIffl

w

a

CAMBRIDGE UNIVERSITY PRES

F. GERRISH

PURE MATHEMATICSAN

cm

>

>Hn

-

VOLUME II

ALGEBRA

TRIGONOMETRY

COORDINATE GEOMETRY



two-volume text-book on Pure Mathe-

cs has been designed to cover the whole of

mathematics required for Part I of the

don B.Sc. General Degree and for any first-

degree courses containing subsidiary pure

hematics. It is the only book of its kind

essed primarily to the General Degree

nt. It also covers most of the Pure Mathe-

cs required by the recently introduced

loma in Mathematics of the Mathematical

ciation.

This book does not assume a complete mastery

earlier work, and much of it is suitable for

pure mathematics required by grammarol pupils preparing for scholarships in

ral Sciences, as well as for the mathematics

lists. However, though Mr Gerrish has

d to supply a real need of examination

nts at these levels, he has treated his

ct in a serious fashion and this book is in no

a cram book for examinations. Thus he

a little beyond the requirements of the

ination syllabus.

By a natural division the subject-matter falls

niently into two volumes which, despite

ional cross-references, can be used inde-

ently as separate text-books:

alculus and diiVerential equations, with appli-

s to topics such as areas, arc-lengths,

s of gravity, moments of inertia, and

metry of plane curves.

Algebra and convergence, trigonometry and

inate geometry of two and fhree dimen-

in which calculus methods are illustrated

n instructive.

nally the book contains many sets of



-8fsm

-a my -mi

-3. SEP.

NAME

This book is due for return on or before the

last date shown above.

-0 4^



8EDGLEY PARK COL

PRE8TWIGH.

MANGHE8T



PURE MATHEMATICSVOLUME II



Uniform with this Volume

VOLUME I: CALCULUS



PUREMATHEMATICS

A UNIVERSITY ANDCOLLEGE COURSE

BY

F. GERRISH, M.A.

VOLUME II

ALGEBRA, TRIGONOMETRYCOORDINATE GEOMETRY



PUBLISHED BYTHE SYNDICS OF THE CAMBRIDGE UNIVERSITY PRESS

Bentley House, 200 Euston Road, London, N.W, 1

American Branch: 32 East 57th Street, New York 22, N.Y.

©CAMBRIDGE UNIVERSITY PRESS

1 <><><>



V

CONTENTSGeneral Preface p a g e xy ji

Preface to Volume II xi x

References and Abbreviations xxi

Chapter 10. Algebra of Polynomials 363

10.1 The remainder theorem and some consequences 363

10.11 Long division; identities,p. 363. 10.12 The remaindertheorem, p. 364. 10.13 Factorisation of a polynomial;

'equating coefficients ', p. 366

Exercise 10(a) ggg

10.2 Polynomials in more than one variable 370

10.21 Extension of the preceding results, p. 370,

10.22 Symmetric, skew and cyclic functions, p. 371

Exercise 10(6) 373

10.3 Polynomial equations : relations between roots andcoefficients 374

10.31 Quadratics: a summary, p. 374. 10.32 Theory of enbicequations, p. 375. 10.33 Quartic equations, p. 377

Exercise 10(c) 37 g

10.4 Elimination 378

10,41 Further examples, p. 378, 10.42 Common root of twoequations, p. 380. 10.43 Repeated roots, p. 381

Exercise 10 (d) 382

10.5 The h.c.f, of two polynomials 382

10.51 The h.c.f. process, p. 382. 10.52 An important algebraictheorom, p. 385. 10.53 Theory of partial fractions, p. 386

Exercise 10(e) 388

Miscellaneous Exercise 10(f) 388



VI CONTENTS

11.2 Determinants page

11.21 Determinants of order 2, p. 393. 11.22 Determinants of

order 3, p. 394. 11.23 Other expansions of a third-order

determinant, p. 395. 11.24 Properties of A, p. 396.

11.25 Examples, p. 397

Exercise 11(a)

11.3 Minors and cofactors

11.31 Definitions and notation, p. 401. 11.32 Expansion by-

alien cofactors, p. 403

11.4 Determinants and linear simultaneous equations

11.41 Cramer's rule, p. 404. 11.42 The case when A = 0:

inconsistency and indeterminacy, p. 406, 11.43 Homogeneouslinear equations; solution in ratios, p. 407

Exercise 11(6)

11.5 Factorisation of determinants

Exercise 11(c)

11.6 Derivative of a determinant

11.7 Determinants of order 4

Exercise 11 {d)

Miscellaneous Exercise 1 1 (e)

Chapter 12. Series

12.1 The binomial theorem

12.12 Properties of the binomial expansion, p. 420.


Exercise 12(a)

12.2 Finite series

12.21 Notation and definitions, p. 422. 12.22 General methodsfor summing finite series, p. 423. 12.23 Series involving

binomial coefficients, p. 424. 12.24 Powers of integers, p. 426.

12,25 'Factor' series, p. 428. 12.26 'Fraction' series, p. 429.

12.27 Some trigonometric series, p. 431. 12.28 Mathematical

Induction, p. 433

Exercise 12(5)



CONTENTS vii

12.4 Series of positive terms page 43^

12,41 Comparison tests, p. 439. 12.42 d'AIcmbert's ratio test,

p. 443. 1243 Speed of convergence of a aeries, p. 446.

12.44 Infinite aeries and infinite integrate, p, 447Exercise 12(e) 442

Exercise 12(f) 446

Exercise 12(g) 450

12.5 Series of positive and negative terms 451

12.51 Alternating signs (theorem of Leibniz), p. 451,

12.52 Absolute convergence, p. 452, 12.53 The modified ratio

test, p. 456. 12.54 Rogroirping and rearrangement of terms of

an infinite series, p. 457

Exercise 12(A) 459

12.6 Maclaurin's series 460

12.61 Expansion of a function as a power series, p. 460.

12.62 Expansion of c x, sinx, cosx, log(l +x), (1 +x) m

, tan -1a?,

p. 461. 12.63 Note on formal expansions, p. 464

Exercise 12 (i) 465

12.7 Applications of the series in 12,62 466

12.71 Binomial series, p. 466. 12,72 Exponential series, p. 470.

12.73 Logarithmic series, p. 472. 12.74 Gregory's series andthe calculation of n, p, 476

Exercise 12(j) 468

Exercise 12(A) 472

Exercise 12(1) 475

12.8 Scries and approximations 476

12.81 Estimation of the error in s = s nt p. 476. 12,82 Formalapproximations, p. 479. 12.83 Calculation of certain limits,

p. 480

Exercise 12 (m) 481

Miscellaneous Exercise 12 (w) 482



viii CONTENTS13.14 Third stage: logical development, p. 489.

13.15 Importance of complex algebra, p. 492. 13.16 Further

possible generalisations of 'number', p. 493

Exercise 13 (a) page 4

13.2 The modulus-argument form of a complex number 4

13.21 Modulus and argument, p. 494. 13.22 Further

definitions, notation, arid properties, p. 490. 13.23 The cube

roots of unity, p. 497

Exercise 13(6) 4

13.3 Applications of the Argand representation 4

13.31 Geometrical interpretation of modulus and argument,

p. 499. 13.32 Constructions for the sum and differ once of z v z vp. 500. 13.33 The triangle inequalities p. 501.

13.34 Constructions for the product and quotient of & XJ z, £ , p. 502.

13.35 Harder examples on the Argand representation, p. 504

Exercise 13(c) 5

13.4 Factorisation in complex algebra 5

13.41 'The fundamental theorem of algebra,3

p. 507.

13.42 Boots of the general polynomial equation, p. 507.

13.43 Principle of equating coefficients, p. 509. 13.44 Repeatedroots and the derived polynomial, p. 509. 13.45 Equationswith 'real' coefficients; conjugate complex roots, p. 510

13.5 Relations between roots and coefficients 0

13.51 Symmetrical relations, p. 512. 13.52 Unsymmetricalrelations, p, 512. 13.53 Transformation of equations, p. 513

Exercise 13 (d)5

r

13.6 Factorisation in real algebra 5

13.61 Roots of the general polynomial equation, p. 515.

13.62 Location of roots in real algebra, p. 510

Exercise 13(e) 5

13.7 Approximate solution of equations (further methods) 5



CONTENTS ix

Chapter 14. de Moivre's Theorem astd someApplications page 528

14.1 de Moivre's theorem 528

14.12 The values of (eos# + i sin p. 529, 14.13 Examples,p. 530

Exercise 14(a) 532

14.2 Use of the binomial theorem 534

14.21 oos m sin fl # in terms of multiple angles, p. 534.

14.22 oasnd, sinnfl, tanri.0 as powers of circular functions,

p. 535. 14.23 tan (OL + 2 + ...+& n ), p. 537

Exercise 14(6) 538

14.3 Factorisation 538

14,31 x n -l, p. 539. 14.32 8*+l,p. 539.

14,33 £e2n -2a; costta+ 1, p. 640. 14.34 sin n(9, p. 542

Exercise 14(c) 544

14.4 Roots of equations 545

14.41 Construction of equations with roots given trigono-

metrically, p. 545. 14.42 Results obtained by using relations

between roots and coefficients, p. 546

Exercise 14(d) 547

14.5 Finite trigonometric series: summation by C + iS 548

14.51 Cosines and sines of angles in A. P., p. 548. 14.52 Otherexamples, p. 549

Exercise 14(e) 550

14.6 Infinite series of complex terms. Some single -valued

functions of a complex variable 550

14.61 Convergence and absolute convergence, p. 550. 14.62 Theinfinite c,.i\, p. 551. 14.63 The exponential series, p. 552.

14.64 The modulus -argument form of expz. p. 554.

14.65 Euler's exponential forms for sine, cosine, p. 554.



X CONTENTS

Chapter 15. Survey or Elementary CoordinateGeometry page 5

15.1 Oblique axes 5

15.11 Advantage of oblique axes, p, 565. 15.12 Cartesian andpolar coordinates, p. 565. 15.13 Distance formula, p. 566.

15.14 Section formulae, p. 566, 15.15 Gradient of a line, p. 567.

15.16 Area of a triangle, p. 568

15.2 Forms of the equation of a straight line 5

15.21 Line with gradient m through [$ v jfj, p. 570.

15.22 Gradient form, p. 570. 15.23 General linear equation

Ax + By + C = 0, p. 571. 15.24 Intercept form, p. 571.

15.25 Line joining P lt P 2 , p. 572. 15.26 Parametric form for

the line through [v lr y t ) in direction 6 t p. 572. 15.27 Generalparametric form, p. 573. 15,28 Perpendicular form, p. 573

15.3 Further results 5

15.31 Sides of a line, p. 573. 15.32 Perpendicular distance of

a point from a line, p. 575

Exercise 15(a) 5

15.4 Concurrence of straight lines 5

15.41 Lines through the meetof two given lines,

p.576,

15.42 Condition for concurrence of three given lines, p. 577

Exercise 15(6) 5

15.5 Line-pairs 5

15.51 Equations which facte rise linearly, p. 579. 15.52 Thelocus ax 1 + 2hxy + fc^ 2 = 0, p. 580. 15.53 The general line-pair

$ = 0, p. 581. 15,54 Line-pair joining to the meets of the line

Ix + my — 1 and the locus s = 0, p. 584

Exercise 15 (c) 5

15.6 The circle 5

15.61 General equation of a circle; centre and radius, p. 586.

15.62 Circle on diameter I\P 2r p- 586. 15.63 Tangent at P vp. 587. 15.64 Chord of contact from P 1? p. 587.

15.65 Examples ; polar, p. 589. 15.66 Orthogonal circles, p. 591

Exercise 15(d) 5

15 7 Conies 5



CONTENTS X

Chaptee 16, The Parabola page 603

16.1 The locus y1 —4ax 003

36.1 1 Focus -directrix property, p. 603. 16.12 Parametric

representation, p. 603Exercise 16(a) go

16.2 Chord and tangent 605

16.21 Chord P x P 2 > p. 605. 16.22 Chord t^, p. 606.16.23 Tangent at F 1} p, 607. 16.24 Tangent at the point t,

p. 607. 16.25 Tangency and repeated roots, p. 008.16.26 Examples, p. 609

Exercise 16(6) 610

16.3 Normal 612

16.31 Normal at the point t, p. 612. 16.32 Conormal points,

p. 612

Exercise 16(c) 614

16.4 Diameters 615

16.41 Corporal definition,p. 615. 16.42 Diameters of aparabola, p. 615

Exercise lG(d) 616

Miscellaneous Exercise 16(e) 616

Chapter 17. The Ellipse 619

17.1

17.11 &P = e.PM, p. 619. 17.12 Focus- directrix property ofx*la* + y*jb 2 - l

t p. 620. 17,13 Focus-directrix property ofx s fa?-ij z

fb2 = 1, p. 620. 17.14 Second focus and directrix,

p. 621. 17.15 Further definitions, p. 621. 17.16 Form of theellipse a 2 /a z + i/

2/&

2 = I, p. 621. 17.17 Circle and ellipse, p. 622

17.2 Other ways of obtaining an ellipse 623

17 21 A ili F l



Xii CONTENTS

17.4 Chord and tangent page

17.41 Chord PjPj, p. 630. 17.42 Chord 0$, p. 631.

17.43 Chord p. 63 L 17.44 Tangent at P„ p. 632.

17.45 Tangent at 0, p. 632. 17.46 Tangent at t, p. 632.


Exercise 17 (b)

17.5 Normal17.51 Equation of the normal, p. 636. 17.52 Conormal points,

p. 636

Exercise 17(c)

17.6 The distance quadratic

17.62 Chord having midpoint I\, p. 638. 17,63 Diameters,p. 639. 17.64 Conjugate diameters, p. 639

Exercise 17 (d)

Miscellaneous Exercise 17 (e)

Chapter 18. The Hyperbola

18.1 The hyperbola -g— fj — 1; asymptotes

18.1 1 Form of the curve, p. 645. 18.12 Asymptote: generaldefinition, p. 645. 18.13 Asymptotes of the hyperbola, p. 646.

18.14 The bifocal property: SP-S'P = ± 2«, p. 647

18.2 Properties analogous to those of the ellipse —+ 1tir D2

18.3 Parametric representation

18.31 Hyperbolic functions, p. 649. 18,32 The point t, p. 650.

18.33 The point 0, p. 650. 18,34 Another algebraicrepresentation, p. 650

18.4 Chord, tangent, and normal

Exercise 18(a)

18.5 Asymptotes : further properties

18.51 Lines parallel to an asymptote meet the hyperbola on yonce, p. 653. 18.52 The equation of the tangent at Pi tends to



CONTENTS Xlii

18.7 Asymptotes as (oblique) coordinate axes page 6o7

18.71 xy = c 2, p. 657. 18.72 Parametric representation, p. G5S.

18.73 The rectangular hyperbola, p. 6S9

Exercise 18(c) 659Miscellaneous Exercise 18 (d) 661

Chapter 19, The General Conic; s = Jcs' 664

19.1 The locus s = 664

19. U Scheme of procedure, p. 664. 19.12 Notation, p, 664,

19.13 Chord P,P 2 of s = 0, p. 666

19.2 Joachimsthal's ratio equation. 666

19.21 The ratio quadratic for $ = 0, p. 666. 19.22 Sides of aeonic, p. 666. 19.23 Tangent at P v p. 667. 19.24 Pair oftangents from P t ,p. 667. 19.25 Chord of contact from P vp. 668. 19.26 Examples; polar of P l wo s = 0, p. 668.

19.27 Chord whose mid-point is P Jf p. 669. 19,28 Diameters,

p. 669. 19.29 Conjugate diameters, p. 670

19.3 The distance quadratic 670

19.4 Tangent and normal as coordinate axes 671

Exercise 19(a) 672

19.5 Number of conditions which a conic can satisfy 673

19.6 The equation s = ks' 674

19.61 Number of possible intersections of two conies, p. 674,

19.62 s = ks' t p. 674. 19.63 Degenerate cases, p. 676.

19.64 Examples, p. 677. 19.65 Contact of two conies, p. 679

19.7 Equations of a type more general than s = ks' 680Exercise 19(&) 681

Miscellaneous Exercise 19(c) 682

Chapter 20. Polar Equation of a Conio 684

20.1 The straight line 684

20.11 Distance formula, p. 684. 20.12 Line joining two points,



XiV CONTENTS20.3 Conies: pole at a focus page

20.31 Polar equation of all non-degenerate conies, p. 688.

20.32 Tracing of the curve Ijr = 1 + ecos0, p. 690.

20.33 Examples, p. 691. 20.34 Chord and tangent, p. 693

Exercise 20(6)

Miscellaneous Exercise 20 (c)

Chapter 2L Coordinate Geometry in Space:the Plane and Line

21.1 Coordinates in space

21.11 Rectangular cartesian coordinates, p. 700 21.12 Othercoordinate systems, p, 701

21.2 Fundamental formulae

21,21 Distance formula, p. 702. 21,22 Section formulae, p. 703

Exercise 21 (a)

21.3 Direction cosines and direction ratios of a line

21,31 Direction cosines, p. 706. 21.32 Angle between two lines,

p. 706, 21.33 Direction ratios, p. 707

Exercise 21 (&)

21.4 The plane

21.41 Equation of a plane in 'perpendicular form p. 710.

21.42 General linear equation Ax + By + Cz+D- 0, p. 710.

21.43 Conditions for two linear equations to represent the sameplane, p. 711. 21.44 Plane having normal I twin and passing

through P t , p. 711. 21.45 Plane P^Pj, p. 712.

21.46 Intercept form, p. 713. 21.47 Angle between two planes,

p. 713

Exercise 21 (c)

21.5 The line

21.51 Line through Pj^ in direction l:m:n, p. 714.

21.52 Parametric equations of a line, p. 715. 21.53 Line of

intersection of two planes, p. 717. 21.54 Distance of a point

feom a plane, p. 718. 21,55 Areas and volumes, p. 719

Exercise 21 (d)



CONTENTS XV

21.7 Skew lines page 727

21,71 Geometrical introduction, p. 727. 21.72 Length of thecommon perpendicular, p. 728. 21.73 Equations of the commonperpendicular, p. 729. 21.74 Alternative method, p. 729.

21,75 Standard form for the equations of two skew lines, p. 731

Exercise 21(f) 732

Miscellaneous Exercise 21 (g) 733

Chapter 22. The Sphere; SphericalTrigonometry 733

22.1 Coordinate geometry of the sphere 736

22.11 Equation of a sphere, p. 736. 22.12 Some definitions andresults from pure geometry, p. 736. 22. 13 Tangent plane at P up. 737. 22.14 Examples, p. 738

Exercise 22(a) 740

22.2 s = hs' 742

22.21 The general principle, p. 742. 22.22 Spheres througha given circle, p. 742

22.3 Surfaces in general 744

Exercise 22(b) 745

22.4 The spherical triangle 747

22.41 Some definitions and simple properties, p. 747.

22.42 Sides and angles, p. 748. 22.43 Polar triangle;

supplemental relations, p. 749. 22.44 Area, p. 750

22.5 Triangle formulae 751

22.51 Cosine rule, p. 751. 22,52 Sine rule, p. 752.

22.53 Supplemental formulae, p. 754. 22.54 Triangb on thegeneral sphere, p. 755

Exercise 22(c) 755

Miscellaneous Exercise 22 (d) 756

Answers to Volume II (25)

Index to Volume II xxiii



xvii

GENERAL PREFACE

This two-volume text-book on Pure Mathematics has been designed

to cover completely the requirements of the revised regulations for the

B.So. General Degree (Part I) of the University of London. It presents

a serious treatment of the subject, written to fill a gap which has long

been evident at this level. The author believes that there is no other

book addressed primarily to the General Degree student which covers

the ground with the same self-contained completeness and thorough-

ness, while also indicating the way to further progress. On theprinciple that 'the correct approach to any examination is from above',

the book has been constructed so that those students who do not

intend to take the subject Mathematics in Part II of their degree

course will find included some useful matter a little beyond the

prescribed syllabus (which throughout has been interpreted as anexamination schedule rather than a teaching programme); while

those who continue with Mathematics will have had sound prepara-

tion. As it is the author's experience that many students who begin

a degree course have received hasty and inadequate training, a com-plete knowledge of previous work has not been assumed.

Although written for the purpose just mentioned, this book will

meet the needs of those taking any course of first-year degree work in

which Pure Mathematics is studied, whether at University or Tech-

nical College. For example, most of the Pure Mathematics required

for a one-year ancillary subject to the London Special Degrees in

Physics, Chemistry, etc. is included, and also that for the first of the

two years' work ancillary to Special Statistics. The relevant matterfor Part I (and some of Part II) of the B.Se. Engineering Degree is

covered. The book provides an introduction to the first year of anHonours Degree in Mathematics at most British universities, andwould serve as a basis for the work of the mathematical specialist in

the Grammar School. Much of the material is suitable for pupils



XViii GENERAL PREFACEor else split up into two or even three courses of reading in Calcu

Algebra-Trigonometry and Geometry taken concurrently. Throughoit has been borne in mind that many students necessarily work withmuch direct supervision,

andit is

hoped that those of even modeability will be able to use this book alone.

A representative selection of worked examples, with explanatorremarks, has been included as an essential part of the text, toget

with many sets of 'exercises for the reader' spread throughoutchapter and carefully graded from easy applications of the bookwto 'starred' problems {often with hints for solution) slightly aboveultimate standard required. In a normal use of the book there

not be time or need to work through every 'ordinary* problemeach set; but some teachers welcome a wide selection. To each cha

is appended a Miscellaneous Exercise, both backward- and forwa

looking in scope, for revision purposes. Answers are provided a

end of each volume. It should be clear that, although practice

solving problems is an important part of the student's training,

sense is this a cram-book giving drill in examination tricks. Howevthose who are pressed for time (as so many part-time and even

students in the Te clinical Colleges unfortunately are) may havepostpone the sections in small print and all 'starred' matterlater reading.

Most of the problems of 'examination type* have been takenPinal Degree papers set by the University of London, and I am gr

ful to the Senate for permission to use these questions. Othersbeen collected over a number of years from a variety of unrecorde{and hence unacknowledged) sources, while a few are home-made.

It is too optimistic to expect that a book of this size will bepletely free from typographical errors, or the Answers from mamatical ones, despite numerous proof readings. I shall be grateful

readers will bring to my notice any such corrections or other sugg

tions for possible improvements.

Finally, I thank the staff of the Cambridge University Pressthe way in which they have met my requirements, and for the



xix

PREFACE TO VOLUME II

Although for convenience of reference the pagination and section

numbering are continued from Volume I, this does not imply depend-

ence of the present volume on the first. Apart from occasional back-

ward references, Volume II is a self-contained text-book on Algebra,

Trigonometry and Coordinate Geometry of two and three dimensions,

in which calculus methods are illustrated when instructive.

Beginning with a chapter which leads from revision of previous

algebraic work to an introduction to formal algebra, an early starton determinants can be made in the next, thereby assembling all the

equipment necessary for the subsequent geometry (which throughout

is real and euclidean) . Certain widely-used general theorems on systems

of linear equations have received more explicit statement and

emphasis than is customary at this level.

In Chapter 12 the passage from finite to infinite series lays the

foundations of4

convergence \ a subtle subject so often misunderstood

and mishandled by beginners. It is hoped that the many somewhatnegative cautionary remarks will stimulate rather than shake the

reader's confidence.

Chapter 13 sets complex numbers on a logical footing by first briefly

retracing the historical steps in their development from * mystery 5

through 'diagram' to the concept of 'ordered number-pair*. Thesubject is often approached from only one of these standpoints, but

the present inclusive treatment combines the advantages of all. The

opportunity is also taken of setting up a general theory of factorisation

and polynomial equations. Despite some repetition, it is felt that this

can be appreciated only after the provisional nature of the con-

clusions in 10.13, 10.3 has been realised. The following chapter, con-

cerned with trigonometrical applications of the preceding algebraic

theories, also introduces some genuine functions of a complex

variable.

A l b d



XX PREFACE TO VOLUME II

methods in which anything is fair') only confusion of principles

arise by incautiously mixing the real and the complex, especially

calculus techniques.

Since many students come to the course regarding Coordinate

Geometry as merely a mixture of 'graphs', Pure Geometry, and

Calculus, an attempt in Chapter 15 has been made to review

'known' parts of the subject more as an illustrated account of lin

and (real) quadratic algebra. Use is made of oblique axes wappropriate.

In the next three chapters, which contain a fairly detailed tre

ment of the parabola, ellipse, and hyperbola by means of th'standard' equations, the emphasis is on parametric methods

the consequent elegant applications of the (real) theory of equations

given in Chapter 10. Joachimsthal's ratio equation and the distanc

quadratic are also used incidentally. Owing to the algebraic analog

between ellipse and hyperbola, the discussion of the latter is centr

mainly on properties of the asymptotes.

The unifying influence of a systematic treatment of 'the gener

conic' by Joachimsthal's method is too valuable to omit. Thisthe powerful

e

s = ks' principle' are the themes of Chapter 19.

A short chapter on polar equations, first revising the straight

and circle in this form, develops in more detail their application to

conic, and thus gives the geometrical complement of the calculu

methods illustrated for various 'polar' curves in Volume I.

In principle, Chapter 21 returns to the fundamentals of cartesian

coordinate geometry, this time for three dimensions. The treatment,

designed to emphasise whenever possible the analogies betweentwo- and three- dimensional cases, while also pointing out important

contrasts, revises the methods of linear algebraic geometry, and mwell be left until fairly late in the course. Finally the sphere is brief

treated^ first geometrically by methods resembling those already u

for the circle in Chapter 15, and then trigonometrically.



xxi

REFERENCES AND ABBREVIATIONS

In a decimal reference such as 12.73(2),

12 denotes chapter (Ch. 12),

12.7 denotes section,

12.73 denotes sub-section,

12.73(2) denotes part.

(ii) refers to equation (ii) in the same section,

ex. (ii) refers to worked example (ii) in the same section.

4.64, ex. (ii) refers to worked example (ii) in sub-section 4.64.

Ex. 12 (6), no. 6 refers to problem number 6 in Exercise 12 (b).

wo means with respect to.

In the text, matter in small type {other than 'ordinary' worked

examples) and in ' starred ' worked examples is subsidiary, and may be

omitted at a first reading if time is short.

In an exercise

no. 6 refers to problem number 6 in the same Exercise.

a 'starred' problem either depends on matter in small type

in the text, or on ideas in a

later chapter;

or is above the general standard of

difficulty.

matter in [ . ..

] is a hint for the solution of a problem,

matter in ( . .. ) is explanatory comment.





363

10

ALGEBRA OF POLYNOMIALS

10.1 The remainder theorem and some consequences

10.11 Long division; identities

The reader will be familiar with the process for dividing one poly-

nomial by another of lower degree; we use it here to lead up to an

important algebraic theorem.

Example

Find the quotient and remainder obtained by dividing ax 2 + bx + cby x —k.

x —k )ax 2 + bx +c ( ax + (b+ak)

ax 2 —akx

(6 + ak) x + c

(b + ak) x —k{b + ak)

ak 2 + bk + c

Observe that the remainder is the original expression with x replaced by k.

If only the remainder is required in the above example, we can

proceed as follows. Let the quotient be Q{x) (a function oix) and the

remainder be R (a constant). Then we have

ax2

+ bx + c = {x-k)Q{x) + R. (i)

If1

we now put x = k in this, we obtain

ak 2 + bk + c = R (ii)

since the term involving Q(k) is zero.

Remarks



364 ALGEBRA OF POLYNOMIALS [10.12

so that the right-hand side has the factor x —k. If the other factor

denoted by Q(x), then

(ax 2 + bx + c)- (ak 2 + bk + c) = (x-k) Q(x),

which is equivalent to (i) with R given by (ii).

(ft) We may regard the 'long division ' as a process for constructing

from the given polynomials x —k and ax 2 + bx + c another polynomial

Q(x) and a constant R which satisfy (i) for all values of x. (A similar

principle, justified in 10.13, ex. (v), holds for any pair of polynomials,

although when the divisor is of degree n > 1, R will be a polynomial

of degree n —1 or less.)

The relation (i), which holds for all values of x, is called an identity

in x (in contrast to an equation in x, which holds only for certain special

values of The sign = between two polynomials in x is used to denote

'equal for all values of and is read 'identically equal to'. Thu(i) would be written

ax 2 + bx + c = (x- k) Q(x) + R. (i

The result of substituting a; = k would still be written with the ordinary

= sign because (ii) is a relation between special numbers only. However, we may continue to use the sign = even when the expressions

are in fact identically equal, unless we specially wish to emphasise

their identity.

10.12 The remainder theorem

We now obtain the remainder when the general polynomial

p(x) = p x n +p 1 x n ~ 1 + ... +p n - X +p n

is divided by x —k. We have

p(x) -p(k) = p (x n - n) +p 1 (x n ~ 1 - k 71 - 1

) + ... +p n - 1 (x - k). (ii

By direct multiplication we can verify that, for any positive integer

m and any x and k,

x m -k m = (x- k) {x™- 1 + x m~ 2 k + x m~ s k 2 +...+ k m~ x).

(When x 4= k this is equivalent to the sum-formula for the g.p.



10.12] ALGEBRA OF POLYNOMIALS 365

Hence the remainder when the polynomial p(x) is divided by x —k is

the number p(k) obtained by substituting k for x in this polynomial.

Also from (iv) we have

The factor case. If p(k) = 0, the polynomial p(x) has x —k for afactor. Conversely, if x —k is a factor, then p(k) = 0.

Examples

(i) Factorise 2x s -llx 2 +nx- 6.

Calling the polynomial p(x), we seek numbers k for which p(k) = 0. Onlythose numbers k which are factors of the last term 6 need be tried, for anyother number could not be the constant term in a factor of the polynomial, f

p(l) = 2-11 + 17-6 #: 0, /. a:- 1 is not a factor.

p{2) = 16-44 + 34-6 = 0, /. x 2 is a factor.

To find the other factor, divide the polynomial by x —2; the quotient is

2x*-lx + 3, so

p(x) = (x-2)(2x*-lx + 3),

= (a:-2)(a;-3)(2a?-l)

on completing the factorisation by inspection.

(ii) Find the valuesof a and b if 6x* + ax

2

+ bx—

2 is divisible (i.e. exactlydivisible, without remainder) by 2x 2 + x— 1.

Since 2x 3 +x~l = (x + l)(2x-l), both x+1 and 2x-l are factors of thegiven polynomial p(x). Hence p( - 1) = andp(J) = 0, which give

= a-b-8, = £a + £&-£,and so a = 7, b = —1.

(iii) Find the remainder when p(x) is divided by (x —a)(x —b), a 4= 6, wherep(x) has degree greater than 2.

Since the divisor is quadratic in x, the remainder will in general be linear in x,

say Ax + B. lfQ(x) is the quotient, then (cf. 10.11, Remark (yff))

p{x) = (x-a)(x- b) Q(x) +Ax + B.

Put x = a: p( a ) = Aa + B.

Put a; = 6: p(b) = Ab + B.

Solving these equations for A and B, we find

_p(a)-p(b) ap(b)-bp(a)A , IS =,a b a b




10.13 Factorisation of a polynomial; 'equating coefficients'

We continue to write

p(x) = p x n +p 1 x n ~ 1 + ... +p n ,

where p #= 0.

Theorem I. Ifp{x) is zero when x has any one of the n distinct values

a v a 2 , . .. , a n , then

P(x) = Po( x ~ a i) (x-a 2 )...(x- a n ). (v

Proof. Since p(a x )— by hypothesis, therefore x —a x is a factor

of p(x). Let the quotient when p(x) is divided by x —a x be Qn -x{x)\

it will be a polynomial in x of degree n —1 whose first term is p^x 11 *

Thenp(x) = (x- a t ) Qn ^{x)

.

Since p{a 2 ) = by hypothesis, we have

= ((h-c^Qn-M-

As a 2 4= a x (also by hypothesis), hence Qn -i(a 2 ) = and therefore

x-a 2 is a factor of Qn ^ x {x). The other factor Qn _ 2 ( x )> obtained b

division, is a polynomial whose first term is

px n - 2

; thus

Qn-ifr) = (x-a 2 )Q n - 2 (x),

and so p(x) = (x - a x ) (x - a 2 ) Qn - 2 {x).

We can continue step by step to remove the factors x —a 3 ,

until we reach x —a n , after which the other factor will be Q (x) whose

only term is^ a; = p . The result (v) follows.

Corollary I (a). A polynomial of degree n in x cannot be zero fo

more than n distinct values of x.

Proof. Expression (v) cannot be zero when x takes any value

different from a v a 2 ,...,a n ; for no factor would then be zero, an

p Q=(= by hypothesis.

Theorem II. // p x n +p 1 x n ~ 1 + ...+p n is zero for more than

distinct values of x, then p = p t = ... = p n = 0, and so p(x) = 0.

Proof. Either all of p ,Pi, --,Pn are z ^ TO >

or there is a first p which is not zero ' say p k (k < n




given that it is zero for more than n values of x. Our second alternative

thus leads to a contradiction, and so only the first alternative is

possible.

Finally, when p = p x = ... = p n = 0, the given polynomial is zerofor all x, i.e. p(x) = 0.

Coeollary II (a). // p x n +p 1 x n ~ 1 + . . +p n = 0, then

Po=Pl = ••• = 2>» = 0.

For since the polynomial is identically zero, it is zero for more than

n distinct values of x.

Corollary II (6). //

p x n +p 1 x n ~ 1 + ...+p n = q x n + q^- 1 +...+q n

for more than n distinct values of x, or for all values of x, then

Po = qo, Pi = $v Pn = q n >

For we have

(Po-q )x n + {p 1 -q 1 )x^- 1 +... + (p n -q n ) = 0,

and the results follow from Theorem II or Corollary II (a).

Corollary11(6)

is the basis of the principle*) of 'equating coeffi-

cients' in an identity between polynomials. It asserts that if two

polynomials of degree n are equal for all values of x (i.e. if the poly-

nomials have identical values for corresponding values of x), then they

agree term by term (i.e. they are identical in form). Without this

result it would be conceivable that two quite distinct polynomials

might take the same values for the same x, as x ranges over the real

numbers.

Corollary II (c).

//

p x n +p 1 x n ~ 1 + ... +p n = q x m + q^™- 1 + ...+q m

(where m > n)for more than m distinct values of x, or for all values of x,

then „ „= q , = q m_ n _ 1 , p = q m_ n , p n = q m.

In particular, the polynomial on the right has degree n.

The proof is like that of Corollary II (b).




The right-hand side

= a(n 8 + 3n 2 + 2n) + b(n z +n) + cn+d

= an 3 + (3a + b)n* + (2a + b + c)n + d.

This is identically equal to n z if and only if

a = 1, 3a + 6 = 0, 2a + 6 + c = 0, d = 0,

i.e. a = 1, b = —3, c = 1, a* = 0.

(ii) Show that x % cannot be expressed in the form

a(x+l)(x + 2) +b(x + 3)(x + 4:).

The expression = a(x s + Bx + 2) + b(x 2 + Ix + 1 2)

= (a + 6) x z + (3a + 76) x + (2a +126),

which is identical with x % if and only if

a + 6=l, 3a + 76 = 0, 2a+126 = 0.

The last two equations give a = 0, 6 = 0, and these values fail to satisfy t

first. The three conditions cannot be satisfied, and so x 2 cannot be written

the form stated.

(iii) If a,b, c are all distinct, prove that

(x —b)(x —c) (x —c)(x —a) (x —a)(x —6) _(a —6)(a —c)

(6 —c)(6

—a)(c

—a)(c —

6)

—

and deduce relations between a, b, c by equating coefficients of the various power

of x.

When x = a, the left-hand side reduces to 1. Hence the polynomial

(x —b)(x —c) (x —c)(x —a) (x —a)(x —b)

(a —b){a —c) (b —c)(b —a) (c —a)(c-b)

is zero when x = a. Similarly, it is zero when x = 6 and when x — c. Thus th

quadratic in x vanishes for three distinct values of x. Hence it vanishes for

values of x.

By equating coefficients of x2

, x and the constant terms, we obtain

1 1 1+ + - o,(a-6)(a— c) (6-c)(6-a) (c —a)(c —6)

6 + c c + a a + b _(a —b)(a —c) (b —c)(b —a) (c —a)(c —b)

be ca ab _(a —6) (a —c) (6 —c)(6 —a) (c —a)(c —6)

(iv) If ax s + bx % + cx + d contains (x+ l) 2 as a factor obtain relations betwee




By eliminating h from the first and third, and from the second and third,

b = d + 2a, c=2d + a.

*(v) Given two polynomials f(x), g(x) of degrees m, n (m > n), prove that there

is a unique pair of polynomials Q(x), B(x) such that E(x) has degree less thann

and

f(x)=g(x)Q(x) + B(x). (a)

The process of dividing/(a;) by g(x) gives a quotient Q(x) of degree m—n anda remainder B(x) of degree less than n such that (a) holds for all values of xexcept possibly those for which g(x) = 0. Thus the polynomial

f(x)-{g(x)Q(x) + B(x)},

whose degree certainly does not exceed m, is zero for more than m values of x.

Hence by Theorem II it is identically zero, which proves (a). Compare 10.11,

Remark (fi).

It is conceivable that, had we proceeded in a different way, we might haveobtained another pair of polynomials q(x), r(x) such that

f(x) = g(x)q(x)+r(x), (b)

where r(x) has degree less than n. Subtraction of (b) from (a) gives

B(x) - r(x) = g(x) {q(x) - Q(x)}. (c)

The left side of (c) has degree < n, while (unless q(x) —Q(x) = 0) the right

has degree ^ n (the degree of g{x)); and by Corollary II (c) this is impossible.

Hence q(x) —Q(x) = 0, and therefore by (c), B(x)—r(x) = 0. This proves theuniqueness.

Exercise 10(a)

Factorise

1 2x 3 + Sx 2 -l. 2 6a 8 -a; 2 -19:c-6. 3 x* + 2x 3 + x 2 - 4.

4 Solve x 3 —4:X 2 + x + 6 = 0.

5 Find the values of a and 6 if 6a; 8 + 7a; 2 + ax + b is divisible by 2x —1 and byx+1.

6 The remainder when (x—l)(x —2) divides x* + ax 3 + b is a;+l. Find

a and b.

7 A polynomial gives remainder 2x + 5 when divided by (x —1) {x + 2). Findthe remainders when it is divided by x —1, x + 2 separately.

8 A polynomial gives remainder 2 when divided by x + 1 and remainder 1

when divided by x —4. Find the remainder when it is divided by (x + 1 ) (x —4).

*9 A cubic polynomial gives remainders 5x —7, 12a;— 1 when divided byx 2 —x + 2, x 2 + x —1 respectively. Find the polynomial.

* 10 When divided by x 2 + 1 a polynomial gives remainder 2x + 3, and by x 2 + 2gives remainder x + 2. Find the remainder when it is divided by (x 2 + 1) (x 2 + 2).




2x + l . , „ a b14 Express — - m the form H .* (x-l)(x + 2) x-1 x + 2

5 . a bx + c15 Express — ——- m the form H—- —-.* {2x+l)(x* + D 2x+l x*+l

If a,b,c are all distinct, prove

^ a(x —b) (x —c) b(x —c)(x —a) c(x —a)(x —b)16 _i L± - + - — + - — = x.

(a —b)(a —c) (b —c)(b —a) {c —a)(c —b)

(a + b + x) (b + c + x) (b + c + x) (c + a + x) (c + a + x)(a + b + x) _^

(b —c)(b —a) (c —a)(c —b) {a —b)(a —c)~

18 (i) Prove that there cannot be two different quadratic expressions in

which take the values A, B, C when x has the distinct values a, 6, c respectively,

(ii) Verify that

A(x-b)(x-c)

[ B(x-c){x-a

) c{x-a)(x-b)

(a —b)(a —c) (b —c)(b —a) (c —a)(c —b)

has the required properties, and deduce from (i) that this is the only sucquadratic.

*19 Write down the (unique) cubic polynomial in x which takes the values

A, B, C, D when x has the distinct values a, b, c, d respectively.

20 (i) If x 3 +px + q contains a factor (x —a) 2, prove 4p 3 + 21q % — 0.

(ii) By writing q = 2a 3, prove the converse of (i).

21 (i) If a;4 + px + q contains a factor (x —a) 2

, prove 27p 4 = 256q 3, and express

q in terms of a.

(ii) Prove the converse of (i).

10.2 Polynomials in more than one variable

10.21 Extension of the preceding results

If p(x, y) is a polynomial in the two variables x, y, we may arrange

it according to powers of x, say as

%nPo(y) + x^Piiy) + • • • +p n (y)>

and consider it to be a polynomial in x whose coefficients are poly-

nomials in y. The preceding theories can then be applied, f Similar

considerations hold for more than two variables.

Examples

(i) If ax 2 + 2hxy + by* + 2gx + 2fy + c = a'x* + 2h'xy + b'y* + + Wv+




Equating coefficients, we have for all y:

a = a', 2(hy + g) = 2(h'y + g'), by 2 + 2fy + c = b'y 2 + 2f'y + c'.

By equating coefficients of like powers of y in the last two identities, we obtain

all the results stated.

(ii) Show that 2x 2 + 5xy —Zy 2 -x + lly-6 has linear factors, and find them.

Since 2* 2 + 5xy - 3y* = (x + Zy) (2x - y),

try putting

2x 2 + 5xy-Zy 2 -x+lly-6 = (x + Zy + a){2x-y + b),

where a and 6 are constants to be determined.

The right-hand side is

2x 2 + 5xy - Zy 2 + (2a + b) x + ( 36 - a) y + ah.

This will be identical with the given polynomial if and only if a, 6 can be chosento satisfy „ „, 2a + b = -\, 36-a=ll, ab = -6.

The first two equations give a = —2, 6 = 3; and these values do satisfy the

third. Hence the given polynomial has linear factors

(x+Sy-2)(2x-y + Z).

(iii) Factorisexy(x + y) + yz(y + z) + zx(z + x) + 2xyz.

Regarding this as a quadratic polynomial in x whose coefficients are poly-

nomials in y andz,

put x——

y:the expression becomes

+ yz(y + z) - yz(z -y)- 2y 2 z - 0.

Hence x + y is a factor. Similarly, x + z is a factor.

Next, regarding the expression as a polynomial in y whose coefficients are

polynomials in x, z, we may put y = —z and show that y + z is a factor.

Hence (y + z) (z + x) (x + y) is a factor. Since this product and the given

polynomial each have total degree 3, any other factor h must be numerical,

£LHCL SOxy(x + y) + yz(y + z) + zx{z + x) + 2xyz = k(y + z)(z + x) (x + y).

To obtain k, we may either substitute numerical values for x, y, z (e.g. x = 0,

y = 1, z = 1) in both sides, or compare coefficients (e.g. of x 2y). We obtain k = 1.

10.22 Symmetric, skew and cyclic functions

In 1.52(4) we defined a homogeneous function of two or more

variables, and we now give further useful definitions.

(1) A function (not necessarily a polynomial) of two or more

variables is symmetrical in these variables if it is unaltered by the




Bemarks

(a) The only symmetrical function of (x, y, z) of the first degree is

constant multiple ofx + y + z. For if Ix + my + nz is unaltered by inter-

change of x, y, thenIx + my + nz = ly + mx + nz,

i.e. (l —m)x+(m —l)y = 0,

so I = m. Similarly, we find I = n, and the linear function reduces

l(x + y + z).

(/?) The most general symmetric polynomial in (x, y, z) consisting

second-degree terms can likewise be shown to have the form

k(x 2 + y2 + z 2

) + l(yz + zx + xy).

(2) A function of two or more variables is skew or alternating if

interchange of any two of them changes only the sign of the function.

For example, x —y is a skew function of (x, y) ; and

(b —c) (c —a) {a —b),

xy(x -y) + yz(y -z) + zx(z - x)

are skew functions of their respective variables.

The reader should satisfy himself that the product of two symmetric

or of two skew functions is a symmetric function, while the product of

symmetric and a skew function is skew.

Considerations of symmetry, skewness and homogeneity often sav

algebraic manipulation, and will be helpful in Ch. 11.

Example

Factorise x 3 + yz + z 8 —3xyz. (Also see Ex. 10(/), no. 3.)

Regarding this as a polynomial in x, we find that when x = —(y + z) t

expression is zero. Hence x + y + z is a factor. The other factor could be foundirectly by long division, or as follows.

The given expression and the factor x + y + z are both symmetric in (x,y,z),

therefore so is the remaining factor.

Since the given polynomial is homogeneous of degree 3 and x + y + z is homo



10.22] ALGEBRA OF POLYNOMIALSWe can find the constants k, I by substituting numerical values:

x = 0, y = 0, z — I give 1 = k;

x = 0, y=l, 2 = 1 give 2 = 2{2k + l}, :. l = - 1.

Hence xP + y^+z 3 —3xyz = (x+y+z) (x* + y2 +z 2 —yz —zx —xy).

(3) Cyclic expressions. Consider again

xy(x - y) + yz(y -z) + zx(z - x),

and suppose the letters x, y, z are arranged around the circumference

of a circle as shown. When the first term xy(x —y) is given, the next

can be obtained from it by replacing each letter bythe one which follows it on the circle. Repetition of

the process gives the third term, and further repeti-

tion leads back to the first term. The given sumthus consists of the term xy(x —y) together with the «

two similar terms obtainable from it by cyclic inter-

change ofx, y, z\ it can be written

y

Lxy{x-y).

The number of letters involved has either to be stated explicitly, or

must be clear from the context; otherwise the S-notation is ambiguous.

Observe that the complete sum is unaltered if we replace x by y,

y by z, and z by x; for this cyclic interchange merely alters the order

in which the three terms occur. The expression (y —z) (z —x) (x —y) has

the same property.

Definition. An expression in x, y, z which is unaltered by cyclic

interchange of these letters is cyclic in (x, y, z).

A similar definition can be given when there are more than three

letters. For convenience of reference and comparison, it is desirable

to write expressions with their terms in cyclic order whenever possible.

Thus we prefer to write bc + ca + ab rather than ab + ac + be.

Remark (y). To test an expression for symmetry or skewness weinterchange letters two at a time. A cyclic interchange involves change

of all the letters. Thus a function may be cyclic but not symmetric,

373




3 Find the values of a for which 2a; 2 —5xy —3y 2 —x + ay —3 has linear

factors. [The factors must be of the form (x —3y + 3k) (2x + y—llk).]

4 Show that x 2 + Axy + 3y 2 + 2x —2y + 6 does not possess linear factors.

Using the remainder theorem, together with considerations of degree, symmetryand skewness when helpful, factorise

5 (x + y + z) 3 -(x 3 + y 3 + z 3).

6 yz(y 2 — 2) + zx(z 2 —x 2

) +xy(x 2 —y2

).

7 x(y-z) 3 + y(z-x) 3 + z(x-y) 3.

8 (x + y + z) 5 -{x 6 + y5 + z 5

).

9 Write the expressions in nos. 6, 7 in the S-notation.

10 Write in full the following expressions, assumed to involve three letters:

(i) Zx2

; (ii) lLx2

y2

; (hi) S6c(6-c); (iv) Sa2

6c; (v) 26c2

.

[In (v), cyclic interchange gives bc 2 + ca 2 + ab 2, which is only half the numbe

of terms implied by S, meaning 'sum of all terms of the type letter x another

letter squared' .]

11 Prove that (i) S(6-c) = 0; (ii) Xbc(b-c) = Sa 2 (6-c);

(iii) £a(6 2 -c 2) = (b-c)(c-a)(a-b).

12 Prove S(6 - c) 3 = 3(6 - c) (c - a) (a - 6).

13 Prove Sa 8 (62 -c 2

) = -(b-c)(c-a)(a-b)(bc + ca + ab).

14 Prove that Sa n (6 —c) contains the factor (6 —c) (c —a) (a —6) for apositive integer n ^ 2.

*15 (i) By taking x = 6 —c, y = c —a, z — a —b in the example in 10.22(2).

duce no. 12. (ii) Similarly, deduce the factors of Sa 3 (6 —c) 3. (iii) Use the result

no. 8 to factorise (6 - c) 5 + (c - a) 5 + (a - 6)5

.

16 Prove that a symmetric or skew function of three variables is also cyclic.

[If /(a, 6, c) is symmetric,

f(a,b,c) = f(b,a,c) =f(b,c,a);

if it is skew, f(a. 6, c) = -f(b, a, c) = - { -/(&, c, a)} = /(&, c, a).]

10.3 Polynomial equations: relations between roots and coeffi-

cients

10.31 Quadratics: a summary

If the equation ax 2 + bx + c = has roots a, fi (possibly equal), t

reader will know that ^ •

ca+/3 = —, a/? = -.

a a




10.32 Theory of cubic equations

If the distinct numbers a, /?, y satisfy

ax 3 + bx 2 + cx + d = 0, (i)

then by Theorem I of 10.13,

ax 3 + bx 2 + cx + d = a(x-a){x-fi)(x-y). (ii)

If the unequal numbers a, ft satisfy (i), then by the Remainder

Theorem x-cc and x-fi are factors of the left-hand side, and by

division the other factor is seen to be linear and of the form a(x —y).

Bythe statement are the roots of (i) ' we mean that

y=

fi,so

that the factor x-fi appears twice in (ii). Similarly, by 'a, a, a are

the roots of (i) ' we mean that the right-hand side of (ii) is a(x - a) 3.

Thus, if a, /?, y (not necessarily distinct) are the roots of (i), then

(ii) holds in all cases. Expanding the right-hand side by direct

multiplication,

ax 3 + bx 2 + cx + d = ax a -a(a+/3 + y)x 2 + a(fiy + ya + afi)x-aafiy.

Equating coefficients,we

find

a + p + Y = -^, PY + Y« + «P = ^ a PY = P)

Conversely, the cubic having roots x = a, ft, y is

(x-a)(x-fl){x-y) = 0,

i.e. x 3 -(a+fi + y)x 2 + (py + ya, + afi)x-otfiy = 0,

in whichthe coefficient of x 3 is +1,

the coefficient of x 2 is —(sum of roots),

the coefficient of x is (sum of the products of the

roots taken in pairs),

and the constant term is —(product of roots).

Observe the sequence -| 1—of the signs.

Remark The relations (iii) do not help us to solve the cubic equation,




Examples

(i) If a, /?, y are the roots of pa? + qx +r = 0, express in terms of p, q, r:

(a) Sa 2, (6) S^, (c) Sa 3

, (a*) Sa 5, (e) S/? 2 ?-

We have Sa = 0, S/?y = -, a^y = --.

(a) Sa 2 = (Sa) 2 -2S/?y = 0-2?- =

a a/>y r

(c) .Firs* method. Since a is a root of the given equation,

pot 3 + q<x + r = 0;

there are similar relations for /?, y. Adding,

pSa 3 + gSa + 3r = 0,

.-. Sa 8 = .

PSecond method. By the example in 10.22 (2),

a 8 +/ff 8 + y 8 -3a/£y = (a+/? + y) (a 2 +/# 2 + y 2 -/?y-ya-ayff)

= 0, since Sa = 0.

3r.-. Sa 3 = 3a#y = .

P(d) From the given equation, £>a 8 + qoc + r — 0, and hence

pa. 6 + qu z + ra 2 = 0.

Adding this to the two similar relations for fi, y,

2>Sa 8 + gSa 3 + rSa 2 = 0.

p pby using (a) and (c), and so Sa 6 = 5qr/p 2

.

(e) Consider (Sa)(S/tfy). A term like y#2y occurs only once, as the produ

ft .[iy. The terms a/?y arises in three ways, from a./ffy, fi.ycc, y.afi. Hence

(Sa)(S/?y) = S^ + Sa/Jy.

3r Sr:. = S/?^ , and Efi*y = -.

p p




In the given equation put y = — I —x, i.e. x = —1 —y:

(-l-2/) 3 + (-l-2/)2-24(-l-2/)-16 = 0,

i.e. y* + 2y* - 2Sy - 8 = 0. (A)

The values of y which satisfy this are related to the values of a; which satisfy

the given equation by the formula y = —1 —x. Since these values of a; are a, fi, y,hence the values of y are —1 —a, —1 — —1 —y; therefore (A) is the required

cubic. (We could also say that

x s + 2x*-2Zx -8 =

is the required equation, because the letter used for the unknown is immaterialwhen the roots are given.)

(b) fiyfoc = afiyloc* = 16/a 2. Put y = 16/x*, i.e. x 2 = 16/y:

\y ) y

so by squaring, 64cc a ^-3j = 16 2 ^1-^ ,

*(?-.) =Ma

,

y\y I \ y)

which reduces to y* —38t/ a + 4% —16 = 0.

The argument used in (a) shows that this is the required equation; but see theRemark at the end of 13.53.

10.33 Quartic equations

By reasoning as in 10.32 we find that if a, /?, y, 8 are the roots of

ax* H-foe 3 + cx % + dx + e = 0,

then a + fi + y + 8 = --, i.e. Sa = --,a a

afi+a,y + a8+fiy+fiS+y8 = -, i.e. Hafi = -,

fiy8+y8a + 8<xfi + afiy = --, i.e. Ea^y = --,Cb Qi

and aBy8 = -.

a



378 ALGEBRA OF POLYNOMIALS [10

Exercise 10(c)

1 If a, ft are the roots of x 2 —5x + 3 = 0, calculate

(i) a 2/? + a/? 2

; (ii) + (iii) (a-l)(/?~l); (iv) a 2 +/? 2;

(v) (a-/?) 2; (vi) ^ + ^; (vii) a 3 + /ff

3; (viii) a 4 + /ff*.

2 If a, /? are the roots of 3a; 2 + 2* —4 = 0, construct the equation whoroots are

(i) i/ a , (ii) a 2, /? 2

; (iii) a + 1//?, /?+ 1/a.

3 Write down the conditions for both roots of ax 2 + bx + c = to be positive.

4 If a, /?, y are the roots of x z + 2 —14a; —24 = 0, calculate

1 0y(i) (a + l)(/?+l)(y+l); (ii) (iii) Sa 2

; (iv) S^;

(v) Za 3; (vi) 2a 4

; (vii)

5 If a, /?, 7 are the roots of x 3 + 4a; 2 + x —2 = 0, construct the equation

whose roots are

(i) P+y, y + a, a+fi; (ii) -j^-L, ^; (iii) a(/? + 7), /% + <*), 7(«+A);

(iv) a 2, /ff

2, 7

2; (v) a 2

,/?~ 2

,7~ 2

.

6 If a, /?, 7 are the roots of 6a;3

—x2

—6x —2 = 0, form the equation whoroots are a + \, /? + \, y + \. Hence solve the given equation.

7 If x 3 + BHx + G = has two roots equal, prove G2 + 4H 3 = 0. [Let

roots be a, a, /?; then 2a +/? = 0.]

8 If the roots of x z + 3Hx + G = are in a.p., prove G = 0. [a + 7 = 2y#.]

9 If a;3 + px 2 + qx + r = 0, find the condition for the roots to be in (i) a.p

(ii) g.p.

10 If a, /?, 7 are the roots of a;3 + px 2 + qx + r = 0, construct the equation who

roots are a 2 —fiy, P 2 —ya, y 2 —ocfi, and comment on the special cases (i) p =(ii) q — 0. Also write down the value of (a 2 - Py) (fi

2 - yet) (y2 - a/?).

r „ r a 3 + r —pct 2 —q<x, v

~|

I a 2-fiy = a 2 + - = = — —= -(pa + q). I

J_ a a a _|

11 If x i + px 3 + q = has roots a, /?, 7, 5, construct the equation whose ro

are a + ji + y, ft + y + 8, y + d + ct, 8 + a + fi.

12 If a;4 + px 2 + qx + r = has three equal roots, prove that p 2 + 12r =

9g 2 = 32pr; and show that the value of the repeated root is —3g/4p.




There may be more than one such relation. Each is called an eliminant

of the system of equations. Owing to the importance of elimination,

especially in coordinate geometry, we give some further examples.

Nogeneral

methodscan be laid down; but see 11.43.

(i) Use of an identity.

Eliminate x, y, z from

x + y+z = a, x 2 + y2 +z 2 = b 2

, x 3 + y3 + z 3 = c 3

, xyz = d 3.

From the identity (10.22, ex.)

x 3 + y3 + z 3 -3xyz = (x + y + z)(x 2 + y

2 +z 2 -yz-zx-xy)

we have c 3 - 3d 3 = a(b 2 - Hyz)

and (Sz) 2 = Ea: 2 + 22yz, so Xyz = %{a 2 -b 2).

Hence 2(c 3 - 3d 3) = a(36 2 - a 2

).

(ii) Use of theory equations.

Eliminate m, nfrom (y-mx) 2 = a 2 m2 + b 2, (y —nx) 2 = a 2 n 2 + b 2

, mn = —1.

The first two equations show that m and n are the roots of the quadratic in t

(y-tx) 2 = a 2t

2 + b 2,

i.e. (a 2 - x 2) t 2 + 2xyt + (6

2 - y2

) = 0.

The third equation shows that the product of these roots is —1, so

b2

-y2

= 1a 2 —x 2

i.e. x 2 +y 2 = a 2 + b 2.

For the geometrical interpretation see 17.47, (ii).

The method illustrated in ex. (ii) can sometimes be used to solve a system of

equations as in ex. (iii) following. See also Ex. 10(d), nos. 12-14.

(iii) Solve for x, y,z:

x + ay + a 2 z = a*, x + by + b 2 z = b i, x + cy + c 2 z = c*.

These equations show that a, b, c are three roots of the quartic in t

t*-t 2 z-ty-x = 0.

Since the coefficient of t 3 is zero, the sum of the roots is zero, and hence the

remaining root is —(a + b + c). Then

—x = product of roots = —abc(a + b + c)

y = sum of the products of the roots taken in threes

= abc —(a + b + c)(bc + ca + ab);

= f h d f h t k i i




10.42 Common root of two equations

(1) The necessary and sufficient condition that the quadratics

ax 2 + bx + c = 0, a'x 2 + b'x + c' =have a common root is

Necessary. If there is a value of x which satisfies both equations,

then by treating them as simultaneous equations in x 2, x and elimin-

ating first x, and then x 2, we obtain

(ab' —a'b) x 2 = be' —b'c, {ab' —a'b) x = ca' —c'a.

{ab'- a'b) {be' -b'c) = {ab'-a'b) 2 x 2 = {ca' -c'a) 2,

so that the condition follows.

Sufficient. If ab' —a'b =(= 0, the condition (i) can be written

(ca'-c'a\ 2 be' -b'c

Put A = {ca'-c'a)l{ab'-a'b)\ then A2 = {be' -b'c)/ {ab' -a'b), and

{ab' - a'b) A2 = be' - b'c, {ab' - a'b) X = ca'- c'a.

{ab' - a'b) {aX 2 + bX) = a{bc' - b'c) + b{ca' - c'a)

——c{ab' —a'b) on simplifying.

Hence {ab' —a'b) (aA 2 + bA + c) = 0; and similarly

Since ab' —a'b #= 0, these show that x = A satisfies both quadratics.

If ab' —a'b = 0, then the given condition (i) shows ca' —c'a =Hence a:b:c = a' :b' :c', and so the quadratics are not independent.

In this case the result holds trivially.

Remark. Direct calculation will verify that

4{{ab' - a'b) {b ' - b' ) - { ' - ' ) 2

{ab'- a'b) {be' -b'c) = {ca' -c'a) 2.

{ab'-a'b) (a'A2

+ 6'A + c') = 0.




(2) When the given equations are of degree higher than the second,

we ehminate the highest powers step by step until two quadratics

with a common root are obtained. For example, if

/ = ax 3 + bx 2 + cx + d =

and g = px 2 + qx + r = (iii)

have a common root, then this root also satisfies p . /— ax . g = 0, i.e.

( pb —aq)x 2 + (pc —ar)x+pd = 0. (iv)

Conversely, a common root of g = and p.f— ax.g = is also a

common root of g = and / = 0. Hence / = 0, g = have a commonroot if and only if (iii) and (iv) have a common root, and a condition

like (i) will express this fact.

10.43 Repeated roots

With the notation of 10.13, suppose that

p(x) = (x-ct) r g{x) (1 < r < n),

where g{a) 4= 0. Then by the Remainder Theorem x —ocia not a factor

of the polynomial g(x). We say that the polynomial p(x) has a repeated

factor x —a of order r; and that the equation p{x) = has an r-fold

root x = a, r equal roots a, a root x = a, of multiplicity r, or a root x = a

of order r. If r = 1, we call # —a a simple factor of p(x), and x = a a

simple root ofp(x) = 0.

// p(x) = has a root x = a of order r, then x = a is also a root of

p'(x) = 0, of order r—l.

Proof.#'(a:) = r(x —a) r_1 g(x) + (x —a) r g'{x)

= (x- a)'- 1 {rg(x) + (x - a) gr'(z)}.

Since ^(a) =# 0, the contents of the last bracket are non-zero whenx = a. Hence a; = a is a root of p'{x) = 0, of order exactly r—l.

Conversely, if p'{x) = has a root x = a of order r—l, then provided

x = a satisfies p(x) = 0, it is a root of p(x) = of order r.

Proof Suppose x = a is a root of p(x) = of order s Then the



382 ALGEBRA OF POLYNOMIALS [10

Exercise 10(d)

Eliminate t from

1 x = t-ljt, y = t2 + l/t?. 2 x = a(l-t 2 )/(l + t

2 ),y = 2bt/(l + t 2)

3 Eliminate x : y from x + y = a, x 2 + y2 = b 2

, x* + y* = c 4.

4 If a 2 + x 2 = b 2 + y2 = ay —bx = 1, prove a 2 + 6 2 = 1.

[(a 2 + a:2

) + (&2 + 2/

2 )-2(GM/-&a:) = 0, i.e. (x + b) 2 + (y-a) 2 = 0, ;. x = -b aV = «•]

5 If a 3cc + 6 3

i/ = xy, b 3 x —a 3 y = x 2 —y2

, and x 2 + y2 = 1, prove a 2 + b 2 =

[Solve the first two equations for a 3, b z

, using the third.]

6 Eliminate x, y, z from xy = a 2, yz = b 2

, zx = c 2, x 2 + y

2 + z % = d 2.

7 (i) Eliminate x, y from y + Ix = al 3 + 2aZ, i/ + ma; = aw3 + 2am, a

y + nx = an3

+ 2an. *(ii) Interpret geometrically.8 (i) Eliminate m, n from

m2x —my + a = 0, n 2 x —ny + a — 0, m—n = c(l +mn).

*(ii) Interpret geometrically.

9 Eliminate Z, m, n from the equations in no. 7 and Im = b.

10 Eliminate A, fi from

a 2 + A 6 2 + A a 2 +^ 6 2 +/t

1 1 Solve the following equations for x. y, z:

x + ay + a 2 z = a 3, x + by + b 2 z = b 3

, x + cy + c 2 z = c 3.

12 Solve a + b + c = 5, bc + ca + ab = 7, abc = 3.

13 Solve a + b + c = -2, a 2 + b 2 + c 2 = 30, abc = -10.

14 Solve x + y + z = 2, x 2 + y2 + z 2 = 14, x 3 + y

3 + z 3 = 20.

15 Prove that the necessary and sufficient condition for

x 3 + 2px 2 + 2qx + r = and x 2 +px + q =

to have a common root is r 2 —Spqr + p 3 r + q3 = 0.

Find the necessary and sufficient condition for the following to have a double ro

16 x 3 +px + q = 0. (Cf. Ex. 10(a), no. 20.)

17 x*+px + q = 0. (Cf. Ex. 10(a), no. 21.)

18 Find the common root condition for ax z + bx + c = 0, px 3 + qx + r =[Eliminate x, x 3 in turn.]

*19 If b 2 # c and (x 2 + 2bx + c) r_1 is a factor of p(#) and of p\x), prove th

(x 2 + 2bx + c) r is a factor of p(x). Explain why the restriction b 2 4= c is needed.




We now show how to find the highest common factor (h.c.f.) of any

pair of polynomials f(x), g(x). It will be convenient to write deg/

for 'the degree of f(x) and so on.

Suppose deg/ = m and deg g = n, where m ^ n. Then we can dividef(x) by g(x), obtaining the quotient q x (x) and remainder r x (x), wheref

f(x) =g(x)q x (x) + r x (x). (i)

The degree of r x (x) must be less than that of g(x), and hence

degr x ^ n —1. We can therefore divide g(x) by r x (x), getting quotient

q 2 (x) and remainder r z (x), where

g(x) =r x {x)q 2 (x) + r 2 (x) (ii)

and degr 2 < degr 1? i.e. degr 2 ^ n —2.

Similarly, dividing r x (x) by r 2 (x) gives quotient q 3 (x) and remainder

rJx), where . . , \ , \ , , \ n~\W r x (x) =r 2 (x)q 3 (x)+r 3 (x) (m)

and deg r z ^ n —3.

Proceeding thus, we eventually obtain

V-aO*) =r s - 1 (x)q s (x)+r s (x) (s)

and r s _ x (x) = r s (x)q s+1 (x) + r s+l (X), (8+1)

where r s+1 (x) = 0. The stage is reached after at most n+lapplications:]: of the successive division process

First, suppose f(x) and g(x) have a common factor h(x), so that

f(x) =f x (x)h(x), g{x) = g x (x)h(x).

The identity (i) shows that

r x (x) = f{x) - g(x) q x (x) = h(x) {f x (x) - g x (x) q x (x)},

so that h(x) is also a factor of r x (x). Identity (ii) then shows that h(x)

is a factor of r 2 (x), and so on. Finally, identity (s) shows that h(x)

is a factor of r s (x). Hence

f l f




a factor of r 8 _ 2 {x) ; etc. Thus from (ii) we see that k(x) is a factor o

g(x), and hence by (i) it is a factor off(x). Therefore

any factor of r s (x) is a common factor of f(x) and g(x). (B

From (B) it follows that r s (x) itself must be a common factor

f(x) and g(x); and (A) shows that every common factor off(x) and g(x

must also be a factor of r s (x). Consequently,

(a) if r s (x) == constant, then f(x) and g(x) can have no algebraic

common factor, and we say that f(x), g(x) are coprime polynomials;

(6) if r s (x) is a polynomial, then r s (x) is the (algebraic) highest

common factor f off(x) and g(x). It has been found as the last non-zero

remainder in the process of successive division just described (the h.c.f

process or Euclid's algorithm).

Examples

(i) Find the h.c.f. of

2a; 4 - 4a; 8 + 5a;2 + 2a; -3 and Xs - x* + x* + 4a; 2 - 2x + 3.

We begin by dividing

2(x 6 -x i + x 3 + 4x*-2x + S) by 2a; 4 - 4a; 8 + 5a; 2 + 2a; - 3,

in order to avoid introducing fractional coefficients. The introduction

removal of numerical common factors at any stage is permissible since thi

does not alter the algebraic factors, f

g = 2x l -4x 3 + 5x 2 + 2x- 3

2x* + 2x 3 - 6a; 2 + 18a;

-6a; 3 +lLr a -16a;- 3

-6x 3 - 6a; 2 + 18a; -5417a; 2 -34a; + 51

Remove factor 17: a;2 — 2x+ 3 = r 2

2a; 5 - 2a; 4 + 2a; 8 + 8a; 2 -4a; + 6 = /2a; 6 - 4a; 4 + 5a; 3 + 2a; 2 - 3a;

2a; 4 - 3a; 8 + 6a; 2 - x + 6

2x* - 4a; 8 + 5a; 2 + 2a; -3

x a + x 2 -3x + 9 = r 1

x 3 -2x 2 + 3x

3a; 2 -6a; + 9

3a; a -6a; + 9

= r,

The K.c.v. is the last non-zero remainder r a , viz. x 2 —2x + 3.

(ii) Test the equation




of / and /' = 5(4a; 4 - 16a; 3 + 15a; 8 + 4x - 4). We therefore begin by applying theH.C.F. process to/and

g s 4a; 4 - 16a; 3 + 15* 2 + 4a;-4.

g = 4a; 4 -16a; 3 +15a; 2 + 4a;-4

4a;4

- 14a;3

+ 8a;2

+ 8a;

- 2ar*+ lx 2 - 4a; -- 2a; 3 + 7a: 2 -4a;-4

= r 9

4a; 5 - 20a;* + 25a; 3 + 10a; 2 - 20a; - 8 = /4a;

B

-16a;4

+15a;3

+ 4a;2

- 4a;

- 4a; 4 + 10a; 3 + 6a; 2 - 16a;- 8

- 4a; 4 +16a; 3 -15a; 2 - 4a; + 4

- 6a; 3 + 21a; 2 -12a;-12Remove —3:

2a; 3 - 7a; 2 + 4a; + 4 = r x

The h.c.f. is 2a; 3 —7a; 2 + 4a; + 4. By using the Remainder Theorem, we find

that it has a factor x —2; hence

H.CB-. = (x -2) (2a; 2 -3a; -2) = (a;-2) 2 (2a;+ 1).

From 10.43 it follows that x —2, 2x+l are repeated factors of orders 3, 2in/, and that x = 2, x = —\ are roots of/ = of orders 3, 2. Since/ has degree 5,

there can be no other roots.

10.52 An important algebraic theoremThe identities (i), (ii), ... in 10.51 show that

ri = a f+b o g, where a = 1, b = -q x ;

r a = (hf+brf, where a x = -q 2 , b t = l + q 1 q a ;

and so on. We thus see thatr s = af+bg, (C)

where a, b are polynomials in x.

Writing/ = r s <J>, g = r t f, then by (C)

1 = a<j> + bxjr,

where 0, \Jr have no algebraic common factor. Since a<fi + bijr has no algebraicfactor (being identically 1), hence a, b are coprime.

(a) // r„ = constant, then we may divide (C) by r s and get the importantTheobem. If f, g are coprime polynomials, then other coprime polynomials

A, B exist for which Af+ Bg = 1.

(6) If r s is a polynomial in x, then we have:

The h.c.B . of two polynomials f, g can be written in the form af+bg, where a, bare coprime polynomials.

Definition. A polynomial is said to be irreducible when it has no algebraicfactor of lower degree than itself.

Cobollaby. If f is irreducible and g, h are polynomials such that f is not afactor of g but f is a factor of gh, then f must be a factor of h.

Proof. ^ Since / is not a factor of g, the degree of any common factor of/, gis less than the degree of / and is therefore zero since by hypothesis / has nosuch factor. Hence /, g are coprime and, by the theorem, polynomials A, Bexist such that




Remarks

(a) The polynomials a, 6 in identity (C) are not unique, for clearly

a, = a-\ — and 0, = o

(where c is any polynomial) would also satisfy (C).

(/?) There are arithmetical results (important in the Theory of Numbers)analogous to the preceding; they can be formulated by replacing 'polynomial',

'irreducible', 'degree' by 'integer', 'prime', 'magnitude' respectively.

10.53 Theory of partial fractions

An application of the theorem in 10.52 is to prove the possibility of resolving

a rational function into partial fractions. In 4.62 we stated and illustrated th

practical methods for doing this in any particular example, but we did nattempt a general justification of our statements.

Definitions. The rational function f/g, where / and g are polynomials in a;,

called (a) irreducible if /, g are coprime; (6) proper if deg/ < deggr.

Iff, g have H.c.s\ h (found as in 10.51), then / = hf x and g = hg x , so tha

(except for those values of x which make g = 0) f/g = f x /g x ; emdf 1 ,g 1 are c

prime. We shall therefore always assume that the rational function is irreducible.

If f/g is not proper, we can express it as the sum of a polynomial and a proper

fraction in the form q + r/g, by finding the quotient q and remainder r whe/ is divided by g.

Theorem. If p x , p 2 are coprime polynomials, the (irreducible) rational function

f/(PiPz) can oe expressed uniquely in the form

r x r 83 + —+ -,

Pi Pi

where qis a polynomial and r x /p x , r 2 /p 2 are irreducible proper fractions.

Proof. Since p x , p 2 are coprime, polynomials A, B exist such that

APi + BP2 = •

. _/_ J(A Pl + Bp 2 ) = Bf+

Af^

PiPz PlPi Pi Pi'

By division, Bf = q x p x + r x and Af = q 2 p z + r 2 ,

where deg r x < deg p x and deg r 2 < degp 2 . Hence

—= ?i + ?2 +—+—»

PiPi Pi Pi

where r x /p x , r 2 /p 2 are proper. They are also irreducible', for if (say) r x /p x reduces

to r x /p x , then result (i) shows that//(p 1 p 2 ) is equal to a fraction with denomi-' h d < d di h h th i




Since p 2 is a factor of the left-hand side, therefore p 2 is a factor of (r2 —r 2 )p x .

As p 2 is prime to p x by hypothesis, it follows from the corollary in 10.52 that

p 2 must be a factor of r 2—r 2 . This is impossible unless r

2—r 2

= 0, for p 2 has

higher degree than r'2 and r 2 , and so certainly degp 2 > deg(r 2 — 2 ). Thus

r 2 = r 2 , and similarly we can show r' x = r x . It then follows thatq'

= q.Corollary. If p lt p 2 , p n are polynomials every two of which are coprime,

then the (irreducible) rational function f/(p x p 2 ...p„) can be expressed uniquely

in the formq-\ 1 1- ... ^

,

Pi Pi Pn

where qisa polynomial and r x \p x , r n /p n are proper and irreducible.

This is proved by repeated applications of the preceding theorem: first to

Pi(Pi---Pn) to give ,

gi + -+ J1;

Pi Pz---Pnthen to p 2 (p 3 . . . p n ) to give

/i ~ ^*2 f 2= ?2+— +;p 2 ...p n p 2 P 3 ..-Pn

and so on. The uniqueness is proved by the same argument as in the theorem.

Remark. For their application to a given rational function f/g, the preceding

results depend on the factorisation of the denominator g into irreducible factors

PiP 2 ...p n (of. 4.63). In 13.61 we shall prove that a polynomial g(x) can be

factorised into the product of linear and irreducible quadratic factors like

(x —a) r, {(x —b) 2 + c 2

}*. Assuming this, the above corollary shows that

f- = q + X—̂— +S- 1*. , (ii)

g (x-a) r {(x —6)2 + c 2 }*

where q and the r x , r 2 are polynomials and degr x < r. degr 2 < 2s.

By division, any polynomial $(x) of degree m can be written in the form

<f>(x) = (x —a)$ 1 [x)+A a ,

where A is constant and<f> x (x) has degree m—1. Similarly

$ x (x) = (x-a)<f> 2 (x)+A x ,

and so on. Combining all these results, we havef

<f>(x) = A +A x (x-a)+A 2 {x-a) 2 + ...+A m(x-a) m.

Hence the proper fraction r x /(x —a) r can be expressed as

Aft J^L-t —1 . . ,

—a)r

(a; —a)

Similarly, any polynomial $(x) can be written

= ( ) A B




if <j>(x) has degree 2s— 1 or less. Hence the proper fraction rj{(x —b) 2 + c

can be decomposed as

A x + B A l x + B 1 A 3 ^x + B s . x

{{x-b) 2 + c 2}' {{x-b^ + c 2 }*- 1 (x-b) 2 + c 2 '

K

The argument already used in proving the theorem shows that each of threductions (iii), (iv) is unique.

By combining ( ii ), ( ii i) and (iv), we completely prove the statements madin 4.62 (assuming, of course, the facts in 13.61; but see the Remark in tha

section).

Exercise 10(e)

Find the h.c.f. of

1 x 3 + 3x z -8x-24, x 3 + 3x 2 -3x-9.

2 2x3

+ 7a;2

+ lOx + 35, 2xi

+ 7x3

- 2x2

-3x + 14.

3 2a: 3 - a:2 + 4a: + 15, a;

4 +12a;-5.

Test for repeated factors, and hence factorise completely

4 a:4 - 9a: 2 + 4a; +12. 5 x* - x 3 + 4a; 2 - 3x + 2.

6 x« - 3x 5 + 6a;3 -3x 2 -3x + 2.

Test for repeated roots, and hence solve completely

7 12a: 4 + 4a: 3 -45a: 2 + 54 = 0. 8 x 6 - 5x 3 + 5a: - 2 = 0.

*9 Find polynomials A, B of least degree such that

^(2a; 3 -3a; 2 + 4a;-l) + B(a: 2 + 2a:-3) = 1.

Miscellaneous Exercise 10(/)

1 Prove a(x 2 —y2

) —2hxy always has linear factors.

2 If 3x 2 + 2Axy + 2y 2 + 2ax— 4y+ 1 has linear factors, prove that A mussatisfy A2 + 4aA + 2(a 2 + 3) = 0. Is this sufficient?

3 Establish the identity Hx z —3xyz = (Sa;) (Lx 2 —1>yz) as follows:

(i) Expand {x + y)3 to show x 3 + y

3 = (x + y)3 —3xy(x + y).

(ii) Deduce that x 3 + y 3 + z 3 —3xyz = (x + y) 3 + z 3 —3xy(x + y + z), and factorise the sum of two cubes.

4 (i) Verify that x 2 + y2 + z 2 —yz —zx —xy = %{(y —z) 2 + (z —x) 2 + (x —y)

2}.

(ii) If x + y + z > 0, prove x 3 + y 3 + z 3 > 3xyz unless x = y —z.

(iii) If x + y + z = 0, prove x 3 + y3 + z 3 = 3xyz.

5 If a: = b + c —a, y = c + a —b, z = a + b —c, prove (using no. 4 (i))

x 3 + y3 + z 3 —3xyz = 4(a 3 + 6 3 + c 3 —3abc).

6 Factorise (i) S(6 3 + c 3) (b-c); (ii) Sa(6 4 - c 4

).

7 U th id tit 2 = 2 + 21



ALGEBRA OF POLYNOMIALS 389

11 'The equation (x— l) 2 = A(a; —2/i) (x —4) has equal roots.' (i) If /t has agiven value, prove this statement holds for A = and one other value, (ii) If

A has a given non-zero value, prove the statement is true for two values of

ft only if A < 1. If A = —15, find the two sets of equal roots.

12 Find the condition for the roots of ax 3 + bx + c = to divide the distancebetween the roots of a'x % + b'x + c' = internally and externally in the sameratio.

*13 If {a 1 x 2 + b 1 x + c 1 )/(a i x* + b 2 x + c 2 ) takes the same value when x has the

values given by a a x 2 + b a x + c 3 = (62 > 4a 3 c 3 ), prove that

a 1 a 2

&l &jj 6 3 = 0.

C\ C2 Cg

[If thevalue taken

isk, then (a 2

'k —a 1 )x

3

+(b

2 k—

b 1 )x +(c

2k —c

1 ) = must havethe same roots as a a x 3 + b a x + c 3 = 0. Hence corresponding coefficients are

proportional.]

14 What can be said about the coefficients in an equation whose roots are

(i) a,fi,—cc—fi; (ii) 0,a,,fi,y; (iii) a/b,b/c, c a; *(iv) tan d v tan a , tan d s ,

where 6 1 + 6 2 + 6 a is an integral multiple of 7r?

15 If a line cuts the curve x = at 3, y = at? ia. three points for which t has

the values t v t 2 , t z , prove Ti 2 t a = 0.

16 If one root of x* + ax + b = is twice the difference of the other two,

prove that the roots are

-136/12a, 136/3a, -136/4a, and that 144a 3 + 21976 s = 0.

17 If a, ft, y are the roots of x 3 +px + q = 0, express a i+fi

i + y t in termsof p, q.

18 Obtain the equation whose roots exceed by 3 the roots of

x* + 12x* + 49a; 2 + 78x + 42 = 0,

and hence solve the given equation.

19 If a, ft are the roots of ax 2 + 2hx + b = 0, find the quartic equation whoseroots are

±1/tx,

±1//3.

20 EUminate A, fi, v from

x 2 y* z 2 x 2 y 3 z 2 x 2 y 3 z 2

+ -?—+ -=1, + e— + = 1, -+^—+ = 1a + A 6 + A c + A ' a+/t b+/i c+/i ' a + v b + v c + v

and \/iv = abc.

21 If x s + Sax 3 + Sbx + c = has a repeated root, prove that this root also

satisfies x 3 + 2ax + 6 = 0. Hence show that the repeated root is (c —a6)/2(a 2 —6).

22 Prove that x* +px + q = cannot have a repeated root of order 3.



390 ALGEBRA OF POLYNOMIALS*25 Determine the values of m and c for which the line y —mx + c is (i)

inflexional tangent, (ii) a double tangent, to the curve y = x 2 (x 2 + 4x —1

[In (ii), the equation x 2 (x 2 + 4x —18) —mx —c = has two double roots.]

26 If 2ac' + 2a' 'c = Wand if ax 2 + bx + c, a'x 2 + b'x + c' have a common factor,

prove that at least one of these quadratics is a perfect square. [See 10.42

Remark.]

27 If a, b, c, d are constants such that ad 4= be, and /, g, p, q are polynomials

in x such that p = af+bg, q = cf+dg, prove that /, g and p, q have the samh.c.f. What happens if acZ = bet

*28 If p{x,y) is a symmetrical polynomial in {x,y) having factor x —y, provthat actually (x —y)

2 is a factor. [p(x,y) = Ha rs (x ry

s + x sy

r). The conditions

for p to be zero when x = y are the same as those for dp/dx to be zero whx = y.]

*29 Cubic equations : Cardan's method^ of solution. Every cubic can be reduced

(see Ex. 13 {d), no. 15) to the standard form

x* + BHx + G = 0.

The identity in no. 3 shows that */ y, z can be chosen so that

y3 + z z = G and yz = -H,

then x = —y—z will be a root of (a). Prove that y3

, z 3 must be the roots of

t2 -Gt-H 3 = 0.

*30 (i) If O2 + 4Z? 3 > 0, show that there are distinct numbers y, z satisfying

so that x = —y —2 is a root of (a). Use no. 4 (i) to show that there are no oth

roots.

(ii) If C 2 + 4H 3 = 0, show that the right-hand side of the identity

no. 3 becomes (x + 2y) (x —y)2

, so that (a) has three roots, two of whichequal, viz. - 2y, y, y.

(iii) If Q2 + 4H 3 < 0, numbers satisfying (6) do not exist. The following

trigonometrical method can then be used.

Put x = kcosd in (a), and choose h so that k 3 :3HJc = 4: —3. Show th

(a) then becomes cos 30 = Q/{2H*J( —H)}. Verify that condition (iii) ensure

that 3d can be found from this, and hence that three values of cos 6 = x/k ar

obtainable.

f So-called, although discovered by Tartaglia.



391

11

DETERMINANTS AND SYSTEMSOF LINEAR EQUATIONS

11.1 Linear simultaneous equations

11.11 Two equations in two unknowns

The usual method of solving

a x x + b x y = c x ,

a 2 x + b 2 y = c 2

by elimination shows that

(a x b 2 -a 2 b x )x = c x b 2 -c 2 b x and (a x b 2 -a 2 b x )y = a x c 2 -a 2 c x . (i)

If a x b 2—a 2 b x 4= 0, these give unique values for x and y.

When a x b 2—a 2 b x = 0, we may assume throughout that a x and b x

(and likewise a 2 and b 2 ) are not both zero; for if a x = = b x , the first

equation becomes = c x , which is either false or trivial. Two cases

arise.

(a) If at least one of c x b 2—c 2 b x ,

a x c 2—a 2 c x is not zero, (i) gives a

contradiction. The given equations therefore have no solution, and

are said to be inconsistent.

(b) If a x b 2—a 2 b x = c x b 2 —c 2 b x = a x c 2

—a 2 c x = 0, then (i) gives no

information. Since a x ,b x are assumed to be not both zero, suppose

a x #= 0. The first of the given equations can then be solved for x in

terms of y. When y is given the value y , let the value obtained for

x be x n ; then ,

X°~ a x•

Since a 2 x + b 2 y - c 2= CT^ Cl

+ b 2 y - c 2a x



392 DETERMINANTS, LINEAR EQUATIONS [11.12

shows that corresponding coefficients are 'proportional', so that th

equations are not distinct.)

Geometrically, the given equations represent straight lines.

a x b 2 —a 2 b x 4= 0, the lines are not parallel, and the above solutionrepresents their unique point of intersection. In case (a), the lines a

parallel, while in (b) the equations represent the same line.

11.12 Three equations in three unknowns

Here we obtain only the general form of the results corresponding

to (i), and postpone discussion of details until 11.4.

Elimination of z from the first and second of

a x x + b x y + c x z = d x ,

a 2 x + b 2 y + c 2 z = d 2 ,

a s x + b 3 y + c 3 z = d 3

gives (a x c 2 -a 2 c x )x + (b x c 2 -b 2 c x )y = d x c 2 -d 2 c x ,

and ehmination of z from the second and third gives

(a2

c3

-a3

c2

)a; + (& 2 C3-&3C a )^ = d 2 c 3 -d 6 c 2 .

Now ehminate y from these last two equations:

{(b 2 c 3- 63 2 )

(a x c 2 - 2 c x ) - (b x c 2- b 2 c x ) (a 2 c 3 - 3 c 2 )} x

= (b 2 c 3- 63 2 ) (d x c 2 - 2 c x ) - (b x c 2

- b 2 c x )(d 2 c 3 - 3 c 2 ).

On simphfying we find that the coefficient of x is

c 2 (a x b 2 c 3—a x b 3 c 2 + a 2 b 3 c x

—a 2 b x c 3 + a 3 b x c 2 —a 3 b 2 c x ),

and that the right-hand side reduces to c 2 times a similar expression.The solution for x therefore has denominator

a i h c 3 - a i &3 c 2 + a 2 &3 c i -a 2 b x c 3 + a 3 b x c 2 - a 3 b 2 c x .

We should find similarly that this is also the denominator in th

expressions for y and z.

11.13 Structure of the solutions



11.2] DETERMINANTS, LINEAR EQUATIONS 393

(/?) The suffixes of the letters a, b, c in each term form one of the six

permutations of the numbers 1, 2, 3. For example, consider a z b x c 3 : if

we interchange the suffixes in pairs until they are in natural order

1, 2, 3, we find that the number of interchanges is odd. Similarly, forthe term a 3 b x c 2 the number is even. It is easily verified that all the

terms for which the number is odd are preceded by the sign —, while

those for which it is evenf have + . (For a given term it can be provedthat the number of interchanges is either always odd or else alwayseven, no matter how the rearrangement to natural order is carried out.)

Hence the sign of each term in (ii) is decidable by the parity of the numberof interchanges of its suffixes from the actual to the natural order.

1 1 .2 Determinants


The symbolA =

a x b x

a 9 b 9

called a second-order determinant, is defined to mean a x b 2—a z b x . The

following five properties^ of A are easily verified.

(1) The value of A is unaltered by interchange of rows and columns.

That is, if

A' =a-, a„

W b 2

then A' = A. We call A' the transpose of A.

By use of this result, any property proved for rows extends at onceto columns, and conversely. The following will therefore be stated for

rows only.

(2) Interchange of two rows alters only the sign of A.

That is, a 2 b 2

a x b x

= -A.

(3) // two rows are identical, then A = 0.

(4) Multiplication of any one row by k multiplies A by k.




(5) Addition to any row of a multiple of another row does not alte

the value of A.

For example, by adding k times the second row to the first row

we obtain a x + ka 2 b x + kb a

a 9 b 9

= A.


The symbol

A =a x

a u

a« b

2 b 9

called a third-order determinant, is defined to mean

c 2 a 2 Co+ CX

a % b 2-&1

h c 3 a s c 3 a 3 b 3

which (by using the definition of the second-order determinants)

equal to

a i(h c 3 ~ h c z) ~ b i( a 2 c 3 ~ %c 2) + Ci(a 2 6 3 - a 3 b 2 )

= a 1 &2 C3-a 1 63C 2 + a 2 63C 1 -a 2 6 1 c 3 + a 3 6 1 c 2 -a36 2 c 1

on rearranging. This is the expression (ii) of 11.12, and is sometimesreferred to as the expansion of A.

The numbers a x ,b x ,

c v a 2 ,c 3 are called the elements of A; wit

our notation, the suffix denotes the row and the letter denotes t

column in which a particular element lies.

The remarks in 11.13 show that the expansion of a third-order

determinant consists of 6 terms, each of which involves an element fro

each row and each column but no two elements from the same row or

same column. The sign before each term is determined by the suffixes

according to the rule in 11.13 (/?). We may write the expansion

A shortly as ^±a i bj

c k .

The definition expressed by (i) above can be readily generalised to

define determinants of fourth and higher orders (see 11.7). Observ

how the second-order determinants in (i) are constructed from A: th

coefficient of a x is the determinant obtained from A after deleting t




The diagonal running from top left to bottom right is called the

leading diagonal, and the product a x b 2 c z of the elements in it is called

the leading term in the expansion of A.

Examples

(i) 4 5

2 4

3 6

= 44 7

6 2

2 7

3 2+ 3

2 4

3 6

= 4(8-42)-5(4-21) + 3(12-12)

= -51.

a h 9

h bf = a

9 f c

b f

-h

h f h b

+ 9 9 fc 9 c

(ii)

= a(bc -/•) - h(ch -fg) + g(hf- bg)

= abc + 2fgh - a/ 2 - bg 2 - ch\

The reader should now do Ex. 11 (a), nos. 1-10.

11.23 Other expansions of a third-order determinant

We may arrange the expression (ii) of 11.22 according to elements

from any row of A and the corresponding second-order determinantsformed by deletion as described above. Thus, grouping by elements of

the second row,

A = -a a (6 1 c 3 -63C 1 ) + 6 a (a 1 C3-a 3 c 1 )-C2(a 1 63-a 3 6 1 )

(iii)ha x c x a x b x= -a 2

b 3 c 3 a s c 3 a z b z

which is the expansion of A from the second row. Similarly,

A = a z {\ c 2 - 6 2 c x ) - b z {a x c 2 -a 2 c x ) + c 3 (a x b % -a z b x )

b x c x a x c x a x b x

b 2 c 2-h + c za 2 c 2 a 2 b 2

(iv)

the expansion from the third row.

Likewise, we may arrange (ii) by elements of any one column ; e.g.

A = a x (b 2 c z -b z c 2 )-a 2 (b x c z -b z c x ) + a z (b x c 2 -b 2 c x )




11.24 Properties of AWe now show that the properties (l)-(5) in 11.21 hold for third-

order determinants. In view of the first, all subsequent propertieshold for columns as well as for rows.

Proof of (I).

A' =«i « a a z

b x b 2 b 3

Ci Ca Co

= a.&3

+ «3b x &2

^2 C3 c 3 C l C 2

by definition,

= a^&gCg - 6 3 c 2 ) - a 2 ( &i c 3 ~ &s c i) + %( &i c 2 - &2 c i)

= A by the line preceding (v) in 11.23.

Proof of (2). Interchange of two rows is equivalent to interchanging

two of the suffixes 1, 2, 3. Results (i), (iii), (iv) show| that the sign

the determinant is changed in every case.

Alternatively,we may

start from the definition; e.g. interchange

of the first and third rows gives

= a»b x Cj

-6 9

= « 3 (&2 C l_ &1 C 2) - Ma 2 C l - a l C t) + C3 ( a 2

- a i 6 2)

= «i(& 3 c 2 - b 2 c 3 ) - h(a 3 c 2 - « 2 c 3 ) + c i( a 3 &2 - Hh)

on arranging by elements of the first row of A; and this is clearly —A

A similar direct proof would hold for other row interchanges.

Proof of (3). By interchange of the two identical rows we obtain

—A, by (2). However, interchange of identical rows clearly leaves th

same determinant A as before. Therefore —A = A, so A = 0.

Proof of (4). Since each term in the expansion of A contains exactly

one element from each row, multiplication of each element in a given




has not been mentioned. Thus, if we add k times the second row to the

first row, we geta x + ka 2 b x + kb 2 c x + kc 2

(vi)

The row 'not mentioned' is the third; using an equation like (iv) to

expand from it, we see that the new determinant is equal to

b t + kb 2 c x + kc 2

60 c«

a x + ka 2 c x + kc 2

= a 9

b9

+ c.

a x + ka 2 b x + kb 2

a 2 b 2

CL2

C2

by using Property (5) for each of the second-order determinants.

Again by equation (iv), this expression is A.

Remarks

(a) There are many extensions of Property (5). Three are indicated

in Ex. 1 1 (a), no. 22; but see the Remark about random manipulations

after no. 29.

(/?) The operation by which determinant (vi) above is obtained

from A can be denoted by r 1-> r x + kr 2 . With this notation, properties

(4) and (5) can be combined in a single statement:

// r t -^k 1 r 1 + k 2 r 2 + k 3 r 3 , then A -> k t A (i = 1 or 2 or 3).

11.25 Examples

Properties (2)-(5) give ways of simplifying a determinant whendirect application of the definition would be clumsy owing to the large

numbers or heavy algebra involved,

(i) Evaluate

By r s-> r 3 - r a , we get

By > thi b

35 29 86

36 31 87

38 32 89

35 29 86

36 31 87

2 1 2




The last determinant is now easily expanded, according to the definition,

2 1 1 1 1 235 -29 + 86

I 2 2 2 2 1

= 35.3-29.0 + 86(-3)= - 153.

(ii) Prove

= 0.

b + c c + a a + b

a b c

I 1 1

By r x-> l* x + r 2 , the determinant is equal to

a+b+c a+b+c a+b+c

a b c

1 1 1

1 1 1

= (a + b + c) a b c

1 1 1

= since the last determinant has two rows identical

(iii) Evaluate

Direct expansion would be easy, but the following method has the advantage

of giving the result in factorised form.

By c 2 c 2—c l5 followed by c 3

-> c 3—c x , the determinant is equal to

on removing the factor (a + b + c)

from the first row,

1 1 l

X y z

Xsy

% z 3

y-x z —x

z z —x z

= (y-x)(z-x)

Xs y 2 + xy + x 2 z 2 + xz + x 2

(y-x) (z-x)

by removing the factor y —x from column 2, and z —x from column 3.

pandingt from the first row, the only non-zero term is

1 1

y2 + xy + x % z* + xz + x 2



11.25] DETERMINANTS, LINEAR EQUATIONS(iv) Without expanding either determinant, prove

1 a 2 a 3

399

be a a 2

ca b b 2

ab c c 2

1 6 2 6 3

Introducing a factor a in row 1, 6 in row 2, and c in row 3 of the left-hand side,

we have . . .

1 a a ae a a 2

ca b b z

ab c c 2

1

abc

abc

abc1 6 2 6 s

abc a 2 a 3

abc 6 2 6 3

a6c c 2 c 3

on removing the factor abc from the first column of the middle determinant,

(v) 'Triangular' determinants.

a x

o 2 6 2

&3 Ca

6 2

6 3 C8

the product of the elements in the leading diagonal. Similarly, a determinant

whose elements below the leading diagonal are all zero is thus readily evaluated.

Remarks

(a) In expanding a determinant from a given row (or column), the amountof calculation is reduced if row- and column-operations can be used to introduce

one or more zeros into that row (or column), or indeed elsewhere also.

(ft) A determinant having a complete row or column of zeros has the value 0.

Exercise 11(a)

Evaluate the following determinants by direct expansion.

Ill4 3 2

5 6 5h g

-h f-9 ~f

7 Show that

1

4 1

3

2 1

Xi a? a

2/i 2/2

1 1

-2

4

3

x s

cos(w— l)x coanx cos(n+l)a;

3

4

-2a

c

b

2

-3

1

= (1 —2acosa5 + a 2 )sina?.



400 DETERMINANTS, LINEAR EQUATIONS

10 Verify that a + A h g

h 6 + A /

9 f c + A

= \* + (a + b + c)X 2 + (bc + ca + ab-p-g*-h*)\ + {abc + 2fgh-af i -bg i -ch*).

Using the properties of determinants, evaluate the following.

11

14

17

13 5

2 6 10

5 7 9

13 14 15

6 7 8

12 3

a— b a+b a

b—c b+c b

c—a c+a c

12

15

18

6 9 15

16 3

3 12 5

101 19 1

102 20 2

103 20 2

111x x 1

X2 X*

13

16

19

1 1 I

4 3 2

5 6 7

b —c c—a a —bcabb c a

x x 1

1 Xs

20 Prove a b c

cab = {a + b + c)(a 2 + b 2 + c*-bc-ca-ab).

b c a

Deduce an identity by comparing this result with that of no. 6.

21 Prove

22 Prove

1 1 1

a b c

a 2 6 2 c

A EE

b x Cj_

a 2 o 2

= (6 —c) (c —a) (a —6).

a x b x + la x c 1 + mb 1 + na 1

a 2 6 2 + la 2 c t + m&2 + na 2

a 3 6 3 + la s c 3 + m&3 + na 8

(This shows that to each column we may add multiples ofthe preceding columns.)What are the values of

(i) a 1 + Xa 2 +/ia i bj^ + Xb^ + fiba Cj^ + Xc^+flCg

K + vb.

(ii)

b,

c 2 + vc s2 + ra 8

aj + A&i & C +/t&i

a 2 + A6 2 6 2 c 2 +/t& 2



11.3] DETERMINANTS, LINEAR EQUATIONS24 Without expanding, prove that

1 be bc(b + c) a 1 b + c

1 ca ca(c + a) = abc b 1 c + a

1 ah ab(a + b) c 1 a + b

and hence evaluate the first determinant.

401

= (b-c) (c-a){a-b)(x-y).

25 Prove a + x b + x c + x

a+y b+y c+y

a a 6 2 c a

[Use the result of no. 21.]

26 JfA,B,G are the angles of a triangle, prove that

—1 cos C cos Bcos G —1 cos A =0;

cos B cos A —1

and by expanding, obtain a relation between cos A, cosB, cosC. [Usea = b cos C + c cos B, etc.]

a. &8

+2/i7 Prove c^ + a^ 6i + 2/x ^ + 2

a 2 6 2 Cjj

a 3 6 3 c s

^tote <Ais reswft cw a general property.28 (i) If each element of A consists of the sum of two terms, prove that A is

equal to the sum of 8 deteraainants. [See no. 27.]

(ii) If each element of A consists of the sum of three terms, how manydeterminants are there in the expression for A as a sum of determinants?

29 By using no. 28 (i), show that

a 1 + A6 1 fti+ztCj c 1 + va 1 a x 6 X c x

D= a 8 + A6 2 6 2 +/tc 8 c 2 + m2 =(l+\/iv) o 2 6 2 c 2

a 3 + A6 8 &3 +/*c 3 c 3 + va z a 8 6 S c 3

Remark. We may be tempted to say that D is formed from A by the opera-tions c t

-> c x + Ac 2 , c 2 c 2 + /tc 3 , and c 3 c 3 + w^. A choice of A, [i, v such that

Xfiv = —1 makes D = 0, and appears to prove that A = 0; but the compoundmanipulations of columns just described are a misuse of Property (5) and havenot left the value of A unaltered. It is essential in applying Property (5) or its

extensions to leave at least one row or one column unaltered at each step.

11.3 Minors and cofactors




second-order determinant obtained from A by deleting the row a

column which contain that element, and prefixed by the sign + or

In some cases the signs run — + — instead of the standard H

We now introduce a notation which will make all this systematic.

Definitions

(a) The minor of an element of A is the second-order determinant

obtained from A by deleting the row and column in which that elemen

lies. For example, the minors of b x ,a 2 are respectively

a 2 c 2

a z c 3

>

h c 3

(b) The cofactor of the element in the ith. row and jth column

the minor of that element multiplied by (- 1)<+ *. For example,

cofactor of b x , which is the element in the first row, second column,

(_ 1)1+2

that of c z is (-1) 3+3

^3

a 2

a z c 3

a x= +a 2 &2

Cofactors are 'signed minors'.

Notation. We denote the cofactors of a x ,b v c 1? a 2 ,

c 3 in A by

corresponding capital letters A x ,Bv Clf A 2 ,..., Cz . The various expan

sions of A can now be written concisely and uniformly: those from

first, second and third rows are

(i

(i

A = a-Â-^ + b-^B^c^

= a 2 A2 + b 2 B 2 + c 2 C2

= a 3 A s + b s Bz + c z C3

and for columns we have

A = a x A x + a 2 A 2 + a z A z

= b x B x + b 2 B2 + b z Bz

= c x Cx + c 2 C2 + c z Cz .




11.32 Expansion by alien cofactors

Consider the determinant

a 2 b 2

a 9 b 9

a 3 ^3

It has two rows identical, and consequently is zero. By expanding

from the first row according to the definition, we have

c 2 a 2 c 2

+ c 2

«2= « 2 -b 2

c 3 a 3 a z

= a 2 A x + b 2 Bx + c 2 C1 ,

where Ax ,Bx ,

Cx are cofactors in

A = u-2 u 2

a* bo

In this way a total of 12 expressions each equal to zero can be formed by

taking the sum of the elements in any row (or column) each multiplied bythe cofactor of the corresponding element from a different row (or

column). These results, which may be called 'expansions' of A by'alien ' cofactors, are useful when employed in conjunction with those

of 1 1 .3 1 for expansion of A by ' true ' cofactors.

Examples

(i) a h g

A= h b f

9 f c

If we interchange rows and columns, we obtain the same determinant; A is said

to be a symmetric determinant, meaning that it is identical with its transpose.

The cofactors are

A = bc-f 2, B = ac-g 2

, C = ab-h 2,

F = gh-af, G = hf-bg, H =fg —ch.

Owing to the symmetry we shall get only three different expansions by true




This determinant arises frequently in coordinate geometry, and the use o

the above relations often saves much algebraic manipulation. The reader i

invited to verify some of them by direct substitution (see 11.22, ex. (ii)).

(ii) The proof of Property (5) in 11.24 can be expressed more concisely in the

notation for cof actors.

a x + ka 2 6 X + Jcb z c x + Jcc s

CT<j 6 2 Cjj

a Z ^3 C3

= (a 1 + ka 2 ) A x + (6 X + kb %) Bx + (c t + fcc 2 ) Gx

= (a 1 A 1 + b 1 B 1 + c x Cj) + Ha^ + 6 a B x + c 2 Cx )

= A + k.O = A.

No. 27 of Ex. 11 (a) can be done in a similar way.

11.4 Determinants and- linear simultaneous equations

We now return to the subject of systems of linear equations (11.1),

and apply our knowledge of determinants to the problem of solving

them.

11.41 Cramer's rule

Consider again the system

a x x + b x y + c x z = d x ,

a 2 x + b 2 y + c 2 z = d 2 ,

a 3 % + b 3 y + c 3 z = d 3 .

Multiply these equations by the cofactors of a x ,a 2 ,

a 3 in

a x b x c x

A =

3 ^3

and add:

(a x A x + a 2 A 2 + a 3 A3 ) x + (b x A x + b 2 A 2 + b 3 A 3 ) y + (c x A x + c 2 A 2 + c 3 A 3 ) z

= d x A x + d 2 A 2 + d 3 A 3 .

The coefficient of x is A, while by alien cofactors the coefficients of

y, z are zero. Hence




column by the elements d v d 2 , d z appearing on the right-hand sides

of the given equations. We may therefore write

Az = A(1),

where A(1) denotes A with its first column modified as just indicated.

Multiplying the given equations by the cofactors of b x ,b 2 ,

b s in Aand adding gives ^ = A(2)

.

and similarly Az = A(3).

If A =|= 0, the solution of the given equations is unique and isf

A(1) A< 2 > A(3)

x = -— , V = Z =A ' * A ' A'

The case A = will be considered later (11.42).

Example

Solve

Here 2 4 3

A = 3 8 6

8 2 9

= 6x 7

9 7

2x + 4y-Sz = 5,

3x-8y + 6z = 4,

Sx-2y-9z = 12.

2

3

8

= 2 x 3 x

2

-4-1

-12

-3

by c x -^Cj-Cg,followed by c 2

->• c a + 2c„

andA<» =

= 6x49;

54

12

= 12 x

4-8

-2

7

2

12

-36

-9

-2

-1

= 2x3x2x5 22 -2

12 -1

1

-3

= 12x7(6+1) 12 x 49.

A




Similarly

and

A(2) =

A(3) =

2 5-33 4 6

8 12 -9

2 4 5

3-8 4

8 -2 12

= 3 x 49, so y - h

= 2 x 49, so z-

11.42 The case when A = 0: inconsistency and indeterminacy

The proof of Cramer's rule shows that, if the given equations are satisfied

(x,y,z), thenAa; = A(1)

, Ay = A< 2>, Az = A< 3 >.

When A = it follows that, if the system possesses a solution, we must have

A<« = A< 2 > = A< 3 > = 0.

If one or more of the determinants A(1), A(2)

, A(3) is non-zero, there can be

common solution; the equations are inconsistent.

Assuming that the consistency requirements (i) are satisfied, two cases arise.

(1) At least one cof actor in A is non-zero.

Suppose C8 4= 0,f and consider the two equations in two unknowns J x, y:

a^ + ^y = d x—c x z,'

a 2 x + b 2 y = d 2 —c 2z.

Since C3 4= 0, these can be solved for x and y (as in 11.11) in terms of z. Whz is given the particular value z , let the values obtained for x, y be x , y .

We may actually solve (ii) for x and y, and show by direct substitution th

the values x Q , y , z do satisfy the third equation of the system. § More elegantly,

however, we can prove this as follows.

Since (x ,y ,z ) satisfies (ii),

a 1 x + b x y + c x z -d 1 =

and a z x Q + bfl y + c. 2 z —d 2 = 0.

Suppose that a 3 x + b s y + c s z —d a = u.

To show that these values satisfy the third equation of the system we have

prove u = 0.

Multiply the above equations by the cofactors of the elements in the la

column of

A(3) =a x 6 X d x

a 2 6 2 <2 a

a 3 6, dn3

and add:




the coefficients of x and y being zero by 'alien cofactors', and A = byhypothesis. Also C3 4= by hypothesis, and A<3) = by the assumed consistency

of the equations. Therefore u = 0, which proves that the solution of the first

two equations for x, y in terms of z will be also satisfy the third.

Since the value of z is arbitrary, we say that the solution is indeterminatewith one 'degree of choice', or that there are oo 1 solutions.

(2) All cofactors in A are zero, but at least one element of Lis non-zero.

We may suppose a 1 4= 0. Then the first of the given equations can be solved

for x in terms of y and z. When y, z are given the values y , z , let the value

obtained for x be x ; then_ î-fri2/o- c i g o

Xq ,

and so

a 2 x + b 2 y + c 2 z —d 2 =a *(di - &i y - Ci z ) + a i 6 2 Vo + a i c 2 2 o - a i d 2

( a 2 d 1 —a 1 d 2 ) + G3 y —B3 z

a 2 d x —a 1 d 2

by hypothesis. Similarly

a 3 dj —a1 d 3

The given equations will therefore be inconsistent unless

a 1 d x = =a x d x

a a d 2 a 3 d 3

If both conditions are satisfied, the solution of the first equation (with y and z

arbitrary) will satisfy the second and third also. The solution is indeterminate

with two 'degrees of choice'; we may say there are oo 2 solutions.

The results just proved are intuitively evident; for when all cofactors in Aare zero, the coefficients in the left-hand sides of the given equations are

'proportional'. Clearly the equations will not be consistent unless the right-

hand sides are in the same proportion; and then the three equations are equi-

valent to only one.

In 21.62 we shall illustrate the results of this section geometrically.

11.43 Homogeneous linear equations

(1) When the numbers on the right-hand sides are all zero, weobtain the homogeneous system




Theorem I. If the homogeneous system is satisfied by values

x, y, z which are not all zero, then A = 0.

Proof. Proceeding as in 11.41, we find that any set of values (x, y,

which satisfies the given equations must also satisfy

Ax = 0, Ay = 0, Az = 0.

Since x, y, z are not all zero, we must have A = 0.

Corollary 1(a). If A =j= 0, the only solution is x = 0, y = 0, z —

In the particular case z = 1 we obtain the non-homogeneous equations

a x x + b x y + c x= 0,

a 2 x + b z y + c 2 = 0,

a 3 x + b 3 y + c 3 = 0.

If these have a common solution, say (x, y ), then the corresponding

homogeneous equations have the solution (x, y , 1) which is certainly

not all zero, whatever x , y may be. Hence

Cobollary I (b). If the non-homogeneous equations possess a commosolution, then A = 0.

A direct proof of this result is indicated in Ex. 11 (b), no. 14.

Corollary 1(c). If A 4= 0, the non-homogeneous equations are

inconsistent.

For by Corollary I (a), the only solution of the homogeneous system

is (0, 0, 0), so (x, y, 1) can never be a solution for any choice of x, y.

A is sometimes called the eliminant of the non-homogeneous

system; for A = is the necessary and sufficient relation between th

coefficients in order that the three equations in two unknowns x, y cabe satisfied simultaneously (cf. 10.41).

Geometrically, the results of Corollaries 1(6), (c) are associated

with concurrence of lines in a plane (15.42).

As a converse to Theorem I we have

Theorem II. 7/ A = 0, then the homogeneous equations are satisfied

by values of x, y, z not all zero.

h h h A h




(1) Suppose that at least one cof actor in A is non-zero, say Bz=j= 0. Then

x = XAZ , y = AB3 , z = AC3

isa solution not all zero, where A

isnon-zero but

otherwise arbitrary.

Fora x A z +b x B z + c x Cz = (alien cofactors),

a 2 A z + b 2 Bz + c i Cz — (alien cofactors)

and a z A z + b z B z + c z Cz = (A = by hypothesis).

(ii) Suppose all cofactors in A are zero, but that at least one element

of A is non-zero, say c x =# 0. Let x, y have arbitrary non-zero values

x = Ac 1? y = /ic x (so A 4= and /i =# 0).

Choose z to satisfy the first equation:

z = -(Aaj +/£& ).

With these values, the left-hand side of the second equation is

a i Xc 1 + biix& x —c %Xa x —c^pib x

= X{a % c x -a x c^+iL$ 2 c x -b x c^

= XBz -fiA z

=

by hypothesis. Similarly the third equation is satisfied by the above

values.

(2) Two homogeneous equations in three unknowns : solution in ratios.

If Az ,Bz , Cz are not all zero, then the equations

a x x + b x y + c x z = 0, a 2 x + b 2 y + c 2 z =have a non-zero solution x = XAZ , y = AjB 3 , z = AC3 ; this is clear byalien cofactors, or by proceeding as in 11.11. The solution can be

writtenx y_

B«

i.e.




Exercise 11(6)

Solve the following systems of equations by vise of determinants.

1 2x-y+

5z = 2, 2 x-y-z = 2, 3 Zx+

y- 4z = 13,

x + ly-IOz = 1, x-2y-3z = 1, 5x-y + Zz = 5,

a? + f/ + z = 2. Zx —y + 5z = 4. oj + y —z = 3.

*4 If a, 6, c are all different, solve

x + y + z = 1, aa; + 6t/ + cz = (2, a 2a; + b 2 y + c 2 z = d 2

.

[Use Ex. 11(a), no. 21.]

5 Prove that the following system of equations is inconsistent:

Bx + y-z = 0, x-4y + 2z = 9, 4x-Zy + z = l.

[Solve the first two for x, y in terms of z, and substitute in left-hand side

the third.]

6 Show that the solution of the system

x —y —z — 0, 3x + Zy-z = Q, 2x + y-z = 3

is indeterminate, and can be written

x — 1 + 2X, y= 1-A, z = 3A,

where A is arbitrary. [Solve the first two equations for x, y in terms of z.]

7 Show that the solution of

3x + y —z = 0, x —±y + 2z = 0. 4x —3y + z =

can be written x = 2A, y = 7A ; z = 13A.

8 Find the values of A for which the equations

\x + y + j2z = 0,

x + Xy + j2z = 0,

V2« + V22/ + (A-2)z =

have a solution other than x = y = z = 0. Find also the ratios x-.y.z whicorrespond to each of these values of A.

9 Find the values of A for which the equations

3x + Xy = 5, Ax —3y ——4, 3x —y = —l

are consistent. Solve the equations when A has these values.

10 Solve (if possible) the equations

x + y + kz = 4k,

+ k + = 2




13 If the quadratics ax 2 + bx + c = 0, px z + qx + r = 0, lx 2 + mx + n = havea common root, show that

a b c

p q r

I m n

= 0.

[If a is the common root, then the non-homogeneous system of equationsax + by + c = 0, px + qy + r = Q, lx + my + n = has the solution (<x 2 ,a); use11.43, Corollary 1(6).]

14 Find from first principles the condition that the equations

a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0, a 3 x + b 3 y + c 3 =

are satisfied by the same values of x and y. [Assuming a 2 b 3 —a 3 b a 4= 0, solve

the last two for x, y in the form

X -y 1

b 2 c 2 a 2 6 2

b 3 c 3 a 3 c 3 a 3 6 3

and substitute these solutions in the first, getting A = 0.]

15 If the homogeneous system

a 1 x + b 1 y + c 1 z = 0, a 2 z + b 2 y + cji

z = 0, a 3 x + b 3 y + c 3 z =

has a solution other than x = y = z = 0, show by means of no. 14 that A = 0.

[If z 4= 0, the equations can be written a^x/z) + b x(ylz) + c x = 0, etc.]

16 Verify in detail that when A #= 0, the solutions given in 11.41 actually dosatisfy the equations. [The solution is given by

Ax = dî + dzAz + dsAs, Ay = d 1 B 1 + d 2 B 2 + d 3 B3 ,

Az = d 1 C1 + d z Cz + d 3 C3 .

HenceA(a 1 x + b^y + ^z) = d 1 (a 1 J. 1 + 6 1 B1 + c 1 C ,

1 ) + c?2(a 1 ^4 2 + 6 1 j5 2 + c 1 C ,

2 )

+ d 3 {a 1 A 3 + b 1 B3 + c 1 C3 )

= eÂ

by true and alien cofactors.]

11.5 Factorisation of determinants

In 11.25, ex. (iii) and Ex. 11 (a), no. 21 we have examples in which a

determinant is expressed as a product of factors without expanding

it directly. Here we consider various methods of factorising a deter-

minant. Direct expansion should be used only when other methods




If A were expanded, we should obtain a polynomial in a, 6, c which is homogeneous of degree 3 (for each term in the expansion consists of a product

factors taken one from each row and one from each column). We may regard

this polynomial as a quadratic in a whose coefficients are polynomials in 6 and

When a = 6, A = since it has two columns identical. Hence byapplying

the Remainder Theorem to A regarded as a polynomial in a, we see that a —is a factor.

Similarly 6 —c, c —a are factors. Hence

A = k(b —c) (c —a) (a —b),

where k must be numerical because A and (6 —c) (c —a) (a —b) are both poly-

nomials in a, b, c of total degree 3. To obtain the value of k, either comparecoefficients of a particular term, say 6c 2

; or substitute particular values f

a, 6, c, say a = 0, 6 = 1, c = - 1. We find k = + 1, and so

A = (b-c)(c-a)(a~b).

(ii) Considerations of symmetry and skewness.

Factorise 1 1

A= a 2 6 2 c

a 3 6 8 c

As in example (i), the Remainder Theorem shows that b —c,c —a,a —b a

factors. The determinant is a homogeneous polynomial of degree 5 in a, 6,

while the product (6 —c) (c —a) (a —6) is homogeneous of degree 3; hence th

remaining factor P must be a homogeneous polynomial of degree 2 in a, 6, c

Since A and (6 —c) (c —a) (a —6) are both skew functions of a, 6, c, hencemust by symmetric in a, 6, c.

Hence by 10.22, Remark (/?), P must be of the form

k(a z + 6 2 + c 2) + l(bc + ca + ab),

where k, I are numerical.

Clearly the expansion of A contains no term in a 4, while the factorised form

does unless k = 0. Hence

A = l(bc + ca + ab) (6 —c) (c —a) (a —6).

By comparing coefficients (e.g. of 6 2 c 3), or substituting particular values f

a, 6, c, we find I = + 1. Hence

A = (6 —c)(c —a)(a —b) (bc + ca + ab).

(iii) Use of row- and column-operations to make known factors appear

explicitly.

One example has already been given in 11.25, ex. (iii). As another

we factorise A in ex. (ii) above by this method.

The Remainder Theorem indicates the factors 6 —a and c —a. By c 2 c 2 —and removal of 6 —a, followed by c 8 -» c 3 —c x and removal of c —a, we have



= (b-a)(c-a)(b-c)

= (b —a)(c —a) (b —c)

by c x c x —c 2 andremoval of 6 —c,

11.6] DETERMINANTS, LINEAR EQUATIONS1 c + a

b + c+a c 2 + ca + a 2

1 c + a

b + a a 2

= (6 —c) (c —a) (a —6) (6c + ca + a6)

on expanding and rearranging cyclically.

413

by r 2 r 2-° r i

Exercise 11(c)

Factoriae the following determinants.

1

*9

1 1 1

a b c

6c ca a6

111a b c

a 3 6 3 c 8

1 b + c 6 2 + c 2

1 c + a c 2 + a 2

1 a + b a 2 + 6 2

1 1 1

6 4

1

6

*8

a b c

a 2 6 2 c 2

6 + c c+a a + 6

a 6 c

a 6 2

be ca ab

1

6 + c

(6 + c) 2

a 2

(6 + c) 2

6c

1

c + a

(c + a) 2

6 2

(c + a) 2

ca

1

a + 6

(a + 6)2

c 2

(a + 6) 2

ab

a* 6* c*

<SoZ«e the following systems of equations by use of determinants.

10 x + ay + a2

z = az

, 11 ax + by + cz = 1, *12 a?/z + 6za; + ca^/ = 0,

x + by + b 2 z = 6 3, a 2

a; + 6 2?/ + c 2 z = 1, ?/z + zsc + a*/ = ajyz,

* + cy + c 2 z = c 8. a 3

a; + b z y + c 3 z = 1 . a 2 /a; + b z/y + c 2 /z = j9

2.

11.6 Derivative of a determinant

If the elements of A are functions of x, then dAjdx is equal to the sum

of the three determinants obtained from A by deriving one row at a time

d db d



414 DETERMINANTS, LINEAR EQUATIONS [11

Proof. It is convenient to use the notation (mentioned in 11.22)

A = I ±a i bj

c k

for the expansion of A. ThencZA „, da, ., „ dbi . dc k

^ = S±rfx

6^ + S±a ^ C*

+ S±a ' 6^-The first sum is the expansion of the determinant obtained from

by replacing the top row a x , \, c x by dajdx, dbjdx, dcjdx; the othe

two sums are interpreted similarly. This proves the theorem.

A similar result holds with 'rows' replaced by 'columns'.

Example

If three rows of a determinant whose elements are polynomials in x becom

identical when x — a, then (x —a) 2 is a factor of A.

First solution. dAjdx —sum . of three determinants each having two ro

the same as A. Therefore when x — a, dA/dx = because each determinant

the sum has two identical rows. Hence x —a is a factor of dA/dx. Since a

A = when x = a, (x —a) 2 is a factor of A by 10.43.

Second solution. Subtract any one of the rows from the remaining two row

When x = a, each element in each of these new rows vanishes; hence by

Remainder Theorem x —a is a factor of each row, so (x—

a)2 is

afactor of A.


The symbol d x

d 2

d 3

d.

called a, fourth-order determinant, is defined to mean

&2 c 2 a 2 c 2 d 2 a 2 &2 d 2 a 2 c 2

K d 3 a 3 d 3 + <a a 3 h d 3 -d, a 3 c 3

&4 c 4 d, «4 c 4 d, a 4 K d* a 4 h c 4

This is a direct generalisation of the definition of 'third-order determinant'

11.22.fWe mention fourth-order determinants because they arise occasionally

the coordinate geometry of three dimensions. It can be shown that

(i) the properties (l)-(5) in 11.21 continue to hold;




The proofs of these statements are not always easy, and for determinants of

order four or more it is better to approach the subject from a more advancedpoint of view than we can consider here.

Write down the derivative of

Exercise 11 00

1 1 2 3 2 1 1 X

X X2 Xs 1 2 X2

1 2x 3x 2 1 3 X3

log a;

1/x

ljx*

x*

2x

-2

4 If u' = du/dx, etc., prove

u v

dx

ww'

w

u

u'

u'

ww'

5 If u, v, w are polynomials in x whose degrees do not exceed 3, prove that

U V wA = w

w

is also a polynomial in x of degree not greater than 3. [Derive A wo x four times,

rejecting zero determinants at each stage.]

6 Prove

7 If

A =

a 2

XdxdyX2

1 1 1

X y z

X2y

2 z 2

= 2(y-x).

prove3A SA 8A

ox oy oz

[Derive by rows, regarding 8/8x + 8/8y + d/8z as operating on the rows.]

8 If

A = find8A 8A 8A

8x+

8y+

8z'x s y* z A

Evaluate the following fourth-order determinants.

*9 1 5 3 *10 1 2 3 *11 2 3 7 5

2 1 2 1 4 -1 7 8 2 9

3 1 2 -3 6 2 4 3 6 7 4

4 6 1 8 1 2 3 3 9 2

*12 Prove

h




*13 Solve by determinants:

3x + 2/ -5z = o,

X + 2z + Zt —

y + Bz- = o,

x + 2y — z- -2f = 14.

Prove (i) 1 1

1 r

1 1

r 3

1 r 2 r 4=

1 r 3 r* r 5

(ii) 1 1 1 1

1 r

1 r 2 r 4

r 3

r 6= r* (r- l) 3 (r 2

1 r 3 r 6 r 9

Factorise 1 1 1 1

A =x a

x % a 2

6

ft2

c

c 2

a;8 a 3 6 3 c 8

Obtain identities by considering the cofactors of a?, x 2, x in A and the coeffi

cients of Xs, x 2

, x in the product.

Miscellaneous Exercise 11(e)

1 Without expanding the deteraiinant, explain why

=

is the equation of the line joining the points (x u yj, {x it y a ).

Without expanding, prove

2 be c 2 6 2 a c b

2 2 = b b a



DETERMINANTS, LINEAR EQUATIONS 417

Factorise

4 1 be b + c

1 ca c + a

1 ab a + b

a a 2 1+a 8

b 6 2 1 + 6 8

c c 2 1+c 8

1

6 2

(6 + c) 2 (c + a) 2 (a + 6)

&2 c 2 + a 2 d 2 bc + ad 1

c 2 a 2 + &2a

2 ca + fta 1

a 2 6 2 + c 2 d 2 a& + ca 1

8 Show that a and a + 6 + c are factors of

(6 + c) 2 6 2 c a

(c + a) 2 c 2

6 2 (a + 6)2

and hence factorise it completely.

9 By factorising the determinants, prove that

2 a + 6 a 2 + 6 2 1 1 1 1 1

a + b a 2 + 6 2 a 8 + 6 8 a 6 X a 6 c

1 c c 2 1 a 2 6 2 c 2

10 Prove that sin 6, cos 6, sin —cos 6 are factors of

cos d sin 20 cos 2

sin 6 sin 2(9 sin2

6sin sin 2 cos 2 6

and find the remaining factor.

11 Express ap + br aq + bs

cp + dr cq + ds

as the sum of four second-order determinants, and hence show that it is equal to

a b P 1X

c d r 8

(This gives a rule for multiplying two second-order determinants.)

Solve

12

13

x+1

x

1

X X2

b 6 2

2x

Sx-2

x

a 3 — s

a s -6«

1

2x

x

= 0.

= (assuming that a, b, c are non-zero and distinct).




15 Show that g,f, c cannot be found so that the equation

x 2 + y2 + 2gx + 2fy + c =

is satisfied by the values (1, 1), (0, - 2), (2, 4) of (x, y). Interpret geometrically.

16 Show that the equations2x + 3y — 4, 3x + Xy = —1, Ax-2y —c

can be consistent whenever c satisfies 4c 2 —156c —439 > 0.

17 Discuss the equations

a; + 2/ + 3z = 4, a; + 2?/ + 4z = 5. a-y + az = 6

when (i) a = 6 = 2; (ii) a = 1, 6 = 2; (iii) a = 6 = 1.

18 Find the conditions which A, /i must satisfy for the following system

equations to have (i) a unique solution; (ii) no solution; (iii) an infinity

solutions: x + y + z= e, x + 2y + 3z = 10, x + 2y + Az = p.

19 Prove that, for two values of A, the equations

(\ + 2)x + 4y + 3z = 6,

2x + (\ + 9)y + 6z = 12,

3x+12y + (X+10)z = k,

have no solution unless k is suitably chosen. When k is thus chosen, find

general solution for each of the values of A.

20 If/, g, h are functions of x which satisfy the differential equations

dL==f+g + 2h,

d

^ = 2g +2h,

8^=

7/+8sr +24fc,

find all solutions of the form/ = a e Aa, g = b e Xx

, h = ce Xx (where A is independent

of x), giving all the possible values of A and the corresponding ratios of

constants a, b, c.

21 If x y z

W(x,y,z)= x y z

x y z

where x, y, z are functions of t and x = dxjdt, etc., prove that

W(x,y,z) = xa

W(l.ylx,zfx).22 Write

V1 =

a 3 6 3 c 3 a 8

and x + a x + b x + c

f(x)= (x + a) 2 (x + b) 2 (x + c) 2

(x + a) 3 (x + b) 3 (x + c) 3

V

1 1 1 1 1 1 1 1 1

a 2 6 2 c 2 a b c , v 3 = a b c

a 3 6 3 c 3 a 3 b 3 c 3 a 2 b 2 c 2



419

12

SERIES

12.1 The binomial theorem

The reader should already be familiar with the work in this section;

we include it for revision and completeness.

12. 11If

n is a positive integer,

(x + a) n = x n + n Cx x n ~ x a + n C2 x n ~ 2 a 2 + . . + n Cr x n -â r +...+a n,

where n Cr denotes the number of selections ofr different objects from n, viz.

n(n— 1) ... (n —r+ 1)

r\'

Proof. Consider the product (x + a t ) (x + a 2 ) ... (x + a n ). To expand

it we must multiply each term in each bracket by the terms in the

other brackets. This can be done systematically as follows.

If we multiply the x from each bracket, we obtain x n.

If we multiply the a-term from one bracket and the a;-terms from

the other brackets, and do this in all possible ways, we obtain

(a 1 + a 2 +...+a n )x n - 1.

Next, multiplying the a-terms from two brackets and the a>terms

from the rest, and doing this in all possible ways, we obtain

(ax

a2

+ ax

a3

+ . . + a2

a3

+ . ..)x n ~ 2,

where the coefficient of x n ~ 2 consists of the products of different en-

terals taken two at a time. The number of terms in this coefficient is

therefore n G2 .

Similarly, multiplying a-terms from three brackets and ar-terms

from the rest in all possible ways gives

{a x a 2 a z + a 1 a 2 a i + . ..)x n ~ 3,

~



420 series [12.12

Finally, the product obtained by multiplying a-terms from all

brackets is a x a 2 ... a n .

We now put a 1 = a 2 = ... = a n = a. The original product then

becomes (x + a)n

. In the expansion, the coefficients of £cn_1

, xn ~ 2

x n ~ z, x n - r

, ... become

na, n C2 a 2,

n Cz a z,

n Cr a r,

and the constant term becomes a n. This proves the result stated.

12.12 Properties of the binomial expansion

(1) The expansion of (x + a) n consists of n + 1 terms.

(2) The (r + l)th term isn

Cr ar

xn - r

and is called the general term.(3) The coefficients of terms at equal distances from each end o

the expansion are equal.

For the (r+ l)th term from the beginning is n Cr a r x n ~ r, while the

(r+ l)th term from the end is n Cn _ r a n -^x r; and

n C = —— = n C .r r\{n-r)\ w~ r

This fact saves work in numerical calculations.

(4) If n is even, there is a middle term, given by r = \n.

If n is odd, there are two (equal) middle terms given by r = %(n ±1).

(5) We can obtain the expansion of (x + a) n in ascending powers of

x by interchanging x and a in the result of 12. 1 1

{x + a) n = a n + n Cx a n - x x + n Cz a n ~ 2 x 2 + . . + x n.

This is useful in approximations when x is small compared with a.

(6) If we write —a for a in 12.11, we obtain

(x - a) n = x n - n Cx ax 71 - 1 + n C2 a 2 x n ~ 2

+ ( - l) r n Cr a r x n ~ r +... + (- l) n a n.

12.13 Examples

(i) Find the coefficient of x° in the expansion of (2x —Bjx) ls.

The general term is

i3Cr ^_£^

r

(2£c) 18 -' = 18a '( -3)r213 ral8 ~ ar

-



12.13] series 421

where the unwritten terms all contain no power lower than x*. On ignoring all

such terms, we have

32 + 5. 16(x- 3a; 2) + ^ 8x 2 (l - Qx) + ^^4z*+ ...

= 32 + 80* - 160a 8 - 440« 8 + . .

after collecting like terms.

(iii) Find the numerically greatest term in the expansion of(l —3*)' if x = J.

The terms in this expansion are alternately positive and negative, but thenumerically greatest term will be the same as the corresponding term in theexpansion of (1 + 3a;) 7

. If the latter expansion is

u 1 + u i x + u 3 x 2 + ...+u B x 7,

7 / 7then = _ _ _ (toy / (3*)*- 1

u r r (7-r)v

' / (r- 1)1 (8-r)v '

8-r= 3xr

= .- if x = l.r 4

Now u f+1 ^ u r according as u r+1 ju r ^ 1, i.e. according as

3(8 -r)

4r1, i.e. 24- 3r gi 4r, i.e.

3f $r.

Hence if r ^ 3, u r+1 > u r , i.e. w4 > u a , u a > u 2 and u. 2 > u v If r > 4, thenu r+1 < u r . i.e. m5 < u it m6 < w6 , etc. These inequalities show that w4 is greater

than the other terms. Thus the 4th term is the greatest. Its value is

7 Cs(-i) 3 = -35x(f)3.

(iv) Prove n Gr _ Y + n Gr = n+1 Gr .

This can be verified directly, or obtained from the identity

(l + x)(l+x)n

= (l+x)»+\in which the coefficient of x T on the right is n+1 Cr . On the left, terms involving

x r will arise from 1 x n Cr x r and x x C^ x r ~ x, so the total coefficient is n Gr + n GT -x.

The result follows.

When the coefficients n Gr in the expansion of (1 +x) n have been calculated,

the above relation shows how to obtain those in (l+x) n+1. The following

scheme, in which each number is formed by adding the two immediately aboveit, is called Pascal's triangle:

(l+x)° 1



422 SERIES [12.2

Exercise 12(a)

Write out the expansions of

1 (3*-2) 4. 2 (x+l/x) 5

. 3 (l-x){l+x) i.

4 Expand ( 1 + 2x —x 2)

6 in ascending powers as far as the term in x l.

Write down and simplify the

5 coefficient of x s in (2 —a;2

)6

. 6 coefficient of a;10 in ( 1 \x 2 + x) 19

.

7 coefficient of x~ 20 in (x 3 - 1 /2x 2)

1B.

8 term independent of a; in (2a; 2 —1 fx)12

.

9 6thtermin(3£c+l/£c) n. 10 coefficient of x 3 in (2 + x - 3a; 2

)7

.

11 Evaluate correct to 4 places of decimals (i) (1-04) 5;

(ii) (0-98) 5.

12 Find r if the coefficient of x r in ( 1 + x) 20 is twice the coefficient of x f ~ x.

13 Find the greatest term in the expansion of (5 + 4a;) 11 when x = f

.

14 Find the greatest coefficient in the expansion of (5 + 4a;) 11.

15 Obtain an identity by equating coefficients of x r in

(l + 2a; + a;2 )(l+a;) n = (l+a;) + 2

.

16 Consider the coefficient of x r in ( 1 + x) m( 1 + x) n = ( 1 + x) m+n to prove

mCrn C + mCr _ x

n C1 + ...+ mG n Cr = ™+n Cr (r^m,r^n).

17 By considering the coefficients of x n in ( 1 + x) n( 1+ x) n = ( 1 + x) 2n

, prove

c2

+ c2

+ c2

+...+c2

= ^L,where c r denotes n Cr .

18 Prove

(1 +x) n+1 -x n+1 = (l+x) n + x(l + x) n - 1 +...+x r {l +x) n ~ r + ... +x n.

Deduce an identity by equating coefficients of x r.

19 Show that (2 + ^3) 5 + (2 —JS) 5 is rational, and find its value.

20 Prove that

vn(n— 1) 7 n x

n(n—l)(n —2) /

x + n(x + y)+v

t

'

(x + 2y)+ -(x + 3y) + ... to (w+1) terms

= 2 n ~\2x + ny

12.2 Finite series

The reader will already have met arithmetical and geometrical

progressions, and perhaps other simple series. We now extend thi

work.

12.21 Notation and definitions



12.22] series 423

If u r = a r x r, then the series is called a power series in x.

Later we shall write s = lim s n provided this limit exists, and call7l->00

s the sum to infinity of the infinite series T,u r . For example, the sum of

the first n terms of the geometrical progression

1, x, x 2, x 3

, ... (x 4= 1)

l-x n

*» = t^;and if \x\ < 1, then x n -> when n-âo (2.72), so s n -> 1/(1 —a).

12.22 General methods for summing finite series

Given a series, our problem is to find a formula for s n , the sum of

the first n terms. The methods (l)-(3) listed below will be illustrated

in this section.

(1) Derivation or integration of a known finite power series and its

sum-function (followed perhaps by substitution of some particular

value for x).

(2) The difference method. If a function /(r) can be found so that

u r ^f{r+\)-f{r),

then by taking r = n, n —1, n —2, . . ., 2, 1 in turn we have

u n -i =f(n)-f(n-1

)>

«/(»- )-/(» -2),

« 2 =/(3)-/(2),

and %=/(2)-/(l).

By ddi g IX f( + l) f(l)



424 series [12.23

12.23 Series involving binomial coefficients

As our standard form of the binomial theorem we take

(l+x) n = l+ n C1 x + n C2 x 2 +...+ n Cr x r +...+x n .

Used thus, the numbers

1 _ nfj nf] nf) 1 _ nfl

are called binomial coefficients, and are sometimes written

. C). 0' G) (:>

and when the index n is evident from the context and is the same f

all, they are abbreviated to

c 0> c l> c 2» '••» Cn-

Thus (1 + x) n = c + qa; + c 2 a;2 + . . . + c n x n

.

The following methods are useful in dealing with series involving

binomial coefficients:

(a) Express the given series as a combination of two or mor

binomial expansions.

(6) Obtain the given series by derivation or integration of

identity based on the standard binomial expansion (i).

(c) Construct a function in which the given series is the coefficient

of a certain power of x, and evaluate this coefficient in another way.

Examples

(i) Sum c + 2c 1 x + 3c i x 2 + ... + (n+l)c n x n.

Method (a)

The sum = (c + c 1 x + c 2 x 2 + ...+c n x n) + (c 1 x + 2c z x i + ...+nc n x ).

The first bracket = (1 + x) n. The second is

n(n— 1) „ n(n—\){n —2) „nx + 2-±- —-x* + 3 £ -x* + ...+nx n

{ (n-l)(n-2) . ,)= nx\\ + {n-l)x + ^- xt+.^+x - 1)



12.23] series 425

Derive both sides wo x:

c + 2c 1 x + 3c 2 x 2 +... + (n+l)c n x n = (1 +x) n + nx{l+x) n ~ 1,

which gives the same result as before.

(ii) Sum c + 2c 1 + 3c 2 +... + (n+l)c B .

This is not a power series., but it can be obtained by putting x = 1 in ex. (i)

c + 2c 1 + 3c a + ... + (n+l)c B = 2 - 1 (n + 2).

(iii) Sum c + ic x + ic 2 + ... +—̂ — c„.n +

Method (b)

Integrate both sides of identity (i) from to 1

[ c x + fax* + $c t x* +... + - — c n a;n + 1 l

1

= [— j—(l +*)»+ 1 ~|1

,

L n + 1 Jo [n + 1 Jo

i.e. c + £c x + Jc 2 + . .. c n = -j- (2»+» - 1).n+1 n+1

Method (a)

_ n n(n— 1) n(n— 1)...2.1The given sum = 1 + h-^-r —-+...+ — ^ ——6 1.2 1.2.3 1.2...n(n+l)

1 (n+1 (n+l)n (n+l)n(n-l)\=

nTT\ l~ + ~r2~ + 1^3 +

= J_{( l + 1) »+i_l

}n+l1

(2»+ 1 -l).n+1

(iv) Sum 0^ + 0^ + 0^ + ... + c n _ 1 c B .

Method (c). Consider

(c + c 1 a; + c 2 a;2 + ... + c B a B )(c + c 1 :c+...+c B a;

n) = (l+x) n (l + x) n

= (l+a;) 2 .

Coefficient of x n ~ x on the left-hand side is

*o n— i I i n— 2 + • • • + c n _i c = Cq Cj + Cj c 2 + . . . + c B _j c B because c r — <?„_,..

Coefficient of x n_1 on the right is

(2n)

(n-l) (n+l) '

which is the value of the required sum.Ex. 12 (a), no. 17 is another illustration of Method (c).



426 series [12.2

3 Prove c x + 2c 2 x + 3c 8 x 2 + . . . + nc n x n ~ x = n( 1 + x) n_1, and deduce the sum

c 1 + 2c 2 + 3c 3 +...+nc n .

4 Find CJ-2C2 + 3C3 -... + ( -l) n_1 nc n .

5 Prove c + 3c x + 5c 2 + ... + (2w + l)c„ = (n+ 1) 2 n . [Method (a); or wr

backwards and add, as for summing on a.p.; or derive x(l + x 2)

n.]

6 Prove 2 + 3c x o; + 4c 2 « 2 + . . . + (n + 2) c n x n = (1 +«) - 1 {2 + (n + 2)a;}.

7 Prove c c r + CiC r+1 + c 2 c f +2 + . . . +8 Using the identity (l+x) n (l —x) n = (l-a; 2

) , prove that when w is ev

What is the value if w is oddl

9 Prove „ . „ , , „ , , , _ a _ (2n-l)\cl + 2cl + Scl+...+ncl = {(n-1) } 2

10 Sum c c 1 + 2c 1 c 2 + 3c 2 c 3 + ...+nc n _ 1 c n .

11 Find a;2 x z x*

r2 Co+ n Ci F4 Ca°

(w'

and deduce that

Co <h c 2 . , , 1U 2»+ 2 -n-3—+ —+ —+ • ..to(n+l)terms =1.2 2.3 3.4 (n+l)(n + 2)

*12 Prove

Ci_

K_ ...

+ (

_ i)n-i Cn = A +J+

i. + ... + I.

[c x + c 2 x + . . . + c n x n ~ x = {(l+x) n - l}/x; integrate from - 1 to 0.]

13 (i) Find the sum of the coefficients in the expansion of (1 + 3a;) 6. [Writ

(l + 3#) 6 = a + a 1 x + ...+a 6 x 6, and put a? = 1.]

(ii) Calculate a + a 2 + a 4 + a 6 and a i + a z + a s- LSee no - 2

*14 Let (l + a; + £t;2

)

n = a + a 1 x + a 2 x 2 + ...+a 2n x 2n.

(i) Deduce other expansions by writing ljx, —a; in place of x, and pro

a r = a 2n -r-

(ii) Deduce theexpansion of ( 1

+x 2

+# 4

)

n.

(iii) Use the identity ( 1+ x 2 + x*) = ( 1 + x + x 2) ( 1 - x + x 2

) to prove that

al-a 2 + a\-...+a 2n = a n .

(iv) Calculate a + a 3 + ...+ a 2n -i«

( v) Calculate a + 2a x + 3a 2 +... + ( 2n + 1 ) a 2n .

12.24 Powers of integersn



12.24] series 427

n(2) 2> 2

- First method. From the identityr=l

(2r + 1)3 - (2r - l) 3 s 24r 2 + 2,

we have by putting r = 1 , 2, . ..

, w and adding

(2rc+l) 3 -l 3 = 24f> 2 + 2w,r=l

/. 24 S r 2 = Sn z + 12n 2 + 4n,r=l

and 2> 2 = + 1) (2n + 1).

r=l

Second method. Using

(r+l) 3 -r 3 s 3r 2 + 3r+l

and summing for r = 1, 2, . . ., n, we have

(rc+l) 3 -l 3 = 32r 2 + 3£r + rc.

r=l r=l

From (1) this becomes

n 3 + 3n 2 + 3n = 3£r 2 +fw(tt + l) + n,r=l

which leads to the result just found.

(3) Sr 8. We have

r=1+ 1)}

2 - {(r - 1) r} 2 = 4r 3.

Summing this for r = 1, 2, . . ., n, we get

{n(n+l)} 2 -0 2 = 42^,r=l

so S»* = i»a (»+l) a

.

r=ln

Observe that this result is the square of 2r=l

The sum could also be obtained by using the expansion of



428 SERIES [12.2

12.25 'Factor' series

(i) Sum 1.2 + 2. 3 + 3. 4 + ...+w(n+l).

The general term is u r = r(r+ 1). Consider

r(r+l)(r + 2)-(r-l)r(r+l) = r(r + l){(r + 2)-(r- 1)}

= 3r(r+l) = Zu r .

This is a suitable difference relation, and by summing for r — 1, 2, ...,n,

32r(r+l) = w(ra+l)(n + 2)-0,r=l

i.e. £ r(r+l) = £n(n+ 1) (w + 2).

r=l

(ii) Sumton terms 1.2.3 + 2.3.4 + 3.4.5 + ....

The general term isf wr = r(r + 1) (r + 2). Consider

r(r + 1 ) (r + 2) (r + 3 ) - (r - 1 ) r(r + 1 ) (r + 2)

= r(r+l)(r + 2){(r + 3)-(r-l)}

= 4r(r+l)(r + 2) = 4w r .

nHence 4 2 r(r+ 1) (r + 2) = n(w+l)(w + 2)(ra + 3)-0,

r=l

i.e. |; r ( r+ i)( r + 2) = in(n+l)(w + 2)(w + 3).

r=l

The same method can be used for any series in which the terms consist

the same number of factors and the first factors in each term form an a

having the same common difference as the successive factors in each term

To obtain a difference function f(r +1), insert an extra factor at the end of

rth term. Observe in exs. (i), (ii) how the sum can be written down fromform of the general term.

Other series may be reducible to this type.

(iii) Sum 1 .2.3 + 2.3. 5 + 3.4.7 + ... to n terms.

u r = r(r+l)(2r+l)

= r(r+l){2(r + 2)-3}

= 2r(r+l)(r + 2)-3r(r+l).

/. 2 ** r = 22 r(r+l)(r + 2)-32 r(r+l)r=l r=l r—1



12.26] series 429

n n n n

r=l r=l r=l r=l

= 2 x £n 2 (n + 1)2 + 3 x \n{n + 1) (2n + 1) + |n(n + 1)

by the results in 12.24, and this reduces to the expression just found.The second method is inconvenient if u T is of degree higher than 3 in r, unless

nwe know formulae for 2 f

-4* etc.

r=l

12.26 'Fraction' series

(1) Examples.

o illi) tfwm —+ —+ —+... tow terms.

1 . ^5 Z . o o . 4

1 1 1

r(r+l) r r+1by partial fractions, and this decomposition is of the form f(r) —f(r + 1). Thedifference method is therefore applicable, so that on putting r— 1,2, ...,n

and adding, 11 ny\u t = = —

.

r=i 1 n+1 n+1

Sum T^ + ^ + z^ +-

tontervm -

l

u r = r(r+l)(r + 2)

Partial fractions are less suitable here because their use would lead to three

fractions. We may decompose u r into two fractions by omitting the last factor:

1 l= {r + 2)-r

r(r+l) (r+l)(r + 2) r(r+l)(r + 2)*11 n

Hence = 2 T\ u„1.2 (n + l)(n + 2)

f

and the required sum is .

H 4 2(n+l)(n + 2)

The terms of the series just considered are the reciprocals of those illustrated

in 12.25. The same method as in ex. (ii) here can be used to obtain a difference

function f(r), viz. write down u r and omit the last factor in the denominator.This gives u r in the form /(r) —f(r + 1 ) which, although not exactly the difference

considered in 12.22 (2), does enable the method to be used.

Other series may be reducible to the above.



430 series [12.26

Since 1 1 _ _ 1 1

rT3~r + 4 ~ Vr' ^ r ~~4~n + l'

a- 1 1 2Since = = 2w r ,

(r + 2)(r + 3) (r + 3)(r + 4) ( r + 2)(r + 3)(r + 4)

n 1 ihence 2Y.w r = .

r _l 3.4 (n + 3)(n + 4)

After reduction, we find

V _ v _ V 5 2n + 5

riP'r-l' r=l^ 24. 2(n + 3)(n + 4)'

(2) Direct use of partial fractions. This method will sum all ' fraction

series summable by the difference method, although often less easily.

Example

(iv) Sum T^ +2̂ r5 + si^ + '- t0ntermS -

2r+l ABC- + —T + -

r r(r+l)(r + 3) r r+1 r + 3'

and by the usual method (4.62) we find A = J, B = C = —f.

2 3 5

6w r =-

H ,r r+1 r + 3

andn n i w 1 n 1

62*. = 22- + 32 --5Sr=i r-ir r =ir+l r rir+3

n 1 n+1 l n+3= 22- + 3S i -5S -

r=l*' r =2* r=i r

/ n 1\ / n 1 1 \

*( , + * + * + ^r) + »(* + * + Sr +S+i)

/ » 1 1 1 1 \

\ r = 4 r + n+l + n + 2 + n + 3/

3 5 5 52(l + i + l) + 3(i + i) + -— — — —

n+1 n+1 n+2 n+337 2 5 5

n+1 n+2 n+3» 37 1ZuU r — —- —-

r=1r

36 3(n+l) 6(n + 2) 6(n + 3)

To sum this series by the method of 12.25 we should write u T as



12.27] series 431

Thus 2 3 2r

(r + 2)(r + 3) (r+ 1) (r + 2) (r + 3) r(r+ 1) (r + 2) (r + 3)'

and each fraction can be dealt with separately.

Success of the 'partial fraction' method in this example depends on the fact

that the sum of the numerators of the three partial fractions is zero, so that then J

major part 2 - °f tne sum cancels out. This will always be the case when ther=4 r

degree of the numerator of u r is at least two lower than that of the denominator.

(3) Sum to infinity. The series in 12.24, 12.25 clearly have no

limiting sum, but those in exs. (i)-(iv) above do possess such a sumto infinity (12.21), viz.

(i) i, (ii) i (iii) a, (iv) ft.

12.27 Some trigonometric series

(1) Series of sines or cosines of angles in a.p.

(i) C = cosa + cos(a+/tf) + cos(a + 2yff) + ...+cos{a + (n-l)/ff}.

Multiply both sides by 2 sin then since

2 sin £/?tt r = 2 sin %/3 cos {a + (r - 1) ft}

= sin {a + (r - J) fl} - sin {a + (r - f ) 0}

by using one of the formulae for products into sums, and this expression is of

the form/(r + 1) —f(r), we have by summing for r — 1, 2, . . ., n:

2 sin \p G = sin {a + (n - £) fi}- sin {a -

= 2cos{a + £(w-l)/?}sin£n/?

by converting the difference into a product.

sin-JnyffC = coa{* + $(n-l)/3}

sin£/ff

(ii) S = sina+sin(a+yff)+sin(a + 2/

ff) + ...+sin{a + (w-l)^}.

The result S = sin{a + £(n- l)yff}Sm

^fsinf/J

can be obtained in various ways:(a) Put a —£7r for a in ex. (i). (This shows that the multiplier 2sin£/ff for

finding G must also suit S.)

(b) Multiply both sides by 2sin£/? (the same factor as for G).

( ) Derive the result of ex (i) wo



432 series [12.27

(2) Other useful differences. Results like

tan0 = cot0-2cot20,

cosec 26 = cot 6 —cot 26,

tan 6 sec 26 = tan 26 —tan 6

are easily verified, and can be used to sum suitable trigonometric

series.

Example

(xv) Sum tan - sec + tan - sec - + tan - sec - + ...ton terms.2 4 2 8 4

6wr = tan-sec —.

If we write 0/2 r for 6 in the last difference just mentioned, we see that

6 6u. = tan - —- —tan —

.

r2 r_1 2 r

n Q:. 2 u r = tan 6 —tan —

.

r =1 2

The sum to infinity is tan d.

Exercise 12(c)

Find a formula for each of the following sums.

1 l 3 + 2 3 + ... + (2n + l) 3. 2 l 2 + 3 2 + 5 2 + ... + (2w-l) 2

.

3 l 2 —2 2 + 3 2 —4 2 + ... to n terms if n is (i) even; (ii) odd. Give a formulawhich includes both cases.

4 Prove that the sum of the products in pairs (without repetitions) of tfirst n integers is -^n(w 2 —1) (Bn + 2). [If s n is the required sum, then

(n \2 n

Z') =2*„+2>2

.]

r=l / r=l

5 Expand (r + l) 4 —r 4. Using the formulae for Sr, 2r 2

, deduce 2 rS-

r=ln

6 (i) By writing r 2 = r(r+ 1) —r, obtain 2 r2 by using 12.25.r=l

n(ii) Writing r 8 = r(r 2 —l)+r = (r— l)r(r+ 1) +r, obtain 2 f 3

-

r=l2/i

7 Calculate 2 * (*•+ 1).l



12.28] series 433

Sum the following series to n terms, and find the sum to infinity when it exists.

10 l a .2 + 2 2 .3 + 3 2 .4 + ....

11 3.5.7 + 5.7.9 + 7.9.11 + .... [Method of 12.25.]

I s l 3 + 2 3 l 8 + 2 8 + 3 8

12 1.4.7 + 2.5.8 + 3.6.9 + .... 13 - + —̂—+ +....

1 1 1 1 1 114 —+ —+ ^^ + --- 15 ir-^ + ^r^+-,

1.4 4.7 7.10 3.5.7 5.7.9 7.9.11

„ „ 2r+l 1 4 716 2 . 17 + 1- h....

r(r+l)(r + 2) 3.5.7 5.7.9 7.9.11

Use partial fractions to find the following sums, and also the sum to infinity,

n i n r +\18 S 7~T~o\* 19 Sr f i r(r + 2) r ti r(r + 2) (r + 3)

>S«m the following to n terms.

20 cos sin 20 + cos 20 sin 30 + cos 30 sin 40 + ...

[cos rd sin (r + 1) = £{sin (2r + 1) + sin 0}.]

21 cos 2 + cos 2 20 + cos 2 30 + . . .. [cos 2 = 4 + £ cos 20.]

22 cosec 20 + cosec 40 + cosec 80 + . . .

23 tan0 + 2tan20 + 4tan40 + 8tan80+....

24 sec 2 0+4sec 2 20+16sec 2 40 + 64sec a 80+.... [Use no. 23.] .

25 sec sec 20 + sec 20 sec 30 + sec 30 sec 40 + .... [tan(r+ l)0-tanr0 = ....]

26 Prove that tan -1 (r + 1 ) —tan -1 r = cot -1( 1 + r + 2

) , and hence sum

cot 1 3 + cot- 1 7 + cot 1 13 + . . . + cot- 1( 1 + n + ra a

).

What is the sum to infinity?

12.28 Mathematical Induction

This is a general principle for proving a given statement whichinvolves a positive integer n.

LetTC

be such a statement. For example,<f> n might be 'n(n+ 1) is

always divisible by 2 or 'the sum to n terms ofl.l + 2.2 + 3.3 + ...

is (n+l)\-V.If by assuming the truth of <j) k we can prove the truth of

<f> k+1

(i.e. if the statement for any particular integer always implies the

corresponding statement for the next integer), and if <$> x is known to



434 SERIES [12.28

Examplesn

(i) Prove 2 r% —Wn + 1) (2n + 1).

r=l

Suppose that, for some integer k,

k£r 2 = i*(Jfc+l)(2Jfe+l>,

r=l

This is the induction hypothesis.

fc+ 1 fc

Then £ ^ 2 = > 2 + (&+l) 2

r=l r=l

= + l)(2fc+l) + (&+l) 2 by the induction hypothesis,

= £(fc+l){fc(2fc+l) + 6(fc+l)}

= £(fc+l){2fc 2 + 7fc + 6}

= $(&+l)(fc + 2)(2A; + 3),

which is similar to the induction hypothesis, but with k + 1 instead of k. Hencif the result holds for n = k, then it also holds for n = k + 1.

When n = 1, the result is true because the left-hand side is l 2 = 1 and thright-hand side is $ . 1 . 2 . 3 = 1.

Hence by Mathematical Induction the statement is true for all positive

integers n.

(ii) Prove that 3 2n + 7 is always divisible by 8.

Write f(n) — 3 2n + 7, and consider

f(k+l)-f(k) = (32 *+ 2 + 7)-(3 2 * + 7)

— 32fc+2_ 32*

= 3 2 *(3 2 -l) = 8.3 2 *.

Iff(k) is divisible by 8, this relation shows that f(k+ 1) is also divisible byAlso/(l) = 3 2 + 7 = 16, which is in fact divisible by 8. Hence the result follows

by Induction.

(iii) Prove the binomial theorem by Induction.

Suppose the theorem holds for some particular value n = k:

{x + a) k = x k + k C1 x k - 1 a + ... + k Gr x k - f a r +...+a h.

Then in the expansion of (x + a) k+1 = (x + a) (x + a) k, the coefficient

a-fc-r+iig( forr _ 1,2,...,^;)

k Cr a r + k Cr _ 1 a r = k+1 Cr a r

by 12.13, ex. (iv). This expression is also correct when r = or k+ 1. Hence



12.28] series 435

Remark. Mathematical Induction applied to the summation of a

series is equivalent to the difference method. For to prove that, for

all positive integers n,^ + ^ + + ^ _

mwe assume that, for some k,

+ ...+U k =f(k),and then show that

u x + Uz+ ...+u k+1 =f(k+l).

Since u 1 + u 2 + ...+u k+1 = (u 1 + u 2 + ...+u k ) + u k+1 =f(k) + u k+1 ,

this is equivalent to showing that

«*« =/(*+ )-/(*).

so that/(r) is a suitable difference function.

Exercise 12(d)

Prove the following results by Mathematical Induction.

1 1 + 3 + 5 + .. . + (2n-l) = n 2.

2 1.1 + 2. 2 + 3. 3 + . ..ton terms = (n+l) -l.3 l 2 + 4 2 + 7 2 + ... + (3n-2) 2 = %n(Gn 2 -3n- 1).

, 8 12 16 , . mx , 14 ——-——+-—-... to n terms = ^ + ( —1)

n_1

3.5 5.7 7.93 v

2ra + 3

5 (i) n(n + 1) (w + 2) is divisible by 6; (ii) n 3 + 2n is divisible by 3.

6 (i) 9 - 1 is divisible by 8; (ii) 9 n - 8n - 1 is divisible by 64.

7 a?2 -1 + ?/2n-i jg divisible by x + y.

8 If 2m x = a + 6, 2u 2 = b + u 1 , and 2u n+2 = u n +u n+1 (w ^ 1), prove

3w„ = a{l - ( - m+ 6{2 + ( - £)»},

and find limw„.n->oo

9 Prove

cos sec + cos 20 sec 2 + . . . + cos n0 sec =s

|

n^

1 * ^_ j

sin 6/ cos



436 SERIES [12.

12.3 Infinite series

12.31 Behaviour of an infinite series

n(1) Convergence. If s n = 2 u r an( * um exists and has the value

r=l n-><»

we defined s to be the s«m to infinity of the infinite series T,u r (12.21).

We also say that the series Hu r converges (or is convergent) to s, anwrite

00

s = y£u rr=l

or 5 = ^ + ^2 + %+....

It will be clear from the definition that s is not a 'sum' in th

ordinary sense of 'the result of adding a number of terms', but is th

limit of such a sum. If we attempt to treat infinite series like finite

ones (i.e. without consideration of convergence), then paradoxical

results may arise.

For example, suppose x > and consider the series

1 « 1 o 1

l+x +-

+ x* + —+xa

+ - + ....X Xz XT

One is tempted to regard this as the sum of two infinite g.p.'s

„« ,111l+x + x 2 + x*+... and - + - + —+ ...,X Xs X3

whose sums to infinity are (see end of 12.21)

1and

llX 1

1 —x 1 —1/x x—\'

which add up to zero. We thus appear to have a series of positive terms whosum to infinity is zero. The explanation is simply that the first g.p. converges(i.e. possesses a sum to infinity) only when \x\ < 1, while the second convergesonly when

|

l/x\ < 1, i.e. \x\ > 1; there is no value of # for which both converge,and hence the original series has no sum to infinity.

Another example is

i+£+i+i+i+*+-= (i+i+£+ -) + £(i

from which + = +



12.32J series 437

(2) The infinite g.p. 1 + x + x 2 + x z + ....

If x =f= 1, then s n = (1 -x n )/(l -x). Hence by 2.72

(i) if —1 < x < 1, then x n -> when n->oo, and so sTC

-> 1/(1 —

(ii) if x>

1, then x n -» oo and so sw

->+

oo; we say $n

is properly

divergent to +oo;

(iii) if a; < —1, then # n -> + oo when n is even and x n -+ - co when

% is odd : s n oscillates infinitely;

(iv) if x = —1, the series is 1 —1 + 1 —1 + ..., so that s n = if n

is even and s n = 1 if n is odd: sTC

oscillates finitely (between and 1).

If x = 1, the formula for s n is meaningless, but the series becomes

1 + 1 + 1 + . . ., so that s n = n and s n -> + oo when n->oo:s n is properly

divergent to+

oo.

(3) Divergence. In this book we shall not give a detailed study of the

behaviour of infinite series, and it will usually be sufficient if weemploy the term 'divergent' to mean 'not convergent'. Thus our

'divergence' will cover proper divergence to +oo, proper divergence

to —oo, finite oscillation and infinite oscillation.

We abbreviate 'S« r is convergent' totyLu r c'; similarly, '£w r d'

means ( Hu r is divergent (i.e. not convergent)'.

12.32 General properties

(1) IfZu r converges, thenMmu n = 0.n—><x>

For by hypothesis s n -+s when n -> oo; and since u n = s n —s n _ vtherefore u n -> s —s = when n->oo.

(2) The converse of this result is false : if u n -> 0, the series 1>u r maynot converge.

Example

2(l/r) is divergent.

We have i + i > i + i =

i+£+*+i>£+i+£+i = h

i+^+..-+^>i% = i, etc.,



438 series [12.32

Given a positive integer n>\, a positive integer p can be found so tha

2 J '- 1 < n < 2 P, and therefore (since all terms of the series are positive)

S^p-i < *n ^ sa*»-

When n -> 00, also 2p

00, i.e. p ^ cc; hence s n -> cx> because Sg*- 1 °o« Thshows that the series uiiuu.sometimes called the harmonic series, is (properly) divergent (to + 00), althoughu n — 1/n -> when n -> 00.

Remark. We have shown that %n -> does not imply convergence

of T,u r ; but */ u n -+> wAew -> 00, 2w r 6e divergent (otherwise (

would be contradicted).

Thus Sr/(r + 1) d because ww

= n\{n + 1) -> 1 when «- -> 00.

(3) Convergence or divergence of a series is unaltered by

(a) removing a finite number of terms from the series;

(b) multiplying every term by a non-zero constant.

These facts are clear from the properties of limits (2.3). Property

(a) is useful when the first few terms of a series behave irregularly,

and also shows that any test for convergence which we shall give need

apply only 'for all n ^ m\ where m is a fixed number; i.e. from som

definite term of the series onwards.

(4) If1iU r ,Sv r both converge, to sums s, t respectively, then T,(au r + bv

converges to as + bt, a and b being constants.

For if _ » _ »

r=l r=ln

then 2 (a>u r + bv r ) = as n + bt nr=l

-> as + bt when n -> 00, by 2.3, (ii).

Similarly, if one of 2v r converges and the other diverges, then

H(au r + bv r ) diverges. However, if both diverge, T,(au r + bv r ) mapossibly converge, e.g. if u r = r + 2~ r

, v r — r, a = 1, b = —1; roughly

speaking, the divergent parts may cancel out.

(5) If s n steadily increases when n increases, then T,u r either converges



12.4] series 439

The behaviour of the series considered in 12.2 can be determined

directly because a formula for s n can be found; cf. 12.26 (3). Whenno such formula is known, we resort to tests for convergence or

divergence. Although these may showthat the series converges, they

do not help us to evaluate its sum to infinity.

We recall that (2.71) the statement 'lims n = s' means that if any»->co

positive number e however small is given, then a number m (in general

depending on e) can be found such that, for all n ^ m,

s —e < s n < s + e.

12.4 Series of positive termsWe suppose all terms of our series are positive in this section. Series

whose terms are all negative can be included by first removing the

factor - 1 (see 12.32 (3) (6)).

12.41 Comparison tests

In these we compare the given series 2w r with a series T>v r whose

behaviour (c or d) is known. Roughly, a series which is less term by

term than a convergent series is also convergent, and one greater than

a divergent series term by term is divergent.

(1) Test for convergence.

If 2iV r c, and

either (a) u n ^ cv n for all n ^ m, where c is a positive constant,

or (6) lim {ujvj = 1^0,n—>oo

then also lLu r c.00

Proof of (a). If 2 v r = t, then since all the terms v r are positive,r=l

v 1 + v 2 + ... +v n < t for all n.

By hypothesis,

Wm + Um+1+---+ Un < C{v m + Vm+1 +...+V n )

< c(v 1 +...+v m +...+v n )



440 series [12.4

Proof of (b). Given e > however small, there is a number m suthat u

l —e < —< l + e for all n ^ m.v n

The right-hand inequality shows that

u n < (1 + e) v n for all n ^ m.

Taking c = l + e in (a), the result follows.

(2) Test for (proper) divergence.

If 1>v r i>, and

either (a) u n ^ cv n for all n ^ m, where c is a positive constant,

or (b)lim (ujv n ) = I > (but not I = 0),fn-*-<x>

then 1>u r d.

Proof of (a). Since T>v r d, then (see 12.32 (5))

v m + v m+1 + . . . + v n oo when n -» oo.

By hypothesis,

«fn+^m+i + •••+«» > c(» m + t; m+1 +...+f;ft )

-> oo when n -> oo.

.*. s n = ^ + . . . + u m + . . . + u n -> oo when n->co,

i.e. Sw r d.

Proo/ of (b). From the left-hand inequality in the proof of (1) (b),

u n > (l —e)v n for all n ^ m.

If Z > 0, then Z —e > for all positive e sufficiently small; we mtake c = I —e in (a), and the result follows.

(3)Standard

comparison series are(i) the g.p. Zaf, which c if < x < 1 and d if x ^ 1 (12.31 (2));

(ii) the harmonic series S(l/r) which d (12.32 (2)).

Another is E(l/r p); for various values of p this gives a whole family

of series, which we now consider,

(hi) X(l/r p) c if p > 1 and d if p ^ 1.

/ X 111 1 , .W 5 « =i^

+ 2^ + 3^ + ••• +^» there bein g n terms,



12.41] series 441

the inequality is clear since the first, second, third, . . . terms in s n have

been replaced by brackets containing 1, 2, 4, . . . positive terms. Making

an upper estimate for the contents of each bracket, we have whenp > 0:

*n < Jt+ (p+p) + (&+ - + p) + - (» brackets)

1 2 4 8, x=

I?+ p +

4^ +8^

+ - (n terms)

= 1 +2^1+ (2î)

2

+ (g^l)3+ - <» terms >

={l

- (î)

71

} /{* ~ 2^}0n summin

gthe G ' p

1

l-(i) 2 '- 1for all n if > 1,

since then the common ratio (£)p_1

is less than 1.

Hence when p > 1, s n is bounded; so by 12.32 (5) the series c.

(b) The case p = 1 gives the harmonic series already investigated

in 12.32 (2).

(c) If p < 1, then mP < n when n > 1, so that IjnP > 1/n. Henceby comparison with S(l/r), S(l/r p

) d whenp < 1.

Alternatively, if0<p< 1 then (without referring to the harmonic

series)

11 1 1 ,

s„ = - —h- —h ... H > nx —= -> 00 when w -> co.n IP 2 p n v n p

If p < 0, then clearly s n -» 00. Hence S(l/r p) d when p < 1.

Examples

(i) Examine the series

5 7 9

Here

1.3.4 2.4.5 3.5.6

2n+3n(n + 2)(n + 3)

[When n is large, « n = 2n/(n.n.n) = 2/n 2; so for large n, «„ behaves like 2/n 2

.

It i therefore likely that Ew c and we try the comparison test for convergence ]



442 series [12.41

Alternatively, using the 'limit' form of the test, and taking v n = l/w a

nôo v n n-yoo \n(n + 2) (n + 3)/ n 2/

before,

i

Hence Z« B c because- Jjv n c.

„ . 1 1 + 2 1 + 2 + 3(n) Mixamme —-\ —H \-....x

'1

2 8 3 3

_ l + 2 + ...+w _ £n(n+l) _ n+1n w3 w8 2n 2

[When n is large, wn = n/2n 2 = l/2n, so that probably 2w„ d.]

We have ,n ltt - > 2^

=&i

f ° rallM '

so if v,, = 1/n, then u n > %v n . Therefore 2w n diverges with S« n .

^ftemotM* ^ /n+l/l\,hm —= lim I——- /-) = \,

TC_oo w» n->oo \ 2n 2/ n/

and the same result follows.

Exercise 12(e)

Show that the following series do not converge because u n -+> 0.

i f+f+ H.... 2 l±f -l^ + l^?.a; a;2

a;3

3 S1

V(r+1)-Vr*

Using the result of 12.41 (3) (iii), state which of the following series converge.

6 » + ++*+.... 7 -1+- -+-^+....

Use the comparison test to ascertain the behaviour of the following series.

s i+i+^+^+^+.-. 9 ^ + ^ +jJL+....

1 2 3 4 5 610 1 1 h-... 11 1 1 K...

2 aT

3 2 4 2 1.2.3 2.3.4 3.4.6

n 13 S „*

. 14 s£tl.r+1 V( r 2 -2r) 3 r +l



12.42] series 443

1 8 If Sw r c, prove that Ew a and 2{w r /( 1 + u r )} also c. [Use the comparison test.]

19 IfZw*c, prove Z(w r /r) also c. r(w„--\ ^ 0, therefore 2—r

|_\ ?*/ w J

11,0 (i) Verify that - < — when r > 2.

r 2 (r— l)r

» 1 w 1 1Deduce that s n = £ - < 1 + £ 7 —= 2 —< 2.

r=lf a r=2 (r-l)r n

(This shows that 2(l/r 2)

converges and has sum to infinity a < 2.)1111,(ii) Prove —< < — when r ^ 1,v

' (r+1) 2 r r+1 r*

and deduce that

1 1 1 » 1 „ 1 111 + 2i+--- +

(Î)i+

»<

r S^<1 +

25+ --- +

^-(This shows that l/(n+ 1) < s-s n < 1/n. Taking n = 100, we see that the first

hundred terms give s correct to only two decimal places. It can be proved that

8 = ^n 2: see 14.34, ex. (ii).)

*21 If Sw r c, prove u n+1 + u n+2 + . . . + u n+v -> when n -+ 00, where p is any

fixed positive integer.

12.42 d'Alembert's ratio test

The tests in 12.41 required an auxiliary series Ev r ; that now to

be stated involves only the given series itself.

1iU r is convergent if

either (a) u n+1 /u n < k < 1 for all n ^ m, where k is constant,

or (b) lim (u n+1 ju n ) = Z < 1.

n-*-ao

HtU r is (properly) divergent if

either (a') u n+1 ju n ^ 1 for all n ^ m,

or (&') lim (u n+1 /u n ) = 1>1.n->ao

Proofs

(a) By hypothesis,

u m+l ^ k u m>

u m+z < ku m+1 s$ k 2 u m from the previous line;

u m+3 ^ k u m+2 ^ from the previous line;



444 series [12.4

which converges since < k < 1 by hypothesis. Therefore Sw r co

verges, since the first m terms u 1 + u 2 + ... +u m do not affect

behaviour.

(b)Given

e> 0,

there isa number m such that

l-e < < l + e for all n ^ m.

Since I < 1, we have l + e < 1 for all e sufficiently small. Taking

k = I + e in the right-hand part of the inequality, the result follow

from (a).

(a') By hypothesis,

um+l ^

um>

um+2 ^

um+l ^

um>

e ^ c -

Hence for n ^ m we have u n ^ u m , so that i* n -I* when n ->

proving non-convergence of The series.must actually be properly

divergent since it consists of positive terms (12.32 (5)).

(&') Since I > 1, we have Z —e > 1 for all e sufficiently small. Tleft-hand part of the inequality in the proof of (6) then shows

U^±1>1 for all n > m,

and the result follows by (a').

Remarks

(a) The complete condition (a) is essential : the result may be fals

if we merely have^±1<1 for all n > m.

Thus although E( l/r) d,

* /i <1 foralln.u n n+l/ n n+1

No fixed k independent of n can be found for which

î- 1 < k and k<l

because n/(n + 1) can be made as close to 1 as we please by taking



12.42] series 445

when the 'limit' form is inconclusive. For a power series it is usually

best to treat this exceptional case from first principles.

{y) No conclusion can be drawn from non-existence of the limit of

u n+il u n- Thus for the seriesf

1 1 i i 1 12

+42

+2 3 +

4 4 +2 B +

4«+ * ''

successive values of u n+1 /u n are

Jl 9 _L 93 JL 95 I 9723' » 2 5 ' 2 7 ' 2 9 '

* '

so that un+1

jun

oscillates infinitely. The series is nevertheless con-

vergent (because, roughly, it is the sum of two convergent q.p.'s):

I 1 1 1 1 12 ^42 ^2^^ 4* ^ ^

2 2n_1 i 2 ™

a-royi-an2 2 4 2

->f+T£ when n -> co;

andj ^+1 = ^ + 22^1^1^+° when ™~>°0-

Hence the series converges, and has sum to infinity Observe also

that for tljis series u n does not steadily decrease to zero when n -> 00

Example

Examine the series 1 + 2x + Zx 2 + 4a; 3 + . . . for all positive x.

Ifx = 0, the series is 1 + + + . . ., which converges to the sum 1.

If x 4= 0, then since u n = nx n ~ x,

w„+i n+1= x -> x when n -> 00.u n n



446 series [12.43

When x = 1 , the test gives no information; but then the series is 1 + 2 + 3 +which clearly d.

Hence the series c when < x < 1, d when x ^ 1. (See Ex. 12 (/), no. 12

a method of finding the sum to infinity.)

Exercise 12(/)Test the following series.

2 2 3 2 4 2 2 2 2 2 3

1 1+Tl

+2i

+3i

+ '- 2 * + * + * +

3 If a and 6 are positive, prove that

q+1 (a+l)(2a+l) (a+ 1) (2a+ 1) (3a+ 1)

6+1+

(6+1) (26+1)+

(6+1) (26+1) (36 + 1)+ '

converges if a < 6 and diverges if a ^ 6.

Test the following, assuming x to be positive.

4 1+r.

+ii

+Si

+ 5i-2

+ +374

+ --

a;2

a;3 x* 1 2 3

8 a;+2 3a;

2 + 3 3a;

3 + 4 3a:

4 + ....

9 l + (l + 3)a: + (l + 3 2 )a: 2 + (l + 3 3 )a: 3 + ....

a;2

10 + +a+l a: + 2 x + %

1.3 1.3.5 1.3.5.711 (i)ix + -—x 2 + -———x3 + ———-xi + ... (aj+l).

2.4 2.4.6 2.4.6.8 '

*(ii) By showing that

1.3.5... (2n-l) 1.2.4... (2n-2) _ 1

2.4.6... (2n)_ >

2.4.6... (2n) ~ 2n'

prove that the series in (i) diverges when x = 1.

*12 By deriving the identity1

l+a; + a;2 +...+a: n =

1—x 1—x

obtain a formula for 1 + 2a? + 3# 2 + . . . + nx n ~ x, and hence verify from fir

principles the conclusions of the worked example in 12.42.



12.44] SERIES

is rapidly convergent when \x\ is small, since

\—x n x n

447

and|x

|

n is small for small n. It converges less rapidly when x is just

less than 1. On the other hand, Ex. 12(e), no. 20 (ii) shows that

1,(1 /r 2) converges slowly.

These notions are necessarily vague since the terms ' a good approxi-

mation', 'fairly small n' have not been specified; they are relative

and not absolute notions, and can become precise only after we have

selected some definite series by which to fix our ' standard of rapidity '.

However, they are useful in a descriptive way.

The essence of the comparison test for convergence (12.41 (1)) is

that if 1u r is term by term less than a convergent series cLv r , then

1u r will converge. We may say that( 1u r converges at least as rapidly

as Hv r; ' and if lim (ujv n ) = 0, that '1u n converges faster than 1v n \

Similar language would describe relative rates of divergence.

d'Alembert's test for convergence consists fundamentally of com-paring the given series 1u r with a g.p. If Hu r happens to converge less

rapidly than any g.p., the test is ineffective; this is the case with

Z(l/r 2) (and indeed with any of the series T,(l/r p

) for p> 1), and is

usually so when lim (u n+1 /u n ) = 1.

Tests of greater delicacy than d'Alembert's can be formulated, but

no finality can be attained since it can be shown that, however

delicate the test, a series can be constructed for which that test is

ineffective. In 12.44 we give a test which is of no fixed standard of

delicacy, and which deals particularly easily with series like 2 (l/r p),

S{l/r(logr) p} for which the ratio test is indecisive.

12.44 Infinite series and infinite integrals

The definitions of 'sum to infinity of a series' (as the limit of a

certain 'finite' sum, 12.21) and 'value of an infinite integral' (as the

li i f h di l h t



448 series [12.44

(1) The Maclaurin-Cauchy integral test.

If f{x) is continuous and steadily decreases to zero for x ^ 1, theoo /*oo

2 /(**) an d f(t) dt either both converge or both diverge. When they conr=l

JlVer 9*> /-co co /-co

S/(r)< f(t)dt+f(l).Jl r=l Jl

Proof. Since /(a;) steadily decreases to zero, we have f(x) > for

n fxx ^ 1 , so that 2 fix)> f(t)dt are increasing functions of n,x respectively.

(

If r < x < r + 1, then/(r+1) </(*)</(#).

Integrating this form r to r + 1,

/(r+l)<j

V+1/(a )& ; </(r).

Summing for r = 1,2, ...,n—l,

im< \

nf{t)dt < S/(r).

r=2 Jl r=l

The argument is illustrated geometrically by considering the sums

the inner and outer rectangles in fig. 124.

/•oo /*n

(a) Suppose f(t) coexists. SinceJ

/(£) eft is an increasing function

of n, it cannot attain its limit; therefore for each n,



12.44] series 449

nso that is bounded. Hence (2.77) S/(r) converges, and

r-2

S/(r) ^ f /W*+/(l).r=l Jl

It now follows from the right of (ii) by letting n -> co that

r=l

(6) If S/(r) converges, then

V/(r)<r-l r=l

By the right of (ii), P /(f) iB < £ /(r),Jl r=l

so that j f(t) dt is bounded and therefore tends to a limit, say Z, when

«,-> oo.

SincefJ

/(£) <ft is an increasing function of x, then if n —1 < X < w,

rn-i rx rn

I f(t)dt<jJ(t)dt<jj(t)dL

When X->co, also n oo, and soJ

/(£) <ft -> Z, i.e.

J/(f) <Zf exists.

(c) Since the series converges when and only when the integral

exists, it follows that divergence of one implies divergence of the

other.

(2) The function

$(n)= if(r)- {n

f(t)dtr=l Jl

steadily decreases as n increases, and lies between and f(l) for all n.

Proof. By (ii), n rn

0< S/(r)- f(t)dt<f(l);r=l Jl

f This step is necessary because the fact that j f{t)dt->l when n-+ oo {n being1

f°°a positive integer) does not by itself imply the existence of I f(t)dt (the limit of



450 series [12.44

also rn+10(n+l)-0(n) = /(n+l)- f(t)dt <

Jn

by using (i). Hence <p(n) decreases as n increases, but < $(ri) < /(l).

Corollary. It follows that lim <fi(n) exists and is non-negative.

The corollary is particularly interesting when both series anintegral diverge, as in example (i) following.

Examples

(i) Taking/(a:) = ljx in the corollary,

${n) = l

+ £ +£+...+--logn.

n

Hence <fi(n) tends to some limit y (Eider's constant: cf. 4.43 (8)) when n -»

and «S y < 1.

(ii) Discuss the series £ ( 1 /r p) by means of the integral test.

With/(a;) = then if p =j= 1,

f*/(f)* = _I-(l_X-M*)f

while if p = 1,

J/(*) <ft = logX.

If p > 1, the integral tends to l/(p —l), and if oo. Hence S (l/r») c if p > 1, D if p ^ 1 (cf. 12.41(3)).

The general theorem also shows that when£> > 1,

1 ™ 1 pp—1 r =l rP p—1

00 JIn fact y\ —> 1, which is a better lower estimate when p > 2.

Exercise 12(g)

1 Prove 2——̂——diverges. 2 Discuss 2*

rlog(2r) r(logr) J '

3 If a > 0, prove that the sum to infinity of

1 1 1

a 2+

(a+l) 2+

(a + 2)2+



12.5] series 451

5 Sketch the graph of y = logx, and show that

/•« n /•»

I logxdx < ^,logr < log x dx + log n.J l r=2 Jl

Deduce the value of lim {(n )1/n

/n}.n-yoo

12.5 Series of positive and negative terms

The special difficulty with such series is that s n is not necessarily

a steadily increasing function of n, so that the principle in 12.32 (5)

does not apply directly.

12.51 Alternating signs (theorem of Leibniz)The simplest kind of series with terms of mixed sign is that in which

the signs are alternately + , —// (i) the terms are alternately + , —

, say

u i ~ u 2 + %—u * + •••

where each u is positive,

i.e. un

steadily decreases to zero,

(ii) u n+1 oo

then the series is convergent.

Pr °°f s 2n = K - u 2 ) + {u 3- 1* 4 ) + . . . + {u 2n _ x - u 2n ).

By (ii), each bracket is positive. Hence s 2n is a steadily increasing

positive function of n. Also

S 2n = Ul ~ ( U2 ~ Ua) ~ (^4 —Ms)—

• • •—

( u 2n-2 ~ U2n-l) ~ u 2n>

each bracket is positive by (ii), and u 2n > by (i). Hence s 2n oo.

Further, _s 2n+l —s 2n + u 2n+l>

and since lim u 2n+1 = by (iii), thereforen->oo

lim s 2n+1 = lim s 2n = I.

n—>oo n—><x>



452 series [12.52

Example

1 —£ + i + converges because the terms satisfy all three conditions

the theorem. The argument used in the proof shows that the sum to infinity

lies between \ and 1; for s 2n <»-»-*•

Also see Ex. 12 (h), no. 8.

If any one of the conditions (i), ( ii ), ( ii i) is omitted, the series may not con

verge. Thus since the series

i + *+i+i+-,

d, c respectively, the 'difference series'

diverges (12.32(4)); conditions (i), (iii) are satisfied by it, but not (ii).

12.52 Absolute convergence

Consider the pairs of series

i + i+i+i + -;

, iii ,iiii 1 k i-i —i—i—i-

2 2 3 2 4 2 ^ -

'AT

2 a 3 2 4 a

The first of each pair converges, by 12.51; but in the corresponding

series where all the terms are made positive, the first diverges (being

the harmonic series) while the second still converges (12.41 (3)).

It is convenient to distinguish between series which remain con

vergent when all their terms are made positive, and those which do not

Definition. If the series TiU r is such thatf S \u r \ converges, then 1iU

is said to be absolutely convergent (a.c).

Observe that nothing has been asserted in this definition about th

convergence of T>u r itself; it is a different series S \u r \(i.e. Ewr with

all its terms made positive) which converges.

Examples

1 + 4+ i not A b + + + + di



12.52] SERIES 453

The importance of absolutely convergent series is (roughly) that

they behave like series of positive terms. For example, compare

Theorems II, III in 12.54.

Although the definition states nothing about the convergence ofEi* r , we can in fact prove that an absolutely convergent series is also

convergent in the ordinary sense. More explicitly

7/2 \u r \converges, then 2i* r also converges.

Before giving the proof of this, we illustrate the underlying idea.

Suppose that the series 2i* r in question is

Ul + U2—U3 + Ui -U 5

-1*6 + 0-W 8 + U9 -U 10 + U11 + U1Z + ....

Consider the following two series:

%+ i* 2 + + i* 4 + +0 +0 + +u 9 + + i* u + i* ia +

+0 +W3 + O +1*5 + 1*6 + + 1*8 + +i* 10 + +0 + ....

The first is formed by replacing all negative terms in the given series

by 0's; the second is obtained by replacing all positive terms by and

changing the sign of the negative ones.

Compare each of these series of positive terms with the series

%+ 1*

2+ 1*3 + 1* 4 + 1*5 + 1* 6 + + 1*8 + 1*9 + 1*io + %i +

%2 +- >

which is convergent by hypothesis. The comparison test (12.41 (1))

with c = 1 shows that each series converges. Hence by 12.32 (4) their

difference (which is the given series) also converges.

General proof. Consider the series

v 1 + v 2 + v s + 1 + w2 + w3 +...,

where we define

(Thus relative to the given series Si* r ,Ei> r is the non-negative part, and

Sw r is the non-positive part with its sign changed.)

Clearly v r ^ |i* r |and wr \u r \. Comparing each of Sv r ,

2t*? r (which

are series of positive terms) with 2 \u r \which is given to be convergent,

the comparison test with c = 1 shows that they converge. Hence

li{v —w converges; i e since u = v —w for all r, Si* converges.

and



454 series [12.52

converges because it is a.c. However, the test is 'one-sided' because

a series which is not a.c. may yet be c ; e.g. 1 —1 + J+...is not a.c,

but is in fact c. Such a series may be called conditionally convergent

(o.c.) or semi-convergent.

Examples

*(iv) The ' triangle inequality 'ffor infinite series.

If YaU t is A.c, its sum to infinity is numerically less than or equal to the sumSlid:

With the above notation we have u r = v r —tv r , \u r \= v T + wr . Also, since

Su r , Hwr converge,00 00 00

r=l r=l r=l00 00

< 2 *V + S wr sincef both these sums are positive,r=l r=l

= S K|.r=l

The device used in this example can also be employed to prove that

\(

b

f(x)dx\ < (b

\f(x)\dx,\J a | J a

where b > a and f(x) is continuous; cf. Ex. 7 (a), no. 5. Define

-f(x) if f(x)<0,f(x) if /(a)>0,

if f(x) =sj 0, if f(x) > 0.

Then/ 1 (a;),/ 2 (a;) are non-negative functions, and

\m\ =f 1 (x)+f t (x), f(x) =A(x)-Mx).Also

f(x)dx\ = f^dx- f z (x)dxI J a

I \J a J a

rb rb

^1 /i(aj)cfa;+ I fi{x)dx, both integrals being non-negative,J a J a

= (

b

{fi(x)+A(x)}dx = (b

\f(x)\dx.J a J a

.



12.52] series 455

If x > 0, all the terms are positive and we can use d'Alembert's ratio test:

Un+l _ XnI Xn ~ X _ X

u n w /(w— 1) n

thus since ^n+i 2 Xf

or alternatively, since lim = 0,

the series converges for all positive x.

If x < 0, write x = —y so that y > 0. The series becomesf

* 2 3

By the case just considered, this is A.c. for all positive y, and hence it is also c.

The given series therefore converges for all negative x.

Consequently, the series converges for all x.

N.B.— From 12.32(1) it follows that (cf. 2.74)

x n

lim —= for all x independent of n.n-oo n\

(vi) Discuss the series x——+ —- + ...

2 3 4

for all values of x.

w2 y s y*

First consider the series 2/ + ~2 * ~3 * ~4 + *

where y ^ 0. If y = 0, the series is + + + . . . which converges to sum 0.

If y 4= 0, the ratio test gives

u n+i y n+x ly n n-^±1 = /— = y^-V when n->oo;

u n w+l/n n+1so if < y < 1 the series converges.

The y-series therefore converges for < y < 1; hence the given series in xis A.c. for —1 < x < 1, and thus also converges in this range.

If x = 1, the given series is 1 —| +i

+ which c (12.51).

If x = —1, the series is —1 —| —i —i— which is properly divergent to

—oo (it is —1 times the harmonic series).

If x < —1, each term of the series is negative, but its numerical value is

greater than that of the corresponding term of 1 + + J + . . • . Hence the series

properly diverges to —oo.

If x > 1, then x = 1 + c where c > 0, and

x n = (l+c) n = l+nc + in(n-l)c z + ... > £n(w-l)c 2,

since all the terms of this binomial expansion are positive. Therefore



456 SERIES [12.53

Summarising, the aeries c for —1 < x < 1, and d for x ^ —1 or x > \. Trange —1 < x < 1 is called the interval of convergence of this series.

12.53 The modified ratio test

We introduce here a modification of d'Alembert's ratio test whichmakes it directly applicable to series whose terms are not all positive.

// either (a)u, n+l ^ k < 1 for all n ^ m, where k is constant,

or (b) lim = I < 1,

then Hu r converges.For either condition ensures that S \u r \

c, being what is obtained b

applying the original ratio test to this series. Hence lLu r is a.c., antherefore c.

If EITHER (a') ^ 1 for all n ^ m,

or (6') lim 'n+l = I > 1,

then Ymt is divergent (i.e. not convergent).

(At first sight all we can say is that 2 \u r \d, i.e. that Y,u r is not a.c;

this does not prevent 1>u r from converging.)

Condition (a') implies that \u n \ ^ \u m \for all n ^ m, so that Mn -f»0

when n-^co. Given e > 0, condition (&') implies that there is

number m such that \u n+1 \> (l —e) \u n \

for all n ^ m, and the result

follows from (a') as in the proofs in 12.42.

Examples

(i) Discuss the series of 12.52, ex. (vi) in this way.

If x — the series converges to sum 0. Since u n = ( —l) -1a;

n /n, we havwhen x 4= that

|a;| when n 00.n+l

Hence if \x\ < 1 the series c; if \x\ > 1 the series n. Only the cases \x\ =

.



12.54]

If x 4= 0, then from

we find

SERIES

m{m—1) ... (m —n + 1)

457

n + 1

n

-> —re I = \x\ when n oo.

Hence */ |»| < 1 </ie series o; */ |a;| > 1 the series d.

The cases \x\ = 1 cannot be decided by means of the tests at our disposal.

However, it can be proved that

if x = 1, the series c if m > —1

and if x ——1, the series o if m Js

and d if ra < —1;

and d if m < 0.

12.54 Regrouping and rearrangement of terms of an infinite series

When dealing with finite sums it is permissible to group the terms in anymanner by means of brackets, or to rearrange the order of the terms, withoutthereby altering the value of the sum. These properties may not extend to

infinite series, f

Example

(i) The series (1-1) + (1-1) + (1-1) + ...,

i.e. + + + ...,

converges to zero. If we remove the brackets we obtain the series

1-1 + 1-1 + 1-1 + ...,

which diverges since s B oscillates between and 1. If we insert brackets as

l + (-l + l) + (-l + l) + ...,

we obtain 1 + + + . .

which converges to 1.

This example shows that, in general, brackets cannot be removed or inserted

in an infinite series without altering its behaviour or its sum. However, the

following result is easily proved by using the definition of 'sum to infinity'.

Theorem I. // 2« r converges to s, then it will still converge to s when brackets

are inserted in any way.

Proof. Let n x < n a < w8 < ... be any infinite sequence of positive integers,

and writefix n%

= X Mr» h — 2 u r> •••

r—1 r=rii+l

Then we wish to prove that 2^ converges to s.



458 SERIES [12.54

Let n v be the first integer of the above sequence which is greater than or equato N. Then (i) holds for all n 2s n v , and so

Tim

2 u r —s

r=l

< e

for all m > p (since the integers n„ +1 , n„ +2 , • . . are a selection from the set of

integers greater than n P ).

Since terms of a 'finite' sum can be grouped in any way,

Hence result (ii) becomes

Tim m2 u r = 2

r=l i=l

2i=l

< e for all m ^ p;

i.e. Y&i converges to s.

The converse of this theorem is false : if a series with the terms groupedconvergent, the series obtained by removing the brackets may be divergent.

This is illustrated by ex. (i), which also shows that there is no general theoremfor divergent series like Theorem I; but see Ex. 12 (h), no. 14.

Example

(ii) If s n denotes the sum of n terms of the series

W+W+i--,and t n denotes the sum of n terms of

1 _l_l_l_i_l_l-Ll 1 1__L A_2 4=13 6 8 '5 10 12 '7

(formed from the first by taking one positive term followed by two negative ones

then the sum to infinity of the second is half that of the first.

Consider

<.»-(i-i)-i+(i-*)-i+(i-A)-A+.-- + (2^ri--4 n _2) 4n

= *-* + *-*+...+1 1

4n —2 4n

The first series is known to converge, say to s; hence when n -> 00, s 2n ->

and therefore t 3n -> %s. Since also

Wl = *3n + 2^T[-*2 S

^ hn+ *= hn +

2nTl ~ 4^T2



12.54J series 459

Theorem II. A convergent series of positive terms can be rearranged in anymanner without altering the convergence or the sum to infinity.

Proof. Let Sw r be a convergent series of positive terms, with sum to infinity s.

nLet Zw^ be a rearrangement of Sw r , and write s' n = 2 u '

r-

r=lEach term of Zt^ appears somewhere in Zwr . Hence for a given n, there is

na number m such that s' ^ s m. Since 2 u r is a steadily increasing function of w

r=l(all u r ^ 0), hence s m < s and so «^ < s. Since is also a steadily increasing

function of n (because all u'T

3s 0), therefore by 2.77 s' n tends to a limit *' wheres' ^ s. Thus ~Lu'

rcertainly converges, and its sum is s'.

Similarly, since each term of Zwr appears somewhere in Hu'r , the preceding

argument will prove that s < s'. Consequently s' = 8.

N.B. —Observe that we could not prove s ^ s' first because we do not know

whether Hu' r converges or not until we have proved s' n < *.Theorem III. An a.c. series can be rearranged in any manner without altering

the absolute convergence or the sum to infinity.

Proof. If Hu'r

is a rearrangement of 2w r , then certainly 2>u'r

is A.c. since

2 \u'r \

is a rearrangement of £ \u r \, and Theorem II applies.

We use the notation v r , wr oi 12.52, and its extension^, w' rfor the rearranged

series Hu'r

. Then clearly Hv'r , Ett/ are respectively the rearrangements of Sv r ,

TiW r entailed by the rearrangement lLur

of Zwr . Since T,v r , Ewr are convergent

seriesf of positive terms, Theorem II gives

00 00 00 00

S < = 2 Vr and S w'r

= 2 wfr=l r=l r—1 r=l

00 CO 00 00 00 00 CO

r=l r=l r=l r=l r=l r=l r=l

We shall not discuss the rearrangement of infinite series any further; but

enough has been said to caution the reader against treating infinite series

analogously to finite ones.

Exercise 12(h)

Determine the behaviour of the following series.

11111 _ i i i_ i I 2 1 h1 8 + 6 7 + Z1.2 2.3

+3.4 4.5

+

3 1-£V2 + W3-£V4+-- 4 £-f+i-f+-&-H+....5 logi+log|+logf+logf+log^+logi|+.-»

6 State which of the series in nos. 1-3 are a.c.

7 Show that 2 {(cos rx) jr 2} converges for all values of x.

8 Under the conditions of Leibniz's rule (12.51), prove u x —u t <s wherea £ 11.

Lw„ «„ I

(n+l)a„jr=i

rj

n*13 S— — [The cases \x\ = 2 may be omitted.]

1.3.5... (2n+l)L

*14 If u r ^ for all r, prove that (with the notation of 12.54, Theorem I)

m n m+1if n m s£ w < then 2 < £ u r < S

i=l r=l i=l

and deduce that the converse of Theorem I holds for a series of positive terms.

*15 If Ew r is conditionally convergent, prove that (with the notation of 12.5

|~ n nHv r and ?jw r diverge to +oo. Ifs„ = Tiu r ,t n — 2 \u r \, then s n -> sand +oL r=l r=l

n » ~1

also 2 «V = + S wr = i(«n-*n)-r=l r=l J

*16 If the series 1 —\ + %—\ + \ —... is rearranged as

(taking two positive terms followed by one negative term), prove that

sum of the new series is f times that of the original. [Prove t 3n = s in + £s 2 nO

12.6 Maclaurin's series

12.61 Expansion of a function as a power series

We proved in 6.52 that if f(t) satisfies certain conditions of co

tinuity and derivability for ^ £ < x, then

f(x) = /(0) + xf>(0) + ^/ (O) + . . . + ^-_^/(-i)(0) + Bn (x),

where (see 6.54(1))

fxr

•»•(*) =

_ f(n)(0x) (Lagrange's remainder),n

x^l-d) 71 - 1

(n-1) f(n \dx) (Cauchy's remainder)

and 6 is some function of x and n (in general not the same function



12.62] series 461

continuous for O^t^x; then the conditions for validity of

Maclaurin's theorem hold for all orders. Consider the infinite series

/(0) +

«+ ^/'(0) +

...+^^(0)+ .... (i)

We have s n (x) =/(0) + */'(0) + ... + ^£^/<»-i>(0),

= f(x)-R n (x)

by Maclaurin's theorem. If Rn {x) -> when n-+ao (perhaps only for

some range of values of a;, say \x\ < c), then

s n (x) -+f{x) when n -> oo, for \x\ < c;

hence the infinite series (i) converges and has sum-function f(x) when\x\ < c. We write

/(z)=/(0) + */'(0) + ^ (\x\ <c), (ii)

and call this series the expansion off(x) as a power series in x over the

interval \x\ < c.

This idea will now be applied to obtain series for some of the

elementary functions; the results are important, and will be discussed

in detail in 12.7. We use the results on wth derivatives obtained in 6.6;

only when such convenient formulae are known is direct discussion

of Bn (x) possible.

12.62 Expansion of some elementary functions

(1) The exponential series.

If f{x) = e x, then f

in) {x) = e x for every positive integer n, and

Lagrange's remainder is

Rn (x) = ê°* (O<0<1).

Since x n jn\ -> when n->co for any fixed value of a; (2.74), therefore

Rn (x) -> when n -> oo, for all x. As/ (r)(0) = 1 for each r,

e*= 1 + *+^ + ^+.. .+£+... (allx).

(2) The sine and cosine series.



462 series [12.6

Also f®r \0) = and /(^(O) = ( - 1 y- 1 for each r, so

sin* = ^-^T+^t- - • + (- 1 )r ~ l

( 2 r-l)+ •'

(a//

Similarly

cos*= ^Ji + ^T- •+(- 1 )

r 1

(2r _2)+ ' (aW ^*

(3) Tê logarithmic series.

If /(a;) = log(l+z), then f n \x) = (- l) - 1 ^- 1) /(1 + x) n] a

so fin)

{0) = ( —l) n_1 (n— 1) . We are supposing a; > —1, for otherwise

log (1 + would not be defined in part of the range x ^ t ^ 0.

When ^ x ^ 1, Lagrange's remainder

v n 1

W {l + 0x) n

shows that l-S^) |< - -> when n->oo.

When x < 0, —xj(l + 6x) may be greater than 1, so that Lagrange's

remainder . , N _1 / —x \ n

~n\l+6x)

might tend to —oo when n -> oo (2.73); insufficient is known abou

the behaviour of as a function of n.

Cauchy's remainder can be written

r w / l —f) \ ™-i

^» = <- 1 ) - 1

rTfe(rr^) •

\x\ nso that if —1 < x < 0, \R n ( x ) I

<^ j^j

because (1 —6)j{\ + 6x) is a positive proper fraction and (see 1.14)

\l + 0x\ > \-6\x\ > l-\x\.

Hence Rn (x) when n-+oo. (Cauchy's remainder also covers t

case < x < 1 already treated.) Thus



12.62] series 463

(4) The binomial series.

lff(x) = (l+x) m, then

f n \x) = m(m - 1) . . . (ra - w + 1) (1 + x) m~ n.

If m is a positive integer, Maclaurin's formula terminates whenn = m+l, and gives the usual binomial expansion containing m+ 1

terms. If m is not a positive integer, then f{n \x) will be continuous

when n > m only if x > —1.

Cauchy's remainder can be written

When |g| < 1,

(m- 1) ... (m-n+ 1) „ ,

a„ = - —̂—: x n -> when n -> oo,n (n-1)

by 2.75. Also (1 —0)/(l +#r) is a positive proper fraction, and

( (l + bl)™- 1 if m > 1,

(l + ftc) 1 - 1 < J11

[(1-IjcI)™-1 if m < 1.

\Rn

(x)\ < |m|(1 ±

[a; )™— 1 when n-+co.

Hence.„ . . m(m —1) _

(l + *)m = l + mx+—^ —J* 2 *...

+m(m-l).„(m-r + l)^

+ (M<J)

If m is a fraction with even denominator, (1 +x) m has two values

which are equal and opposite. Here we intend the positive value, since

the sum of the series when x = is clearly + 1.

As in (3), Lagrange's remainder would cover only the range

^ x < 1. We omit consideration of the cases x = ± 1.

(5) Gregory's series for tan -1 x.

If f(x) = tan -1 x, then

/< )(») = (-l) - 1 (w-l) (l+* 2 )-* n sin(ncot- 1 a;),

so / (2r)(0) = and /^- (O) = ( - l)' 1 (2r - 2) .



464 SERIES [12,63

Hence , ac8 x? , <xr , * 8r-1

,

tan 1 * = *__ + _-... + ( (|*| ^ l).

The value of tan -1 x in the range —\n to + ^n -

is intended (i.e. the principal

value when —1 < x ^ + 1), since when x = the sum of the series is clearly

zero. No more than the above can be proved about the sum because the serie

converges only for \x\ < 1 : see Ex. 12 (h), no. 9.

12.63 Note on formal expansions

In Ex. 6 (6), no. 15 we considered the function y = tan -1a;. Having

proved that it satisfies the differential equation

(1 + a?) y n+2 + 2(n + l)xy n+1 + n(n+l)y n = 0,

we put x = and showed that the Maclaurin coefficients a r = (y r ) xare given by ^ = - n {n + l)a n .

Since a = and % = 1, all coefficients with even suffix are zero

while by successive application of the above recurrence formula

those with odd suffix, we find that

GW-i = ( ~ I) 1 (2n - 2) (n = 1, 2, . . .).

Assuming that y possesses a Maclaurin expansion, we thus obtainfor it the series „ K » r ,

Since this converges for \x\ ^ 1 (see Ex. 12 (h), no. 9), it is reasonable

to expect this series to have y = tan -1a; for its sum-function whe

\x\ ^ 1. However, as we have nowhere considered Rn (x), we hav

not proved by this argument that the series has sum-function tan -1

nor even that tan-1

x can be expanded at all as a power series.

It can be proved that, within its interval of convergence, every power series

is the Maclaurin series of its sum-function. In othqr words, if Ha r x r converges

to s(x) (for \x\ < c, say), then a n = d n) (0)/nl (n = 0, 1,2,...). Assuming this

what we have shown by the above argument is that, if the series (i) converges

to s(x) when \x\ sS 1, and/(£c) = tan -1#, then

s(0)=/(0), «'(0)=/'(0) *W(0) = /<'>(0), ....

These equations do not prove that s(x) is identical with/(a;) for \x\ ^ 1. Fl



12.63] SERIES 465

tinuous in a certain interval, the Maclaurin series off(x) may not converge at

all for that interval.

Thus, to ensure the existence and validity of the Maclaurin expansion of

f(x) in 12.61, it is essential to prove somehow that B n (x) -> when n -> oo, for

all x in the interval concerned. However, despite the cautionary nature ofthese remarks, it is true that for 'ordinary' functions the formal expansionsobtained as above are valid whenever they converge; and with this assurance

the reader may accept them with confidence.

Exercise 12(i)

State the sum to infinity of

3 Assuming the exponential series, state series for ch x and sh x.

Obtain Maclaurin series, stating the range of validity, for the following functions.

4 e^sina;. [See 6.61, (vi).] 5 e~ a coax. 6 cosa;cha;.

d n

7 Prove —{e x coa * cos (x sin a)} = e x 008 * cos (isma+ na),dec

and deduce that for all x,

x 2 x 3

e XC0BCt cos (x sin a) = l + a;cosa + —cos2a + —cos3a + ....

*8 By showing that Rn (x, h) -> in Taylor's theorem, prove that

coa(x + h) = cosa; —nsmjc ——cosa:+— suiir— ...

2 3

for all x and h. [See Ex. 6 (6), no. 24.]

*9 Obtain a Taylor series for sin (x + h).

10 If y = ch (sin -1 x), prove

(l-x*)yi-xy x -y = and (l-x*)y n+i -(2n+ l)xy n+1 = (n a + l)y n .

If ch(sin _1a;) = a + a l x + a i x % -\-

obtain an expression for a 2n , and prove a an+1 = 0.

11 If a; = cos0 and y = cosn# where n > 1, prove (1 —aj2

)y a—xyi + n 2 y = 0.

Assuming that y = o + a t x + a t x 2 + ..., prove

(k+l)(k + 2) a k+2 + (n 2 —k 2 )a k = (k — 0,1,2,... ).

If n is an integer, show that the assumed expansion is a polynomial of degree

n; and if n is even and greater than 2, prove a 4 = ^ —1)* w2 (n 2 —4).

12 If 2/ = cos log ( 1 + x), prove

(l+a;) a2/ n+8 + (2n+l)(l+a;)2/ n+1 + (n a +l)2/ n = 0.



466 series [12.

12.7 Applications of the series in 12.62

12.71 Binomial series

/-v r™ •

m(m— 1)...

(m—

r+l). ,

,,(1) The expression — '——^ » often abbreviatedf

I , is called a binomial coefficient. The expansion of (a + x) m can

reduced to the standard one in 12.62 (4) as follows:

{a) if

(6) if

(x\ m1 + -I =etc;

(a\ m

1+-I = etc.

Examples

(i) Write down the general term and the first four terms of the expansion

(1 + 2a;)-3 in ascending powers of x, stating when it is valid.

There will be an expansion in ascending powers of a; if \2x\ < 1, i.e. if \x\ <and then the (r + l)th term is

t-»)(-«)-(-»-r+D(2xy

. <-ir«:«-tr + 2)

if Iif\

- ( _ i)r(r +

^+ 2)

2 r x r = ( - l) r 2 y - 1 (r + 1) (r + 2) x r.

1.2

We can obtain the first four terms by writing r = 0, 1, 2, 3 successively

thisformula:{1 + 2x) -s = 1 _ 6x + 24 x *-80x* + ... (\x\ <

(ii) .Find <7&e 9£ft term in the expansion of(l —4x)~ 2 if\x\ > \.

Since [4a?j > 1, the expansion can be made only if we expand in powers ofl/x,

i.e. in descending powers of x.

i-^)- = (i-i) w- s =i i(i-i) '.

and the 9th term in this is (putting r = 8 in the general term)

1 (-2)(-3)...(-9) / 1\ 8 2.3...9/1X 10 9

\4seJ 8 \±x)6a; 2 8 \4xJ 8 \4a:/ (4«)10

(iii) Calculate ^102 correct to four places of decimals by using the binomial series



12.71] series 467

Since we require the result to be correct to 4 places, we must work with

5 places; and owing to the factor 10, (1-02)* must be found to 6 places. Cal-

culating the terms shown, we find

(1-02)* = 1 + 0-01 -0-000,05 + 0-000,000,5+...

= 1-009,950,5...,

the next term clearlyf being too small to influence the sixth place. Hence

V102 = 10-0995.

(iv) Expand ( 1 + x + x 2 + jc3

)-6 as far as the term in x*.

Since 1 + x + x 2 + x 3 = ( 1 - a^)/( 1 - x), the given expression is

(l-a;) 5 (l-a: 4 )~ 6 = (l-5a;+10a; 2 -10a; 3 +...)(l + 5a; 4 +...)

= l-5*+10a; 2 -10a; 3 + ...

on neglecting all terms in x* and higher powers.

( v) Find the coefficient of x r in the expansion ofl/{(2 —x)(l + 3x)}in ascending

powers of x, stating the condition of validity.

By resolving into partial fractions we find

1 11 3 1

(2-«)(l + 3a0 ~ 72-£C+

7 1 + 3*

= ^(i-4^)- 1 +f(i + 3»)- 1.

The first bracket can be expanded in ascending powers of x if \%x\ < 1,

i.e. |a;| < 2; the second can be so expanded if |3a;| < 1, i.e. \x\ < §. Hence both

can be expanded if \x\ < \, and then the coefficient of of is

1 (-l)(-2)...(-r) / IV 3 (-l)(-2)...(-r)

14 r \ 2) 7 r\

= *<*>' k - 3)- = \ - ( - sr *} •

(2) Summation of series reducible to binomial expansions.

In the result

(l- a; )^=l + (-3)(-^) + - : 4^(-^) 2 + 3

1 72

4 3 5(- a; )

3+ -

, n 3.4 , 3.4.5 ,

which is valid when \x\ < 1, let us put x = \. The series becomes



468 series [12.71

which can be written « A „ A K3

3.4 3.4.5

Our result above shows that the sum to infinity of this series

(l-i)-3 = 8.

Conversely, many series of this form (in which the factors in the

numerator and denominator of each term are in a.p., while the

number of such factors is the same but increases as we proceed along

the series) can be reduced to binomial expansions, and hence their

sum to infinity written down.

Examples

(vi) Find the sum to infinity of1 1.3 1.3.5

4+

I78 4.8. 12+

The factors in the numerators form an a.p. with common difference 2, andthose in the denominators have common difference 4. We begin by dividingeach factor in the numerator of each term by 2, and each factor in the denomin-ator by 4: lis. i a &

1_i

(l)+iT2

(l)2 ~r2f3 (l)3+ - -

The factors in the numerators now ascend by the common differencewhereas for a binomial series they descend by 1. This can be adjusted here bintroducing signs as follows:

and this series is clearly the expansion of ( 1 + x) m with x = £ and m = —Hence the required sum to infinity is ( 1 + = ^/f = \ ^6.

, „ t1 1.3 1.3.5 1.3.5.7

(vu) -H 1 1 h...' 3 3.6 3.6.9 3.6.9.12

1 12. lii JL 2. & Z= f(f) + ^(f) 2 + ff|(f)

8 + ^(l)* + ...

= {l+^(-|) + Z |^(-f) 2 + i

1

2

f 3 f(-|) 3 + -.-}-l

= ( - )-*- = V3-L

Exercise 12(J)



SERIES 469

Find the named terms in the following expansions.

8 8th term in (1 - 2a;)- 8 if \x\ <\. 9 10th term in (1 - 3a;)- 2 if \x\ > £.

10 What is the first negative coefficient in the expansion of (3 + 2a;)* if |a;| < f ?

[Consider the sign of ar+1

/ar

.]

* 1 1 If u r is the (r + l)th term in the expansion of ( 1 + x) m, \x\ < 1, show that

u r \ r }

Deduce that when r becomes greater than m+ 1, then u r+1 /u r< according

as a; > 0; and hence that, after a certain stage in the expansion, the terms are

alternately positive and negative if x > 0, but all have the same sign ifx<0.

Find the numerically greatest term or terms in the expansion of

12 (1 +x) 21 ' 3 when x = f . [Use the argument in 12.13, ex. (iii).]

13 (1-x)- 10 whena; = |. 14 (1 +x)~* when x = \.

Use the binomial series to calculate correct to four places of decimals

15 (4-08)*. 16 V98 -

17 Expand ( 1 + x + 2a; 2 )-* as far as the term in x*.

State the condition that the following can be expanded in ascending powers of x,

and find the coefficient of x r in the expansion.

2 + 3a; x18 „ „ , . 19(l-a;)(l + 2a;) 6-a;-a; 2

a;2

1 T _ l-x l

{l-x)*(2-z)' 1 +x + x 2 \_~ 1 -a; 3J

'

*22 (l+a;+a; 2 + ...+a; m- 1)

B if r <m.23 Find the coefficient of x 2r and of x 2r+1 in the expansion of

/j±=. W <I. [Writer -^.]Find the sum to infinity of

1 1.3 1.3.5 , 1 1.4 1.4.724 1 +

8+

8Tl6+ 8Tl04 + -- 25 1 -4 + 4T8-4XT2 + --

4 4.7 4.7.10 JJ_ 3.4 3.4.526

20+

20T30+

20.30.40+

2T4+

2.4.6+

2.4.6.8+ *

28 Verify that when \x\ < 1,



470 SERIES [12.72

*29 By writing Pascal's triangle (12.13, ex. (iv)) as 1

shown here, we observe that the nth vertical column 1 1

gives the coefficients in the binomial series for (1 —x)~ n. 12 1

Justify this rule by proving 13 3 1

(rwr--/)- (r)1 4 6 4 11 5 10 10 5

12.72 Exponential series

(1) Irrationality of e. Taking x = 1 in the series of 12.62 (1),

111 1e=1 + H +

2+

3+ - + H +

From this it is possible to calculate e to any specified degree of

accuracy (see the example in 6.53).

Since this series consists entirely of positive terms,

,1 1 1

1 2 n\

The error in this estimate is

1 1 1+ , . + , , „„ + ...

(n+l)\ (n + 2)\ (n + 3)\

=(n+l)\{

1+n + 2

+(n + 2) (n + 3)

+ )

<* + i)\{

1 +n~+l

+(n+l) 2+ )n +

1

(n+1) l-l/(»+l)

1 n+1 _ 1

(n+l)\ n n\n'

by summing the g.p.,

These results can be used to prove that e is an irrational number,

i.e. that e cannot be expressed in the form p/q where p, q are integers.

For suppose if possible that e = p/q; then from the above inequalities

with n = 11



12.72] series 471

where r , q\ q\

is an integer, and so is p.{q— 1) . Our conclusion is that the integer

p.(q— 1) lies between / and I+l/q, which is impossible since 1/q is

a proper fraction. Hence e cannot be written as p\q for any pair of

integers p, q; i.e. e is irrational.

(2) a x, chx, shx.

If a > and m = log a, then a = e m and

m2 x 2 m3x 3 . „a x = e™* = l+mx + ——H-^r-+... for all a;,

i.e. o* = 1 + a; log a + ^ (log af + —(log a) 3 + . . . (all x).

From the definitions chx = %(e x + e~ x), shx = %(e x —e~ x

) and

ch._ 1+ + + ... + ^_^ + ...,

(aU*).

8h ' = ,+ 3 + 5i + ' W )(3) Series reducible to exponential series. The first step is always to

write down the general term of the given series, and then put it into

a suitable form.

Examples. „ . I 2 2 a 3 a

(l) Sum to infinity —+ —+ —+

.

2 3 4

Here T(r+1)

The numerator r 2 = (r+ \)r —r = (r+l)r —(r+ 1) + 1,

1 1 1and so u. = (-- — if r > 1.r

(r-l)l r\ (r+1)

00 oo 2.00 1 00 1

=



472 series [12.73

(ii) Calculate « (r + 2)2

Since (r + 2)2 = r 2 + 4r + 4 = r(r- l) + 5r + 4,

1 5 4+- if(r-2)l (r-l)l rl

ooJ}2

GO

r=l 1 r=2

= 9 + e + 5(e-l) + 4ê-l-^= 10e-4.

Exercise 12(k)Give the coefficient of x r in the expansion of

1 6 **+ 2. 2 i±^. 3

g, + te + 1. 4 ±f .

e a e x e 2x

5 Write down the series whose sum to infinity is (i) 1/e; (ii) e 2; (iii) (e + 1/e)

.Find the sum to infinity of each of the following series.

32 33 3 4 42 44 466 3 1 1- .... 7 H 1 1

f- ....2 3 4 3 5 7

, 2 3 4 1 1+2 1+2+38 H 1 1 h .... 9 —+ -— + ———+ ....

1 2 3 1 2 3

3 5 7 =° ~'

12 S

2 2.4 2.4.6 r =i(r+l)

r=l(r + 2)r

12.73Logarithmic series

(1) There are several important forms of the result

{](^ tJC^

log(l+*) = *-- + ---+... (-1 <x^ 1).

(a) Replacing x by —x,

x 2 x 3£C

4

log(l-z)



12.73] series 473

. _ 1 + X Xs X5. , _

.

i.e. ^ l ° 8 T^x= x+

s+ ~5 + - (~ 1<x<1 )'

(c) In (6) write , ,

1+x . y —1=y,

i.e.

x= -

—-.

1-x y y+1

The condition —1 < x < 1 becomes y > 0, and

(d) In (6) put1+x p+l . 1— i.e. x =l-# p ' ' 2^ + 1'

Then 1log ( p + 1 ) - log p = 2 {—+ 3 +

, (2i>+l) 5 +

and the left-hand side is defined only if p > 0, which corresponds to

the condition < x < 1 under which (6) certainly holds.

The series (b), (c), (<?) are more rapidly convergent than either the

standard series or (a) separately; (c) is used for calculating natural

logarithms, and (d) for calculating logarithms of consecutive numbers.

Example

(i) Find the coefficient of x r in the expansion of log ( 1 —a; + a;2

) if |*| < 1.

l+x z

log(l-a; + a:a

) = log- = log(l+* 3 )-log(l+ar).1+x

We have log(l+ar») = a?8 - —+ —-... + (- l)

r_1 —+ ...

2 3 r

and log(l+a:) = + -... + (- I)' 1 —- - ....

If r is not a multiple of 3, the coefficient of x r is ( —l) r_1 /r.

If r is a multiple of 3, the coefficient of x r is

(2) Series reducible to logarithmic expansions.



474 series [12.73

(iii) Sum to infinity 111+

275+

377+

r(2r+l)

1 2

r 2r+l

2 2

on using partial fractions,

2r 2r+l'

where the denominators are now consecutive numbers. Then

•^»{»-»+«-*>+- + (s-5Tl)}->

2{1 —log 2} when n->

oo,

since log2 = 1 —£ + £-;£+....

The sum of the series is therefore 2 —log 4.

(iv) Find the sum to infinity of

whenever it converges.

x x* x 3

l72+

273+

374+

x r 1 1 \. = = x r

[ ) .

r{r+l) \r r+ljoo oo/i 1 \

_ » x r 1 « x r+1

r =l r x r=1 r+l

provided that each series converges (12.32(4)), and when —1 ^ x < 1 this

the case because the first is —log ( 1 —x) and the second is —( I fx) {log ( 1 —x) +

Hence oo /l \V u. = |--l)log(l-a;)+l when - 1 s$ x < 1.

r=l \« /

When x = 1,

u. = =, s n = 1 -> 1 when n -> oo,r

r(r+l) r r+1 w+1

and the sum is 1.

It is easily verified that the series does not converge unless —1 < x ^ 1,

the problem has been solved completely.

(v) Sum to infinity 111



12.73] series 475

and we find by the usual method that A = \, B = —\, G — \. Hence

£ 1 *

2r 2r+l 2r + 2

= l/±_ L_V2\2r 2r+lJ 2\2r+l 2r + 2/

1 n / 1 1 \ 1 w / 1 1 \A Sn=

2 r? x \2?~ 2r~+l)~

2 r ?i \27+I~ 2r + 2j

-i(log2-l)- |(log2-l + £) when n -> oo,

= f-log2.

Exercise 12(/)

-Find <Ae coefficient of x r and give the first four terms in the expansions of the

following, together with the conditions of validity.

1 log(l-4a;). 2 log(2 + *). 3 log(l+a;) 2.

1 + x +4 log(l + 5a; + 6a; 2

). 5 (1 -a;)log(l +x). 6 log .

7 log(l-a; + a;2 -a; 8

).

8 Expand log ( 1 + 2x + 3a; 2) as far as the term in x l

.

9 If |a;| > 1, expand log(l+a;)-loga; in powers of 1/x, giving the general

term.10 Taking x = \ in 12.73(6), prove log 2 = 0-6931472 correct to 7 places

of decimals.

11 Taking x = \ in series (6) and using no. 10, prove log 10 = 2-3025851.

12 Use series (d) and no 10. to prove log 3 = 1-098612.

1 3 Prove 2 log 7 + log 2 = 2 log 10 —a, where

<r = 0-02 + i(0-02)2 +i(0-02) 3 + ...,

and deduce that log 7 = 1-945910.

Find the sum to infinity of the following series.Ill 1 _1_ _1_14

T72~2T2 2 +3T2 5 15

173+

373 s +573*

+

x 2 X* x* 111,g >_> + ».... 19 +

1.2 2.3 3.4 1.2.3 3.4.5 5.6.7

i i i r j i i



476 series [12

24 If y — 2x* —1, prove (under conditions to be stated) that

1 1 1 _ 2 2 2

i^*2»* + 3a« + y + 3?/3 + 52/

6 + *'

25 If —J7r < a < prove

sin 2 a + £sin 4 a + ... + -sin 2n a + ...

n. 2 3 sinHa 2 n + 1 sin 2n ia

= 2 2 sinH<x + 7T± -+-- + —+....

2 to

x 3 x 6 I 2x \26 Iff{x) = x + -+- + ... and \x\ < 1, prove/

1

^—^1 = 2/<<r).

12.74 Gregory's series and the calculation of nIf we put x = 1 in the result of 12.62 (5), we obtain

^=1-* + *-*+...,

a relation discovered by Leibniz. This series converges very slowly (see Ex. 12(

no. 4) and is impracticable for calculating n.

A better method, given by Machin, depends on the result

4 tan 1 £ - tan- 1 ^9 = Jar,

which we now prove. Let a = tan -1-J,

so that tana =\

and

2 tan 2atan 4a =

1 —tan 2 2a_ 4 tana IV ( 2 tan a 1

2~|

~ l-tan 2 a/ |_~ \l-tan 2 aj J

which is a number just greater than 1. Hence 4a is just greater than \ir; if

write 4a = \ti + tan -1 x, then1 + x

£|£ = tandzr + tan-1

*) = y-—and a; = ^ ¥ .

Thus on putting x = \, in Gregory's series we have

II 1 1) i

1 1

izr-4J--—

+g gf .

...J | 239 3 23g3+

It can be shown that 6 terms from the first bracket and 2 from the second g

7t to eight decimal places; while 21 and 3 terms from the respective brackets

give ir to 16 places.



12.81] series 477

by an application of Maclaurin's or Taylor's theorem as in 12.61, the

remainder term R gives the error in the approximation (see 6.53);

but owing to the indeterminate nature of the number 6, an estimate

of R has to be made in practice. When R cannot be obtained explicitlybecause of difficulty in calculating f

{n) (x), or when the sum-function

of the given series is not known, the method in 12.72 (1) can be used.

Examples

(i) If we

then the error is (see 12.62 (3))

(i) Ifwetake log(l+z) = x -— + —-... + (- l)n —

2 3 n

and

(n+l)(l + 0a) n+1

x n+lWhen x > 0, 1 + 6x > 1 and so |-R„+i(a;)| < .

Whena;<0, \l + 6x\ > l-d\x\ > 1- \x\ - 1+x,

|i2» + i(^)|< (n+1l

)(

l

1+a;)n+1.

(ii) Ifwetake e x = 1 + x +X

- + ... +X

2 (n-1)

the error is (see 12.62(1)) E n (x) = (x n /nl)e 9x. When x < 0, e 9x < 1 and

|i2 n (a?)| < |#|n

/tt. ; but if a; > 0, e 9x < e x and all we can assert is that

\B n (x)\ <—e x,

n

which is of no help since we are trying to approximate to e x.

The error is „ „ ,

n\+

(n+l)+

(n + 2)\+ '

_ xn

/ , x x * \~n\ \

+ w+l +(n+l)(n + 2)

+ ' /'

and when x > this is less than

x n / x x 2 \ _ x n 1

wl\+ n+l +

(n+l) 2+' J

=nll-xl(; n+1)

if x < n + 1, on summing the g.p. Thus if x > and n > x —1, the error in theapproximation is less than

x n n+1



478 series [12.8

For when n> m,

K+a |<* K+1 |,

K+s| < * K+2I < & K+l|> etC '

The error after n terms is

u n+l + u n+2 + Un+Z + • • • >

which is numerically less than or equal to (cf. 12.52, ex. (iv))

Kn+ll +\u n+2 1 +

|mtc+3| + •*•

< |t Wl |(l+* + *»+...)

1

= U, n+1 1-&

on summing the g.p., since < k < 1.

Another useful result is based on Leibniz's rule (12.51).

If for all n > m the terms of a series alternate in sign and steadily

decrease to zero in numerical value, the error in taking s == s n (n >cannot exceed the modulus of the first term neglected.

Let the series be

u 1 -u 2 + u 3 -... + (-l) m- 1 u m + {-l)û m+1 +...,

where for all n > m, u n > and u n+1 m) is / , x„ r ^( - 1 )» {u n+1 - u n+2 + u n+z - . .

.}.

The argument in the proof of Leibniz's rule shows that, for n fixed

u n+l ~ u n+2 + u n+3 ~ • • •

converges to a limit En , where < E n < u n+1 . Hence the numerical

value of the error does not exceed u n+1 .

Example

(iii) Consider ex. (iii) in 12.71: we wish to show that all terms after

fourth in the expansion of ( 1 + 0*02)* cannot influence the 6th place of decimals.

The general term is

u r+1 = i --^ 1}((H)2r (r;*l).

v\



12.82] series 479

the terms steadily decrease numerically. Hence the error is numerically not

greater than4 —A —£ -S

1(0-02)4

*•-*•-§.-# 3.5

4(0-02) 4 = 6-25 x 10- 9

.

2*4

12.82 Formal approximations

The following examples illustrate methods of obtaining approxi-

mations, but no error estimate is made.

(i) Ifxis small, find the best approximation for log ( 1 + x) which has the form

i \ax

(a) By expanding both sides of

ax x(l+ax)W1 + bx'

{' 1 + bx+cx 2

'

axlog(l + a;) =

1 + bx

when x is small, we have

x-$x 2 + %x3 - ... = ax{l-bx + b 2 x 2 -...)

= ax —abx 2 + ab 2 x z ——This will hold approximately for all small x if a = 1 and ab = \, i.e. b = |.

The expansions differ in their x 3 -terms since ab 2 4= h Hence

2x

log ( 1 + x)== i-

correct to order x2

.

Ji + X

(b) A similar method could be used, but the following is more convenient.

rromx(l + ax) = (l + bx + cx 2 )log(l+x),

x+ax 2 = (1 + bx + cx 2)

{x-%x 2 +\x z -lx l + ...)

= x + (b-$)x 2 + {c-\b+\)x* + {-\c + \b-\)x l + ....

We can obtain three equations for a, b, c by equating coefficients of

a = 6-|, = c-\b + \, = -ic + i6-i

from which a = 6 = 1, c = \. Hence

and since it can be verified that, for these values, the coefficient of x 5 on theright is non-zero, this approximation is correct to order x*.

(ii) Prove(

1+ I) _.(,_£+j

iL + £)).If 2/ = (l + l/n)», then



480 series [12.83

Writing exp (x) = e* for ease of printing, we have

*— { 1 -L + £t +0$))

=e[

i+ {-i +3s-«

+o(^)}

+ ^{-i + i +o(^)F

+o (^)]

= e

[1 -^ + 2^ + °(^)]-

(iii) If p is small, prove that a root of the equation e x + x = 1+p is approxi-

mately \p - -^p 2.

For a first approximation we expand the left-hand side as far as terms in

1 + 2x = 1 +p, .'. x = \p.

For the second approximation, expand as far as x 2 :

l + 2x + $x* = 1+p, i.e. 2x{l + \x) = p.

Using the first approximation x = \p, this becomes

2a*l + ip) =p,

so x = \p{ 1 + 1^)- 1 = ip - &p 2.

12.83 Calculation of certain limits

Example„. , e x -l-xFind lim

0 «--log(l+aO

x*We have e x = 1 + x + —+ a» 8

,

£ I

where a = ^ + ^+| + ....

113 4 5

<31 / \x\ Id 2 \

3 l-iHif 1*1 < 4.

xSimilarly log(l + a?) = x ——+bxz

,

where \b\ < + i T + iir+ —



12.83] series 481

Hence e x -l-x ix* \ l(x 2. J

i + ax, -> 1 when x -+ 0.

The above argument is often given incompletely as follows:

e x —1 —xMm = limx _ »-log(l + a;) x ->o x

i

2i+ -

= lim — = 1.

x-+0 » + •••

The result is easily obtained by two applications of l'Hospital's rules (6.9).

Exercise 12(m)

Prove the statements in nos. 1-4.

1 The error in taking log{(w+ l)/n} == 2/(2n+ 1) is less than

1

6n(n+l)(2ra+l)*

2 The remainder after n terms of

,1 1 1

1+Ii

+2i

+8i

+ -

lies between (H —̂— | — and\ n+l/w \ nf n\

3 If |aj| < 1, the error in taking

„ 1+x x sa;

2 1

i lo g^ =? Z + —+ ... +;1-x ' 3 2n-l

is less than the numerical value of a:2n+1 /{(2w + 1)(1— x 2

)}. Putting n = 2,

prove that log 1 1 == 2 log 3 + 0-20067 correct to 5 places of decimals.

*4 The error in taking

**l-*+*-... + <-D-îs numerically less than l/(2n+ 1). Deduce that the first 50 terms of the series

give \it correct to only two places of decimals.

5 Use the expansion of (1 —-^r)~* to find <J2 correct to 7 places of decimals.



482 SERIES

8 e x/*J(l + 2x) = 1+x 2 —f#

3. Write down the corresponding approximation

for l/{e x j(l-2x)}.

9 log 1 + e x)} = \x + \x 2 correct to order x 3

.

10 ( 1 + xy-+ x = 1 + x + x 2 + ix z. [Take logarithms.]

x x* r4

11 = 1 - - + . [Method of 12.82, ex. (i) (6).]e x -l 2 12 720.

w \ / j

12 Find an approximation of the form a + bx for (8 + 3»)*/{l + ( 1 —x) 6 (4 + #)

x being small.

13 Find an approximation of the form a + bx + cx 2 for 1/{(1 —x)<s/(l+x)},

x being small.

14 Find approximately the small positive value of x which satisfies

(1+x)6 = 1-03(1

-xf.15 Neglecting x*, prove that e^logfl+a;) = —log (1 —x). Hence find

approximation to the root of e x log ( 1 + x) = £ which lies between and 1

16 If p is small, prove that x = p —p 2 is an approximate solution of x e x —17 If p is small, prove that a root of «log£c+a; = 1+p is approximately

1 + \p, and find a closer approximation. [Put x = 1 + h, where h is small.]

Calculate the following lirftits.

_ a;-log(l+a;) trx a x —l,

18 lim

^'

. 19 lim (a > 0).

x-+0 x a:->0

Miscellaneous Exercise 12(h)

n 1 2 n

1 Prove y = —r= or (n —r) «.

2 Prove (c + c 1 + ... + c„) 2 = 1 + »»01 + 2 C 2 + . . . + 2 <7 2B .

3 Calculate eg + + . . . + c 2, and prove

c 2 + 2c 2 +...+rac 2 = £n(c 2 + c 2 +...+c 2 ).

4 (i) Verify the identity

(l+x}» = l+x{(l+x) n - x + (l+x) n - 2 + ... + (l+x) + l},

and use it to prove n Gr+1 = »- 1 Cr + n ~ 2 Gr + n ~ 3 Gr + . . . + r Cr for < r < n.

*(ii) Conversely, assuming this relation, prove the binomial theorem.

5 For the series (l) + (3 + 5) + (7 + 9+ll) + (13 + 15+17 + 19) + ...,find

(i) the number of terms in the first r— 1 brackets;

(ii) the first and last terms of the rth bracket;



SERIES 483

Find the sum to n terms and to infinity of

1 1 1 3 56 1 + T+2 +

l + 2 + 3+ ' * 7

2.3.4+

3.4.5+

4.5.6+

8 If r > 0, prove tan-1

(l/2r8

) = tan-1

(2r+ 1) - tan1

(2r- 1). Hence find

9 If n straight lines are drawn in a plane so that no two of them are parallel

and no three concurrent, the number of regions into which the plane is divided

is denoted by f(n). Prove f(n+ 1) —f(n) = n+l, and deduce the expression

for/(n).

10 Prove that (l + 2cosa: + 2cos2a; + ... + 2cosna;)sin£a; = sin.(n + %)x.

^ , , fw sin(n + i)a5 ,Deduc^thatt :

—: dx = it

Jo sm$xwhen n is a positive integer or zero. What is the value when n is a negative

integer?

11 (i) Prove

sin no: (1— = 2cos(n— l)a; + 2cos(n —3) a: + ... + (

suix [2 cos x

according as n is odd or even. Deduce that

%7T sin nx•71

Jo

dx = n or

q 8W.Xaccording as n is an odd or even positive integer.

(ii) Prover n /sin na;\ 8 _

Jo \sina;/

for all positive integers n. [Square the relation in (i), and use Ex. 4 (Z),no. 30 (i).]

sin0 6 6 612 Prove that = 2»cos-cos- ... cos-.

Deduce the values of

n 6 « 1 6 . « 1 6(i) Slog cos- ; (ii) S ^ tan ^5 ( m) S ^tan-.

r=l ^ r=l z z r=l z z

Use Mathematical Induction to prove the following (nos. 13-16).

13 l 4 -2 4 + 3 4 -... to w terms = ( - l) n 1 in(n 3 + 2n 2 - 1).

d n

14 (x*- 1 e 1 ) = ( - l) n eW«/af»+ 1.

dx n

1 1 1

15 If U» = nTi + nT2 + - +*n>



484 SERIES

16 If u n+% —(a + /?) u n+1 + <xfiu n — for w > 1 , then

(a -/?)«„ = (a - 1 - yff - 1) w2 - ayff(a»- 2 - /?»- 2

) u v

17 If s n denotes the sum of n terms of

x 2x 2 4a* 8a 8 16a; 16

+ ^ + ^—: + t— + r— r. + • • •

,

1 + a 1+a; 2 1+cc* 1+* 8 1+a; 1

prove that s n = xj{\ —x) —mx mj{\ —x m) where m = 2 and x =t= 1. Deduce that

the series converges when |a| < 1. [Either use Induction, or take logarithms

and derive the identity obtained from

1 _ 1+x _ (l+a:)(l + a;2

) _1-x ~ 1-x 2 ~ 1-x 1 J

18 Assuming that 1 + —+ ^ + ... - ~,2 a 3 Z 6

•provethat

1 ^ ,111W 1 + 3i +52

+ - = ¥ ; (U) 1 -2 2+ 3i-ii + - =

12-

19 (i) Find the coefficient of a:4 and of x* in the expansion of

(l+x+x st +x 3 + x i)

a.

(ii) How many solutions in non-negative integers has the equation

(a) a + b + c = 4; (6)a + 6 + c = 7?

20 In how many ways can a total of 10 be thrown with three dice of different

colours, the faces being numbered 1, 2, 3, 4, 5, 6? [We require the numberpositive integral solutions not greater than 6 of the equation a + b + c — 10

This number is the coefficient of x 10 in the expansion of

(x + 2 + x 3 + x* + x 6 + a;6

)3

-]

21 If a;2 +px + q = has roots a, /?, prove that (for a range of x to be stated),

-log(l+px + qx 2) = {<x + P)x + \{a?+p 2 )x* + \(a*+P 3 )x 3 +....

22 A circular arc subtends an angle 6 at the centre of the circle, and a, b,

are the lengths of the chords of the arc, off of the arc, and of \ of the arc, respec-

tively. Find numbers x, y, z independent of d, a, b, c such that the length of thearc is approximately ax + by + cz, correct to order 6 e

. [Use the series for sin x.}

Find the sum to infinity of

„_ , 3 3.9 3.9.15 nA _ r + 223 l + r + r-rr + r-rr-^7+-— 24 S-

8 8.16 8.16.24 r(r+ l)3 r

^ 2 4 6 8 ,23425

3+

6l+

7+

9+ -' 26 ^Tl + irzi^'

9 915 9 15 21



SERIES 485

32 S- — (-l^x^l).r(r+l)(r + 2)

/v»4 /y»8 /yil2 /y*3 /yi7 /yillJU JU »V •*/ *Mj U/

33 M 1+35+51+15 + -' < >3 + 7l

+lTl

+ --

34 If p > 1, prove

2 2 , »+l 2 2

Taking p = 26, 31, 49, calculate log 5 to 3 places of decimals. [Choose a, 6, c so

that a log f£ +6 log |K clog f£ = og5.J

35 Denoting by nP r the number of arrangements of n unlike things taken r

nat a time, prove 1 + 2 ^r = the largest integer less than e (n ). [Use Ex. 12 (m),

r=lno. 2.]



486

13

COMPLEX ALGEBRA AND GENERALTHEORY OF EQUATIONS

13.1 Complex numbers

13.11 Extension of the real number system; <J(—1)

Consider the six equations

If we knew nothing about fractions or negative numbers, we should

only be able to assert that (i) and (iv) are both satisfied by x = 3, anthat the others have no solution (i.e. that we could find no natural

numbers which satisfy these).If we are subsequently acquainted with fractions (i.e. if our lis

of numbers is extended by admitting rational numbers), we can sa

that (i) and (iv) are satisfied by f , while (ii) is satisfied by f . The others

would remain insoluble.

Not until signed numbers are introduced can we solve (iii) bx = —f ; and then (iv) is found to have two solutions ± f . We still have

no solutions for (v), (vi).

To enable us to say that (v) has a solution (the necessity for doingthis was indicated in 1.11), we introduce the number *J2 which ca

be proved (see 1.11) to be non-rational. Even then, (vi) has no solution.

It will now be clear that what can be said about solutions of equa-

tions (i)-(vi) depends entirely on what we are prepared to call

'number'. In other words, the possibility of solving a given equation

depends on how far the concept of ' number ' has been extended. Fractional

and negative numbers enable all linear equations with integral

2x = 6,

2x = 5,

2x + 3 = 0,

(i)

(ii)

(iii)

x 2 = 9,

x 2 = 2,

x 2 +l = 0.

(iv)

(v)

(vi)



13.12] COMPLEX ALGEBRA 487

the meaning of 'number', so that a question like 'Does *J(—1) exist?

is futile until we specify what kind of 'numbers ' we are going to workwith.

Hence we have two alternatives: either (a) we leave ^( —l) as ameaningless symbol; or (b) we generalise still further the concept of

'number' to include such expressions as this. It is clear that, unless

we choose (6), we can never have any complete theory even of quad-

ratic equations. To arrive at our generalisation we follow the historical

sequence of development; we use the symbol i as an abbreviation

(introduced by Euler) for ^( - 1) in 13.12, 13.13.

13.12 First stage: formal development

At first i was regarded as something mathematically disreputable

whose use, nevertheless, gave results quickly and simply to problems

which could be solved more laboriously by other (respectable) means.

The procedure was formal: i was treated like an ordinary algebraic

symbol (with the peculiarity of satisfying i2 ——1), and was used in

the spirit of 'press on and see where we get'. Cardan (1501-76) was

probably the first to do this.

Accordingly we should have (a, b, c, d being 'ordinary numbers'):

(a + bi) + (c + di) = (a + c) + (b + d)i (i)

by regrouping and taking out the factor i;

(a + bi) (c + di) = ac + adi + bci + bdi 2

= (ac —bd) + (be + ad) i (ii)

by first multiplying out the brackets, and then using i 2 = —1 and

grouping.

Also, if a + bi = c + di, then a —c = (d —b)i. Squaring both sides

gives (a —c) 2 = —(d —b) 2, which would assert that a positive number

is equal to a negative number and is a contradiction unless both are

zero, i.e. a —c = and d —b = 0. Hence

if a + bi = c + di, then a = c and b = d. (iii)



488 COMPLEX ALGEBRA [13.13

ordinary numbers and expressions weref called real when contrast

was desirable.

This intuitive approach, however unsystematic it may appear, di

in fact lead to many advances because it produced results whichcould be verified subsequently by orthodox means.

13.13 Second stage: geometrical representation

In the spirit of 13.12 we see that if any real number a is multiplied

by i, we obtain an imaginary number ia; multiplying again by i, wget iia = i

2 a = —la = —a, which is again a real number. Thus tw

successive multiplications by i convert a real number into its negative.

— 4 \ J. ^-a O a x

Fig. 125

Consider now the geometrical representation of this result. Suppose

for definiteness that a is positive, and let OP be the corresponding

segment of the a;-axis. Then two multiplications by i have the effect

of rotating OP, originally along the positive asaxis, through two right-angles (say in the counterclockwise sense, to agree with trigonometrical

conventions) to position OP'. It is therefore reasonable to interpret

one multiplication by i as turning OP counterclockwise through on

right-angle. Hence the imaginary number ia is represented by a poin|;

P at distance a, measured along Oy (fig. 126). Similar considerations

apply when a is negative.

This geometrical representation was published in 1806 by Argand,

although others had the idea a little earlier. In it we are regardingnot as a number at all, but as an operator which rotates lines Ocounterclockwise about the origin O through a right-angle. (Observe

that ordinary numbers can also be interpreted as operators in the

process of multiplication: the 2 in 2a as the operator which doubles

the length of OP and preserves the sense; the —2 in —2a as the

operator which doubles the length of OP and reverses the sense;

and so on )




length a along Ox, followed by a turn counterclockwise through one

right-angle and a step of length b parallel to Oy, we end up at what

we should refer to in graphical work as the point (a, b), say Q (fig. 127).

Although the interpretation of * as an operator and Argand'srepresentation of a + bi both have important applications, the signi-

ficant thing at this stage is that we are led away from the vague

'imaginary expression' a + bi towards the familiar and precise idea

P

ia

\\

a P X a

Fig. 126 Fig. 127

of the coordinates of a point (a, b) in a plane, i.e. a pair of numbers

a, b taken in a definite order. It is from this notion that we can re-

construct the theory of 'imaginary' numbers on a logical basis.

13.14 Third stage: logical development

(1 ) Our aim now is to make a fresh start by setting up and developing

an algebra of ordered number-pairs^. That is, we begin with the definite

idea of a pair of numbers, and as far as possible try to retrace our

steps to get at * indirectly. This approach is contained in the work of

Gauss (1777-1855) and others.

We write our number-pairs, usually called complex numbers, as

[a, b] where a, b are ordinary (real) numbers. As we are about to invent

a new kind of algebra, we can make the rules as we please provided

they do not contradict each other or lead to contradiction. However,

our purpose is to make the new algebra as much like ordinary algebra

as possible; and so, with a view to preserving the structure of the formal

results (i)-(hi) of 13.12, we define the fundamental operations with




(b) Addition of number-pairs.

[a, b] + [c, d] is to mean [a + c, b + d],

(c) Multiplication of number-pairs.

[a, b] x [c, d] is to mean [ac —bd, be + ad].

Since subtraction is the opposite of addition, we can define

[a, b] —[c, d] to mean the number-pair [x, y] such that

[x,y] + [c,d] = [a,b].

By (6) this gives [x + c, y + d] = [a, b],

which by (a) implies

x + c = a and y + d = b;

hence x = a —c, y = b —d,

so that [a, b] —[c, d] = [a —c, b —d].

Similarly, the quotient [a, b] -r- [c, d] can be defined in terms

multiplication to be the number-pair [x, y] such that

[x, y] x [c, d] = [a, b].

By (c) this gives [xc —yd, yc + xd] — [a, b],

and so by (a) xc —yd = a and yc + xd = b.

If c, are not both zero, we can solve these simultaneous equations

for x, y: , , , 7* ac + bd _bc —adX =

~& + ¥' V =~c? + d?'

Hence if [c, d] # [0, 0],

r n r Vac + bd bc-ad~][a,6] + fcfl-L ?Ta? ,^ T? J.

(2) By using the above definitions we can show that complexnumbers obey all the laws of ordinary algebra. For example,

*/ [ a > x t c > ^] = [0j °L ^ eri t a> ^] or t c

> ^] or w*W &e [0, 0].




Hence at least one of the real numbers a 2 + b 2, c 2 + d 2 must be 0; and

this implies a = = b, or c = = d, or both. The result follows.

See also Ex. 13 (a), nos. 19-25.

(3) Application of definitions (b), (c) to number-pairs of the form

[a,0],[c,0]gives [M ] + [c,0] = [a + e.0],

[a, 0] x [c, 0] = [ac, 0].

Apart from the presence of the brackets and the second part in

each pair, these results are exactly like the laws of addition and

multiplication of the ordinary numbers a, c: the laws of combination

of complex numbers [a, 0] and real numbers a are structurally the same.fOwing to this fact, we agree to abbreviate [a, 0] to a, and in particular

we write 1 for [1, 0]. (There is a precedent for this: we agree to write

2 for +2, etc., because the (unsigned) natural numbers and the

positive signed numbers behave exactly the same algebraically.)

In definition (c) let us now choose a = 0, c = 0, b = l,d = 1. Then

[0, l]x[0, 1] = [-1,0],

i.e. [0,1? = [-1,0];

and according to our agreement about abbreviation of the right-hand

side, this can be written r _ , n9 ,

[0, l] 2 = - 1.

Thus in the algebra of number-pairs there is an element whose square

is equal to —1, viz. [0, 1]. Let us abbreviate [0, 1] to i; then we have

number-pairs i and —1 for which i2 = —1.

By definition (6),

[a, 6] = [a, 0] + [0,6]

= [a, 0] + [b, 0] [0, 1] by definition (c),

= a + bi by our abbreviations.

If we now rewrite definitions (a), (b), (c), replacing [a, b] by a + bi,

etc., we recover the results ( ii i) , ( i) , (ii) respectively in 13.12; but now




Since [b, 0][0, 1] = [0, 1][6, 0], as is easily verified, we can write

[a, b] as a + ib instead of a + bi when convenient. Our conventions

amount to writing a for a + Oi, i for + li, and bi for + bi.

(4) Conclusions. In the algebra of real numbers ('ordinary algebra

V( —1) does not exist. In complex algebra (the algebra of ordered

number-pairs) there are pairs abbreviated to i, —1 which do satisfy

i 2 = —1, so that this pair i has the property which*J(

—1) would hav

(if it existed) in real algebra.

Correct results in complex algebra are obtained by applying th

laws of ordinary algebra to the number-pairs in their abbreviated

form a + bi, with the replacement of i2 by —1 whenever it occurs.

That is, the intuitive work of 13.12 is correct, provided we realise

that we are handling abbreviations for pairs.

Remark. The relations 'less than', 'greater than' are not defined

complex algebra; e.g. it is meaningless to write 3 + 2i < 4 + 5i.

particular, the terms 'positive', 'negative' cannot be applied

complex numbers.

13.15 Importance of complex algebraComplex algebra provides a method for proving results of real algebra.

This is so for two reasons.

(i) A single equation in complex algebra is equivalent to twequations in real algebra, by definition (a) in 13.14. Hence pairs

results can be obtained from one calculation, thus making for economy.(ii) From any general relation between complex numbers there c

be obtained a special case by taking the second parts of all numbers

to be zero. Because of the exact correspondence between complexnumbers [x,0] and real numbers x (13.14(3)), we deduce a relation

in real algebra.

These remarks will be illustrated in this chapter and the next.

Complex algebra offers generality and completeness. For example, wcan assert that in complex algebra every quadratic equation has twroots (possibly equal).

Since complex numbers arose from the attempt to solve all quadratic




has n roots (i.e. the roots are themselves complex numbers). Thuscomplex algebra has a finality which the real number system lacks,

and no further generalisations can be obtained in this direction.

13.16 Further possible generalisations of 'number'

(1) Ordinary algebra is 1 -dimensional, i.e. any real number can be repre-

sented by a point on a line, say the axis of x.

Complex algebra is 2-dimensional, since a complex number is representedby a point in a plane, with reference to two axes.

Is there a 3-dimensional algebra —an algebra of ordered triplets [a, b, c]?

Hamilton (1805-65) considered this problem and showed that no genuine'algebra' could be constructed, although a system of elements (which may becalled vectors) of the form ai + bj + ck where a, b, c are real numbers and i, j, kdenote [1,0,0], [0,1,0], [0,0,1] would obey some of the usual laws, and is

useful in 3-dimensional mechanics., for example.However, Hamilton did construct an algebra of ordered quadruplets [a, 6, c, d]

or quaternions, in which all the usual laws are obeyed except that xy =# yx. Wecould similarly think of an 'algebra' of n-vectors [a x ,a 2 , ...,a„] or even of

infinite sequences [a v a 2 , a 3 , . . .].

(2) Further, we may consider arrays with more than one row, e.g.

where the elements are real numbers. With suitable laws of combination, thealgebra of rectangular arrays or matrices would be obtained. Although such analgebra was originally investigated for its own sake, it is now widely used,

e.g. in theoretical physics and much geometry. Nor is there need to restrict

the elements to be real.

The reader must seek elsewhere for development of these suggestions.

Exercise13(a)

Simplify and express in the form X+Yi:1 (2 + 7i) + (3 + 4*). 2 (2 + 7*)-(3 + 4i). 3 (2 + 6t) + (5-3t).

4 (2 + 6i)-(-5 + 3*). 5 (3 + 5t)(2 + 3i). 6 (4 + 5i) (4- 6»).

7 (3 + 5i)-j-(2 + 3»). 8 (6 + 5*)/(3 + 2*).

9 ( 1 + i) 2, and hence ( 1 + i)

1, ( 1 + i)«.

10 (l+*)(l + 2i)(l + 3i). 11 (cos0 + *sin0)(cos0 + isin^).

12 (cos0 + isin0)/(cos0 + *sin0).




14 If Z = X+ Yi and Z = (z + l)l(z—\) where z = x + yi, express X, Yterms of x and y.

15 Complex numbers z u z 2 are given byi

z1

=R 1 +

i(oL, z2

=R 2

——g,Ill

and z is denned by - = —I—

.

Z Zi z%

Find the values of o» for which z is of the form x + 0*.

16 If J(a + bi) = x + yi and x > 0, prove x % —y % = a 2 and 2xy = 6. Henceexpress ^/(3 + 4i) in the form x + yi.

17 Express (x + yi) 3 in the form X+ Yi. Write down the correspondingexpression for (x —yi) 3

.

18 Prove {£( - 1 ± i V3)}3 = 1.

B?/ direct appeal to the definitions (a)-(c) in 13.14(1), verify that complexnumbers satisfy the following algebraic laws.

19 [a, b] + [c, d] — [c, d] + [a, 6] (commutative law for addition).

20 [a, 6] x [c, c?] = [c, d] x [a, 6] (commutative law for multiplication).

21 ([a, 6] + [c, cZ]) + [e,/] = [a, 6] + ([c, cZ] + [e, /]) (associative law for addition).

22 ([a,b]x[c,d]) x[e,f] = [a,b] + ([c,d]x[e,f]) (associative law for multi-

plication).

23 ([a, 6] + [c, d]) x [e, /] = [a, 6] x [e, /] + [c, rf] x [e, /] (distributive law).

24 [a, 6] + [a, 6] + . . . tow terms = [a, 6] x [w, 0], each being equal to \na, nb].

25 By using the multiplication rule, verify that —i = [0, —1] satisfies

x*+l = 0.

26 Consider ordered number -pairs {a, 6} of integers (positive, negative

zero, but with 6 4= 0) which satisfy the following laws:

(a) {a, b) = {c, d} if and only if ad = be;

(b) {a, 6} + {c, d} = {ad + be, bd};

(c) {a, 6} x {c, d} = {ac, bd}.

Obtain expressions for {a, 6} —{c, d} and {a, 6} + {c, d}. Verify that this algebrais isomorphic to (i.e. the same in structure as) the algebra of rational numbers(fractions). (We could define fractions abstractly in terms of integers in thi

way.)

13.2 The modulus-argument form of a complex number

13.21 Modulus and argument

We i 13 13 h h l b + + b




When x + yi is so expressed, and r is restricted to be positive (cf. 1.62),

then r is called the modulus and 6 the argument of x + yi, respectively

written \x + yi| , arg (x + yi)

.f We may think of x + yi as ' the number r

in direction 6\ a generalisation of the idea of 'directed number' as

presented in elementary algebra.

It is customary to write z forx + yi, and to write r = |z|, = argz;

z may be called the number of the point P in Argand's diagram, and

sometimes P is called the affix of z.

Frequently it is necessary to express a complex number, given in

the form x + yi, in terms of its modulus and argument. From equa-

tions r = +J{x* + y*) (iii)

since r is restricted to be positive; and

cos0:sin0: 1 = x:y: + j(x 2 + y2

), (iv)

which gives a unique value of 6 in the range —it < 6 < + tt. Any value

differing from this by any positive or negative integral multiple of

2n would give the same point P and consequently is a possible value

of argz. Thus argz is a many-valued function, and we define its

principal value to be that in the range —ttkO^tt. (The single

equation tan# = yjx does not determine argz because it gives twovalues for 6 in this range.)

Examples

(i) Express (5 + i)/(2 + Zi) in modulus -argument form.

5 + i _ (5 + i)(2-U) _ 13-13t _2 + 3*

~(2 + 3i)(2-3i) ~ 4 + 9 ~ *'

We require r, such that rcos0 = 1, rsin0 = —1. Hence r = +<J2, and is

given by cos0:sin0:l = 1: —1:^/2; the principal value of the argument is

therefore —\n. The required result is

{cos ( —\n) + i sin ( —In)}.

(ii) Express 1 —cos —i sin in modulus-argument form.

1 —cos0— *sin0 = 2sin 8 £0 —2isin£0cos£0

= 2 sin £0 (sin £0 —*cos£0)

= 2sin£0 {cos ( \d - \n) + * sin (|0 - \n)}.




In numerical cases like ex. (i) the modulus and argument can usually

be written down easily after drawing a sketch.

We remark here that the complex number z, represented by t

point P(x,y),

can also be associated with the displacementOP

in t

sense from towards P. Given the axes, there is an exact correspond-

ence between the complex number x + yi, the point P(x,y), and t

vector OP. Also see 13.32.

13.22 Further definitions, notation, and properties

(1) We agreed in 13.14 (3) to abbreviate a complex number z of t

form x + Oi to x. It is customary to go further and call z ' real ',

although

this double use of 'real' is most unfortunate. Similarly a numberthe form + yi is shortened to iy and called purely imaginary.

If z is 'real', then it is represented by a point on the x-axis; aargz = or tt according as z (i.e. a;) is positive or negative.

If z is purely imaginary, it corresponds to some point on the y-axis

and arg z — ± \n according as y < 0.

(2) Real and imaginary parts. In the complex number z = x+yi

x is called the real part (or first part) and is sometimes written x = 8%{zand y is called the imaginary part (or second part), written y = J(z)

The process of deducing from the relation z x —z 2 that x x= x % a

Vi = 2/2 i s called equating real and imaginary parts, respectively.

If z lf z 2 are expressed in modulus-argument form as

r(cosd + ismd), s(cos0-Msin^),

then from r(cos 6 + i sin 6) = s(cos )

it followsby equating real and imaginary parts that

r cos 6 = s cos <j> and r sin 6 = s sin <j>.

Squaring and adding gives r 2 = s 2, so that r = s since r, s are positive.

The two equations now become

cos d = cos<f),

sin 6 = sin <j>,

so that in general 6, <f> differ by some integral multiple of 2ir. If 6,

denote the principal values of argz argz 2 then 6 =




(3) Conjugate complex numbers. Given a complex number z = x + yi,

the number x —yi obtained by changing the sign of i is called the

conjugate of z and is written z (or sometimes z* when the previous

notation is required to denote 'average value'). Conjugate complexnumbers have the following properties.

(i) The product zz = (x + yi) (x —yi) = x 2 + y* = |z|2 is 'real'.

(ii) The sumz + z = (x + yi) + (x-yi) = 2x = 20t(z) is 'real'.

(iii) The difference z —z — (x + yi) —(x —yi) = 2yi = 1iJ(z) is purely

imaginary.

(iv) \z\ = |z|; but (for principal values) argz = —argz, except

when z is 'real and negative', i.e. of the form x + Oi where x < 0.

In the a;y-plane the points which represent z, z are images in the #-axis.(v) If z = z, then z is 'real

'; if z = —z, then z is purely imaginary.

If z x = z a , then x x = x 2 and y x = y % \ hence x 1 —iy 1 = x 2—iy 2 , i.e.

z x —z 2 . The process of deducing from z x= z 2 that z x

= z 2 is called

taking conjugates.

Finally, the definition of 'conjugate' shows that z x + z z= z x + z 2 ,

(z x z 2 ) = z x z 2 , (z x \z 2 ) = z x \z 2 , (z) = z.

(4) Properties of\z\. From the relation \z\ 2 = zz in (3) it follows that

I ^1 ^22 = Z X Z 2^ Z l Z i) ~ Z 1 Z \' Z 2 Z 2

=|

Z l|2,

l^l2.

Hence\

z i z %\ =|

z i|-| z 2|>

since both sides are positive by definition. Similarly, if z 2 #= 0,

Zl|

2

r= (Zl

) (Zl

) = Zl * 1 = lZl

l

2

Z 2| \ Z 2/ \ Z 2/ Z 2 Z 2 \

Z z\2>

IZ

\Z I

— =z 2\ |

Z 2|

13.23 The cube roots of unityf

If z is a number whose cube is + 1, then z 3 —1 = 0, i.e.

(z-l)(z 2 + z+l) = 0;

hence either z - 1 = or z 2 + z+ 1 = by 13 14 (2) From the latter




Also, cosfTr = —| and sin \-n = —\ ^3, so the second root

cos f 71 + i sin f n. It is also w, as is clear from the original formula for

^ ow(jfi — (cos f 77 + i sin f it)

2

= cos 2f 7T —sin 2 §7r + 2i sin §7r cos §7r

= cos jn + i sin f /r.

Hence the roots of z 2 + z + 1 = can be written z = to, w2. There a

thus three cube roots of + 1 in complex algebra, viz. 1, (o, a>2

. Threlations „ , 2 . , , A 2 _

often simplify calculations.

Example

Direct expansion shows that

a b c

c a b = a 3 + 6 s + c 8 —3abc.

b c a

The operation c x c x + c 2 + c 3 reveals the factor a + b + c.

Since the expression on the right can be written

a 3 + (bw) 3 + (ceo 2)

3 - 3a(b(o) (co 2),

similar reasoning reveals the factor a + bo) + co) 2. Likewise, since the expression

is also a 3 + (6w 2)

3 + (cw) 3 —3a(6w 2) (ceo), a + 6w 2 + ceo is a factor.

We now have three complex linear factors, so that any further factor mube numerical. Comparison of the coefficients of a 3 shows that it is + 1. Hence

a s + b 3 + c s -3abc = (a + b + c) (a + bo) + cco 2) (a + ba> 2 + c(o).

By comparing this result with

a s + b 3 + c 3 -3abc = (a + b + c)(a 2 + b 2 + c 2 -bc-ca-ab)

(see 10.22 (2), example; or Ex. 11 (a), no. 20), it follows that

a 2 + b 2 + c 2 -bc-ca-ab = (a + bw + cio 2) (a + b(t)

2 + co)).

Exercise 13(6)

Express in modulus -argument form:

1 l+iy/3. 2 1-*V3- 3 4 1-i.

6+iJ3 2+i5 (l + + 6 V

7 (l+i)^




12 If|

(z - 1 )/(z + 1 ) |= 2, prove that the point P which represents z lies on the

ciTcle ^+^+^+1 = 0.

13 If w = z/(z + 3) where w = u + iv and z = x + iy, and if z lies on the circle

(x + 3)2

+ y2

= 1,

find the locusof

14 Prove that 1 a) to2 1 1 1

to 2 1 to = 0, 1 to w2

0) w2 1 1 to 2 to

15 Prove to2n + to + 1 = 3 or according as n is or is not a multiple of 3.

16 Express x 3 + y3 as the product of three linear factors, and deduce the factors

of x 2 -xy + y2

. [Method of 13.23, ex.]

17 Prove (a + <ob + w2 c) 3 - (a + to 2 6 + wc) 3 = - 3i • (6 - c) (c - a) (a - 6). [Use

factors of x 3 —y3

.]

18 Prove that (a 2 + ah + 6 2) (a;

2 + a*/ + «/2

) can be written in the form

[Use tf + ab + b 2 = (a- <ob) (a - (o 2 b), etc.]

*19 Prove that (a 3 + b 3 + c 3 -3abc)(x 3 + y3 + z 3 -3xyz) can be written in the

form X3 + Y3 + Z 3 —3X YZ, where either

(i) X = ax + by + cz, Y = cx + ay + bz, Z = bx + cy + az,

or (ii) X - ax + cy + bz, Y = cx + by + az, Z = bx + ay + cz.

[In (i), (a + ba) + <xo*)(x + y<o 2 + z(o) = X+ Ya) 2 + Z(o, etc.]

*20 If a r (r = 1, 2, 3) denote the cube roots of - 1, prove

(l+xx^x^Dil+x^ + x^Dil+x^+xâl) = (l+x 3 )K

[A typical factor is ( 1 - tc8 a 3

)/( 1 - xa) = ( 1 + « 3)/( 1 - aa).]

21 If the equations x 3 = 1, gkc s + 6a; + c = have a common root, prove

a 3 + b 3 + c 3 -3abc = 0.

[One of the three roots 1, to, to 2 of a:3 = 1 satisfies the other equation, so

(a + b + c) (aw 5 + bo + c) (aw 10 + ba> 2 + c) = 0.]

13.3 Applications of the Argand representation

13.31 Geometrical interpretation of modulus and argument

If z is represented by P, then \z\ = J(x2 + y

2) is represented by OP.

Also .

=|

(a? x —» a ) + - y 2 ) |




If is the principal value of arg z, then it is given by

cos 6 : sin 6 1 = x : y : *J(x2 + y

2)

and is the angle of turn from the direction Ox towards the direction

OP, which can be written L {Ox, OP). Similarly, arg (z x—z 2 ) is the angle

defined by

cos 6 : sin 6 : 1 = x x - x 2 : y x - y 2 : <J{(x x- x 2 )

2 + (y x - y 2 )2

},

and represents the angle of turn from Ox towards P2 PX , written

L(Ox,P %Px ).

Examples (some loci)

If P x is the fixed point z x and P is the variable point z, then

(i) the locus \z —z x \

—cis the circle with centre P x and radius c ;

(ii) the locus arg(z —z x ) = ex, is the half -line P X P for which /_(Ox,P x P) is

(The other half is given by arg (z —z x ) = n + a.)

(iii) If P x , P 2 are fixed points z x ,z 2 , the locus \z —z x \

= \z —z 2 |is the perpen-

dicular bisector of the line P X P 2 - For PP X = PP 2 , and the result follows by purgeometry. Alternatively, it can be proved algebraically from

(x-xJt + iV-Vi)* = (z-*a) 2 + (2/-2/ 8 )a

by simplifying.

(iv) If P, Q are the points z, 2z + 3 —4i and P moves on a circle of centre anradius r, find the locus of Q.

Write w - 2Z + B-U. Then tt> - 3 + 4» = 2z, so

|w-(3-4*)| = \2z\ = 2r

since \z\ = r. Therefore Q lies on the circle of centre (3, —4) and radius 2r.

13.32 Constructions for the sum and difference of z lf z 2

(1) Sum. Let Px , P2 correspond to z x , z 2 , and complete the parallelo-gram P2 OPx P3 . Then

x 3 = projection of OP3 on Ox

= projection of OPx + projection of PX P3

= projection of OP\ + projection of OP2 f



COMPLEX ALGEBRA 5013.33J

Observe that the construction for P3 is the familiar parallelogram

law for compounding ('adding') two displacements OP v OP 2 . Tothe sum of two complex numbers corresponds the sum of the associated

vectors.

(2) Difference. Since z x—z 2

= z x + ( —z 2 ), we first construct the

image P 2 of P2 in 0, which represents —z 2 . The above sum-construc-

tion performed with Px ,P 2 will give P4 ,

corresponding to z x—z % .

y,

^ /L

//

//

X

.J/

p*

p*

Fig. 129

As OP± is equal to, parallel to, and in the same sense as P2 Pl5 wecan represent z x

—z 2 by the displacement P 2 P v

13.33 The triangle inequalities

By applying the theorem that 'the sum of two sides of a triangle is

greater than the third side' to triangle OPx P3 , we have

OP3 ÔP x + Px P3 ,

i.e.f OP3 ÔP x + OP2 ,

and hence |z, + z 2 \ ^ \z t \ + \z 2 \.

From this we deduce the following, exactly as for real numbers

\z x -z 2 \ > \\z x \-\z 2 \\,

|z 1 + z 2 | ^ ||^l| I^H'




with equality when and only when z is ' real and positive i.e. of the form x +where x 2s 0. (Similarly J{z) < |z|.) By 13.22(3),

h + Z 2 |

2 = (Zl + Z2)(2l+2 2 )

= %\Zx

-\- Zg

+Zj Z

2-f- Z

2Z

2

= Iz^ + ^z^R |z 2 |*

since z x z 2 and Z\Z % are conjugate complex numbers. As M{z x z 2 ) < |ziZ 2 |,

KI 2 +2M 2 | + N 2

= (Nil + W) 2» since ^1 = . |z 2 |

= \z t \. |z 2 |.

Therefore \z x +z 2 | < ja^J + |z 2 |,

positive signs being chosen because the modulus of a complex numberpositive by definition. Equality occurs if and only if z x z 2 is 'real and positive',

i.e. zx

= a 2 z2

where a is 'real'.

The other results can also be established similarly.

13.34 Constructions for the product and quotient of z„ z 2

It is convenient to use the modulus-argument form.

(1) Product. Let z x — r(cos 6 + i sin 6), z 2 = s(cos<f> + i sin (j>), so that

Px ,P2 have polar coordinates (r, 0), (s,

<f>).Then

zx

z2

= rs(cos 6+

i sin6)

(cos(p +

i sin <j>)

= rs{(cos 6 cos <j> —sin sin 0) + *( sm # 008 ^ + 008 # sin (j>)}

= rs{cos (0 + <p) + i sin ((9 + <j>)}.

Therefore z x z 2 is represented by the point which has polar coordinates

{rs,d + (f>).

To construct this point, let A be the point whose polar coordinates

are (1,0). Draw triangle OAPx , and construct triangle OP2 P3 directly

similar to

OAP1(fig. 130). Then

§i=

6J'80 op s = o Pl .OP^ rs .

Also x6p 3= #ap 2 + p 2 6p 3 = a*5p 2 + ^0Pi = <p + e.

Hence P3 is the required point.

The above calculation verifies the result \z 1 z 2 \= Iz-J . |z 2 |, and also

shows that (cf. the end of 13.22 (2))



13.34] COMPLEX ALGEBRA

and the principal values are

503

arg Zi = arg z 2 = f n, arg {z x z 2 ) = - § n.

Weobserve that multiplication of a complex number z

2by

r(cosd + i&md) turns the corresponding vector OP 2 through angle

and multiplies its length by r.

(2) Quotient. If z 4 = z 1 /z 2 , then z x = z 4 z 2 , and so we can use the

product-construction in (1), making Px take the role of P3 there.

We therefore construct triangle OPx P2 , and then make triangle

OPÂ directly similar to this (fig. 131).

Alternatively, the construction can be obtained from

Zj _ r cos 6 + i sin

z 2 <scos0 + isin0

_ r (cos# + isin#) (cos0 —isin^)

s (cos <j> + i sin $) (cos<f>

—i sin ^)

r (cos cos <j> + sin sin 0) + i(sin cos —cos sin <j>)

~ s cos 2) + i sin (0 - (j>)}.

s






It is easy to prove that, if the centre is O, the radius OH of the circle is given by

OH* = OA. OB.

For since AH/HB = AK/KB, it follows that AHjAK = HB/KB, and so

OA-OH OH -OB . 20 _ 2 OHOA + OH~ OH+ OB'

i e* 20H~ 20B

by properties of equal ratios, which gives the result.

Fig. 134

We can now show that the circles in (a), (b) cut orthogonally (see 15.66(1)).

If T is a common point, then OT 2 = OA . OB. By the converse of the tangent-

chord theorem, OT touches the circle (a). Hence the radius GT of (a) is per-

pendicular to OT, and so touches (6) at T. Thus the tangents to (a), (6) at Tare perpendicular, i.e. the circles are orthogonal.

*(ii) Prove that the triangles P^P^P^, P 4 P S P 6 are directly similar if and only if

Zi 2 2 Z 4 Z s

(a) If the condition holds, then by taking moduli, P 1 PJP 3 P 2 = P4PJP 6 P 6 ;

and by taking arguments and using the Remark in 13.34,A A

•f8-P 2-^*1

=P 6-P &P

and are in the same- sense. Hence by the test 'common angle and containing

sides proportional', triangles P^^P^P^, P 4 P 6 P 6 are directly similar.

(6) Conversely, if the triangles are directly similar, then the above tworelations hold, so that (z 1

—z i )((z 3 —z 2 ), (z 4 —z 5 )l(z 6—z 5 ) have the same modulus

and the same argument. They are therefore equal.

The condition can be written

2„-2

= 0, i.e.

z 1—z 2

Z4-Z5 Z 6- Z 5 = 0.



506 COMPLEX ALGEBRA

Exercise 13(c)

1 Indicate in a diagram the points representing £(z x + z 2 ), z x + 2z 2 , z x—2z 2

2 If collinear points Px

, P2

, P3

are such that PxP

2= 2P

2P

3, find the relation

between z x , z 2 , z 3 .

3 Write down the complex number represented by the point dividing P X

in the ratio k : I.

4 What is the locus of P if 2 < |z + 1 - 2i\ < 3?

5 Find the greatest and least values of (i) \z —4| if \z\ ^ 1; (ii) |z + l|

|z-3| ^ 5.

6 Prove that |z + z 2 |

2 + |zi-z 2 |

2 = 2 {| z i|2 + \

z s\2

} (i) geometrically, (ii) alg

braically. [For (i) use the theorem of Apollonius on triangle OP 3 P 4 in fig. 129

7 (i) If |zj. + z 2 | = \ z i~ z a|» prove that argz x and argz 2 differ by \it or(principal values intended), (ii) If arg{(z 1 + z 2 )/(z 1 — 2 )} = \n, prove |Z]J = |z

[Treat geometrically.]

8 Verify the following construction for the roots of z 2 —2az + b 2 = 0, whea, 6 are real and < a < b. With the origin O for centre, draw a circle of radius

From the point A on Ox at distance a from O draw a perpendicular to Ox cutting

the circle at P, Q. Then P, Q represent the roots.

9 Verify the, following construction for *Jz. Let A be (1,0); produce PO to

so that OB = 1. Through O draw a line parallel to AB to meet the circle PAin Q and R; these are the required points.

10 Given the points representing z x , z 2 , construct those representing the twvalues of V( 2 i z 2)-

11 P x is any point, and on the circle on OP x as diameter points P 2 , P 8

chosen so that PxOP 2 = P 8 OP 8 = <}>. Prove z\ cos 20 = z x z z cos 2 0.

12 If G is the centroid of triangle P X P 2 P 3 and 4z x + z 2 + z 3 = 0, prove that

bisects P x (?. [Use Ex. 15 (a), no. 1.]

13 If A is real and z = 4A + 3*(1 —A), prove that P lies on a straight lin

As A varies, prove that the least value of |z| is 2-4, and interpret geometrically.

14 If z x — 2 = z 4 —z 3 , prove that P x P t P i P l is a parallelogram and that

point \{z x + z 2 + z 3 + z 4 ) is its centre.

15 Show that the point P 2 , where z 2 = (l+AiJz and A is real, lies onperpendicular at P x to OP x . Find A if argz 2 = argZj^ + i^r, principal value

being intended.

16 If z x , z 2 are the roots of z 2 —az + a? = 0, where a is a complex number, prov

that the points z x , z 2 are the vertices of the equilateral triangles drawnopposite sides of the line 0.4.

17 ABO is equilateral, with sides of length 1 and centroid at O. If A repre

t what d B C t? Show th t the sum of th three numbers




20 Interpret geometrically the equation

*21 On the sides of triangle ABC are drawn triangles BOX, CAY, ABZdirectly similar to each other. Prove that the centroids of ABC, XYZ coincide.

[By 13.35, ex. (ii) and properties of equal ratios,

x —c y —a z —b So; —Sa

b —c c —a a —b '

so Hoc = Sa.]

*22 For any complex number z prove that the triangle with vertices zz x , zz 2 ,

zz z is directly similar to the triangle with vertices z lt z 2 , z 3-

23 If P1P3 and P 2 P 4 are equal and perpendicular, the sense of rotation from

P X P S to P 2 Pi being clockwise, prove z 1 -iz 2 -z z + iz i = Q, and conversely.

[|z 4 -z 2 | = Iza-Zilandargftzs-ZiVî-Sa)} = + \tt* Hence (z 3 - Zi)/(z 4 - «,) = <•]

24 On the sides of a plane convex quadrilateral, squares are drawn externally.

Prove that the centres of these squares are the vertices of a quadrilateral whose

diagonals are equal and perpendicular. [Let the vertices in clockwise order be

Zi, z 2 , z s , z 4 . The centres of the squares are £(z x + 2 ) + |i(z 2 - z x ), etc. Use no. 23.]

25 By considering the modulus of the left-hand side, prove that all the roots of

z n sin na + z n_1 sin (n —l)a + ... +zsina = 1

lie outside the circle |z| = -J.

[1 = |z n sinwa + ...+zsina| < |z|»+ ^l -^ ... + |z|. If a root z satisfies

|z| < \, then for this z, 1 < (£)n

+ (i) -1

+ . . .

+ & = 1 - {\)n

, a contradiction.]

13.4 Factorisation in complex algebra

13.41 'The fundamental theorem of algebra'

We remarked in 13.15 that complex numbers cannot be generalised

by attempting to solve equations of degree higher than 2. This is a

consequence of the following theorem.

In complex algebra the polynomial equation

p(z) = p z n +p 1 z n ~ 1 + . . . +p n = (p Q *0) (i)

has AT LEAST ONE TOOt.

This result, known also as d'Alembert's or Gauss's theorem, has

not yet been proved by any strictly algebraical argument. We shall

assume it here; the reader may later study a proof which depends on

the complex integral calculus.




In complex algebra the equation (i) has exactly n roots.

Proof. By the fundamental theorem there is a root z —<x x , i.e.

The equation (i) is therefore equivalent to

Po(*n ~ «?) +Pi(z n ~ 1 - a? 1

) + . . . +p n - 1 (z - cc x ) = 0;

i.e. since z —ol 1 divides each bracket, to

(z-<x 1 )f 1 (z) = 0,

where f x (z) is a polynomial of degree w-1 in z, whose first term

PqZ 71-1 (as is clear by starting the division).

Again by the fundamental theorem, f x (z) = has a root z = a 2 (n

necessarily distinct from ol x ), and by similar reasoning

/i(z) = (z-oc 2 )f z {z),

where 2 (z) is a polynomial in z of degree n —2, beginning with p z n

Continuing thus, we find

p(z) = (z- ai )(z-a 2 ) ... (z-a^J/^z),

where n-1 (z) is a polynomial of degree 1 in z, and begins with p z;

must therefore be of the form p z + k, or p (z —cc n ) say. Therefore

P(z) = Po( z ~ a i) (2 - a 2 ) . . . (z - a J.

This shows that p(z) has n factors linear in z; or equivalently, tha

the equation p{z) = has n roots a l5 a 2 ,...,a n (which may or m

not all be distinct). Also, (ii) shows that these are the only values o

which make^>(z) zero; for if /? is such a value, then

i>o(A-ai)(A-a 2 )-(/?-aJ = 0,

and since p 4= by hypothesis, at least one of the factors fi—cx,^

f}—cc 2 , fi —a n must be zero (13.14(2)), i.e. /? coincides with an

Hence p(z) = has exactly n roots.

If the roots are not all distinct, (ii) takes the form

p(z) = p^z - tj** ( - a )»« ( - a ) * (i




the notation has been chosen so that fi x = oc x , /? 2 = a 2 , then (iv) can be

wittenp(z) = p (z - a x

) »* (a - a 2)«* . . . (z - a*)**, (v)

It now follows that mr = n r (r = 1,2, ...,&)• E.g. if 7% < n x , then from (iii) and

(v) we should obtain, after dividing outp

(z —oc x )mi

:

(z - oc x )n^ (z - a 2 )

n « . . . (z - a ft )n * = (z - a 2 )

TO > . . . (z - a fc )m*,

which is impossible since the left-hand side is zero when z = cc x but the right-

hand side is not.

13.43 Principle of equating coefficients

We can now deduce results for complex polynomials corresponding

to Theorem II and its corollaries in 1 0. 1 3. The enunciations and proofs

are the same. In particular, the 'principle of equating coefficients' is

valid for polynomials in complex algebra.

Example

Solve the following three simultaneous equations in z x , z a , z 3 , where a x , a %, a 3 are

all different:

^rr + -^r+-^T- 1 =(y = 1 ' 2 ' 3 )-

c^ + A, a^ + Af a 8 + A,

Consider the expression

When expanded, it is a polynomial in A of degree 3, with leading term —A3.

The given equations imply that this expression is zero when A = Alt A2 , A8 .

Hence it must be identical with

-(A-A 1 )(A-A a )(A-A 3 ).

Putting A = —a x in the identity, we have

(a 2- a x ) (a 3 - a x ) z x

= ( - 1)4 (a x + Ax ) (a x + Aa ) {a x + A3 ),

which gives z x . Similarly the substitutions A = —a 2 , A = —a 3 give z a , z 3 .

13.44 Repeated roots and the derived polynomial

The definitions of 'repeated factor', 'repeated root' in 10.43 are

retained for complex polynomials. Continuing to write

p(z) ss p z n +p 1 z n ~ 1 + ... +p n ,

we define the derived polynomial of p(z) to be




where h is complex (say h = g+irj) and h means that £ and n\ tend to ze

separately and in any manner.' There are difficulties underlying this procedure;they can only be fully examined in dealing with general functions of a complexvariable, a subject which may be studied at a later stage.

From the definition given it is straightforward to verify the usua

rules of the differential calculus, and in particular the product rule

in this 'differential calculus of complex polynomials'. The theorems

on repeated roots given in 10.43 remain valid in complex algebra

since their proofs are exactly as before.

13.45 Equations with 'real' coefficients; conjugate complex roots

(1)// the coefficients

p,

p ly p nin the polynomial p(z) are 'real',

and if a + fii (where fi =f= 0) is a root ofp(z) = 0, then its conjugate a—is also a root.

Proof. The condition for a fii to be a root is p(a + fii) = 0, i.e.

Po(a+Pi) n +p 1 {oc + /3i) n - 1 + ...+p n „ 1 (cc+j3i)+p n = 0.

On expanding the binomials and replacing i 2 by —1 whenever

occurs, this equation can be written

P + Qi = 0,

where P, Q are real polynomials in (a, fi).

This condition is equivalent to the separate equations P = 0, Q =Hence also P —Qi = —0* = 0, which (because p(z) has real coefficients)

is equivalent to the statement p(a —fii) = 0, i.e. a— fii is a root

p(z) = 0. See Ex. 13 (d), no. 7 for an alternative proof.

If the coefficients in p(z) are not all real, the argument breaks dowat the last step. For example, if p(z) is the linear polynomial z —(1 +then clearly z = 1 + i is a root of p(z) = 0; but 1 —i is not a root since

p(l-i) = (l-i)-(l + i) = -2i # 0.

The correct statement in the general case is that a fii is a root

p(z) = 0, where p(z) is the polynomial obtained from p(z) by changing

the sign of i in all the coefficients. For if p(oc + fii) = P + Qi, then wmust still get a true result by changing the sign of i everywhere on both

sides f (because the equality was obtained by using only the property




Conjugate complex roots occur to the same order.

For if p{z) = has an r-fold root z = a (yff 4= 0), then it also has

a root oi-fii. Remove the factor (z-a,-fii)(z-oc+fii) from p(z),

obtainingf x

(z), say. Then/^z) = has an (r- l)-fold root a+fii, and

therefore also a root <x—fii; and by removing the corresponding factors

from/ 1 (z) we obtain 2 (z), say. Proceeding thus, we arrive at a poly-

nomial f r (z), where a +fli is not a root of f r {z) = 0; hence a—fii cannot

be a root (otherwise its conjugate a +/3i would be a root). Thus a fti

is also an r-fold root of p(z) = 0.

Example

Prove that the equationft* 6 2 6 a

x —a x x —a 2 x —a n

has n roots in real algebra if a l9 a 2 , . . ., a n are all different, the Vs are non-zero, and4= 0.

Consider the corresponding equation in complex algebra, the a's, 6's and k

all being real. Denote any rootf by a + fii; then a —/?* is also a root, and hence

r=1r

\oc—f}i —ar a+/3i —a r )

n*

%=1 (a-a r )a +^ 2

Since the sum is non-zero, we have /? = 0. Hence all the roots are of the form

ct + Qi. In real algebra there are thus n roots.

(2) There is a theorem in real algebra on surd roots of an equation

with rational coefficients which can be proved by an argument like

that in (1).

If the polynomial equation p(x) = (in real algebra) has rational

coefficients and if x = a + b*Jc is a root (where ^c is a surd), then also

a —b^jc is a root.

Proof. The condition for a + b Jc to be a, root of p(x) = is, whensimplified, of the form P + Q ^jc = 0, where P, Q are rational numbers

or zero. This implies that P = and Q = 0, for otherwise if Q =f= the

surd yjc would be equal to the rational number —P/Q, which is a

contradiction. Since








By inspection, the coefficients will be integers when h = 3 (this is the smallest

such value). The new equation is then

2/4 -42/ s + 54y+135 = 0,

where y = 3x.

(ii) Find the equation whose roots are the squares of the roots of

3x 3 -x* + 2x-B = 0.

Write the given equation as

x(3x 2 + 2)=x* + 3,

and put y = x*: yfy(3y + 2) = y + 3.

Squaring, y(9y* + 12y + 4) = y* + 6y + 9,

i.e. 9y a + 1 It/ 2 - 2y - 9 = 0.

Remark. Strictly, we should show that every root of equation (c) is the squarof a root of equation (a),f as follows. If y satisfies (c), then (by reversing

working) it satisfies one of the equations

±*Jy.(3y + 2) = y+S.

Thus either +*Jy or —*Jy satisfies (6), i.e. the given equation (a). In eithe

event, y is the square of a root of (a).

Further examples of the transformation process are given in 10.32, ex.

and Ex. 10(c).

Exercise 13(d)

The algebra is complex.

1 Prove that the equation

z 2 z 3 z1+i +

2i+ 3T + - + ^=

has no repeated roots.

2 Solve z 4 —4z 2 + Sz —4 = 0, given that 1 + i is a root.

3 Given that 2 + ^/3 is a root of x 9 —2a; 2 —7x + 2 = 0, solve the equationcompletely.

4 Prove that w is a repeated root of 3a; 6 + 2a; 4 + x 3 —6a; 2 —5x —4 = 0, ahence solve the equation.

5 One root of x« + x 5 -$x*-10x 3 -9x 2 +x+l = is *J2 + J3. Find all

roots.

6 ' The linear equation az + b = with real coefficients a, b has a root aand therefore also a root <x~fii: two roots for an equation of degree 1.' Explainthis apparent paradox.

7 (i) Verify that (z - a —/?i) (z —a (ii) = z 2 —2az + a 2 + ft2

, where a, /? are re

L + b b h i d i d h d d d




(iii) If a is a root of p(z) = 0, show by putting z = a in the identity f

p(z) = (z 2 - 2<xz + a 2 + /? 2) q{z) + az + b

that act + 6 = and aft = 0. If /# 4= 0, deduce that a = and 6 = 0, and hence

that z —a—fiiia also a root of p(z) = 0.

*8 Prove the theorem about conjugate surd roots in 13.45(2) by a methodsimilar to that of no. 7. Also prove that such roots occur to the same order.

9 Ifa,# y, S are the roots of x* +px 3 + qx 2 +rx +s = 0, find Zafiyia+fi+y).

10 One root of a;4 +px 3 + ga; 2 + rx + s = is equal to the product of the other

three. Prove ( ps + r) 2 = s(q + s + 1 )2

.

* 1 1 Find the sum of the cubes of the roots of x 5 + x 2 + x —1 = 0.

Solve the follouring equations.

12 x 3 —5a; 2 —16* + 80 = if the sum of two roots is zero.

13 xi

+ 3x 3 -5xi

-6x-S = if the sum of two roots is -2.14 x* - 2x z - 2 1* 2 + 22a; + 40 = if the roots are in a.p.

15 Increase the roots of x s + Sax 2 + Bbx + c = by k, and find for what value

of k the new equation contains no quadratic term. (This process is called

'removing the second term from the given equation'.)

16 Find two substitutions of the form y = x—k which remove the third term

from x 4 + 4a; 3 —1 8a; 2 —100a; —112 = 0, and use one of them to solve the equation.

17 Find a substitution y = kx which will transform the equation

8a; 4 + 8a; 8 -18a; 2 -16a; -3 =

into one with integral coefficients of which that of the leading term is + 1.

18 If a, f$, y are the roots of x 3 +px 2 + qx + r = 0, find the equation whoseroots are (i) a(fi+y), /3(y + a), y(x+P); (ii) (a- l)/a, (/?- (y- l)/y.

19 If the roots of x n —1 = are 1, a x , a 2 . . n _ 1? prove that

{l-a 1 )(l-cc 2 )...(l-cc n ) = n.

[Construct the equation whose roots are 1 —1, 1 —a lf ....]

*20 If a, /?, y are the roots of x 3 +px + q = 0, form the equation whose roots are

1,7 7* *fi7 p ay pa

[y + 2 = (fi* + y 2 + 2/?7)//?7 = a 2 /^ = - a 3/q = pa\q + 1.]

13.6 Factorisation in real algebra

13.61 Roots of the general polynomial equation

(1) We have proved that complex roots of an equation with real

coefficients occur in conjugate pairs (13.45 (1)). Since

(z-a-j3i)(z-a + fii) = (z-a) 2 + /?2

,






of generality we can assume that h, k are integers with no commonfactor and that k > 0. Then

so ^ = - (pJi 11 - 1 + p 2 h n ~ x k + ... +p n k n ~ 1)

k

= an integer.

Hence h n jk is an integer, so that k = 1 . Thus the rational root is in fact

an integer h. Since - ,„ ,, ,

~

/. p n = - hih -1

+ Pl hn

-* + ... +p n - 1 ),

and so h is a factor of

(2) Change of sign of a polynomial.

If the polynomial p(x) is non-zero when x = x x and when x = x 2 , then

the number of roots ofp(x) = between x x and x 2 is

odd ifp(x^), p(x z ) have opposite signs,

and even or zero if p{x^), p(x 2 ) have the same signs.

These results are 'intuitively obvious' properties of all continuous

functions (cf. 2.65). For a polynomial p{x) they can be proved as

follows.

By result (vii) in 13.61,

2>(a) = p (x - a x ) {x-a 2 )...(x- a k ) g(x),

where g(x) is a product of factors of the form (#-&)2

+ c2

, or else (in

the case when p(x) consists entirely of linear factors) is 1. In either

event, g(x) is positive for all values of x.

If p(Xj), p(x 2 ) have opposite signs, the linear factors cannot be all

absent, and (a? x—a x ) (x x

—a 2 )... {x 1 —a k ), {x 2—a ± ) {x 2 —a 2 )... (x 2 —a k )

must have opposite signs. Now x t —a,x 2—a have like signs unless a

lies between x x and x 2 . Hence an odd number of a 1} a 2 , . .

.

, a k lie between

x and x




Examples

(i) Let p s be the numerically greatest coefficient (or one such coefficient

two or more are numerically equal) in

p(x) = x n +p 1 x n ~ 1 +... +p n .

Then Ip^-i + .-.+p^x+Pnl ^ \p 1 \.\x\»- 1 +... + \p^ l \.\z\ + \p n \

< \p s \{\x\ n -i + ... + \x\ + l}

If |a;| > 1 it follows that

p(x) -1x n

Given any positive proper fraction, e.g. \, we shall have |i? g |/{|a;| —1} ^ § if

is such that \x\ > 2| p s \ + 1 = K', say. Hence if K > K\

p{x) = x n (l + d) for all \x\^K, where \0\ < J.

This shows that p(x) has the same sign as x n for all x ~$t K and all x < —i

we say that p(x) is dominated by for large (positive or negative) x. Thup(if), p( —K) have opposite signs if n is odd, and the same sign if n is evensince K can be as large as we please, we have proved the theorem in 13.61 (2).

The symbols p{co), p{ —<x>) are customarily used to denote '#>(i£), p(—Kwhere K is arbitrarily large'.

(ii) Solve the example in 13.45 (1) by a 'change of sign'' argument.

We may suppose the notation to be chosen so that a x < a % < . . . < a n . Consider the polynomial

p(x) = (x - a x ) (x - a 2 ) . . . (x - a n ) f(x)

= bl{x-a 2 ) ...{x-a n ) + b\{x-a 1 )(x-a s ) ... (x-a n ) + ...

+ b%(x - a x ). . . (x - a n _ ) + k(x - a x ) (x - a 2 ) . . . (x - a„).

When x takes the values

—00 a x a a ... 0 n-\ a n + °°»

p(x) has the same sign as

k(-l) n (-1) 1 (-1) - 2... -1 +1 k.

In this sequence there are exactly n changes of sign, whether k is positive

negative. Hence p(x) = has at least n roots. Since p(x) is of degree n whek =|= 0, it cannot have more than n roots. The number of roots is therefore




Examples

(iii) Find the number of roots of x* - 2a; 3 + 6a: 2 - 10a; - 8 = in real algebra, andlocate them between consecutive integers.

Writing p(x) =x l - 2x* + 6a; 2 - lOx - 8,

then p\x) = 4a; 3 - 6a; 2 + 12x - 10 = 2(x - 1) (2a; 2 - x + 5),

the factor x —1 being discovered by trial.

Hence p'(x) = has one root in real algebra. By 6.23(3), p(x) = cannot

have more than two roots; and if p(x) = actually has two roots, the root x = 1

of p'(x) = must lie between them. We therefore consider the sign of p(x)

X —CO 1 +00

Sign of p(x) + - +

Thus p(x) = has two roots, one less than 1 and the other greater than 1.

By trial we find that p(0) < 0, p( —1) > 0, so there is a root between and —1.

Similarly we find p(2) < 0, p{3) > 0, so the other root lies between 2 and 3.

(iv) Find the range of values of k for which the equation

p(x) = x*-2Qx 2 + 4:Sx-k =has 4 distinct roots.

p'{x) = 4(a; 8 -13a;+12) = 4(a;-l)(a;-3)(a; + 4).

Possible roots of p{x) = must be separated by roots of p'(x) = 0, viz. —4, 1, 3.

x |-oo —4 1 3 +oo

Value of p(x)\

+oo -352-A; 23-A; -9-k +oo

In order that p(x) = shall have 4 distinct roots we require there to be four

changes of sign in the above sequence; the signs must therefore be

+ - + -+.This will be so if and only if -9 < k < 23.

The reader should sketch the graph of y = x* - 26a; 2 + 48a;, which has a maxi-

mum at x = 1 and a minimum at x = 3 (and at x = —4). The line y = k cuts the

curve in four distinct points if and only if this line lies between the levels of thesetwo turning points.

(v) Prove that the equation d^x^—l^/dx 1 = has 4 distinct roots which all

lie between —1 and + 1.

Putting f(x) = (a;2 -l) 4

, then f(x) = has roots +1, —1 each four-fold.

Hence/'(a;) = has roots +1,-1 each three-fold (by 10.43) and a root x = abetween +1,-1 (by Rolle's theorem). Then f(x) = has double roots +1,-1,a root ft between —1 and a, and a root y between a and + 1. Similarly, f '{x) =has simple roots +1, —1, and roots between —1 and fi, ff and y, and y and + 1



520 COMPLEX ALGEBRA

Exercise 13(c)

The algebra is real.

Find the rational roots (if any) of the following equations.

1 3a^-lla; 2 + 9a;-2 = 0. 2 2a: 4 + 5a; 8 + 14a; 2 -3a;-54 = 0.

3 3a; 4 - 4a; 3 + 6a: + 5 = 0.

Determine the number of (real) roots of the following equations, and locate e

between consecutive integers.

4 x* + 2x 2 + Sx-l = 0. 5 x«-x 3 + x 2 -2 = 0.

6 « 5 -2a: 3 + a;-10 = 0.

7 If a t < a 2 < a 3 < a 4 < a 6 < a 6 , prove that the equation

(x —a 1 )(x —a 3 )(x —a 5 ) + k 2 (x—a 2 )(x—a i ) (x—a e ) =

has 3 distinct roots for any real k.

8 If a > 0, 6 > and p < q, prove that al(x—p) + b/(x —q) = x has thr

roots a, ft, y such that a. <p < ft <q<y.n i

9 If a x , a z , a n are all different, prove that the equation 2 =r=l x-a r

exactly n—1 roots (i) by considering changes of sign of

n ip(x) = (x —a x ) ...(x —a n ) S ;

r=\ x —a r

(ii) by applying Rolle's theorem to f(x) = (x —a t )... (x —a n ).

[We may assume a x < a 2 < . . . < a n without loss of generality.]

10 (i) If p n < and n is even, prove that the equation

p(x) =p x n +p 1 x n - 1 +...+p n =

has at least one positive and at least one negative root.

(ii) Prove that the number of positive roots of p(x) = is odd if and onif p n < 0.

Determine the number of roots of the following equations by first finding the ze

of the derived polynomial ; and locate each between consecutive integers.

11 x i + 4x 3 -Sx 2 -l = 0. 12 8a; 8 -5a 4 -40a; 8 - 50 = 0.

13 Prove that x z + 3a; —3 = has only one root a, and hence that

x* + 6a; 2 -12a; -9 =has just two roots.

14 If a > 1, prove x 5 —5a*ar + 4 = has 3 roots. What happens when a <




has 3 roots if r lies between pq— 2(p2 —q)a and pq —2(p t —q)fi. [By long

division, f(x) = (x—p) (x* —2px +q) + 2(q —p 2) x + (pq —r), so

/(a) = 2(q -p*)* + {pq - r), etc.]

1 8 Prove that x* + ax + b = has 3 distinct roots if and only if 276 a + 4a 8 < 0.

*19 Use the argument in ex. (i) of 13.62 to prove that p(x) = has every root

numerically less than 1 + \p g \. [If aisaroot, —x n = p 1 x n ~ 1 + ... +p n ; by taking

moduli, \x\ n < \p t \{\x\«- l}l{\x\ - 1} if \x\ 4= 1. If \x\ > 1, the right-hand side

< b«|-M n /{H-l}> so 1 < |p,|/{|a;| - 1}, i.e. \x\ < l + \p t \. Thus any root

numerically greater than 1 must be numerically less than 1 + \p,\.]

*20 Replacing 'numerical value' by 'modulus', verify that the results of

13.62, ex. (i) and no. 19 hold in complex algebra, even when the coefficients

p r are not real.

13.7 Approximate solution of equations (further methods)

We have already explained Newton's method (6.73), which applies to equa-tions in general. The following method also has this advantage, although it

usually does not give an approximation of a required order as quickly as

Newton's.

13.71 The method of proportional parts

For a straight line, the increment of the ordinate is proportional to theincrement of the abscissa. For a curve, a small arc will usually not depart far

from the chord joining its ends, so that over this arc the ratio

(increment of ordinate)/(increment of abscissa)

will not differ much from the gradient of the corresponding chord.

These geometrical considerations lead to the ' principle of proportional parts '

the increment of any continuous function is approximately proportional to the

increment of the variable, the range of this

variable being small. An analytical formula-

tion and an estimate of the error are given

in Ex. 6(e), no. 15.

Suppose now that numbers a, b have beenfound such that /(a) = A > 0, f(b) - B < 0.

Geometrically, the points (a, A), (6, B) are onopposite sides of Ox, and preferably close to

Ox (a wise choice of a, b will secure this). Theline joining these points has equation (b,B)

AFig. 135




ExampleFind the root of x z —2x —2 = correct to 3 places of decimals.

By trial we find/(l) = —3,/(2) = 2, so there is a root between 1 and 2.

this root is 1 +h, then by 'proportional parts'

/(2)-/(l) J(l + h)-f(l)

2-1 ' (l + h)-l'

and since /(l +h) = 0, this gives 5 = S/h, i.e. h = 0-6.

A better approximation is therefore x = 1-6. Since

/(l-6) = -l-l, /(l-7) = -0-487, /(l-8) = +0-232,

the root actually lies between 1-7 and 1-8. If the root is 1-7 + h, then

/(l-8)-/(l-7).

/(l-7 + ft)-/(l-7 )

1-8-1-7 ' (l-7 + fc)-l-7'

0-719.

0-487

so that h = 0-0677, and a better approximation is x = 1-768.

Applying the process once again,

/(l-8)-/(l-768) ^ /(l-768 + ft)-/(l-768)

1-8-1-768 h0-241 ^ 0-009

1,e 032 h

since /(1-768) = 0-009. Hence h = 0-001195, and x = 1-76919 = 1-769 to thrplaces of decimals. (Taking a = 1-768, 6 = 1-8, c = 1-769 in Ex. 6(e), no.

shows that the error is less than 0-0002.)

13.72 Horner's method

Suppose that the equation p{x) = has a root 2-76 .. . and has no other rwhose integral part is 2. Thenp(2), p(3) have opposite signs; this fact, discovered

by trial, gives the first figure of the root.

The first figure 2 is now removed from subsequent calculations by diminishingthe roots of p(x) = by 2. The new equation then has a root 0-76 .. .. To avodecimals, we multiply the roots by 1 0, so that the new equation has a root 7 • 6

The figure 7 is discovered by showing that the last equation has a rbetween 7 and 8. The roots are then (iiminished by 7, and next multiplied byThe figure 6 is then found by trial. The process is repeated until the required

number of decimal places has been found.

Example




By trial, this has a root between 7 and 8. Diminish the roots by 7 by puttingz = y —1, i.e. y = z + 7; we get

2 3 + 51z 2 + 6672 -487 = 0.

Multiply the roots by 10:

2 s + 510z 3 + 66,7002 - 487,000 = 0.

By trial this has a root between 6 and 7. Diminish the roots by 6, putting

t = 2 —6, i.e. 2 = t + 6; we obtain

t 3 + 528* 2 + 72,928* - 68,224 = 0.

Multiply the roots by 10:

t 3 + 5280* 2 + 7,292,800* - 68,224,000 = 0.

The process can be continued; but since the numbers in the last two terms are

large incomparison with those

inthe other terms,

clearlya good

approximationis given by6 * 7,292,800*-68,224,000 = 0,

i.e. t = 9.

Hence the required root is x = 1-769.

Remarks

(a) The method used in the last step gives a rough estimate for the trial root

when applied at the previous stage: 66,7002 —487,000 = gives 2 = 7. Actually

this is too large; writing

0(z) = 2 3 + 5102 2 + 66,7002 - 487,000,

we should find that 0(7) and 0(8) have the same sign. This indicates that weshould calculate 0(6), which is found to have the opposite sign. Although this

estimate is rather rough, it is better than none, and often saves much futile

trial. Estimates thus made increase in accuracy the further we have got in theprocess; in fact we may approximately double the number of significant figures

already obtained by using this 'division estimate '.

(/?) Horner's method will also give rational roots. Since these can always betested for exhaustively (13.62(1)), they should be removed from the equationbefore applying the method. The same applies to repeated roots: see 10.51,

ex. (ii).

(y) To approximate to a negative root, first change the sign of all the roots (by

putting y = —x), and then approximate to the corresponding positive root of

the new equation.

(S) Case of nearly equal roots. If two roots lie between consecutive integers

n and n + 1, then p(n) and p(n +1) have the same sign, and detection of these

roots is difficult; more refined methods than 'change of sign' are required.

Assuming that these roots have been approximately located, Horner's methodcan still be used.

(e) Although the method applies in principle to equations in general, its




Given a cubic x 3 + ax 2 + bx + c = 0, whose roots are a, /?, y (say), we construct

the cubic whose roots are a 2, /?

2, y

2 by putting y = x 2: the equation becomes

*Jy{y+*>) = -(cm/+c),

and by squaring,y

3

+(26 - a 2

) y2

+(6

2 - 2ac)y

- c 2 = 0,

i.e. ^-f^ + By + C = 0,

where A = 26-a 2, B = b 2 -2ac, O = -c 2

.

If the root-squaring process is applied n times, we obtain an equation whroots are a 2 *, ytf

2 , y2 . Suppose that |a| > |/?| > |y|; then for n sufficiently larg

a 2 will be very large compared with /?2 * and y 2 , so that a 2 is approximately

'minus the coefficient of the quadratic term ' in this equation.

Since /?2 is large compared with y

2 , the coefficient of the linear term will

approximately (a/?) 2 . Similarly the constant term is approximately —(a/?y)

Example

Solve x 3 —2x —2 = in complex algebra.

Here a = 0, 6 = —2, c = —2. Proceeding step by step, we construct

following table.

A = 26-a 2 B = b 2 - 2ac G = -c 2

a 2 -4 4 -4a 4 -8 -16 -16a 8 -96 -256a 16 -9216 -49,152 -65,536

By trial the given equation is found to have a root a near 1-7; and this is

only real root (e.g. see Ex. 13(e), no. 18). Since ctfiy = 2, it follows t

yffy = 2-r 1-7 = l«2,i.e.r 2 = 1-2 where/? = r(eos# + *sin0), y = r(cos0 —isin#)

hence r == 1-1. Thus a > r = |/?| = |y|.

From the theory just explained,

a 18 = —J. = 9216, whence a = 1-7692;

and a 16 (r 2)

16 = -G = 65,536, so r = 1-0632.

Referring now to the given equation and considering the sum of its roo

a + 2rcos0 = O, from which cos 6 = -0-83215 and 6 = 146° 19'. Henr(cos 6 ± i sin 6) can be found. The three roots are approximately

1-7692, - 0-8847 ± 0-5896*.

Exercise 13(/)

The algebra is real.

Solve the following equations approximately, correct to at least 2 places of decimals



COMPLEX ALGEBRA 525

8 x = e~ x [three places].

9 10* = 20x [2 roots, each to three places].

*10 x x = 1*5. [This has a root between 1 and 2; write the equation asa; log a; = log 1-5.]

*11 Find the smallest positive root of x = tan -1 a:, correct to four places ofdecimals.

*12 Find the smallest positive root of sin a; = tha;, correct to two places ofdecimals.

* 1 3 If e is small, prove that 6 + sin 6 cos 6 = 2e cos 6 has a small root which is

approximately e —£e s. [Use Newton's method; assume the series for sin x, cos x.]

*14 Show (graphically or otherwise) that f(x) = 1 + x~ 2 —tana; = has a rootnear x = K, where K = ^n + nn and n is a large positive integer. Prove thata better approximation is K+ A where f(K) + \f'(K) = 0, and that A = 1/2K 2

.

Find an approximation correct to terms of order l/K*.

*15 Prove that large roots of seca;= 1 + 1/a; are given approximately byx = 2rm where n is a large positive integer. Show that a better approximation is

x = 2nn ± \j^J{mr). [Assume the series for cos 6.]

Miscellaneous Exercise 13(g)

1 If c is real, and the number (1 + i)/(2 + ci) + (2 + 3*)/(3 +i) is represented

in the an/-plane by a point on the line y = x, prove c = —5±*J21.2 If c 2 + s 2 =l, prove (1 + c + *s)/(l +c —is) = c + is, and write down a

similar result for(

1

+ s +ic)/( 1

+ s —ic)

3 If Z 2 + ra 2 +n a =l and m+ m = (l+l)z,

l + im 1+iz

4 If z = r(cos0+isin0) and a = p(cosa+tsina), calculate \z —a| 2 in termsofr, p, 6, a. Deduce that |1 -az\*- |z-a| 2 = (1 -r 2

) (1 -/> 2).

5 If |zi-«,| < £|z 2 |, prove (i) |af x | > i|z 2 |; (ii) l^ + a, > f |z 2 |.

6 The number z is represented by a point on the circle whose centre is 1 + 0*

and radius is 1.

(i) Represent the number z —2, and prove (z —2)/z = i tan (argz).(ii) Construct the point representing z 2

, and prove that

(a) |z 2 —z| = |z| and (6) arg(z— 1) = arg(z 2) = f arg(z 2 —z).

7 (i) If P x , P 2 represent z x , z 2 , and the line P x P a is turned through a positive

right -angle about P t into the position P-iP^ prove that P 3 represents

Zl + ^Zjj-Zx).

(ii) Two opposite vertices of a square are 1 + 2i, 3 —5i. Find the numbersof the other vertices.

8 (i) If Jc is a real constant, interpret z + k(z 2—z ) geometrically.



526 COMPLEX ALGEBRA

10 Interpret the following as loci:

(i) |z + 3*|2 -|z-3i| 2 = 12; (ii) |z + K| 2 +|z-H| 2 = 10fc 2

(Jfe > 0);

(iii) arg —- = Jtt; *(iv) |z-z x |- |z-z 2 |

= 1.

z — 2

1 1 The numbers z lf z 2 are connected by the equation z x = z 2 + l/z 2 .

(i) IfP 2 describes a circle of radius a 4= land centre O, show that P 1 describes

the ellipse %1 y2 x

(l+a 2)

2+

(l-a 2)

2=

^ 2'

(ii) If z 2 = r(cos0 + isin0), determine the locus of P x when 6 — \tt andvaries.

12 liu + iv = ajz where a is real, show that the curves in the ay-plane whicorrespond to u = constant, v = constant are (in general) orthogonal system

of circles.

13 If {(z + c)/(z —c)} 2 = (w + 2c)j{w —2c) and c is real, prove that whz = c(eos 6 + i sin 0), then w = 2c cos 0. Hence show that if z describes the circ

|z| = c, then w describes the segment of the a?-axis between ± 2c, once in eadirection.

14 If z, w are represented by P, Q and zw + w —z+1 = 0, prove that whz = cos 6 + i sin 6, then \w\ = ±tan£0 according as =S 6 < 71, —n^dÔ.P describes the circle \z\ = 1, prove that Q describes the y-axis, and indicate

corresponding directions of motion.

*15 An ellipse has foci ( + ae, 0) and z x> z 2 correspond to the ends of conjugate

semi-diameters. Prove z 2 +z| = a 2 e 2. [Using eccentric angles, if

z x —a cos (j) + ib sin ^ then z 2 = ± (a sin<f>

—ib cos 0).]

* 16 If w = (z 2 + az + 6) /(z 2 + cz + d) and a, 6, c, d are real, prove that the point

z for which u is real lie either on the real axis or on a circle whose centre is onreal axis. [Clear fractions, equate real and imaginary parts, and eliminate w

In the following, a> and w2 denote the cube roots of + 1, w 4= 1.

17 If x = a + b, y = a + bco, z = a + b(o 2, prove that Lx = 3a, ILyz = 3

a*/z = a 3 + 6 3, Sa 2 = 3a 2

, Sa 3 = 3(a 3 + 6 3).

3n18 (i) If f{x) = Ea r af, what does f(x) +f((ox) +f(o) 2 x) represent?

r=l*(ii) Calculate similarly f(x) + (of((ox) + (o z

f(u)2 x) and f(x) + w2/(waj) + o>/(w

19 If 2 + p x z n - x + ... +p n = (z —Oil) (z —a 2 ) . . . (z —a„), prove that

(l + a 2 )(l+a 2 )...(l + a 2) = (l-p 2 +p 4 -...) 2 + (î-p 8 +P6--) 2

.

[l+p 1 <+i? 2 «2 + ...+p„« n = (1-<M) ... (l-a n £). Put<= ± i, then multiply.]

* 20 Find conditions for z 2 + (a + &*) z + (c + a*) = to have a real root, whea, 6, c, a are real.

21 If = + i a root of 3 +pz + = real) prove a is a real root





528

14

DE MOIVRE'S THEOREM ANDSOME APPLICATIONS

14.1 de Moivre's theorem

14.11 If n is an integer (positive or negative), then

(cos d + i sin d) n = cos nd + i sin nd ;

if n is rational, then cos nd + i sin nd is one of the values~\ of

(coad + ismd) 11.

Proof, (i) Let nbea positive integer. By direct multiplication,

(cos d + i sin 0) (cos <j> + i sin<f>)

= (cos 6 cos $ —sin d sin<f>) + i(sin d cos <p + cos sin 0)

= cos(0+ 0) +

isin(0+ 0).

Similarly,

(cos d x + i sin d x )(cos 2 + i sin 2 ) (cos 3 + i sin 3 )

= {cos (d x + d 2 ) + i sin (^ + d 2 )} (cos 3 + i sin d 3 )

= cos (0J + 6 i + 3 ) + i sin ((^ + 2 + S )

by two applications of the above result. Proceeding step by step

this way, we find

(cos d x + i sin d x ) . . . (cos 6 n + i sin 6 n ) = cos (20) + i sin (20).

Putting d 1=

2= . . . = 6 n = 0, this becomes

(cos 6 + ism d) n = cos nd + i sin nd.

(ii) .Lei nbea negative integer, say = —m. Then

(cos + i sin 0) w = (cos 6 + i sin 0)~ m





530 DE MOIVRE'S THEOREM [14.1

No further values are given by taking other values of k, because a

other value of k differs from one of the values 0, 1 , 2, . .. , q - 1 by som

integral multiple of q.

Consequently, (cos 6 + i sin 6)vla

has exactly q different values, vthose in (i) given by k = 0, 1, 2, q- 1. Any other set of q integral

values of k could be chosen, provided they give q distinct values of

e.g. we often take k = 0, + 1, + 2, ....

The q values (i) are represented in

the Argand diagram by points

on the unit circle \z\ = 1:

£(Ox,OP 1 )=pe/q,

and arcs PXP2 , P^Pz, Pq -iP q sub-

tend angles 'Znjq at the centre 0. Thus

PXP2 PZ ... Pais a regular g-sided poly- Fi 8- 136

gon inscribed in the unit circle.

Since every complex number z can be written in the form

r(cos#-Msin0), where —tt<6^7t,

the above shows that the values of z plq are

cos I —+ + i sin +-2kn

,)

where k = 0, 1, 2, q— 1 and (as elsewhere in this book) r pli denotes

the positive qth root of r p . They are represented by points equallyspaced around the circle \z\ = r plQ

.

Definition. It will often be convenient to abbreviate cos0 + isin#

to cis0.

14.13 Examples

(i) Solve the equation (z+ l) n = z .



14.13] DE MOIVRE'S THEOREM 531

By taking nth roots of both sides, and using the general result (i) in 14.12,

z+1 / 2kn\ . . / 2kn\'

A , ft= cos|0 + |+*sin +) (k = 0,l,2,...,n-l).

z \ n J \ n )

I 2kn . . 2kn\z I cos 1+* sin 1 = 1,

\ n n )

. / „ . a kn n . . kn kn\l.e.y zl —2 sua 2 h2*sm —cos —I = 1,

\ n n n J

n . . hn ( kn . . kn\and 2%z sin —I cos h*sui —I = 1.

n \ n n J

Hence by the theorem (with index —1)

n . . kn kn . . kn „ , „2*zsui —= cos *sin — (k = 0, 1, 2, ...,n— 1).

n n n^„ , ^ 1 / Jen \If k 4=0, 2 = —( cot 1)

2^\ n J

1 / kn\= --ll + icot— I (k = l,2,...,n-l).

These are the n —1 solutions.

Alternatively, instead of quoting the general theory, we may proceed fromstage (a) above by writing

cos 2kn + * sin 2kn',

hence by de Moivre's theorem,

z+1 2kn . . 2kn= cos H*sin ,

z n n

which takes distinct values for k = 0, 1, 2, . . ., n —1; etc.

(ii) Show that the roots ofz n = 1 can be written 1, a, a 2

From z n = 1 = cos2&7r+*sin2&7r,

2kn . . 2knz = cos Msin ,

n n

which takes distinct values when k = 0, 1, 2, . . ., n —1.

The value k = gives 2=1. Write2n . . 2n

a = cos h*sinn n

for the case k = 1; then since

2kncos — . . 2kn I 2n . . 2n\ k

+ *sin = (cos h*sm — =



532 de moivre's theorem [14.13

Compare the brevity of this work with the algebraical treatment of the equation

2 3 = 1 in 13.23. Observe also that

. [2{n-r)7i\ .f 2rn\

a n ~ r = cis |— ———-

—| = cis|27T ——

{2rn\ 2rn . . 2rn —> = cos ism —= oc r

,

n ) n n

and hence a n_1 = a, a n-a = a 2, .... Also see Ex. 14(a), no. 21.

*(iii) If p, q are coprime integers (i.e. have no common factor other than ±1)the q values of (cos 6 + i sin 6) PlQ can be written

cos^(0 + 2for) + *sin^(0 + 2for), (

where k = 0,1,2,

...,q— 1.

These q expressions each satisfy z = cospd+iainpd, and are therefore

values of (cos0+isin0) 1 ' / <1 '. They are all distinct; for if

then as in 13.22(2),

cis -(d + 2k 1 ir) = cis-(0 + 2fc a 7r),

2 3

P(d + 2k 1 n)--(d + 2k z n) = 2mn

? 3

for some integer m, i.e. p(k x—k 2 ) = mq. Since —k 2 \ < q, this shows that

has a factor in common with p, which is impossible becausep,

q are coprime.

Remarks

(a) Since the equation z 9 = cispd has exactly q roots, the set of values of (

must be the same as the set of values of (i), in some order.

(/?) If p, q are not coprime, the function (cos0 + *sin0) 1 ' /9 is understoodmean {(cos0 + isin0) 3>

}1/<r

, and therefore has q distinct values, viz. the values

(i) of (coapd + i smpd) 1 . However, in this case the expression cis {(p/q)(6 + 2kn)

does not take q distinct values, and consequently, does not represent all th

values of (cos0 + isin0) , 'A'. See Ex. 14(a), no. 8.

Exercise 14(a)

1 If x = cis d, y —cis 0, and m, n are integers, prove

1 = 2 cos (md —nd)).yn x m r/

„ . (l + cos0 + *sin0) 8

2 Simplify .. ,, 4•

(cosf —*sin£/)*



DE moivre's theorem 533

8 Write down the six values of (cos + i sin 0)i and the two values of(cos + i sin 0)*. Observe that cis {f(0 + 2ifor)} has only two distinct values, andtherefore cannot completely represent the first function.

9 Simplify (cos0 + isin0) 8 in two ways. By equating real and imaginary

parts, deduce that

cos30 = cos 3 0-3cos0sin 2 and sin 30 = 3 cos 8 sin 6 - sin 80,

and express tan 30 in terms of tan 0.

10 If z = cis 0, express in terms of 0:

(i) \; (ii) z+1-; (iii) z-\; (iv) z + ^; (v) z--~.

Solve completely the following equations.

11 z = -l. 12 (z+l) n + (z-l) n = 0.

13 (z-l) = z». 14 (1+z) = (l-z) B.

15 z 2n —2z n cos ncc + 1 = 0. [First solve as a quadratic in z .]

16 By expanding ( 1 + z) n, where n is a positive integer, and putting z = cis 0,

prove n(2 cos £0) cis \nd = 21 n Gr cis rd.

r=0

n I it\ n

Deduce that S n Cr cos —= 2 n ~* I cos —I .

r= 2w \ 4w/

17 If w is a positive integer, prove ( 1 + *)n + ( 1 —i) n = 2* n+1 cos Jn7r. Writing

(l + a;) = c +c 1 x+c 2 x 2 +...+c n x n,

prove c —c a + c 4 —. . . = 2* cos Jw7T

and c x —c 3 + c 5 —. . . = 2* B sin Jtmt.

18 If a = cisf7r and /? = a + a 2 +a 4, 7 = a 3 + a 5 + a 8

, prove /ff+y = —1 and

/?y = 2. Write down the quadratic having roots fi, 7, and deduce that

. 2n . 4=n . 87r , .sm— +sm —+ sm— = +JV 7 '

7 7 7v

19 If cos0 + cos0 + cos^ = and sin0 + sin0 + sin^ = 0, prove that

cos 30+ cos 3^ + cos 3^ —3008(0 + ^ + ^) = and a similar result for sines.

[Put x = cis 0, y = cis 0, z = cis ^ and use Ex. 10 (/), no. 4 (iii).]

20 What conditions have to be satisfied by z in order that the points repre-

senting all integral powers of z should (i) lie on a circle with centre the origin,




14.2 Use of the binomial theorem

14.21 cos m8 sin B 6 in terms of multiple angles (m, n being positive

integers or zero)This transformation is sometimes needed for the integration

circular functions: see 4.82. Writing z = cos 6 + i sin 6, then

- = cos0 —isin#,z

so 2cos0 = z+-, 2*sin0 = z —.

z * z

Also z n = cosw# + isinn0 and l/z n = cosw0 —iainnd, so that

1 1z n +— = 2 cos n6, z n - —=2i sin nd. (

z z n

By means of (i) and the binomial theorem, the given function ca

be expanded in powers of z and 1/z. By (ii) the resulting expansion

can be expressed in multiple angles.

Examples

(i) Express cos 8 in terms of circular functions of multiple angles.

From (i),

(2cos0) 6

= (z+^y

15 6 1= z« + 6z 4 +15z 2 + 20 + —+ -. + -z

z 2 z 4 z 6

=(

a *4) +6 (*<+ ^ +15

(*s

4) +2 °

= 2 cos 60+

6(2 cos 40) + 15(2 cos 20) + 20 by (ii).

.-. cos 6 = -^(cos 6(9 + 6 cos 40 + 15 cos 20 + 10).

(ii) Express cos 6 sin 4 6 in terms of multiple angles.

(2cos0) 5 (2isin0) 4 = ^z + -^ (z-^j*



14.22J de moivre's theorem 535

We observe that coa n d can always be expanded in terms of cosines

because it depends on the expansion of (z+l/z) n. Since sin m

depends on {z—\jz) m, which involves terms like z r +ljz r if m is

even, and terms like zr

—\\zr if

mis

odd,therefore sin m

canbe

expressed in terms of cosines or sines according as m is even or odd.

14.22 cos /18, sin #18, tan /18 as powers of circular functions (n being

an integer)

We now reverse the process in 14.21

cos n& + i sin nO = (cos 6 + i sin 6) n = (c + is) n, say,

= c n +

^c n ~Hs +

j^jc n -*(is)* +

Qc - 3

(i5)3 +

j^jcîs)* + . .

-(2)

cn v+6)

cn_M - -HI®cn ~ is -

(3cn ~* s * + ••) -

Hence cos nd = c n -Qj

c n ~ 2 s 2 +j^j

c n ~^ - . . . (hi)

and sin nO = c^s -(^j

c 3 * 3 + . . . . (iv)

Taking the ratio of sinnfl to cosnd, and dividing top and bottom

of the right-hand side by c n , we obtain

t*nn0 = U/.

V ;

,, (v)

1-t W[%'-.where £ = tan#. WW

The formulae (iii), (iv) can be transformed by means of the relation

s 2 + c 2 = 1 to show that cosw-0 can be expressed entirely in powers of

cos/9, and so on; see for example Ex. 14 (6), no. 20, and ex. (iii) below.

Examples

(i) Express cos 56 and sin 60/sin 6 in terms of cos 6.

cos 56 = c 8 - Q c 8 * 2 + Q cs*

= c 8 - 10c 8( 1 - c 2

) + 5c( 1 - c 2)

2

= 16c 8 - 20c 8 + 5c.




(ii) Write down the formula for tan 50. What equation is satisfied by tan

if (a) tan 50 = 0; (6) 56 = Jct? Solve each of these equations.

(a) If tan 50 = 0, then 5t- lOt* + <B = 0, so « = tan 6 satisfies

x 6 -10x z + 5x - 0.

Now if tan 50 = 0, then 56 = m and 6 = \rir, so that tan 6 = tan Jnr, andvalues r = 0, 1, 2, 3, 4 give distinct values of tan Jttt. Therefore the roots of

above equation are x = (corresponding to r = 0) and

tanf7r, tan-§7r, tan |tt = —tan f 7r, tanf7T = —tan^7T,

i.e. 0, ±tanf7r, ± tan \tr.

(6) If 50 = \t\, then tan 50 = 1 and so

5t - 10« s + 1* = 1 - 10* 2 + 5**;

hence * = tan satisfies

a 5 -5a:*-10:r 8 +10a; 2 + 5a;-l = 0.

But if tan 50 = 1, then 50 = fyr+rn and = ^n+^rn, so that

(4r+l \

and this takes distinct values for r = 0, 1, 2, 3, 4. The required roots are therefore

a: = tan ( 20~7 (r =

' 1 ' 2 ' 3 ' 4 )-

The equation in (6) could also be solved algebraically by the methodEx. 13 (g), nos. 26-28.

*(iii) Ifnis odd, prove that sinn0 can be expressed in the form

sinn0 = 6 1 s + 6 8 s 8 + ... + b n s n ,

and find b lt b n . Also prove that (r + 1 ) (r + 2) 6 r+a = (r 2 —n 2) b r , and hence find 6

From formula (iv) and the relation c 2 = 1 —s 2,

sinw0 =(^j

(l-s 2 )K«-i) s - f^j (l_ s 2)i(»-3) g 3 + f^j(l- a *)H«-*) 8 *- ....

Since each of n— 1, n— 3, ... is even, this can be expanded as a polynomial i

involving only odd powers of s and having degree n, for

b = ffi i f n




Also /n\& = coefficient of a = II = n.

»By deriving the identity ahind = 2 &r s r twice wo 6, we obtain first

ncosn# = 2 r &r*r_lc

r=ln

and then - n 2 sin nd = 2 0(r - 1) Mr-2 c 2 - r& r * r-1*]

r-=l

= S [r(r - 1) 6 r s r a - {r(r- 1) b r + rb r } a']

= X [r(r - 1) 6 r s r - 2 - r a 6 r s r];

n nhence - n 2 S 6 r * r = S[r(r-l) &r s r_a - r 2 6 r a r

].

r=l r=l

Equating coefficients of « r,

- n 2 6 r = (r + 2) (r + 1 ) &r+a - r 2 6 r ,

from which the required relation follows. Taking r — 1, we have

2.3& 8 = (l-n 2 )n, i.e. 6 8 = Jn(l-w 2).

14.23 tan (et + 6 2 + . . . + 6„)

cos(0 1 + 2 +... + 0J + isin(0 1 + 2 + ...+0J

= (cos 6 X + i sin 6 X) (cos 2 + * sm ^2) • • • ( cos @n + * sm ^«)

as in 14.11, proof (i),

= cos 6 X cos 2 . . . cosTO ( 1 + #2) (1 + it 2 ) . . . ( 1+ it n ) where t r

= tan r ,

= cos X cos 2 . . cos Gn {l + iS x + i 2 2 2 + i 3 S 3 +...),

where 2 r denotes the sum of the products of t lt £ 2 ,...,t n taken r at a

time; this last step follows from the argument used in proving thebinomial theorem, 12.11. Equating real and imaginary parts,

cos(0 1 + 1 +...+0 B ) = cos<9 1 ...cos^ n (l-S 2 + S 4 -...), (vi)

andsin(0 1 + 2 +... + n ) = cos tfj... cos tfJSi- 3 +...). (vii)

By division,

tan(0 + + = ^~^^ 5 (™)



538 DE MOIVRE'S THEOREM [14.

Exercise 14(b)

Express the following in terms of multiple angles.

1 cos*0. 2 sin 4 0. 3 sin 6 0.

4 cos* sin 5 0, 5 cos 6 sin 8 0.

Calculate

'in

6 Jsin 7 0d0. 7 JW0sin 6 0d0.o

sin 4 0cos 6 0d0.

9 Express cos 60 in terms of (i) cos 0; (ii) sin 0.

10 Express sin 70 in terms of sin 0.

1 1 Express cos 70/cos in terms of sin 0.

12 Solve completely the equation cos 50 + 5 cos 30+10 cos 6 = \.

1 3 Solve completely 1 6 sin 6 = sin 50.

14 Prove cos 70/co's = 1 - 2(2 cos 20) - (2 cos 20) 2 + (2 cos 20) 3. Hence prov

that the roots of x 3 - x 2 - 2x + 1 = are x = 2 cos {%(2k + 1) n}, k = 0, 1, 2.

15 If x = 2 cos 0, prove ( 1 + cos 70)/( 1 + cos 0) = (x 3 -x 2 ~2x- l) 2. [Express

in half-angles.]

*16 (i) Give the last terms in formulae (iii), (iv) when n is (a) even; (6) odd.

(ii) State the last terms in numerator and denominator of formula (v) wh

n is (a) even; (b) odd.

17 Write down the expression for tan 30 in terms of t = tan 0. By first solving

tan0 = 1, show that the roots of x 3 —3x 2 —Sx+ 1 = are tan ^77-, tan -^n atan |7T. What are the roots of x 2 —4x + 1 = 0?

18 (i) If a + 2 + 3 + 4 = mr, where n is an integer, state the relation between

h, t 2 ,t 3 ,

£ 4 , where t r = tan r .

(ii) If tan0 1} tan0 4 are the roots of ti + at 3 + bt 2 + ct + d = 0, express

tan (0 X + 2 + 3 + 4 ) in terms of the coefficients. What can be said about t

angles if c = at

19 If tana, tan/?, tany are the roots of ax 3 + x 2 + bx + 1 = 0, prove tha + /? y is an integral multiple of tt.

*20 (i) Prove that cosw0 can be expressed as a polynomial of degree nc = cos0 of the form a„c n + a n _ i c n ~ 2 + ...+a n _ Sr c n ~ 2r + the last term bein

a or a x c according as n is even or odd. (ii) Prove that a n = 2 n_1. (iii) If n is even

put = \n to show that a = ( —l)* n; if w is odd, consider lim (cosn0/cos 0)

prove a 1 = n( —l)^ - 1). (iv) By deriving twice wo and equating coefficients

of c r, prove (r+ 1) {r + 2)a r+2 = (r 2 —n 2 )a r . (v) Hence expand cosn0 (a)

descending powers of c; (6) in ascending powers of c first for n even and the




nwe have used 2/(r) to denote /(l)+/(2) + ... +f(n), here it is con-

venient to write nn/(r)-/(l)/(2).../(n).

r=l

As with the E-notation, the range of values of r can be omitted whenclear from the context.

14.31 x»-l

We have x n —1 = if x n = 1 = cis2?rr, i.e. if x = cis (2r7r/w). For

r = 1 , 2, . ..

, n it follows that x —cis {2m jn) are distinct factors of x n —1

.

Since r = —k and r = n —k give the same value to cis(2nr/%), wemay use r = 0, —1, —2, . . . instead of r = n, n —1 . n —2, . . .

.

Case (i) : n even. Take r = 0, ± 1, ± 2, ± 1), This gives

\n —1 pairs of factors, together with the single factors corresponding

to and \n; in all, there are 2{\n —1) + 2 = n factors.

The factors corresponding to r = ± k are

/ 2kn . . 2kn\ I 2kn . . 2&7r\

a;— (cos h*sm L x— Icos *sm 1,\ n n /' \ w ft /

and their product is

/ 2knyt

/ . 2&7r\ 2. 2&7T

x —cos + sin ) = # 2 —2#cos hi.\ n J \ n J n

The factors corresponding to r = 0, |w are respectively x—l,x + l.

Hence* n_1 / 2kn \

for n even, x n —1 = (x —1) (# + 1) n I # 2 —2a; cos + 1 (i)

Case (»»): n odd. Take r = 0, +1, ±2, ±£(w— 1); this gives

1 + 2{%{n —1)} = n factors altogether. Arguing as before, we find that

*(n-l)/ 2kn \fornodd, x n -l = (x-l) U [x*-2xcos + 1). (ii)




Case (i) : n even. As in 14.31, we wish to pair off conjugate comple

linear factors so as to get a real quadratic product. Now the factors

. 2r-l . 2r'-l

X —CIS 71, X —CIS 71n n

are conjugate if (2r- 1) + (2r'—l) = 0, i.e. if r' = - (r- 1); therefore

with the values , „ ,

r=l, 2, 3, ...,\n

we pair r' = 0, - 1, -2, - {%n- 1).

This gives \n + {l + (\n— 1)} = n factors altogether. The factors

corresponding to r = k, r' = —(k —1) are

/ 2k-l . . 2k-l \ I 2k-l . . 2k-l \X—ICOS 7T + *Sin 7T\, X—|COS TT —l&m 7T\

\ n n J \ n n J

and their product is x* —2x cos {(2k —1 ) njn} + 1 . Hence

i n l 2k —1 \for n even, x n + 1 = II \x % —2xco& tt+ 1 1 .

&=i\ n I

Case (ii) : n odd. We pair off the values

r=l, 2, 3,...,i(n-l)

with r' = 0, -1, -2, -%(n-3).

The value r = $n + 1) gives 2r —1 = n, to which corresponds the r

factor x + 1. Then as before we have

Hn-l)/ 2k -1 \for n odd, x n +l = (x+l) H \x 2 —2x cos n+l).

k=i \ n J

14.33 x^-lx cos na + 1

The equation x 2n —2x n cos + 1 = is quadratic in x n:

(x n —cos na) 2 = —1 + cos 2 not = —sin 2 na,

.'. x n = cos na ± i sin na.

H h 2 l f hi h



14.33] de moivre's theorem 541

The product of the conjugate factors corresponding to r = k is

[X —cos/ 2kn\ . . / 2k7T$( I 2/br\ . . / 2kn\)

„ rt / 2kn\ ,= x* —2x cos I a -\1 + 1.

V n J

n-i i / 2kn\ \

:. x 2n -2x n coanoc+l = ]J lx 2 -2xcoaâ + —\ + l\. (v)

Examples

Various deductions can be made from the result (v).

(i) Putting a = 0, we obtain

n—if 2hn \

(rcn - l) 2 = T\W-1xcos +l},

&=ol n )

and if n is even this is equal toin -II 2kn \*

(x-l) 2 (x+l) z n W~2xcos + 1}fc-i I n )

because the values k = 0, k = \n give the factors {x— l) 2, (x+ 1)

2 respectively,

while factors corresponding to k and n —k are equal. Taking square roots (in

which the positive sign is chosen since in real algebra all the quadratic factors

are positive and x n —1, x % —1 always have the same sign when n is even), weobtain formula (i) of 14.31. If n is odd, we likewise deduce formula (ii).

Similarly, by putting a = n/n in (v), we obtain formulae (iii), (iv).

(ii) Divide both sides of (v) by x n:

1 „ «-M 1 n I 2hn\\B

H 2cosraa = TT {a?H 2cos|a + 1}.

*n

*-ol * \ n J)

is0,

2cosn0 —2cosna = H (2cos0 —2cos (oH )},fc=ol \ n })

n-i i / 2kn\\cos nd— cos net = 2 -1 H {cos0 —cos laH )}.

ft=ol \ n J)

(iii) In (v) put x = 1 and a = 2/?:

n—1 i I 2kn\\

x

Putting x = cis 8,




To decide the sign of the right-hand side, first suppose < /? < rt\n\ th

each factor on the right is positive, and so is sin nfi. Hence the sign + is appr

priate for this range of /?. As /? increases, sinn/? changes sign whenever /? pass

through a value kn/n, and simultaneously one factor on the right changes

Hence the sign is always + :

n—l / w\= 2 1 n sin(/? + —

fc=0 \ »/

(iv) By taking logarithms of the modulus of each side of this last result

assuming that /? is not zero or an integral multiple of 2n,

n—l I /jfc7j-\

I

loglsinwytfl = (n-1) log 2+ Slog sin(^ + —) .

fc=0I \ n /\

Deriving won—l i lvn\

ncotnfi = S cot IAH I

fe=0 \ nj

(v) de Moivre's and Cotes's properties^

of the circle.

A AX A 2 ... A n _ 1 is a regular n-sided

polygon inscribed in the circle of centre

O and radius a; P is a point such that

OP = x and /.(OP, OA ) = 0.

Since sides of the polygon subtend angle

2n/n at O, we have

L(OP,OA r ) = 6 + 2m /n.

Also, by the cosine rule,

PA* = x 2 + a*-2axcos[d +Fig. 137

( + v)-n-ll 1 2rn\

PAl.PA\...PAl_ 1 = II |a 2 -2aa:cos|0 + —j+a 2

= x 2n - 2a n x n cos nd + a 2n

by a slight extension of formula (v). Hence de Moivre's property:

PA . PA X ...PA n _ 1 = V(«2 - 2a n x n cos nd + a 2n

).

In particular, if P lies on OAQ , then 6 = and so

PA .PA 1 ...PA n _ 1 = \x n -a n\.

AIf OP bisects A n _ x OA^ then 6 = njn, and so

PA^.PA x ...PA n _ x = x n + a n.

These last two results are known as Cotes's properties.




Hence for r = 0, 1, 2, ...,n —1, sin 6 —sin (rn/n) are distinct factors of sinntf;

consequently n _ , ,

sin n# = &„ ^ ( sin 6 —sin —I. (vi)

r=l\ w/

This can be expressed alternatively as follows. The n distinct values ofsin (m/n) are given by r = 0, ± 1, ± 2, + 1). The factors which corre-

spond to r = k, r = —k are sin 6 —sin (kn/n), sin 6 + sin (kn/n), and have productsin 2 —aw? (kn/n). Hence

«»- >/ foAsin w# = 6 n sin # JJ (sin 2 —sin 2 —I. (vii)

fc=l \ »/Divide each factor following sin# in (vii) by —sin 2 (kn/n); then

sin nd —A sin/^D/ sin 2

(9 \

\A

[~ am? (kn/n))'

where J. is independent of 6 and can be determined by dividing both sides bysin 6 and then letting 6 -> 0. We obtain

. sinn#A = lim —;——= n,dÔ sin d

i(n-l) / am 2$ \and so sinw# = nsin0 TT 1 1 —r- tt^ —r~> I • (viii)

fc-L1

\ am^ (kn/n) J

Examples

(i) Comparison of series and product.

By identifying the product (viii) with the expansion of sinw# obtained in

ex. (iii) of 14.22, we have for n odd:

i(n-l) / ]c7t\ns n ( 1 - s 2 cosec 2 —) = ns + %n(l -w 2 )s 3 + ... +& B s B

.

ft=l \ n/

Equating coefficients of the various powers of s gives identities involving

cosec 2 (Jen /n); e.g. from those of s 3,

—n 2j cosec 2 ——£n( 1 —n 2),

ft=l »K»-D Jen

i.e. 2 cosec 2 —= A(n 2 —1) for n odd. (ix)

fc=l w

CO J(ii) Proof thm 2 - = |tt 2

.

If < < J7r, then sin 6 < 6 < tan 6 and so

J- < cosec 2 = l + cot 2 < l+^- a .




Writing 11 1s»» = —I h • • • H* p 2 2 m2

and using formula (ix), we have #

7T 2 W2 -1 7r2 n-l

< ^ —< ^ J + **-»Since the series S(l// 2

) is known to converge by 12.41 (3), therefore s m tend

to some limit s when m oo in any manner, and in particular through

sequence —1) for odd n. Letting n -> oo, the above inequality gives

s < |7T 2 ^ s,

so that s = \n*.

Exercise 14(c)

1 Obtain real quadratic factors of x6

—Xs

+1. Byequating coefficients

x B and of x 3, prove that

cos \n + cos | n + cos f 7T = and cos \it cos %n cos ftf = ^.

2 Find the real quadratic factors of x s —4a; 4 +16.

3 Write down real quadratic factors of x 6 + x 5 + x i + x 3 + x i + x+l. [T

expression is (x 7 —l)/{x— 1).]

By substituting ± 1 for x and 2/3 or 2/3+n/n for a in formula (v) of 14.33, pro

the following.n-l l (2r+l)n\

4 cosw/? =2«~ 1

IIsinj^ +-

——].r=0 I 2w )

n—1 I rjjX

5 sinn/? = ( - l)* n 2 n_1 JJ cos —> if n is even.

»-i ( (2r+l)n\6 cosn/? = (-l)*«2 n ~ 1

]J cos{,ff+- ———} if n is even.

7 Obtain a result by deriving no. 4 logarithmically wo /?.

n-l / r7T \

8 Prove 2 cosec 21 /? H—} = w2 cosec 2 n/?.

r=0 \ »/

n_1 f / 2/7r\)9 Prove ch nx —cos na = 2 -1 TT { ch x —cos ( a H }

} . [Divide formular=o I \ w /J

by o;n and then put x = e v

.]

10 Express # n_1/(#

2n —2# n cos noc + 1) as a sum of n partial fractions. [Deriv

(v) logarithmically wo a.]

11 Prove that

1 1 /2r+l \) 1 / lVt 1/ 1 « ™

x n + — = TT {* + —2cos — n\), x n = \ x--\ II |a; + --2cos—x n

r i A\ * \ 2n J) x n

\ x) rJi\ x n




13 (i) Obtain a result from no. 12 (i) by putting = 0.

(ii) By dividing both sides of no. 12 (ii) by sin0 and then letting -> 0,

P rove n-i mVn = +2 - 1

II sin —.

r=l n

14 (i) By deriving no. 12 (i) logarithmically, prove

nt&nnd _ n ~ 1 1

2jsin0 r=: QCOsd —coa{(2r+\)7il2n}'

(ii) What result is obtainable similarly from no. 12 (ii)?

15 With the notation of 14.33, ex. (v), prove that if P lies on the circumference,

then PA . PA X ...PA^^ 2a |sin £w0|

16 Show that the solutions of ( 1 + x) %n + ( 1 - xf n = are

2r-lx = ±i tan n, where r = 1, 2, ...,n.

4nn i (2r—l)n\

Deduce that (l + a;)2n + (l-a;) 2n = 2 FT a;

2 + tan 2 -—r=i \ 4« /

» (2r— 1)ttand hence prove that £ sec 2 = 2n 2

.

r=l 4n*17 Prove that

/ 2r7r\\cosn0 —cosraa = 2 n_1

JJ <cos0 —cos |a + -—}>

r=ol \ w /;

by using Ex. 14 (6), no. 20 and arguing as in 14.34.

14.4 Roots of equations

14.41 Construction of equations with roots given trigonometrically

(i) Form the equation whose roots are cos f n, cos f n, cos § n.

= %n, %Tt, § tt all satisfy cos 30 = — Since cos 30 = 4 cos 3 —3 cos 0, the

equation 4a; 3 —3a; = —£ is satisfied by a; = cos \tj, cos f n, cos f-

. The required

equation is therefore_ + 1 = q

(ii) Form the equation whose roots are cosfTT, cos cos f7T.

Consider the equation cos 40 = cos 30. It is satisfied by

40 = 2mn ± 30,

where m is any integer or zero, i.e. by = \mn (this includes both solutions

obtained from the double sign ± ).

Writing x = cos 0, the equation becomes




(iii) Form the equation whose roots are tan 2 \n, tan 2 § tt, tan 2 \tt, tan 2 %tt.

The equation tan W= is satisfied by 6 = \rm, where n is any integer

zero. Writing t = tan 6, it becomes (see formula (v) of 14.22)

Removing the root t = 0, we see that the equation

<8 - 36* 6 + 126* 4 - 84< 2 + 9 =

has roots i = tan {\mr), where n = 1, 2, . . ., 8, i.e.

+ tan^7T, ±tanf77-, ±tanf7r, ±tan^7r.

Putting x = t 2, it follows that

* 4 - 36# 3 + 126a 2 - 84a; + 9 =has roots tan 2 £7r, tan 2

f7<r, tan 2 \tt, tan 2f7T.

N.B. —Since tan 2 \tt = 3, we may remove the root x = 3 from the

equation, obtaining x s —B3x 2 + 21x—3 = Q, which consequently has ro

tan 2 Aw, tan 2 |7r, tan 2 ±7r.

14.42 Results obtained by using relations between roots acoefficients

If the results of 13.51 are applied to equations like those found

14.41, various trigonometrical relations are obtained.

Examples

(iv) Prove sec f tt + sec %tt + sec fzr = 6, and find sec \tt + sec f tt + sec f 7r.

The equation whose roots are the reciprocals of the roots of that in ex.

will have roots sec f tt, sec %tt, sec § tt. This equation is x 3 —6x 2 + 8 = 0, andsum of its roots is 6.

Since sec \tt ——sec %n, sec %tt = —sec %tt, and sec f tt = —sec f tt, hensec \tt + sec § tt + sec %n = —6.

(v) Calculate tan 2f tt + tan 2

f tt + tan 2f 7T.

From ex. (ii), the equation having roots sec f tt, sec f7r, sec far is

x 3 + 4x 2 -4x-8 = 0.

Putting 2/ = x 2, the equation having roots sec 2

f7r, etc. is

2




(vi) Prove that cos 3a = 4 cos a cos (a + § tt) cos (a + f 7r).

Consider the equation cos 30 = cos 3a, which is satisfied by 6 = oc + f for

where k is any integer or zero. It can be written

4x3

—Zx —cos 3a = 0,

where x = cos 6, and is satisfied by the distinct values x = cos a, cos (a + f tt)

and cos (a + f tt). By taking the product of the roots, the relation follows.

Exercise 14 (d)

1 Form the equation whose roots are cos \tt, cos \ tt, cos ^n.

2 Form the equation whose roots are ± cos \n, ± cos f tt.

[Consider cos 56 = —1.]

3 By considering cos 30 = cos 26, construct the equation whose roots are

gos%tt and cos£7r. Hence find cos 36°, cos 72° in surd form.

2m4 Show that x = 2 cos

-yy(r = 1, 2, . . ., 5) are the roots of

x 6 + x i -4x 3 -3x 2 + 3x+l = 0.

[Consider cos 66 = cos 56; remove the root x = 2 at the end.]

*5 (i) Construct the equation whose roots are ± 2 sin f 77, ±2sin^7T, ± 2 sin f 77

by eliminating x from y = 2^(1 — 2 ) and the result of ex. (ii) in 14.41.

(ii) Verify that 2 sinf 7r, 2 sin f 7r, 2 sin f rr are the roots of a;3 = + ^7 (a;

2 —1).

[The equation in y obtained in (i) can be written

2/8 = 7(2/2- 1)

2, i.e. y

t = ±Jl(y*-l).

Since 2sinf7r > 1 and 2sinf7T > 1, and also 2sinf7T = —2sin$7r lies betweenand —1, these values of y give y

3 and y2 —1 the same sign.]

Use the equations obtained in exs. (i)-(iii) of 14.41 to prove the following.

6 COS § TT COS %TT + COS f TT COS § TT + COS f 71 COS %TT = —f7 tan 2 \tt + tan 2

f tt + tan 2 %tt = 33 and tan \tt tan f 7r tan %tt = + V3 -

8 sec 2 £77- + sec 2f 7r + sec 2 $7r = 36 and sec 4 \tt + sec 4

%tt + sec 4 \tt = 1 104.

9 Calculate sin 2f zr + sin 2

f 7r + sin 2f tt.

10 Prove sin 3a = —4 sin a sin (a + f tt) sin (a + f 7r). [Consider sin 30 — sin 3a.]

1 1 Prove tan a + tan (a + \ tt) + tan (a + f 7r) = 3 tan 3a.

[Consider tan 36 = tan 3a.]

*12 B id i t 0 = h h i




14.5 Finite trigonometric series: summation by C+iSThe method of summing certain trigonometric series which

illustrated in the following examples gives results in pairs, and is als

applicable to infinite series (see 14.66, ex. (i)).

14.51 Cosines and sines of angles in a.p.

Write

C = cos a + cos (a +fi) + cos (a + 2fi) + . . . + cos {a + (n —1) /?}

and 8 = sina + sin(a+/?) + sin(a + 2/?) + ... + s in{a + (n— 1)/?}.

ThenG + iS = cisa + cis (a + /?) + cis (a + 2/?) + ... + cis{a + (w- l)fi}

= cis a + cis a cis /? + cis a(cis /?)2 + . . . + cis a(cis

= cis a{l - (cis^) w}/{l - cisytf}

on summing the g.p., provided cis/? 4= 1, i.e. J3 is not an integral

multiple of 2n. Now|

l-(cis^) n = 1-cisn/?

= 1 —cos nfi —i sin nfi

= 2 sin 2\nfi —2i sin \nfi cos \nfi

= —2* sin î/?(cos + i sin

= —2i sin cis

—2i sin cis \nflHence C + iS = cis a —2i sin £/? cis

sin Aw/? . ,1

. „= . ^^ cisla + ln-l /?}.

Equating real and imaginary parts, we obtain

„ sin \nB sin InB . f ,. , x „

C = \% cos {a + $(11 1)0} # = \ C sin{a + |




Write

andf

Then

14.52 Other examples

(i) Sum 1 + a; cos 0+ a;2 cos 20 + ... ton terms, where x is real.

G - l+a;cos0 + a: 2 eos20+...+a; n- 1

cos(n-l)0

S= a; sin + a;2 sin 20+ ... +a: n - 1 sin(ra- 1) 0.

C + iS = 1 + a; cis + a:2 cis 20 + ...+ a: - 1 cis (w- 1)0

= l+a;cis0 + (a;cis0) 2 + ... + (£ccis0) n - 1

1 —(*cis0) n=

. ^ on summing the g.p., if x cis 4= 1»1 —X CIS

1 —# n cisw0

1 —x cis

_ (1 —05 cis n0) {l-a cis (-0)}~ (l-a:cis0){l-a;cis(-0)}

to make the denominator real,

1 —a; cisn0 —iccis( —0) +£c n+1 cis(n— 1)0

l —2x cos + x 2

since cis 0, cis ( —0) are conjugate. Taking the real part of this result,

_ 1 —a; cos —a; cos n0 + a3n+1 cos (n— 1)0

1 —2a; cos + x 2

N.B. —If |a;| < 1, then x n -> when w -» oo, and the series has a sum to infinity

^ 1—03 008lrm (7 = .

w _kjo 1 —2a; cos + x 2

given by

If \x\ ^ 1, there is no sum to infinity.

(ii) Prove that sin a + ^ sin (a + /?) + ^ sm ( a + + • • • + sin (a + nf})

= (2cos£/ff) n sin(a + ^/?).

Write C = cosa+^ cos(a+/?)+Q^ cos (a + 2/?) + . . . + cos (a + n/?)

and *S' = sina+^ sin(a+/ff) + ^ sin(a + 2ytf) + ... + sin(a + W/fl).

Then C + t£ = cisa +j^j

cis(a+^)+ ^ cis (a + 2/?) + . . . + cis (a + w/?)



550 de moivre's theorem [14

= cisa(l + cis/?) n

= cisa(l + cos/? + *sin/?) n

= cis a (2 cos 2£/? + 2i sin cos £/?)

= cis a (2 cos \fl cis £/?)

= (2 cos \fi)n cis (a + in/?).

.'. S = (2 cos J/?) sin (a + on equating imaginary parts.

Exercise 14(e)

1 Sumsina —sin(a+/?) + sin(a + 2/?) —sin(a+ 3/?) + ... to 2w terms.

S1X1 { Th -J~ X ) $2 Prove 1 + cos0 sec0 + cos20 sec 2 + ... +cosn0 sec 0 = - — —

sin cos

3 Calculate Y. ( | sin 2r0.

r-lW

4 Find (2 cos 0)n -

(^j(2 cos 0) - 1 cos + f^j (2 cos0) - 2 cos 2(9 - ... to (n +

terms.

5 Sum to w terms the series whose rth term is (2r —1) af -1. If a = 2n/n

w is a positive integer, prove that

1 + 3 cos a + 5 cos 2a + . . . + (2n —1) cos (n —1) a = —n.

and 3 sin a + 5 sin 2a + ... + (2n —1) sin (n —1) a = —n cot £a.

6 Find the sum to n terms of sinJ. + (6/c)sin2J. + (6 2 /c 2 )sin3^. + ....

b < c in the triangle .ABC, deduce that the sum to infinity is (c/a) sin G.

7 If is not zero or an integral multiple of n, find the sum to infinity of

sin a + cos sin (a + 6) + cos 2 sin (a + 2/9) + cos 3 sin (a + 30) + . . .

8 If n is a positive integer, prove that

1 + nx cos + x* cos 26+ a:3 cos 30+ ... + x n cosnd = r n cosna,

where r = + A/(l + 2»cos0 + £c2

) and cosa:sina : 1 = (1 +«cos0) :scsin0

Further examples appear in Exs. 14 (/), (g).

14.6 Infinite series of complex terms. Some single-valued fun

tions of a complex variable

14.61 Convergence and absolute convergence

(1) Limit of a complex function of n. If s n = o~ n + ir n , and if o* n ->

d T h > h + h > d i




and so \s n -(<r + iT)\ = \(<r n -<r) + i(T n -T)\ < |<r n -<r] + |t„-t| < e.

Conversely, if this last condition is satisfied for all n 5s N, then since

|o n-o-| < |(o-»-o ) + *(t„-t)| <e,

we have <r B -> cr (and similarly r n -> r).

(2) For the series 2z r , where z r = a;,. + iy r and a; r , y r are real,

n n nSn = 2 z r = 2 «r + * 2 2/r = ° « + * T»> sav -

r=l r=l r=l

By (1), 5 n tends to a limit if and only if <r n , r n both tend to limits whenn -> oo. Accordingly we give the following definitions.

2z r is convergent if 2# r and Sy r both converge.

If YiX r ,Y,y r converge to sums X, Y, then Z = X + iY is the sum to

infinity of 2z r .

(3) If the series (of real positive terms) S |z r jconverges, we say that

2z r is absolutely convergent (a.c).

It follows that if T,z r is a.c, then lLz r is also convergent in the sense

of (2). For since- \x r \ < J{x? + yf) = \z r \, and S \z r \is convergent by

hypothesis, the comparison test of 12.41 shows that S converges;

i.e. 2# r is a.c. and consequently (by 12.52) also converges. Similarly

ljy r converges, and hence by definition 2z r converges.

14.62 The infinite g.p. l + z + z 2 + z 3 + ...

Write sjz) = 1 + z + z 2 + ... + Z - 1.

If we multiply by z and subtract (just as for a g.p. with a real commonratio), we find that, if z =i= 1,

1 z n

1-z l-z

Putting z = r(cos0 + *sin#), where r > 0, we have

z n = r n (cosnd + iainnd).

Ifr < l,thenlimr n = 0;andsince |r n cosw#| ^ r TC and |rn sin?&0| ^ r n

,

therefore both r n cos nd and r n sin w# tend to zero when n -> oo. Hence




smnd oscillate between ± 1 when n^-oo, and again s n (z) does n

tend to a limit. If z = 1, clearly s n (z) = n.

The g.p. l + z + z 2 + ... therefore converges only if \z\ < 1, and its su

to infinity is then 1/(1 —z). We can write

(1-z)- 1 = l + z + z* + z* + ... (\z\ < 1).

The g.p. is absolutely convergent when the series of modu

l + r + r 2 + ... converges. This is the case when and only wh^ r < l,i.e. \z\ < 1. Hence the ranges of convergence and of absolute

convergence are the same.

14.63 The exponential series

Z Z 2 2 3

1 +l

+2

+3

+ -

(1) Write z = r(cos 6 + i sin 6). Then since the series

1 +Ti + 2

+3

+ -

converges for all r, the series

z z 2 z z

1 +T

+2

+3

+ -

is a.c. for all complex numbers z, and hence converges for all z.

Denoting the sum-function of series (i) by expz, or by exp (z) whnecessary, we have

(z z 2 z n \1 + n + iTi + --- + for all z.

1 2 n\)

Remark. The notation is chosen because of the analogy with t

real series » ,

1 + n +2

+3

+ -

whose sum-function e x is occasionally written expo; for convenience

in printing. We usually avoid writing e z for the sum to infinity

k b k) h h z



14.63J de moivre's theorem 553

(2) By using the definition (ii), it can now be shown that expzobeys the functional law

exp z x x exp z 2 = exp (z 1 + z 2 ), (iii)

just as exp x does in real algebra.

Let |zx| = r lt |z a |= r 2 , and write

z z 2 z n

*»W = l + Tl+

2+ - +

nl'

L Z l Z l Z l\ f, 2 2 2 S Z »\*n(Zl)-*n( 2 a) = + —+—+•..+—} {1 + —+ —+...+ —v u n \ a/

^T

iiT

2T

nlf\ 11 2 ra j

2, 2* 2?= 1 + —+ —+ ...+ —

1 2 w

Z2

2X

22

2*Za

g za

1 111 211 nilw 2 « w2 ^.2 «2a 2 z l z 2 z

lz 2 ,ZLZl

2 1 2I 212 n 2

2t* «>2^n rt-w^fi

+ _2 _|_ _J_2 + -i_L + . . . + J—1,

w l n 2 n n\n\

We now add up these terms 'by diagonals ' : the terms of total degree * (s < n)

1 *iJg

2 2 i2g

2 , 2j

s (s-l) l (s-2) 2 s

1 ( 8 , , , 1

and consequently

= ^ (* + *)•»

2*2°*»( z l)*»( z 2)-*n( z l+ z a) = 2 ( iv

)

p,qP ' '

where the last summation is for all p and q for which p + q > n, p ^ n, q ^ n;

for these are precisely the terms not included in the preceding summation bydiagonals.

In exactly the same way we havepPpQ

s n (r 2 ) * n (r 1 + r 2 ) = £ . ( v)

We know that, when n -»> oo, sjfi) -> e r s s n (r 2 ) -> e r», and s n (r 1 + r 2 ) -+ e ri+r *.

Hence the left-hand side of (v) tends to zero, so also y\ - -> 0.

From (iv),




14.64 The modulus-argument form of exp z

By (iii), expz = exp{x + iy)

= exp x exp (iy).

Now by putting z = x in (i),

expz= +^+|| +

= e x for all x.

Also « n (»y) = 1+^+^- + -.+-1 2 n\

y 2 ,y* \ , : L y 3, y a

=l

i -2 + 4 --rn 2/ 3 + ^ '

where the series in the brackets are finite series. Since by 12.61 (2)

, y%

2/4

cost/ =1-1 + ^-...

, y z y 5

and amy = y __+|__...

for all (real) values of y, we have by letting n-+co that

exp {iy) = lim s n {iy)n—> oo

= cosy + isiny.

Therefore expz = e x (cosy + i sin 3;).

This result shows that exp z is periodic, with period^ 2m. For

exp (z + 2m) = exp {x + (y + 2n) i}

= e x {cos (y + 2tt) + i sin (y + 27r)}

= e x (cos y + i sin y) = exp z.

14.65 Euler's exponential forms for sine, cosine

F




First adding, and then subtracting, we find

cos y = f {exp {iy) + exp ( - iy)}, sin y = ^ {exp {iy) - exp ( - iy)},

(vii)

2i

where we observe the factor l/2£, not 1/2, in the last result.

These formulae, often written

cos y = \{e iy + e- iv), sin y = ^ (e iv - er iy

),

are known as Euler's exponential forms. However, they are equivalent

to the infinite series expansions of cosy, siny stated above.

14.66 Examples

(i) Find the sum to infinity of

l+cos0 tan + ^ cos 20 tan 2 6 + ^ cos 30 tan 3 0+....

Write t = tan 0,

tz

t3

C = l+«cos0+ —cos 20 + —cos30 + ...,

2 3

t2

t s

and S= «sin0 + —sin20 + —sin30 + ....

t2 z z t

3 z 3

Then O + iS = 1 +tz+ —+ —+ where z = cis 0,2 3

= exp {tz) for all z and all real £,

= exp (sin + i sin tan 0)

= e sln0 cis(sin0 tan0).

.-. G = e sln0 cos (sin tan 0).

(ii) Trial exponentials: the case of failure. In 5.33(2) we remarked that thetrial method of solving y + ay' + by = by the substitution y — e mx will fail

if the quadratic m2 + am + 6 = has no (real) roots, i.e. if a 2 < 46. Putting

a 2 + 4p 2 = 46, we find that in complex algebra the roots are m = —\a ± ip, so

that by a formal application of 5.32 (2), II, the general solution would be

—e -Ux e ipx + ft g-ipas)

_ e -iax [A(cospx + * sin pa;) + B(cospx —i sinpx)}



556 de moivre's theorem [14.

If u, v are derivable functions of x, and we define the derivative of u + iv

be u' + iv', then

d—{coax + iaiax) = —sinx + icosx = i(coax + ismx),dx

di.e. —le ix

) = ie %x.

dx

More generally, by formula (vi) we have (for real constants a, 6)

e (o+rt) B _ e ax( cos fa _f. I gin fa) —e ax cog fa _j_ 1 g ax 8m fa f

d,\ —{e(a+it)xj — e ax

( a cos fa _ & gin fa) + i e ax (a sin bx + b cos 6a;)

dx= e ax {(a + ib) cos bx + *(a + ib) sin &#}

= (a + ib)4 a+ii)x.

Thus the law d{e mx )\dx = roe M, already known from 4.41 (5) for all real

stants m, also holds for complex ones. Equivalently, we may write

e (a+ib)xd x = —* e (o+t6)a + C)a+ib

where c is a complex arbitrary constant.

This result permits rapid calculation of certain integrals involving cbinations of exponential and circular functions. For example, put

C =j

e ax cosbxdx, S = je ax sinbxdx;

then C + iS = je ax (cos bx + i sin bx) dx

=I e (.a+iU*dx = e (a+ib)x + C

J a + ib

a —ibe ax (cos bx + i sin bx) + c.

a* + b

Equating real and imaginary parts,

„ a cos bx + b sin bx „ asin

bx —b cos bx _G- Z e ax + c L , S = —— e ax + c 2 .

Cf. Ex. 4(e), nos. 28, 29.

A similar method, combined with integration by parts, will apply to

cf. Ex. 4 (m), no. 35.

cos, _

x n e ax fadx;sm

Exercise 14(f)




Find the sum to infinity of the following series.

8 ccsm0 + (x 2 /2 )sm20 + (a:3 /3 )sin30-|-....

9 cos a - cos fi cos (a { ) + ( 1/2 ) cos 2/? cos (a + 2/?) - ...

10 S —see'/? sin r/?. 11 eos0 + —cos30 + —cos50+....r=lT 3 5

12 i^.r=l

13 If w = exp(z/a) where w = u + iv, z = x + iy, and a is real, find the locus

of w corresponding to each of the lines y = 0, y = \Tia in the z-plane.

14 If u + iv = exp (x + iy), find the locus of the point (u, v) in the w-planewhen (i) x = constant; (ii) y = constant. Verify that these two loci cut ortho-gonally.

1 5 Criticise the following argument. ' If y = cos x + i sin x, then

dy—= —sin x + i cos x = *(cos# + isina;) = %y\dx

therefore by 4.41, Remark (a), y = A e ix. Since y = 1 when x = 0, therefore

A = 1. Hence cosaj+isina; = e ix.

i

14.67 Generalised circular and hyperbolic functions

(1) Consistently with formulae (vii), we define

cos z = |{exp + ex P ( ~ * z )}>

sin z = {exp (iz) —exp ( —iz)}, I

and then tanz to be sinz/cosz, etc.

These definitions are equivalent to

cos z + i sin z = exp (iz), cos z —i sin z = exp ( —iz),

(viii)

z 2 z 4 2 6

and also to cos z = 1 —^7 + 77—771 +2 4 o

z z 3 z 5

Sm2=l -3

+ 5 -

(all z). (ix)

Directly from these definitions it can now be shown that the

formulae relating trigonometric functions of a real variable continue

to hold for a complex variable. (The definitions were chosen with a






In any formula relating circular functions of z x ,z 2 , ... we may replace

z l5 z 2 , ... by iz 1} iz 2 , ... and then use results (xii) to translate it into

hyperbolic functions of z x ,z 2 , .... In the course of this translation each

product of two sines becomes i2

multiplied by a product of two hyper-bolic sines. Since the formulae all hold in particular when the variables

are real, we have the justification of Osbom's rule stated in 4.44 (4).

An example of the detailed verification is the following:

ch (A + B) = cos {iA + iB)

= cos (iA) cos (iB) —sin (iA) sin (iB)

= chA chB-i 2 ahA ahB

= ch.A chB + ahA ahB.

Examples

(i) cos (x + iy) = cos x cos (iy) —sin x sin (iy)

= cos a; chy —iainx shy;

sin (x + iy) = sin x cos (iy) + cos x sin (iy)

= sin a; ch.y + icosx shy;

sin (x + iy) sin (x + iy) cos (x —iy)and \mi(x+%y) = ——— = ——r- :

cos (x + ry) cos (x + iy) cos (x —%y)

where we have multiplied top and bottom by the conjugate of the denominator

in order to make the new denominator real. Using the formulae for products

into sums, . . _ . , _sin 2a; + sin 2»2/ sinzaj + ^shzy

tan (x + iy) = —- = —— r——cos 2a; + cos 2*2/ cos2a; + ch22/

(ii) From ex. (i),

|cosz| 2 = cos 2 x ch 2 y + sin 2 x sh 2y

= (l-sin 2 a;)ch 2«/ + sin 2 a;(ch 2 2/-l)

= ch 2 y —sin 2 x

< ch 2 y for all x.

Similarly | cos z| 2 = cos 2 x + sh 2 y ^ sh 2 y for all x.

Thus |sh?/| < |cos2| < chy for all z.

The reader should verify that the same inequalities hold for |sinz|

(iii) Find the equation of the curve described by the point (x, y) when z = x + iy

varies so that sh. z —z is real. Sketch the part of the curve which lies between the






1 3 Expand £(sh z —sin z) in ascending powers of z.

U8eC+iS to find the sum to infinity of

M M sin 26 sin 40 sin 60 n cos 50 cos 96

14 +1

h..- 15 cos0H\-

1-....21 41 61 5 9

Miscellaneous Exercise 14(A)

1 If x —cis a and y = ciafi, prove that

a; -2/ .. , (x + y)(xy-l) sina + sin/?= * tan +(a p) and ,

— — =:—

x+y (x —y)(xy+l) sina —sin/>

2 If the real and imaginary parts of (1 +ix) n are equal, n being a positiveinteger and x real, prove that x = tan {(4r+ l)7r/4n} where r is any integer

or zero. [Put x —cot 6.]

3 Prove

{(cos a + i sin a) —(cos ft + i sin /?)} + {(cos a —i sin a) —(cos fi —i sin /? )}

= An 2 n + 1 sin n |(a-y 5) cos^(a+/?),

where A„ = ( —l)* n if n is even, and An = ( —l)*< n + 1) if n is odd.

4 (i) Solve x 6 + x i +x* + x* + x+ 1 = 0.

(ii) Find the solutions of x6

= 1 for which 1 + x + a;2

=}= 0.

5 If n is a positive integer, show that every root of (z+ l) an + (z— l) 2n =is purely imaginary. If the roots are represented by P x , P a , . . ., P 2n and is the

origin, prove OP\ + OP| + . . . + 0P 2„ = 2n(2n - 1).

6 Show that the roots of (z —l) 5 = 32(z + 1)B are represented by points on a

circle of radius f, and that the roots are

^- 3 + 4i sin ^ ^5 - 4 cos (r = 0, 1, 2, 3, 4).

A I 2r7r\ 33 5 / 2m\Deduce that XI I— 3 + 4isin 1 = —— ]J[ (5 —4cos— —I.r=l \ 6 / 31 r=1 \ 5 /

7 If a = cis (27r/n) and n, are integers, prove that 1 + a* + a 2 * + . • . + a (n-1)fc

is equal to n if is a multiple of n, and is zero otherwise. [Sum the g.p. whena* 4= 1.]

8 Express (1 —t 2 )/{(l —ot)(l —bt)} in partial fractions when the non-zero

constants a, b are (i) unequal; (ii) equal. Taking a = ft-1 = cis0 and a suitable

value of t, deduce that if < <j> < \n,






20 Prove that if a is real, the equation exp z = z + a has no purely imaginary-

root. If x + iy (y + 0) is a solution, prove that x> 0. [Use sixxyjy < 1.]

21 If u + iv = (z-l)exp(-*a) + (z-l)- x exp(»a) where z = x+iy and a is

real, find u and v in terms of x, y, a. Prove that the locus of z when v = consists

of a circle with centre (1,0) and unit radius, and a straight line through thecentre of this circle.

22 If exp(w + *«) = rcistf, prove u + iv = logr + i(d + 2nn), where n is any

integer or zero.

*23 If x+iy = cos (u + iv), prove

(l+x) 2 +y 2 = (ch« + cosw) 2 and (1 -x) 2 +y 2 = (chv-cosw) 2.

If a; = cos 0, y = sin0 (0 < < tt) in the preceding, find cosw and cht> in

terms of cos \0, sin \d, justifying the choice of sign when square roots are taken.

24 If u + = coth (x + iy), prove v = - sin 2y/(ch 2x —cos 2y). Show that

iv = {1 - exp (2a; - 241/)}- 1 - {1 - exp (2x + 2m/)} 1

00

and deduce that if x < 0, then « = - 2 2 e 2ra sin 2ry.r=l

25 If to = z 2, sketch in the Argand diagram the path of w when z describes

the sides of the rectangle whose vertices are ± a, ±a + ia (a real), starting at

O and moving counterclockwise.

26 (i) Sketch the graph of y = thx. Show that the equation thas = Aas has

two real non-zero roots if < A < 1.

(ii) Prove that the equation tanz= Az has two purely imaginary roots if

< A < 1; but that if A has any other real value, all roots of the equation are

real.

Find the sum to infinity of

27 cos 2 a + ^ cos 2 2a + ^ cos 3 3a + . . . (a real)

.

28 a?sin0+— sin30 + |-sin50 + ... (x, 0real).3 5

29 Expand eax

cos bx as a powerseries in x. [Consider e ax cis bx = exp (ax

+ibx)

and put a + ib —r cis 0.]

30 (i) By considering exp (a;) + exp {<ox)+ exp (a) 2 x) where w3 = l,w 4= 1. prove

1 + |.+?L + _ + ... = ^{e* + 2 e-** cos (%x V3)} for all real x.

*(ii) What results are obtained by considering

(a) exp (x) + (0 exp ( (ox) + w2 exp (o) 2 x)

(b) exp (a;) + w2 exp (wa;) + (o exp (w 2a;) ?



564 DE MOIVRE'S THEOREMand solve the latter. Writing z' = ze m

, prove that x' = A cos [it, y' = Bainwhere [i

2 = n 2 + k 2 and A, B are real arbitrary constants.

32 s, ijr are the intrinsic coordinates of a point on a plane curve, and x, ythe cartesian coordinates. Writing z = x + iy, prove that

dz ., d 2 z . .,— = e**, —= ik e 1 ^.ds ds 2

[Assume equations (iii) of 8.12.] Deduce that

K~y\ds 2

)+

\ds 2) ) ~d^ds 2 ~dsds 2

'

(Cf. Ex. 8(d), no. 8.)

33 Verify that the formulae for rotation of axes through angle 6 in 15.73can be written z = z' e id

, where z —x + iy and z' = x' + iy' . Deduce fromthe formulae for the reverse transformation.



565

15

SURVEY OF ELEMENTARYCOORDINATE GEOMETRY

15.1 Oblique axes

15.11 Advantage of oblique axes

Although it is usual to refer cartesian coordinates to two perpen-

dicular axes Ox, Oy, there are problems in which use of oblique axes

is more convenient. For example, in proving properties of the triangle

by coordinate methods, two of the sides can be chosen for coordinate

axes. The coordinates (x, y) of a point

P are then its signed distances from Oy,

Ox measured parallel to Ox, Oy re-

spectively. The angle xOy (measured

counterclockwise from Ox) is denoted

by a).

In sections 15.1-15.5 we revise the

main results about 'the straight line', Fig. 139

and extend them to the case of oblique

axes when this does not introduce any essential complication. The

reader should observe which results remain the same in form whether or

not the axes are oblique', they are marked by the sign {&>}.

Notation. In Chapters 15-19 we use the convention that Px

has

coordinates (x v yj, and so on.

15.12 Cartesian and polar coordinates

We first obtain the relations analogous to those in 1.62 for rect-

angular axes, taking O as pole and Ox as initial line (fig. 140). DrawPN perpendicular to Ox; then

ON OM MN



566 ELEMENTARY COORDINATE GEOMETRY [15.13

Example

If the line PiP 2 is of length r and makes angle 6 with Ox (fig. 141), then

r cos 6 = (a; a - a^) + (y 2 - yj cos w,

rsin# = (y 2 —yi) sin o>.

Fig. 141

15.13 Distance formula

From the above example, or directly from the cosine rule applied

to triangle PX P2 Q (fig. 141),

PX P\ = r 2 = (x 2 - Xj) 2 + (y 2- y x )

2 + 2(x 2- x x ) (y 2

- y x ) cos o).

15.14 Section formulae {to}

These give the coordinates of the point dividing PX P2 in the ratio k

Let P(x, y) be the required point, so that PX P\PP 2 = kjl.

(1) Internal division. With the construction shown in fig. 142

triangles PX PQ, PP 2 R are similar; hence

PP 2 PR

Since PX Q = x —x x and PR = this becomes

from which x=l+k

•



15.15] ELEMENTARY COORDINATE GEOMETRY 567

In each expression observe that the number k, which corresponds

to the segment PX P, is multiplied by the coordinate of P2 ; while I is

multiplied by that of Pv

(2) External division. Lettering fig. 143 (which illustrates the case

k > I) as in fig. 142, we still have similar triangles PX PQ, PP^R,

so that Pil^M~ PR'

i.e.

from which

Similarly

k =I

~

x =

y =

l-k'

l-k'

Fig. 142 Fig. 143

(3) Summary. In both cases above, k and I were understood to be

positive numbers. However, the result of (2) can be written

x = + x zetc.,

l + (-k)

which is formally the same as that of (1) for division in the negative

ratio —k : I. Hence if we agree that PX P : PP 2 shall be reckoned posi-

tive when P divides PX P2 internally and negative when the division is

external, both cases are covered by the same formulae, viz.

iyi+ty



568 ELEMENTARY COORDINATE GEOMETRY [1

definition makes no reference to Oy, we have (as for rectangular a

that if two lines with gradients mx , m2 are parallel, then m1= m

they are perpendicular, then mx m2 = —1; and in the general case

angle between them is given by

tan -1 m1 +m 1 m2

*

For cc = d x —6 2 , where

tan 6 X= mx , tan 6 2 = m2 ,

tan 6 X—tan d 2

and so

tana = m1—m2

1 + tan ^ tan # 2 l + mx m2

' Fig. 144

Example

From the example in 15.12, the gradient of PxP 2 i s

(2/2-2/1) sin wtan0 =

(x 2 - x ± ) + (y 2- y x ) cos ft>

'

15.16 Area of a triangle

(1) One vertex at 0. Let the polar coordinates of P2 , P3 be (r 2

(r 3 , 3 ) ; see fig. 145. Then by 15.12,

r 2 cos d 2 = x 2 + y 2 cos (o, r 2 sin d 2 = y 2 sin &>,

and similarly for P3 .

Area of triangle OP2 P3 = £OP 2 . 0P 3 sin P2 OP3

= £r 2 r 3 sin(0 3 -<92 )

= \r 2 r 3 sin # 3 cos # 2—

^2 r 3 cos 3 sin 6 2

= i(^2 + 2/2 cos w ) (2/3 sin w )

- %{y 2 sin w) (# 3 + y 3 cos

= \ sin &>(a; 2 «/ 3 —x 3 y 2 ) after reduction

_ 1- 22/2

2/3

smw.

B h A di h P P d




(2) Triangle PX P%PZ . We may choose new axes through one of the

vertices, say Px , which are parallel to the original axes Ox, Oy. The new

coordinates of P2 are then (fig. 146)

and of P, are

x % —x % ~ x i> 2/2 — 2/2 Vv

x z= x z~ x i> Vz =

2/3 2/i-

Fig. 145

Fig. 146

By (1), area of P1 P2 -f>

3 is

ft

2/3

sinw = \#2 x \

XZ~ Xl

•Ps—

l

2/2-2/1

Vz-Vi

2/2-2/1

Vz -2/i

sinw

sm<y




Example

The points P x , P 2 , P 3 will be collinear if and only if the area of triangle P 1 Pis zero, i.e. */ and only if

x 2 y 2 1 = 0.

#3 2/s 1

15.2 Forms of the equation of a straight line

15.21 Line with gradient m through (x lt y£Assuming the axes are rectangular, let P(x, y) be any point of th

line. Then the gradient is (y —yJKx —Xj), and hence

00 ~~~~

i.e. . y —y 1 = m(x —x 1 ).

This equation, which is satisfied by the coordinates of any point Pthe required line, is what we mean by 'the equation of the line'.

The result has been used in previous chapters when finding equations

oftangents and normals to a curve. It can be written

y = mx + c,

where c = y x —mx v There is no simple analogue for oblique axes.

15.22 Gradient form

( 1 ) Rectangular axes. The above shows that any line with gradient

has an equation of the form y = mx + c. Conversely, any equation

which can be put into the form y = mx + c (where m, c are constants)represents a straight line of gradient m making intercept c on

For when x = 0, the equation shows that y = c, so (0, c) lies onlocus; also {y —c)j(x —0) = m, so that the gradient is constant aequal to m.

(2) Oblique axes. If the line makes angle 6 with Ox and intercept c on Oy,gradient is tan 6; and if P(x, y) is any point on the line, then by the example15.15 its gradient is .°




and sin 6y = -r—. a\ x + e '

sin (o) —6)

The equation of the line is therefore still of the form y = mx + c, but now mno longer represents the gradient

ofthe line.

If the equation y = mx + c is given, then m = (y —c)jx, and the expression (i)

for the gradient of the join of P{x,y) and (0, c) becomes msinw/(l +wcosw),which is independent of (x,y). Hence this equation represents a straight line

of gradient

1+m cos (o

clearly it makes ^/-intercept c.

Example

Find an angle between the lines y = mx x + c x , y = m2 x + c 2 , the axes beinginclined at angle a>.

The gradients of the lines are

m, sin o) m2 sin a)

1+miCosw l+m 2 cosa)

By 15.15, an angle between the lines is given by tan _1{(/« 1 —/* 2 )/(l +/* 1 /t 2 )}

unless fixfi 2= —1 (in which case they are perpendicular). After reduction this

is found to be , > .

(m 1 —m2 ) sin to

tan -1, — ;

1

+(m

1+ m

2) cos a) + m1 m2

the lines are perpendicular if 1 + («* 1 + m2 )cosw + m1 m2 = 0, and are parallel

if mx = m2 .

15.23 General linear equation Ax + By + C = {to}

If B =|= the equation can be written y = —(AjB)x —CIB, which

by 15.22 is the equation of a straight line.

If B = but A #= 0, the equation can be written x = —C\A y which

is the equation of a straight lineparallel

to Oy.Since these are the only possibilities, it follows that every linear

equation in x, y represents a straight line.

15.24 Intercept form: line making intercepts a, b on Ox, Oy {to}

Let the gradient form of the equation of the required line be

y = mx + c. Since the line passes through the points (a, 0), (0, b), wenave = ma + and b =




Example

ABC is a triangle with sides CA, CB along given lines, and 1/CA + lis constant. Prove that AB passes through a fixed point.

Choose the lines along which CA, CB lie

to be Ox, Oy. Let the coordinates of A, B be(a, 0), (0, 6); then by hypothesis 1/a+ljb = k,

where k is constant.

The equation of AB is xja + yjb = 1. Theabove condition shows that this line passes

through the point (\jk,\jk), which is inde-

pendent of a, b.Fig. 147

15.25 Line joining P v P 2 {to}

(1) For rectangular axes, if P(x, y) is any point on the line, then

V-Vi = Vi-Vzry* /y» /y* rv**As 1 1 2

since each expression is equal to the gradient of the line.

(2) This result is also true for oblique axes. For if the required line

y —nix + c, then t/ x = mx x + c and y 2 = mx z + c, so that by subtraction

y-y 1= m(x-x

1) and

y 1-y

i= m(x

1-x

2);

and the equation is obtained by division.

(3) In either case the result can be obtained as follows. Conside

the equationy

yi

y%

= 0.

It is linear in x and y (as is clear by expanding from the first row),

therefore by 15.23 represents a line. It is satisfied by (x^y^) and(x 2 , y 2 )

(since then two rows become identical). Therefore it represents

the line (This form of the equation

could also be seen from the example in

15.16.)

15.26 Parametric form for the line through

(jtj y ) in direction 8




Hence the coordinates of any point on the line are expressible in terms

of the parameter r as (x x + r cos 6,y x + r sin 6).

Later we shall use this form in the treatment of conies. There is

no simple analogue for oblique axes.

15.27 General parametric form {to}

More generally, the equations x = at + b, y = ct + d (where a, b, c, d are con-

stants of which a, c are not both zero, and t is a variable parameter) always

represent a straight line. For elimination of t shows that

c(x —b) = a(y —d),

which is linear iax,y and therefore represents some line (by 15.23).

15.28 Perpendicular form

Let the perpendicular from O to the line have length p (essentially positive)

and make angle a with Ox. Then the foot N of the perpendicular has polar

B

P/

N

\P(*. y)

—>-A \ \ x

Fig. 149

coordinates (p,cc). Let any other point P of the line have cartesian coordinates

(x,y)

and polar coordinates (r, 6); then

p = rcos(0 —a) = rcosd cosa + rsin# sina

= x cos a + y sin a,

so that the line has equationx cos a + y sin a = p.

Alternatively, the intercepts on the axes are OA = p sec a, OB —p cosec a,

and the result follows from 15.24.




i.e. if h ax,+by x + ct = —̂

r1—•

I ax 2 + by 2 + c

If Plf P2 lie on opposite sides of A (fig. 150), then P divides P\Pinternally and so kjl is positive; hence ax 1 + by 1 + c, ax 2 + by 2 + c hav

opposite signs.

Fig. 150 Fig. 151

Similarly if P\, P2 lie on the same side (fig. 151), kfl is negative anso ax 1 + by x + c, ax 2 + by 2 + c have the same sign.

It follows that the converses of these statements are also true.

Examples

(i) Menelatis's theorem. Let the line ax + by + c = divide the sides P S PP 8 P X , jP]P 2 of the triangle P 1 P 2 P 3 in the ratios k ± : l lf k 2 : Z 2 » k 3 : h' Writing

u n = ax n + by n + c {n = 1,2,3)

we have by (i) above:

Multiplying,

&2

&2 ^3

ij 1% l 3

u 9

'

1.

(ii) Ceva' 's theorem. If lines P 1 QX , P%Qi, P3Q3through the vertices of triangle P 1 P 2 P 3 are con-current at (h, k) and divide P 2 P 3 , P 3 Pi, P\Piin the ratios k t : l x , k 2 : l 2 , k 3 : l 3 , then since the

equation of P X QX is (see 15.25(3))

y

= 0,

Fi 152



15.32] ELEMENTARY COORDINATE GEOMETRYOn multiplying, .

Ki /Cg K3 _ jZj Z 2 Z3

575

15.32 Perpendicular distance of a point from a line

Let d denote the length (essentially positive) of the perpendicular

from Px to ax + by + c = 0, and let the foot be Pa . Then

d^{x % -x x f + {y 2 -y x f. (ii)

Since PX P2 is perpendicular to

we have

ax + by + c = 0,

= -1,

Fig. 153i.e. b{x 2 -x x )-a{y 2 -y x ) = 0. (hi)

Because P2 lies on ax + by + c = 0,

ax 2 + by 2 + c = 0;

and this can be written (analogously to (iii)) in the form

a{x z- x x ) + b(y 2 - x ) = - {ax x + by x + c). (iv)

Square and add (iii) and (iv), using (ii):

d 2 {a 2 + b*) = {ax x + by x + cf.

• a - 1

afti + fyi + ca ~ ±

V(«2 + &2

)

'

where the sign is chosen to make the expression positive: +if

ax x + by x + c > 0, - if ax 1 + by 1 + c < 0. Also see Ex. 15(a), no. 15.

If c 4= and the equation of the line is written so that c is positive,

then a x x + b x y + c will be positive when Px lies on the origin side of

the line, and negative otherwise, by 15.31.

Exercise 15(a)f

1 Prove that the point which divides in the ratio 2 : 1 the median P x Qx of



576 ELEMENTARY COORDINATE GEOMETRY [

3 P is the point (h, k) referred to axes at angle (0. Prove that the line join

the feet of the perpendiculars from P onto the axes is sin (o J(h 2 + k 2 + 2hk co

4 Two fixed lines Ox, Oy are cut by a variable line at A, B respectively;

P, Q are the feet of perpendiculars from A, B onto Oy, Ox. JiAB passes throu

a fixed point, prove that PQ will also pass through a fixed point.

5 If the equal sides AB, AC of an isosceles triangle are produced to E,that BE . CF = AB2

, prove that EF always passes through a fixed point.

Perpendiculars PM, PN are drawn from P onto Ox, Oy. Find the locus of P6 PM+ PN = 2c. 7 OM+ ON - 2c. 8 MN= 2c [use no.

9 From a point P on one side of a fixed triangle, perpendiculars PM,are drawn to the other sides. As P varies on this side, find the locus of the

point of MN.10 Through a fixed point E any two straight lines are drawn to cut one f

line Ox at A, B and another fixed line Oy at G, D. Prove that the locus of

meet of BG, AD is a straight line through O.

11 If a straight line passes through a fixed point, find the locus of the

point of the part intercepted between two fixed intersecting lines.

*12 A line AB of constant length c slides between two given oblique

which meet at 0. Find the locus of the orthocentre of triangle OAB.*13 Using similar triangles, give a proof of Menelaus's theorem (15.31, ex.

« by the methods of 'pure ' geometry.

*14 Using areas, prove Ceva's theorem (15.31, ex. (ii)) by 'pure' methods.

15 Obtain the result of 15.32 as follows. Let ax + by + c = cut Ox, OL, M. Equate \p.LM to the area of triangle P X LM. Verify the result wthe line is parallel to Ox or Oy. (This method could be used when the axes

oblique.)

15.4 Concurrence of straight lines

15.41 Lines through the meet of two given lines {w}

If L,

Uare linear functions of x, y, then the equation

L + kL' =

is satisfied by any point which satisfies both L = and UHence if the lines L = 0, U = intersect, the locus L + kL' = pas

through their common point.

If k is constant, L + kU is linear in x, y, and so L + kU = is t

the equation of some straight fine through the meet of L = 0, L'




Conversely, if the linear equation L = can be written in the form

L l -f lcL 2= 0, where h is independent of x, y and L x , L 2 are linear with

constant coefficients, then L = passes through a fixed point, viz.

the meet of the lines L x = 0, L 2 = 0.

Example

The mid-points of the diagonals of a complete quadrilateral are collinear.

If ABGD is any quadrilateral, let AD, BC be produced to meet at E; andBA, CD to meet at O. The resulting figure is called a complete quadrilateral

and AC, BD, OE are its diagonals.

Choose OAB for Ox and ODC for Oy. Call 4(2o, 0), 5(26, 0), (7(0, 2c), D(0, 2d),

and let Mbe the mid -point of OE. Through Mdraw lines parallel to EA, EB;

E

Fig. 154

by the intercept theorem of elementary geometry these will cut Ox, Oy at the

points A'(a, 0), B'(b, 0), C (0, c), D'(0, d), and their equations will be

* V , x ya a be

Thus ikf lies on the lines

dx + ay = ad and ca; + by = 6c,

and therefore also on(dx + ay —ad) —(cx + by —be) = 0,

i.e. (d— c)x + (a —b)y = ad —be.

The mid-points of BD, AC are (6, d), (a, c), and these clearly lie on this last




If two of these intersect, say the second and third, the commopoint (obtained by solving for x, y) is

a 2 b z -a z b 2' a 2 b z -a z bj

This lies on the first line if

M&2 C3- &3 C2)+M C2 a 3- C3 a 2) + C l( a 2&3- a 3&2) = °>

i.e. if

A =61

2 ^2

3 ^3

= 0.

If no two of the lines intersect, then they are all parallel; this meathat (with the notation of 11.31) Cx = C%

= Cz — 0, and so

A = CiCi + CgOa + CgOg = 0.

Hence */ the lines are concurrent or parallel, then A = 0.

Conversely, if A = 0, the point (i) lies on the first line because the

a f b 2 c z -b z c 2 \[ b

fc 2 a z -c z a 2 \ +ca 2 b z -a z b 2

= 0.

This converse argument fails if a 2 b z —a z b 2 = 0; but a similar argumentapplies to the lines taken in a different order unless also

a z b 1—a 1 b z

= and a 2 b z—a z b 2 — 0,

in which case the three lines are parallel. Hence when A = the lin

are either concurrent or else all parallel.Thus the condition A = is necessary and sufficient for concurrence

or parallelism; see also 11.43, Corollaries I (6), (c).

Exercise 15(6)

1 Find the equation of the line concurrent with 2x —3y —3 =x + Sy— 15 = and passing through ( —2, —1); verify that it passes througthe origin.




5 (i) Show that for all numbers k the equation

2x + 3y-l + k(3x-y + 4:) =

represents a line passing through a fixed point,

(ii) Find the point of intersection of the lines

2x + 3y-l + a(3x-y + 4:) = 0,

b(2x + 3y-l)-c(3x-y + 4) = 0.

6 Find the equation of the line joining the meet of

3x + 4y-l = 0, 2x-3y+l =

to the meet of 3x- 5y + 37 = 0, 5x-2y-8 = 0.

Test the following sets of lines for concurrence.

7 3x-2y + <t = 0, 2x-y-l - 0, 1x-5y+13 = 0.

8 5x + 2y-4: = 0, x-By + 2 = 0, 3x + 8y-6 = 0.

9 (p + l)x + (p-l)y+p = 0, (q-l)x + (q+l)y + q = 0, x = y (p * g).

10 rc-^ + aJ* = 0,x-t 2 y + at* = 0, (^yi + fl-M^ = o(«i + ««) (*i * < 2 )-

11 If the lines x- 2y + 3 = 0, 2x- 3y = 0, ax + y+ 1 = are concurrent, find

the value of a.

12 If the lines ax + 2y+ 1 = 0, bx + 3y+ 1 = 0, ca; + 4?/+ 1 = are concurrent,

prove that a, b, c are in arithmetical progression.

13 Find the condition for concurrence of the distinct fines 3ax + 2y = a 3,

3bx + 2y = b 3, 3cx + 2y = c 3

. [Use theory of equations.]

*14 Find the condition for concurrence of the distinct fines

x sin 3a + y sin a = a, a: sin 3^ + y sin/? = a, xsin 3y + ysm.y —a.

[If (h, k) is the common point, then 6 = <x,fi,y must satisfy h sin 3d + k sin 6 = a,

i.e. 4h sin 3 —(& + 3ft) sin 6 + a = 0. Since the term in sin 2 6 is absent, the roots

sin a, sin/?, siny of this cubic in sin0 satisfy 2 sin a = 0.]

15 (i) If a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0, a 3 x + b 3 y + c 3 = are distinct

lines, and numbers I, m, n none of which is zero can be found such that

l{a1

x+

b-l y +

c1

)

+m{a

2x

+b

2 y +c

2)

+n{a 3 x + b z y + c z ) = 0, (A)

prove the lines are concurrent or else all parallel.

*(ii) If A = 0, use 11.43, Theorem II to show that numbers I, m, n not all

zero exist such that

a^ + a^m + a^n = 0, 6 1 Z + 6 2 TO + &3 W = 0, c 1 l + c 2 m+ c 3 n =

and hence that (A) holds. Deduce the converse of (i).

15.5 Line-pairs



580 ELEMENTARY COORDINATE GEOMETRY [15.

(iii) x 2 + Sxy + 2y 2 + Sx + 5y + 2 = can be written (x + y + 2) (x + 2y + 1)

and therefore represents the pair x + y + 2 = 0, x + 2y+l = 0.

(iv) x 3 + 3x 2 y + 2xy % = 0, i.e. x(x + y)(x + 2y) = 0, represents the three l

x = 0, x + y = 0, aj + 2?/ = 0.

In general, the equation

represents the pair of lines a 1 x + b 1 y + c 1 = 0, a %x + \y + c 2= 0. F

if (x, y ) satisfies either of these equations, then it satisfies the original

one; and conversely, if (x, y ) satisfies the given equation, then eithe

a i x o + bxVo + c x = or a 2 x + b 2 y + c 2 = or both.

15.52 The locus ax 2 + 2hxy + by 2 =

(1) If b = 0, the equation is x(ax + 2hy) = 0, representing a lin

pair.

If b 4= 0, the equation can be written

When (h/b) 2 > a\b, i.e. h 2 > ab, this quadratic has distinct roo

yjx = mx ,ra 2 , and then y = mx x or y = m2 x. Hence if h 2 > ab, th

equation represents two distinct lines through O.

When h % = ab, the equation is (y/x + h/b) 2 = and therefore repre

sents the single line hx + by = 0. We say conventionally that

equation represents two coincident lines, or a repeated line.

When h 2 < ab, the quadratic for yjx has no roots. The original

equation is satisfied solely by the values x = 0, y — 0. Henceh 2 < ab, the equation represents the origin only.

The three general results just given include the case 6 = disposed

of at the start.

The work of 15.5 so far is valid for oblique axes; in the following

shall suppose the axes to be rectangular.

(2) Angle between the lines ax 2 + 2hxy + by 2 = 0. If the separate lin

have equations = m = m then the above quadratic f

(a^ + b^ + Cj) (a z x + b 2 y + c 2 ) =




The required angle is given by

teng=m,-m a 2V(A*-qfe)

b{l+ajb)

2<](h*-ab) a + b

'

provided a + b 4= 0.

If a + b = 0, then Wjmg = —1 and £Ae Zir&es are perpendicular; and

conversely.

(3) Bisectors of the angles between these lines. By expressing equality

of the perpendiculars from (x, y) to the lines y —m1 x = 0,y —mi x = 0,

the equations of the angle-bisectors are

y-m x x _ y-m 2 x

V( +™i) -V( 1 + WD*

Hence the two bisectors have equation

{y-m^f {y-m z x) 2

= Q1+raf l+m|

i.e. (1 +m|) («/-m 1 a;)2 -(l + w&f) {y-m %xf = 0,

i.e. (raf- m|) a;2 —2(m - m2 ) (1 —m1 m2 )xy —{m\ —ml)y'

2' = 0.

If 4= m2 , this reduces to

K + mjj) (z 2 -*/ 2) = 2(l-m 1 ma )a;2/,

i.e. by (i), h(x 2 -y*) = (a-b)xy.

This equation satisfies the perpendicularity condition at the endof (2), as of course it should from elementary geometrical considera-

tions. It can be written

x 2 xy y

a h 6=0.1 1

Th




If s can be factorisedf in the form

s = (A^ + Brf + CJiAzX + Bzy + Cz),

then (i) represents a line-pair. By equating coefficients of the second-

degree terms in (ii),

a = A 1 A 2 , 2h = A X B 2 + Z BX , b = BXB 2 \

so we also have

ax 2 + 2hxy + by 2 = (A 1 x + B 1 y)(A 2 x+B 2 y).

Therefore */ (i) represents a line-pair, then the equation

ax 2 + 2hxy + by 2 = (ii

represents the parallel line-pair through 0.

This fact can be used to calculate the angle between the lines

which is the same as the angle between (iii), found in 15.52 (2).

When h 2 = ab, the lines (i) may be coincident or parallel; for although

this condition implies AX :A 2 —B x \ 2 , these ratios may or may n

be equal to Cx : C2 .

(2) Necessary condition for s = to represent a line-pair.

If (i) represents a line-pair, s can be written in the form (ii); andA x

=j= and A% ^ § (i.e. if a #= 0, since a = AX A 2 ), this means that (

regarded as a quadratic in x, can be solved in the form

Bf C-t Ba Cox =

~A y ~T' x=z ~T y ~T' (lv

Now (i), written as a quadratic in x, is

ax 2 + 2(hy + g)x + {by 2 + 2fy + c) = 0, (

so x = - [ - (hy + g) ± V{(% + g)2 - a(by 2 + 2fy + c)}].

Hence (i) can be solved in the form (iv) if and only if the quadratic

in y under the square-root sign is a perfect square, i.e. if

(h 2 - ab) y2 + 2(gh -af)y+ (g

2 - ac)

is a perfect square. This is so if and only if

2 2 2




Since we are assuming a 4= 0, the condition is

abc + 2fgh - a/ 2 - bg* - ch? = 0. (vi)

If a = but b 4= 0, then equation (i) can similarly be solved as aquadratic in y, and the same condition (vi) is obtained.

If a = 0, b = 0, but h 4= 0, equation (i) is

2hxy + 2gx + 2fy + c = 0,

i.e. v + j. x+ l y + lL = o

which, if factorisable, is

H) H) - °-

so that f = -h

,

i.e. 2fgh-ch* = 0,

and this is what condition (vi) becomes when a = b = 0.

These are the only cases that need be considered, since if

a = b = h =

then (i) would not be of second degree. We have therefore shown that

the necessary condition for (i) to represent a pair of straight lines is (vi).

By ex. (ii) in 11.22, (vi) can be written

= 0. (vii)

a h g

A= h b f

9 f c

The determinant A is called the discriminant of s, or of equation (i).

Remark (a). The necessary condition A = is not sufficient for (i)

to represent a line-pair. For A is zero when

a = b=l, c=f = g = h = 0,

whereas x 2 + y2 has no linear factors. (A sufficient condition is in-




in 10.21, ex. (ii)). For the general case (i), equations giving the poin

of intersection (sometimes called the centre or vertex of the line-pair)

can be found as follows.

Writing (i) in the form (v), we see that to a given value of y thcorrespond in general two values of x; but if y is the ordinate of

meet P, then (v) will give only one

value of x, viz. the abscissa of P.

Now when (v) has equal roots, that

root is (see 1.31 (&))

so the coordinates of P satisfy Fig. 155

ax + hy + g = 0.

Similarly, writing (i) as a quadratic in y, viz.

by 2 + 2(hx +/) y + (ax 2 + 2gx + c) = 0,

(

at P we havey

= hx+f

i.e. hx + by+f = 0.

Equations (viii), (ix) suffice to determine P as the point (GjC, Fj

see ex. (i) in 11.32 for the notation.

Remark (/?). Since (i) can also be written

(ax + hy + g)x + (hx + by+f)y + (gx+fy + c) = 0,

it follows that the coordinates of P must also satisfy

g%+fy+c = 0.

Equations (viii)-(x) are therefore consistent, and Corollary

of 11.43 gives (vii) as a necessary condition for s = to represent

intersecting line-pair. Compare (2) above, where (vii) was shown to

necessary for any type of line-pair, not necessarily intersecting. A

see 15 73 ex (ii)




Consider the homogeneous equation in x, y:

ax 2 + 2hxy + by 2 + 2(gx +fy) (Ix + my) + c(lx + my) 2 = 0.

It is satisfied by the points which satisfy both the equation of the line

and of the locus, and hence it represents some other locus through these

points. Since the equation is homogeneous in x, y, it represents two

lines through O. Hence it must be the equation of the required line-

pair.

Exercise 15(c)

[Rectangular axes.]

1 Find the area enclosed by the lines x2

+ 4xy + 2y2

— and the line x + y = 1

.

2 Show that the area enclosed by the lines

ax* + 2hxy + by 2 = and Ix + my = 1

is <J(h2 - ab)/(am z - 2hlm + bl 2

).

3 Prove that the product of the lengths of the perpendiculars from (x x , y x )

to the lines

ax 2 + 2hxy + by 2 = is ± {ax 2 + 2hx 1 y 1 + by\)IJ{{a-b) 2 + Ah 2}.

4 Find the equation of the lines through which are perpendicular to the

lines ax2

+ 2hxy + by2

= 0. [If (x, y) lies on either of the required lines, then( —y, x) lies on one of the given lines.]

5 Find (i) the angle between, (ii) the equation of the bisectors of the angles

between, the lines 5x 2 + 2xy —4y 2 = 0.

6 If the lines ax 2 + 2hxy + by 2 = 0, a'x 2 + 2h'xy + b 'y2 = have the same

angle-bisectors, prove that h(a' —b') = h'(a —b).

7 Show that any pair of lines which has the same angle -bisectors as the pair

ax 2 + 2hxy + by 2 = can be written ax 2 + 2hxy + by 2 + \(x 2 + y2

) = 0. [Use no. 6.]

Find the equation of the pair of lines, one of which passes through (p, q),

and whose angle -bisectors are x 2 —y2 — 0.

8 Show that the following equations each represent pairs of straight lines;

find the angle between them, and their point of intersection.

(i) 2x 2 + lxy + 3y 2 -4x-ly + 2 = 0;

(ii) 15x 2 + xy - 2y 2 + 3x - y = 0;

(iii) xy-3x + 2y-6 = 0;

*(iv) x 2 + 4xy-2y 2 + 6x-12y-l5 = 0.

9 Write down the equations of those line -pairs through the origin which are

perpendicular to the line-pairs in no. 8. [Use no. 4.]

10 Find the equation of the lines joining O to the meet of x + 2y = 3 with the




15.6 The circle

15.61 General equation of a circle; centre and radius

Referred to rectangular axes, the equation of the circle with centr

C(M) and radius r is{x _ hf + {y

_ kf = ^When simplified, this equation is of the form

x 2 + y2 + 2gx + 2fy + c = 0,

in which the coefficients of x 2 and y2 are equal, and there is no xy-term.

Conversely, by completing the square for the terms involving x a

for those involving y, equation (ii) can be written

(* + <7)2 + (2/+/)

2 = 9'2 +/ 2 -c.

If g2 +f 2 —c > 0, this represents the circle with centre ( —g, —/) a

radius «J(g2 +f 2 —c). If g

2 +f 2 —c = 0, it represents the single poi

( —ff>

—/); and if g2 +f 2 —c < 0, the equation does not represent a

locus.

Example*Referred to oblique axes at angle o), the circle with centre C(h, k) and radius

has equation

(x-h)* + (y-k) 2 + 2(x-h) (y-k) cosw = r a,

which is of the formx 2 + y

2 + 2xy cos (0 + 2gx + 2fy + c = 0.

15.62 Circle on diameter P l P2

Let P(x,y) be any point on the required circle; then PX P ±Pby 'angle in a semicircle'. Since

gradient of PX P = UzMlt and gradient of P2 P = U^ 2

t•C *^

hence Zz* _ lf




15.63 Tangent at P x

Let Px (x x , y x ) be a given point on the circle (ii) in 15.61. Then since

the tangent at Px is perpendicular to the radius CPX , and CPX hasgradient (y 1 +f)l{x 1 + g), the equation of the tangent is

i.e. +/) - y\ -fy x + x(x x + g)-x\- gx x = 0,

i.e. xx x + yy x + gx +fy = x\ + y\ + gx x +fy x .

Since Px lies on the circle,

x{ + yl + 2gx 1 + 2fy 1 + c = 0,

and so the equation of the tangent at Px can be written

xx i + VVi + Qx +fV = ~ 9 x x ~fVi - c >

i.e. xx x + yy x + g{x + x x ) +f(y + y 1 ) + c = 0. (iv)

Remark. This result can be remembered by the following rule of

alternate suffixes. 'Write xx for x z ,x + x for 2x, etc., in the equation of

the circle: , . „, sxx + yy + g(x+x)+f(y + y) + c = 0;

and attach the suffix 1 to alternate variables.'

It should be emphasised that this rule is only an aid to memory,

and certainly does not constitute a proof that (iv) is the equation of

the tangent at Px . Other applications will appear in later chapters.

Example

Find the condition for the line Ix + my = nto touch the circle (ii).

The line will be a tangent if and only if the perpendicular from the centre

( —g, —f) to it is equal to the radius *J(g2 +/ 2 —c)

—Ig —mf—n ,, „

i.e., on squaring, (lg + mf+n) 2 =(fl

,2 +/ 2 —c) (Za + m2

).




For simplicity we consider the circle x 2 + y2 = a 2

, whose centre

the origin. The tangent P2 PX at P2 nas equation

xx z + yy 2 = a 2.

Since Px lies on this line,

#1^2 + 2/12/2 = « 2

which shows that the coordinates x = x % , y = y 2 satisfy the equation

x x x + y x y = a 2,

i.e. that P2 lies on the line xx x + yy x = a 2.

Pi

Fig. 156

Similarly, P3 lies on this same line, which must therefore be

chord P2 P3 . Hence the chord of contact from Px to x 2 + y2 = a 2 is

^1 + 2/2/i = a 2-

The same argument applied to the general circle

x 2 + y2 + 2gx + 2fy + c =

shows that the chord of contact from Px has equation

xx i + 2/2/i + 9( x + x i) +f(y + 2/i) + c = 0,

i.e. (x 1 + g)x + {y x +/) y + (gx x +fy x + c) = 0.

Remark. The equation xx x + yy x = a 2 has two quite distinct

meanings according to whether Px lies on the circle x 2 + y2 = a

outside it. When Px lies on the circle, it represents the tangent at

when Px lies outside, it represents the chord of contact from Px . Simila




As in 15.54, consider the equation-

which reduces to c(x 2 + y 2 ) = (gx+fy) 2 .

It is satisfied by the points common to the circle and line, i.e. by the points

of contact of the tangents from O. It is homogeneous of degree 2 in x, y, andtherefore represents a line-pair. Hence it represents the pair of tangents from 0.

Fig. 157 Fig. 158

(ii) Pair of tangents from P x to x 2 + y2 = a 2

.

When P x is given, let P a be any other point in the plane. Any point on P X P 2

will divide it in some ratio k:

I, and hence will have coordinates

/ lx x + kx 2 ly x + ky 2 \

\ l + k ' l + k )'

This point will lie on the circle if

/ lx x + kx 2 \ 2( ly + ky 2 \ 2 _

\ l + k ) \ l + k J

i.e. (x 2 + y\ —a 2 )k 2 + 2{x x x i + y 1 y 2—a 2 )kl + (x\ + y\ —a 2 )l i = 0. ( v)

This quadratic gives two values for the ratio k : I, which correspond to the points

A, B where the line P X P 2 cuts tne circle; the two values are P X A :AP 2 andP 1 P:PP 2 (fig. 158). Equation (v) is known as JoachimsthaVs ratio quadratic

for the circle x 2 + y% = a 2

.

If P a lies on either tangent from P t , then the points A, B will coincide, andhence (v) will have equal roots k : I, so that

(x 2 + yl-a 2)(xl + y

2 -a 2) = (x^ + y^-a 2

)2

.

This equation shows that the point P 2 (flJ 2 » Vz) nes on * ne locus

(x 2 + y2 -a 2

) (x 2 +y 2 - a 2) = (xx x + yy x - a 2

)2

,




Let the tangents at the extremities A, B of the chord AB through P x mat P 2 . Then by 15.64, AB has equation

xx 2 + yy 2 = a 2.

Since

P xlies on this line,

x x x 2 + y x y 2 = a2 5

and this shows that P 2 lies on the line x x x+y x y = a 2.

Fig. 159 Fig. 160

If P x lies inside the circle, the locus of P 2 is the complete line, whichentirely outside the circle. If P x lies outside, the locus of P 2 is that part of

line which is outside the circle.

(2) A variable chord through P x meets the circle x* + y2 = a 2 at A, B; P

chosen so that P x , P 2 divide AB in the same ratio (one internally, the other exter

ally), f Show that P 2 lies on the line xx x + yy x = a 2.

Since by hypothesis AP 1 :P 1 B = AP 2 :P 2 B, hence also P X A :^4P 2 = P 1 P:so that A, B divide the line P X P 2 internally and externally in the same rat

say ± k' : V






to both sides the previous equation is obtained; hence by the convers

of the theorem of Pythagoras, triangle C'PC is right-angled at P.

The condition is therefore both necessary and sufficient for orth

gonality.

Example

Prove that in general there is only one circle orthogonal to three given circles.

If the given circles are

x 2 + y2 + 2g r x + 2f r y + c r = (r = 1,2,3),

we require to find g, f, c for which

2î + 2//i-c = c x ,

2 99i + 2fA- c = c 2»

2^3 + 2// 3 -c = c 3 .ndIf the determinant

A =91 A92 fi

9z fz

is non-zero, then this system of three equations for the three unknowns g,

can be solved uniquely by Cramer's rule (11.41). The condition A 4= me

that the three centres are not collinear (15.16, ex.) and that no two centrcoincide; this is the situation 'in general'.

Exercise 15(d)

1 Write down the equation of the circle through O whose centre is (pProve that the tangent at O is px + qy = 0. [Use ' tangent _L radius '.]

2 Show that the chord of x 2 + y2 = a 2 whose mid-point is (x x , y x ) has equatio

xx x + yy x = x\ + y\. [If P is any point of the required chord, OP x _L PP\.~\

3 Prove that the mid-points of those chords of x 2 + y2 + 2gx + 2fy + c

which pass through P x lie on the circle (x —x 1 )(x + g) + (y —y 1 )(y+f)[CM±CP 1 .]

4 (i) Find the equation of the chord PxP 2 of x 2 + y2 = a 2

. Deduceequation of the tangent at P x . (ii) Also find the tangent at P x by using Calculu

5 If Ix + my = n touches x 2 + y2 = a 2

, find its point of contact. \xx x + yy x

and Ix + my = n are the same line if x x jl = y x /m = a 2 /w.]

6 Find the condition for y = mx + c to touch x 2 + y2 = a 2

, and deducethe lines y = mx ±a*J(l+m 2

) touch the circle for all values of m.

7 (i) Find the condition for the chord of contact of tangents from P



ELEMENTARY COORDINATE GEOMETRY 593

centres at (A, 0), its equation is y = m(x —A) where m is chosen so that the

perpendicular to this line from either centre is equal to the corresponding

radius.]

Prove that the direct common tangents to

(x - a x f + </2 = r\, (x - a 2 )

2 + y* = r\

have equation {(» ]. —r 2 )x —(a^ —a 2 r x )}2 = {(% — 2 )

2 —(r x — 2 )2}y

2.

Find the equation of the transverse common tangents.

10 If the line of centres of

x* + y* + 2gx + 2fy + c = 0, x* + y* + 2g'x -r 2fy + c' =

cuts the axes at A, B, show that the circle on diameter AB is

(g-g')U-S')^+y 2) = (gf'-rfHig-rtx-V-Dy}-

*1 1 A variable circle passes through the meet O of two given lines, and makesintercepts OP, OQ such that m.OP + n.OQ = 1. Prove that this circle passes

through another fixed point. [Use oblique axes.]

12 A circle passes through (h, k) and cuts orthogonally the circle

x 2 + y* + 2gx + 2fy + c = 0.

Prove that its centre lies on the line

2(h + g)x + 2(k+f)y = fc2 + F-c.

13Find

the equation of the circle orthogonal to x 2

+ y2

+2x —

2y +1 = and

# a + y2 + 4x —4y + 3 = and whose centre lies on the line 3x —y —2 = 0.

14 Find the equation of the circle orthogonal to

x 2 + y2 = 5, x 2 + y

2 + 6x+l = and x i + y2 -<ix-4 ty + l = 0.

15 Prove that an angle between the circles

8 = x* + y* + 2gx + 2fy + c = 0, s' = x 2 + y2 + 2g'x + 2f

,

y + c' =

(supposed to intersect) is given by2gg' + 2ff'-c-c'

COB 6 =2V(fl' 2 +/ 2 -c)V(9'' 2 +/

,2-c')'

16 If 8 = 0, «' = are the circles in no. 15, prove that s + As' = is also the

equation of a circle for any constant A except A = —1 . What is the interpretation

when A = —1? If s, *' intersect, explain why s + As' = passes through their

common points for all A. (As A varies we obtain a system or family of circles.)

17 If a circle or = cuts s = 0, s' = orthogonally, prove that it also cuts

orthogonally each member of the system s + As' = 0. Interpret the case A = —1

by first showing that the centre of a lies on the line s—s' = 0.

18 Prove that the length of a tangent from P to a 2 + % + 2gx + 2f + c =




20 Find the equation of the circle which passes through ( —3, 2) andpoints of intersection ofthe circles 3a; 2 + 3y 2 + 2x—'7y—6 = 0, x 2 + y

2 + y —2

[Use no. 16.]

21 Find the equation of the circle whose diameter is the common chordthe circles x 2 + y 2 -2x + 2y + 3 = 0, 5x* + 5y 2 -x + ly-12 = 0.

15.7 Conies

15.71 Definitions

The locus of a point P in a plane such that the ratio of its distanc

from a given point 8 to its distance from a given line d is constant

called a conic.

The point 8 is called the focus, the line d the directrix, and the cstant (denoted by e) the eccentricity of the conic. According as e =

the conic is called an ellipse, parabola, or hyperbola; e is essentially

positive.

If M is the foot of the perpendicular from P to the line d,

definition is equivalent to the relation

SP = e.PM.

The name 'conic' is an abbreviation for 'conic section'. Originallythe ellipse, parabola and hyperbola were obtained as the sections mby a plane drawn perpendicular to any one generator of right circula

cones with acute, right and obtuse vertical angles respectively. La

they were defined by different sections of the same right circular co

the ellipse, parabola and hyperbola being respectively the curves

section by a plane making an angle with the base less than, equal

or greater than that made by the generators. If the cone is a doub

one, the hyperbola will consist of two separate parts or branches.

A plane parallel to the base will give a circular section; one throug

the axis of the cone will give two intersecting lines (the generators

that plane); one touching the cone along a generator will give a sing

line; while a plane through the vertex but not cutting the cone e

where will give a single point. Hence with this approach, a circl

a line-pair, a single line, and a single point are all conic sections.

I b df th t f each plane which gi an lli i






to a simpler equation than others. Hence, when an equation wo o

set of axes is given, it is often desirable to change the axes in order

simplify the equation. This can be done by(a) changing the origin, leaving the directions of the axes unaltered

(a translation of axes)

;

(6) changing the direction of the axes, leaving the origin unaltered

(a rotation of axes);

(c) both together.

Remark. When a problem involves equations of two or more loc

in general it is possible to simplify only one of the equations in th

way; e.g. see 16.12, ex. (ii).

O'

Fig. 165

(2) Change of origin. If the new origin 0' is taken at the point (h,

then the point P whose coordinates were (x, y) now has coordinates

*',*,') given byx ^ x _ h} y> = y - k .

Thus, to change the origin to the point (h, k) we substitute x' + h for

and y' + kfor y: the locus whose equation wo axes Ox, Oy isf(x, y) =becomes f(x' + h, y' + k) = wo axes O'x', O'y'

.

We usually omit the dashes in the new equation, and say tha

f(x,y) = becomes f(x + h,y + k) = 0.

The above work remains valid for oblique axes ; an example occurred

in 15.16(2).

Examples




Referred to the original axes, this equation becomes

(x 2 + y\- a 2) {(x - x x )

2 + (y- y x )2

} = {x x (x - x x ) + y x (y - y x )}\

i.e. (x 2 + y\ - a 2) {(x 2 + y

2 - a 2) + (a;

2 + y\ - a 2) - 2(xx x + yy x

- o 2)}

= {(xx x + yy x- a 2

) - (x 2 + y\- a 2)}

2,

i.e. {x\ + y\- a 2) (x 2 + y

2 -a 2) = (xx x + yy x - a 2

)2

.

(ii) Point of intersection of the line-pair ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0.

Suppose the lines meet at P x (x x , y x ). Changing the origin to P x , the equation

becomes

a(x + x x )2 + 2h(x + x x )(y + y x ) + b(y + y x )

2 + 2g(x + x x ) + 2f(y + y x ) + c = 0,

i.e. ax 2 + 2hxy + by 2 + 2(ax x + hy x + g) x + 2(hx x + by x +/) y

+ (ax 2 + 2hx x y x + by 2 + 2gx x + 2fy x + c) = 0.

Since the new origin is the meet of the lines, this equation must involve only

terms in x 2, xy and y

2 (15.52 (1)). Hence we must have

ax x + hy x + g = 0, hx 1 + by 1 +f = 0,

and ax 2 + 2hx 1 y x + by 2 + 2gx x + 2fy x + c = 0.

The first two equations deterrnine P x . As in Remark (/?) in 15.53 (3), the third

is equivalent to gx x + fy x + c = 0.

Fig. 166

(3) Rotation of axes through angle 6. Let the point P, whose co-

ordinates wo Ox, Oy are (x,y), have coordinates (x',y') referred to

the new axes Ox', Oy', where




since from triangle OPN', x' = rcos^ and y' = rsin^. Similarly,

y = r sin (0 + <j>) = r sin cos<f> + r cos sin <j>

= x' sin + y' cos 0.

Thus, to rotate the axes through angle we substitute x' cos —y' sin

for x and x' sin + y' cos for y: the locus f(x, y) = becomes

f(x cos —y sin 0, x sin + y cos 0) = 0.

The reverse transformation can be obtained

either by solving the above two equations for x', y', in terms of

or by writing —6 for 6 and interchanging (xr

, y'), (x, y) in tabove equations;

or by considering

x' = r cob {{d + < >)- 6} = y' = rsm{(d + <f>)-6} = ....

Both transformations can be read off from the scheme

x' y'

X

y

cos 6 —sin 6

sin 6 cos 6

(4) Change of both origin and direction of axes. By combining t

substitutions found in (2), (3) we see that, under the most general

change of rectangular axes, the locus whose equation is f(x, y) =becomes », n . „ 7 . n ^ 7 ,

f(x cos a —y sin a + h, x sin a + y cos + k) = 0.

It follows that a change of axes leaves the degree of any poly

nomial equation unaltered. For, instead of the linear function x,we hav

substituted the linear function x cos 6 —y sin 6 + h, and similarly for

to powers and products of linear functions correspond powers aproducts of the same degree.

15.74 Reduction of 5 = to standard forms

By rotating the axes through angle 6, the equation






(vii) reduces to By 2 + 20x = 0,

o .

i.e. r =-~fi

x - (v

(c) If A = = and B ={= 0, equation (iv) becomes

By 2 + 2Fy + G = 0,

which represents two parallel lines, a repeated line, or nothing

according as F 2 = BC.(d) IfB = 0, J. 4= 0, F + 0, or */ j? = ^ = 0, A * 0, then by tran

forming equation (iv) by rotating the axes through £tt, we obta

the cases considered in(6),

(c).

We cannot have A = B = 0, for then (iv) would not be of seco

degree, and hence (see 15.73 (4)) neither would (i). Thus all admissible

possibilities have been considered.

It has now been shown that, by a suitable change of axes,

general equation (i) of second degree can be reduced to one of

following standard forms:

^+^ = -1, ccx*+]3y* = 0,

2/ = a±V(a 2 -/?).

Clearly the fourth equation represents nothing; the fifth represents

an intersecting line-pair, a repeated line, or a single point; andsixth represents a pair of parallel lines, a repeated line, or nothing.

The following three chapters are respectively concerned with

first three loci, which will be proved to be ' conies ' in the sense of

definition given in 15.71: see 16.11, 17.12 and 17.13.

Exercise 15(e)

1 By a change of origin, the points ( —1,3), (4, —2) become (a, 5), (3,/?

find a, ^.

2 Find the new equation of the locus x 2 —xy— 6y 2 —Zx +14y —4 = w



ELEMENTARY COORDINATE GEOMETRY 601

6 Change the origin to (1,0) and then rotate the axes through the acute

angle whose tangent is £ , for the equation

43a; 2 - 48xy + 51y 2 + 86a; - 48y + 1 8 = 0.

7 Showthat the

centroid of a lamina, as defined in7.81, does not depend on

the choice of coordinate axes. [Wo the new axes the centroid is given by

and similarly for y'. Thus (x' ,y') is the point (x, y) referred to the new axes.]

1 If the points {at\,at\), {at\,at\), (at\,at\) are collinear, prove that

M3 + M1 + M2 = 0* Conversely, show that when this condition is satisfied the

three points are collinear. [Use theory of equations.]

2 Find the condition that the four distinct points (kt r ,k/t r ), r = 1,2,3,4,

shall be concyclic. Show also that no three of these points can be collinear.

3 Show that a necessary condition for concurrence of the lines

arcos3a + 2/ —acosa = 0, xcosSfi + y —acos/tf = 0, a;cos dy + y —acosy =is cos a + cos /? + cos y = 0, and prove that this condition is also sufficient.

4 If the line-pairs ax 2 + 2hxy + by 2 = 0, a'x 2 + 2h'xy + b'y 2 = have a line in

common, prove (ab' -a'b) 2 + 4(ah' -a'h) (bh' -b'h) = 0. [Use 10.42(1).]

5 If one of the lines ax 2 + 2hxy + by 2 = is perpendicular to one of the lines

a'x2

+ 2h'xy + b'y2

= 0, prove (aa' -bb')2

+ ±{ah' + b'h) (a'h + bh') = 0.

6 If the lines x 2 —2pxy —y2 = and x 2 + 2qxy —y

2 = are such that onepair bisects the angles between the other pair, prove pq = 1.

7 If lx + my + n = and ax 2 + 2hxy + by 2 = form an isosceles triangle,

prove that h(l 2 —m2) = (a —b) Im. Find also the further condition if this triangle

is equilateral. [The line is parallel to an angle-bisector of the pair, so its gradient—Z/m must satisfy {1 —(y/x) 2 }/(a —b) = (y/x)/h.]

8 If ax 2 + 2hxy + by 2 + 2gx + 2fy + c = represents a pair of lines meeting at

A, and parallel lines are drawn through O to meet these at B, G, find the equa-

tions of the diagonals OA , BC of the parallelogram so formed. If the parallelogram

is a square, prove a + b = and h(g 2 -f 2 ) - fg(a - b). [OB J_ OG and OA J_ BC]9 Prove that the equation m(x 3 —3xy 2

) + y3 —Sx 2 y = represents three

straight lines equally inclined to one another. [In polar coordinates the equation

is m = tan 30. Writing m = tan 3a, the three lines are 6 = a + \m, r — 0, 1, 2.]

10 If A, B, G are fixed and PA 2 + PB 2 + PC 2 is constant, prove that the locus

of P is a circle whose centre is the centroid of triangle ABC. [Choose O at this

centroid; then Hx A = = 2y^.]

1 1 Find the centre and radius of the incircle of the triangle whose sides are

4x-3y + 2a = 0, Bx- 4y+ 12a = 0, 3« + 4y- 12a = 0.

x' = — I x'dA —— (xcoad + yaiad + h)dA = xcoad + yaind + h,

Miscellaneous Exercise 15(/)



602 ELEMENTARY COORDINATE GEOMETRY1 3 Show that the line (x —a) cos 6 + y sin 6 = r touches the circle

{x —a) 2 + y2 = r 2

,

and state the coordinates of the point of contact.

A pair of parallel tangents is drawn to a given circle, and another pairpendicular to these is drawn to a second circle of equal radius. Prove that ediagonal of the square formed by the four tangents passes through a fixed poi

14 All members of a family of circles pass through two given points. Prothat the common chords of these circles and a fixed circle not belonging to

family are concurrent.

15 Write down the equation of the line-pair through P x and P 2 whichparallel to (i) Oy; (ii) Ox. Explain why the locus

(x-x 1 )(x~x t ) + {y-y 1 )(y-y 1 ) =

passes through the vertices of the rectangle formed by these line -pairs,

deduce that this locus is the circle on diameter PiP 2 .

16 If the orthogonal circles

x 2 + y2 + 2g 1 x + 2f 1 y + c 1 = 0, x 2 + y

2 + 2g 2 x + 2f 2 y + c 2 =

have centres A, B and cut at C, D, prove that the circle through A, B, G,

2(x 2 + y2

) + 2( 9 r 1 + g 2 ) x + 2(/ x +/ 2 ) y + (c x + c 2 ) = 0.

If the equation of the circle on diameter CD is written in the form

x 2 + y2 + 2g x x + 2f 1 y + c 1 + X{2( gi -g z )x + 2(/ x -/ 2 ) y + (c x

- c 2 )} = 0,

prove A = —r\\AB 2 where r x is the radius of the first circle.

17 Find the equation of the line -pair joining O to the meets of the li

4a; 2 - 15xy —4ty 2 + 39x + 65y - 169 = and x + 2y = 5. Show that the quadr

lateral having the first pair and also the second pair as adjacent sides is cycli

and find the equation of its circumcircle.

18 Show that the line -pair joining to the meets A, B of the line Ix + mywith the conic ax 2 + by 2 = 1 has equation

(a —I2

) x 2 —2lmxy + (6 —m2) y

2 = 0.

If AOB is a right-angle, deduce that AB touches the circle (a + b) (x 2 + y 2 )

19 Find the equation of the line -pair joining O to the meets of Ix + myand the circle x 2 +y 2 + 2gx + 2fy + c = 0. Hence find the coordinates of

circumcentre of the triangle formed by Ix + my = 1 and the lines

ax 2 + 2hxy + by 2 = 0.

If the lines ax 2 + 2hxy + by 2 = vary, but remain equally inclined to

axes, show that the circumcentre varies on a fixed line through O.

20 Explain why the equation



603

16

THE PARABOLA

16. 1 The locus y 2 = 4ax

16.11 Focus-directrix property

Since the equation y2 = 4ax can be written in the form

(x —a)2

+ y2

= (x + a)2

,

any point P{x, y) on the locus is such that SP 2 = PM2, where 8 is the

point (a, 0) and PM is the perpendicular from P to the line x + a = 0.

Hence by the definition in 15.71, the locus y2 = 4ax is a parabola with

focus (a, 0) and directrix x + a = 0.

It is symmetrical about Ox because

the equation is unaltered by replacing

y by —y; Ox is called the axis of the

parabola, and O is the vertex. Without

loss of generality we may always

suppose a > 0; a; is then positive for

all points on the curve. The parabola

meets Oy where x = and hence y = ;

it therefore touches Oy at 0, and Oy is

called the tangent at the vertex.

The chord LL' through S at right-

angles to the axis is called the latus rectum; its equation is x = a, and

it cuts the curve at the points (a, ± 2a), so that its length is 4a.

Remark. The equation y2 = 4ax also shows that the square of the

distance of P(x, y) from Ox is proportional to its distance from the per-

pendicular line Oy. The locus of a point satisfying this condition is

therefore a parabola.

Fig. 167



604 THE PARABOLA [16

Hence each point on y2 = kax has coordinates of the form (at 2

, 2

for just one value of t. Also each value of t determines exactly

point of the curve. The equations

x = at 2, y = 2at

therefore give a proper parametric representation (see 1.61); t

can also be written ±9 n± ,x:y:a = t

2 :2t : 1.

The point P(at 2, 2at) is briefly referred to as the point t of the curv

t is the parameter of P. By using these coordinates, the condition t

P lies on the curve is automatically satisfied without reference to

cartesian equation; cf. the Remark in 16.21.

Examples

(i) Find the condition for the chord joining the points t lf t 2 to be a focal c

(i.e. to pass through the focus).

Any line through the focus (a, 0) has an equation of the form

l(x —a) + my = 0.

This cuts the curve at points t for which

l(at 2 -a)+m.2at = 0.

It will be the chord t^ if the roots of this quadratic in t are t x and < 2 « Since

product of the roots is —1, the required condition is t x t^ = —1.

This necessary condition for a focal chord is also sufficient. For if =then t x and t 2 are the roots of a quadratic equation of the form

pt 2 + 2qt-p = 0,

which shows that the points t x , t 2 lie on the line px + qy —ap = 0, andclearly goes through the focus (a, 0).

(ii) Concyclic points on the parabola. A circlef

x* + y* + 2gx + 2fy + c =

cuts the parabola at points t which satisfy J

aH* + 4aH 2 + 2agt 2 + 4aft + c = 0,

i.e. aH l + 2a(2a + g) t* + 4aft + c = 0.

Since this equation is quartic in. t, a circle and parabola can intersect i

most four points. In real algebra a quartic equation has four, two, or no ro

(some or all of which may coincide); hence (ignoring possible coincidences)



16.2] THE PARABOLA 605

Suppose there are four intersections, given by t = t x ,t 2 , t a , t v Then these

numbers are the roots of the above quartic; and since the term in t 3 is absent,

WehaV6h + ** + ** + h = 0.

This necessary condition for concyclic points on the parabola is also sufficient.

In a general way this can be seen because three points of the curve determine

a unique circle, and one condition is required for this circle to pass through afourth point of the curve. In detail, given the four numbers t x , t z ,

t 3 ,t if where

Ht x = 0, consider the circle through t lt t 2 , t 3 : it cuts the curve again at a point

t' t (since a quartic with three distinct roots must have a fourth), and so by the

above' t 1 + t 2 + t 3 + t' l = 0.

Hence t' A= < 4 , and so t v t it t z , t t give concyclic points.

Exercise 16(a)' 'The parabola'' in this exercise means the curve y 2 = 4ax.

1 Find the gradient of the chord t x t 2 .

2 If the point t is one extremity of a focal chord PQ, find the length of PQ.

3 If M is the mid-point of a focal chord PQ, prove that the distance of Mfrom the directrix is equal to $PQ.

4 P,Q,R are points on the parabola such that PQ passes through the focus

and PR is perpendicular to the axis. When P varies on the parabola, prove that

the mid-point of QR lies on the parabola y2 = 2a(x + a).

5 The chord PQ subtends a right-angle at the vertex O. Prove that the mid-point of PQ lies on the parabola y

2 = 2a(x —4a).

6 A circle is drawn on a focal chord as diameter, and cuts the axes at

(x lt 0), (a> 2 , 0), (0, y x ), (0, y 2 ). Prove that x x x 2 - y x y 2 = - 3a 2.

7 (i) Prove that the line Ix + my + n = cuts the parabola in at most twopoints when I 4= 0.

(ii) If there are two common points t x and t 2 , prove that

t 1 + t 2 = —2m/l, t 1 t 2 = n/al.

(iii) If there is only one common point, prove that am 2 = In, and conversely.

8 A circle cuts the parabola at A, B, C, D. Show that the chords AB, CDare equally inclined to the axis. [Use no. 1 and ex. (ii) in 16.12.]

9 Prove that the locus of the centre of a circle touching Oy and the circle

x 2 + y 2 —2ax = is the parabola y2 = ±ax. [If P is the centre, show that

SP = PM for a suitable S and a*.]

10 A variable circle passes through a given point A and touches a given line I.

Prove that the locus of its centre is a parabola, and identify the focus anddirectrix.



606 THE PARABOLA [16.2

since y\ = 4ax x and y\ = 4ax 2 . The equation of the chord is therefore

y—Vi = Ix —x-,).

This equation is not symmetrical in the pair of coordinates (x x ,

(x 2 , y 2 ). To make it so, first clear of fractions:

2/(«/i + 2/2) ~ 2/i - 2/i2/2 = 4 «« - 4^1-

By using the condition y\ = ±ax x again, this simplifies to

lax - (y x + y 2 )y + y x y 2 = 0.

Remark. To obtain (ii) we have appealed once to the condition

y\ = 4ax 2 that P2 lies on the curve, and twice to y\ = 4ax x . Contrast

the directness of the work in 16.22, where the use of parameters

automatically ensures that Px ,P2 lie on the curve.

For another way of proving (ii), see Ex. 16 (6), no. 1.

16.22 Chord t xt 2

(1) Since the gradient of the chord is

2at x -2at 2 2at\ —at\ t x + 2

'

the chord has equation2

y-2at x = 7—7- (x-atl),t, l~r t 2

i.e. x-%(t x + t 2 )y + at x t 2= 0.

(2) Alternatively (using the theory of quadratics) let the required

chord be| Ix + my + a = 0. This line cuts the curve at points t for whilt

2 + 2mt+l = 0.

Since the fine is the chord t x t 2> this quadratic in t must have roots

and t 2 , so t x + t 2 = —2m\l and t x t 2 = \\l. Hence

^ — and Tfb —— ^- ?

t x t 2 2 t x t 2

d h i d h d




(3) Alternatively, since the numbers t x ,t 2 are the roots of

t2 -{t x + t

2 )t + t x t 2 = 0, ( iv )

we may use the parametric equations x : y : a = t2

: 2t : 1 to substitutefor t

2, t and thereby construct the linear equation

x - Wi + h)y + at ih = °-

This represents the required chord t x t 2 because it cuts the curve at

the points t given by (iv), i.e. t = t x and t = t 2 .

16.23 Tangent at P x

For curves other than the circle, the tangent is defined as a limit:

the tangent at Px is the limit of the chord PX P2 when the point P2 tends

to Px along the curve; but see 16.25.

(1) The tangent at Px to y2 = Aax is therefore obtainable from

equation (ii) by letting y 2-> y v and is

4ax-2y 1 y + yl = 0.

Since y\ = 4ax v this can be written

yy x = 2a(x + x 1 ). (v)

Observe that this can be written down from the equation y2 = ±ax

by applying the 'rule of alternate suffixes' (see 15.63, Remark).

(2) Alternatively, the gradient of the tangent at Px can be found by

calculating dy/dx when x = x x , y = y x . Thus from y2 = 4ax,

« dy , dy 2a2y = 4a and = —dx dx y

and the tangent at Px has equation

2a.y-yi = —(x-xi)>

which is equivalent to the limit of (i) when y 2-> y x , and reduces to

(v) above.



608 THE PARABOLA [16.

(2) Alternatively, the gradient can be obtained by calculus:

dy _ dy jdx 2a 1

dx=

dtj dt=

2ai=

l'

and then the equation written down as

y —2at = ^(x —at 2).

Geometrically, the parameter t is the cotangent of the angle

made by the tangent-line at the point t with the ic-axis ; for tan i/r =

16.25 Tangency and repeated roots

For a curve whose equation is a polynomial in (x,y), the simu

taneous solution of this , equation and that of the chord P1 P2 leads

equations for x and y containing the factors

(x - x x ) (x - x 2 ), (y - y x ) (y - y 2 )

respectively, or to an equation in a parameter t containing

(*-*i)(*-f.).t

Since the equation of the tangent at Px is the limit of the equation

the chord P1 PZ when P2 tends to P1 along the curve, hence if

equation of the tangent is solved with the equation of the curve,

will give equations for x, y containing the factors (x —x x )2

, (y —yor an equation for t containing (t —t x )

2.

Thus tangency corresponds to repeated roots, and is often dealt w

thus rather than by direct appeal to the limit definition. For examplethe line Ix + my + n = meets the parabola where alt 2 + 2amt + n =and will be a tangent if and only if this quadratic has equal root

i.e. if am 2 = In; cf. Ex. 16(a), no. 7.

More generally, if two loci cut at P1 and P2 , and P2 is madeapproach P2 along one of them, then the limit of the other will tou

the first at Px because they have a common tangent there. For example

if t % t in ex. (ii) of 16.12, the limiting circle will touch the parabola




16.26 Examples

(i) The tangents at t lt t 2 meet at the point {at 1 t 2 , a(t ± + 2 )}.

This result can be verified by direct solution of the equations

x —hy + at l = 0, x —t 2 y + at\ = d.

Alternatively, these two equations express that t t and £ a are the roots of the

quadratic in t

at 2 -ty+x = 0,

y xso that - = U + t z and - = t x

t 2 ,

a a

giving the required intersection (x, y).

The coordinates can be remembered by an ' extension ' of the rule of alternate

suffixes applied at at*, 2at.

(ii) The orthocentre\ of the triangle formed by three tangents to a parabola lies

on the directrix.

The tangents at t lt t 2 meet at (af 1 « 2 .a(<i + * 2 ))- Tiie perpendicular from this

point to the tangent at t z has equation

h x + y —aih + t^ + t^t^,

and this line meets the directrix x = —a where y = a^ + ^ + t^ + t^t^. Thesymmetry of this result shows that the other two perpendiculars meet the

directrix at this same point, which must therefore be the orthocentre of the

triangle of tangents.

(iii) (a) Find the condition for y = mx + cto touch the parabola y2 = 4ax.

(b) If this line also touches the circle x 2 + y* = r 2, prove that

a 2

m* + m2 —- = 0,r 2

and hence find the equations of the common tangents to the parabola and the circle

x 2 + y2 = $a 2

.

(a) The line y = mx + c cuts the curve y2 = 4ax at points for which

(mx + c) 2 = 4ax,

i.e. m2 x 2 + 2(mc -2a)x + c 2 = 0.

The line will touch the curve if and only if this quadratic in x has equal roots, i.e.

(mc-2a) 2 = m2 c 2,

ai.e. c = —

.

mHence for all m 4= 0 the line y = mx + a/m touches y

2 = 4ax.




(6) Similar methods will give the contact condition for the line and circ

but it is simpler to equate the radius r and the perpendicular from the cent

(0, 0) to the line mx —y + ajm = 0:

ajmT = ± V(m 2 +l)'

from which the equation for m follows.

Taking r 2 = \a 2, the equation becomes

mi + m2 -2 - 0, i.e. (m 2 - 1) (m 2 + 2) = 0,

so that m = + 1. The common tangents are thus

y —x + a, y ——x—a.

(iv) Tangents from P x ; the chord of contact.

From a given point P x at most two tangents can be drawn to the parabola.For the tangent at the point t will pass through (x lf y x ) if

x i ~ tyi + a t 2 = 0,

and if y\ > 4mx x this gives two values t = t x , t 2 ; i.e. there are then two poin

the tangents at which pass through P x .f

The argument used in 15.64 will show that the chord of contact of tangentsfr ° mPli8

yy 1 = 2a{x + x 1 ).

(v) Pair of tangents from P x .

The condition for the pointl lx x + kx 2 ly x + JcyA

\ l + Jc ' l + Jc )

to lie on y2 = 4tax reduces to

(yl- 4aa 2 ) h 2 + 2{ Vl y % - <La{x x + x %)} U + (y

2 - AaxJ I2 =

(JoachimsthaVs ratio equation for the parabola).

The argument in 15.64, ex. (ii) shows that the pair of tangents from P x

equation{y 1 y-2a(x 1 + x)} 2 = (y

2 -4ax 1 )(y2 -4ax).

Exercise 16(6)

1 Verify that, when simplified, the equation

{y-yi)(y-y*) = y2 -*ax

is linear in x and y ; and that it is satisfied by the points P lt P 2 on the parabola.

Deduce the equation of the chord PxP 2 .

2 Deduce from equation (iii) in 16.22 that t x t % is a focal chord if and o1



THE PARABOLA 611

5 Find the equation of each of the tangents from ( —2a,— a) to y2 = 4ax.

6 The ordinate of the point P on the parabola is PN, and the tangent at Pmeets the axis at T. Prove that the vertex bisects TN.

7If the tangent at

Pmeets the axis at T, prove SP = ST.

8 (i) If PM is the perpendicular from a point P of the parabola to the

directrix, prove that the tangent at P bisects angle SPM. [Use no. 7 and pure

geometry.](ii) Deduce the property of the parabolic reflector: rays from a source of

light at S leave the surface in a beam parallel to the axis. [Angle of incidence

= angle of reflection.]

9 If the tangent at P meets the directrix at Z, prove SZ _L SP.

10 PQ is a variable focal chord; UP is the tangent at P, and QU is parallel

to the axis. Find the locus of the mid-point oiPU.

1 1 Show that any one of the following statements implies the other two.(i) PQ is a focal chord;

(ii) tangents at P and Q are perpendicular;

(iii) tangents at P and Q meet on the directrix.

12 If P, Q are the points t x ,t v and tangents at P, Q meet at T, prove that

triangle TPQ has area \a 2 (t 2 —t^) 3. If this area is always 4a 2

, and the locus of T.

13 (i) If the chord t x t % subtends a right-angle at the point t, prove that

+ -M a ) + 4 = 0.

(ii) If the circle on chord PQ as diameter cuts the parabola again at Hand K, prove that the chords PQ, HK make equal angles with the axis.

14 A circle cuts the parabola at A, B, C, D; tangents at A, B meet at T, andtangents at G,D meet at T'. Prove that the axis bisects TT'. [Use 16.12, ex. (ii).]

15 A circle passes through the vertex of the parabola and cuts the curve

again at A, B, G. The tangents at B, G meet at T. Find the ratio in which ATis divided by the axis.

16 (i) If a circle touches the parabola at A and cuts it again at G, D, prove that

the tangent at A and the chord GD are equally inclined to the axis.

(ii) If circles touch a parabola at a fixed point, prove that the commonchords not through this point are parallel.

*17 (i) If G -> A in no. 16 (i), then by 8.42 (2) the limiting circle is the circle

of curvature of the parabola at A. If A is the point t lf prove that the remaining

intersection D has parameter —3£ x . What can be said about the inclination of

the tangent at A and the chord AD1(ii) A circle cuts the parabola at A, B, C, D. If the circles of curvature at

these points cut the curve again at A', B' , G', D' respectively, prove the latter

points are also concyclic.

*18 If t x #= t 3 in ex. (ii) of 16.12, and t 2-> t x and < 4

-+3 , the limiting circle

touches the parabola at < and at t (double contact) Prove that the chord of



612 THE PARABOLA [16.

21 Prove that the pair of tangents from T(x l9 y x ) cuts Oy at points A, B suthat the mid-point of AB is M(Q,\y x ). [Use ex. (v) of 16.26.] If P, Q are tpoints of contact of these tangents, prove SM _[_ PQ. [Use the chord of contactfrom T.]

22 If tangents are drawn to the parabola from points of a given line, provthat the corresponding chords of contact are concurrent.

*23 Prove that tangents at the extremities of a variable chord throughmeet on the line yy x = 2a(x + x 1 ). [Method of 15.65(1).]

*24 A variable chord through P x meets the parabola at A, B; P 2 is chosenit so that P x and P 2 divide AB (one internally and the other externally) in t

same ratio. Prove that P 2 ^ es on * ne l me VVi — 2a(x + x 1 ). [Method of 15.65(2);

use the ratio quadratic in 16.26, ex. (v).]

*25 Defining the polar of P x wo y% = 4a* to be the line yy x = 2a(x + x t ), prov

thatif

the polar of P x passes through P 2 , then the polar of P a passes throughWhat is the polar of the focus?

*26 Use the ratio quadratic (16.26, ex. (v)) to obtain the equation of t

tangent at P x . [Since P x lies on y2 — 4ax, the ratio quadratic has one root k =

the other root gives the remaining intersection of PxPa with the curve. If P x

is the tangent at P x , this intersection must coincide with P x , and k = is

repeated root. Hence y x y% —2a(x t + x t ) = 0, which gives the locus of P 2 .]

16.3 Normal

16.31 Normal at the point t

Since the tangent at the point t has gradient l/t, the normal there

has gradient —t. Hence the equation of the normal at t is

y —2at = —t(x —at 2),

i.e. tx + y = 2at + at s.

Example

Prove that the normal at t meets the curve again at the point —t—2/t.

If the other intersection is the point s, then the normal at t is also the chorts, so that

gradient of normal at t = gradient of chord ts,

2i.e. —t =

t + s

from which s = —t —2/t. Also see Ex. 16 (c), no. 4.




curve such that the normals from them pass through (x, y ). Ignoring

coincidences, a cubic has either one or three roots; therefore either

one or three normals can be drawn from a given point. Also see

Ex. 16(c), no. 15.

Suppose that the normals at the points t x ,t 2 ,

t 3 are concurrent at

(x, y ). Then these numbers are the roots of the cubic

at 3 +(2a-x )t-y = 0. . (ii)

Since the term in f2 is absent,

fi + <a + 'a = °- (iii)

Three points on the curve which are such that the normals at them

are concurrent are called conormal points. Hence (iii) is a necessary

condition for the points t x ,t 2 ,

t 3 to be conormal. The point of con-

currence is given by

2a-x Q = dLt x t 2 , y = at x t 2 t z . (iv)

The necessary condition T,t x = Ofor conormal points is also sufficient.

In a general way this is clear because only one condition is needed for

concurrence of three lines. In detail, given t x ,t 2 ,

t 3 , we can define

numbers x and y by equations (iv) above; then, provided

h + t 2 + t 3 = °>

we see that t lt t 2 ,t 3 are the roots of the cubic (ii), which expresses that

the normal at each of the corresponding points passes through (x, y ).

If two roots of the cubic (ii) are equal, then two of the three normals from(x , y ) coincide. If t 2 = t s , then t x = —2t 2 , and (x , y ) is given by (iv) as

x -2a + Zat\, y = -2at\.

By 8.53, ex. (i), these equations show that (x ,y ) lies on the envelope of the

normals (i.e. the evolitte) of the parabola. From any point on this locus only twodistinct normals can be drawn.

Three normals from a point can coincide only when < x = < 2 = t 3 , i.e. when each

is zero (since 2^ = 0); and then (iv) shows that {x ,y ) is the point (2a, 0).




(ii) Find the equation of the circle through the feet of normals which are ccurrent at (h,k). (See also 19.7, ex. (i).)

The feet of the three normals are the roots of the cubic

at s

+ (2a-h)t-k = 0.

If the required circle is x 2 + y 2 + 2gx + 2fy + c = 0, its meets with the curvsatisfy (cf. 16.12, ex. (ii))

aH* + (4a 2 + 2ag) t 2 + laft + c = 0.

Since by ex. (i) t = is one such meet, therefore c = 0, and the other three aregiven by

at* + (4a + 2a) * + 4/ = 0.

The cubics (v), (vi) now have the same roots, and so

4a + 2g = 2a-h, 4/ = —k.

The required circle is therefore

x 2 + y2 - (h + 2a) x - \ky = 0.

(iii) (a) Prove that the normal at P x has equation 2a(y —y x ) + y x (x —x t ) = 0.

(6) Hence prove that the feet of the normals from (h, k) to the parabolaon the curve xy + (2a —h)y —2ak = 0.

(a) By 16.23 the tangent at P t has gradient 2a/y lt so the normal at P x hgradient —yJ2a, and its equation is

y-yi = - y

f{x - x J'

(b) Let P x be the foot of a normal drawn from (h, k) to the curve. Then sinc(h, k) lies on the normal at P x ,

2a(k-y 1 ) + y 1 (h-x 1 ) = 0,

which shows that P x lies on the locus

2a{k-y) + y{h-x) = 0.

Exercise 16(c)

1 PN is the ordinate of a point P on the parabola, and the normal at P cuthe axis at G. Prove that NG = 2a.

2 The tangent and normal at P meet the axis at T, G respectively. ProvT8 = SG.

3 The line through the mid -point Mof a focal chord PQ and parallel to taxis meets the normal at P in V. Find the locus of V.

4 If the normal at t meets the curve again at s, find s in terms of t by sustituting x = as 2

, y = 2as in the equation of the normal and either (a) factorising

the cubic in t or (6) observing that the resulting quadratic in * necessarily h




(iv) Show that normals at the extremities of a focal chord can never meet on

the curve.

6 Prove that the normals at the extremities of all chords of the form

x + ky + 2a = meet on the curve. Show that all such chords pass through a

fixed point, and give its coordinates. [This line cuts the curve whereat 2 + 2akt + 2a = 0, so that t x t 2 = 2.]

7 Find the equations of the three normals from (15a, 12a) to y2 = 4ax.

8 Show that two of the three normals from (5a, 2a) to y2 = 4tax coincide.

9 Prove that the centroid of the triangle formed by conormal points lies

on the axis.

10 PQ is a chord of a parabola drawn in a fixed direction. Prove that the locus

of the meet of the normals at P and Q is a line which is itself normal to the

parabola. [If t lf t 2 are the extremities, then

2/(«i + h) = m, i.e. t 1 + t 2 + {- 2/m) = 0;

hence the normals at t x , t 2 meet on the normal at the (fixed) point —2/m.]

11 If normals at P, Q, B are concurrent, and the chords PP', QQ', BR' are

parallel to QB, BP, PQ respectively, prove that the normals at P', Q', B' are

also concurrent.

12 Prove that the normals at t lt t 2 intersect at the point (a;, y), where

x = a{t\ + t x t 2 + t\ + 2), y = - atitjfa + 2 ).

[Solve directly; or eliminate t 3 from S« x = and equations (iv) in 16.32.]

13 A variable chord PQ subtends a right-angle at the vertex. Tangents atP, Q meet at T, and the normals at P, Q meet at N; Mis the mid-point of PQ.Find the loci of T, N, M.* 14 If Cr is the centre of the circle through the feet of normals from P r , prove

that when P ly P 2 , P 3 are collinear, then so are Glf C2 , G3 , and conversely.

[Use ex. (ii) of 16.32.]

15 Show that y2 = 4ax meets the locus in 16.32, ex. (iii) (6) at points P(x,y)

for which y 3 + 4a(2a —h)y —8a 2 k = 0. Deduce that at most three normals can

be drawn to the parabola from a given point (h, h).

16.4 Diameters

16.41 General definition

The locus of the mid-points of a system of parallel chords of a conic

is called a diameter of the conic.

The parallel chords are sometimes called ordinates to the diameter.

In 17.63 the above definition will be reconciled with the usual meaning




therefore the equation of the diameter bisecting chords of gradient

Thus the diameters of a parabola are straight lines parallel to the axi

Also see Ex. 16 (d), no. 6.

Example

Prove that the tangent at the extremity P of a diameteris parallel to the chords which the diameter bisects.

The diameter y = 2a jm meets y2 = 4ax where

x —aim 2, i.e. at the point P(a/m 2

, 2a/m). The tangentat P (which is the point 1/m) has gradient

1/(1 /m) = m,

and is therefore parallel to the chords of gradient m. Fig. 168

Exercise 16 (d)

1 Write down the equation of the diameter of y2 = x which bisects chord

parallel to 2x— 3y+ 1 = 0.

2 Write down the gradient of the chords of y2 = 2x which are bisected

the diameter 2y + 3 = 0.

3 Prove that tangents at the extremities of any chord of a parabola meetthe diameter which bisects this chord.

4 V is the mid-point of a chord QQ' of a parabola, and TQ, TQ' are tangents.Prove that the parabola bisects TV.

*5 Prove that the circle drawn on a focal chord as diameter touches thedirectrix. [Use no. 3.]

6 By considering the ^-coordinate of the mid-point of the chord y = mxof y

2 = 4ax, show that the locus of the mid-points of chords of gradient mthe line y = 2a /m.

Miscellaneous Exercise 16(e)

1 A point is such that its distance from a fixed line is equal to the length

the tangent from it to a fixed circle. Prove that its locus is a parabola. Locatthe focus and directrix when the given line touches the circle.

2 Obtain the equation of the circle through the points (p, 0), (q, 0), (0, r).

A circle passes through a fixed point, and the chord cut off by it from a givline is of constant length. Prove that the locus of its centre is a parabola.

3 Two parabolas have a common axis and their concavities are in oppositesenses. If any line parallel to the common axis meets theparabolas at P, Q, provthat (provided the latera recta are unequal) the locus of the mid -point of



THE PARABOLA 617

7 G is the centroid of a triangle inscribed in a parabola; G' is the centroid of

the corresponding triangle of tangents. Prove that GQ' is parallel to the axis,

and that the parabola divides GO' in the ratio 2:1.

8 The normal at P meets the parabola again at Q, and the axis divides PQin the ratio k:l. Prove that the ^-coordinate of P is 2ka/(l —k). For what value

of k is P an extremity of the latus rectum?

9 Normals at the ends of a focal chord PQ cut the curve again at P', Q'.

Prove that P'Q' is parallel to QP and that P'Q' = 3QP.

10 If normals at P, Q meet on the curve, prove that the meet of tangents at

P, Q lies either on a line parallel to Oy, or on the locus y\x + 2a) + 4a 8 = 0.

11 A,B,C,D are concyclic points on the parabola; AB is a focal chord; AG is

the normal at A. Show that the axis divides BD in the ratio 1 : 3.

12 Find the locus of a point P(h, k) such that two of the normals drawn from

it to the parabola y 2 = 4ax are perpendicular.13 Prove that three normals cannot be drawn from (h, k) to y

2 = lax unless

h > 2a. [Prove f(t) = at 3 + {2a —h)t —k is steadily increasing for all t if h «S 2a,

so that f(t) = has just one root.]

14 Prove that chords of y2 = 4ax which subtend a right-angle at the point t Q

of the curve all pass through the point (a(t 2 + 4), - 2at ). (Cf. Ex. 19 (c), no. 2.)

*15 A variable triangle is inscribed in y2 = 4ax so that two of its sides touch

y2 = 4bx. Prove that the third side touches y % = lex, where (2a —b) 2 c = ab 2

.

[Use the condition km 2 = In for tangency of Ix + my + n = and y2 = Ikx.]

16 Given a line I and a point £ not on it, a variable line is drawn through Sto meet I at V. A line p is drawn through V perpendicular to SV. Prove that

p touches a parabola having S for focus and I for its tangent at the vertex.

[Choose I for 2/-axis, and the perpendicular to it through S for a;-axis; let these

meet at O and p meet SO at T. If p has equation Ix + my + n = 0, then

OV —-n/rn and OT = w/Z. By geometry OV2 = TO. OS, so am 2 = nl where

a = OS.]

17 Show that the line through P x in direction (15.26) cuts y2 = 4ax at

points for which r is given by

r 2 sin 2 + 2(y x sin 6 - 2a cos 6) r + (y2 - 4ax x ) = 0.

(This is called the distance quadratic for y2 = 4ax because it gives the distances r

from P x of the points of the parabola on the line through P x in direction 6.)

18 Prove that the chord of y2 = 4ax whose mid-point is (x x , y±) has equation

yy x—2ax = y\ —2ax x . [For the required chord the roots of the distance quadratic

are equal and opposite, so y x sin 6 —2a cos 6 = 0, which detemiines cos 6 : sin 6.

The equation x —x x : cos 6 = y —y 1 : sin 6 of the chord becomes

{x-x 1 )jy 1 = (y-yjfia.-]

19 Deduce from no 18 that (i) the locus of the mid-points of chords through



618 THE PARABOLA22 Using no. 17, obtain the equation of the tangent at P v [Two roots r =23 Obtain the equation of the pair of tangents from P x . [Equal roots r.]

*24 Show that (3x + 4y)2 = 54a; —53y + 29 represents a parabola as follows.

(i) Verify that the equation can be written

(iii) If PM, PN are the perpendiculars from P(x, y) to the (perpendicular)

lines 3x + 4=y + 1 = 0, 4x-3y + 2 = 0, (a) shows PM2

= 2PN. By the Remarin 16.11, the curve is a parabola of latus rectum 2, having axis 3x + 4y + 1 =and 4x —3y + 2 = for tangent at the vertex.

(iv) For points on the curve, (a) shows that 4x —3y + 2 ^ 0; hence the curvis on the origin side of 4x —3y + 2 = 0, i.e. lies below this line. Sketch the curve*25 Sketch the parabola (x + 2y) 2 = 56x + 12y— 184, giving the equationthe axis and the coordinates of the vertex.

*26 (i) If ab = h 2, then the terms of second degree in

form a perfect square, say (px+ qy)

2. Prove by the method of no. 24 that, unles

g :f = p :q, the equation represents a parabola whose axis is parallel to px + qy =(ii) In the exceptional case show that the equation can be written

(px + qy + X) 2 = A2 —c where g = Xp and / = Xq. This represents a pair of parallel

lines, a repeated line, or nothing according as A2 ^ c. (Algebraically, a parallel

line-pair is thus a degenerate parabola.)

*27 Show that the curve whose parametric equations are

x - at 2 + 2bt, y = ct 2 + 2dt (ad 4= 6c)

is a parabola. [Eliminate t, and use the result of no. 26 (i).]

28 A curve is given parametrically by x =f(t)/h(t), y = g(t)jh(t). Prove th

the chord joining the points t,t + e has equation

(3x + 4y + k) 2 = (6k + 54) x + (8k - 53) y + (k 2 + 29).

(ii) By choosing k so that the lines

3x + 4y + k = 0, (6k + 54:)x + (Sk—53)y + (k 2 + 29) =

are perpendicular, express the equation in the form

ax 2 + 2hxy + by 2 + 2gx + 2fy + c =

x y 1

f(t) g(t) h(t) =0.

f(t + e) g(t + e) h(t + e)

Deduce that the tangent at the point t has equation

x y 1

f(t) g(t) h(t) =0.



619

17

THE ELLIPSE

17.1 The loci

17.11 SP = e.PMWe now consider the locus denned by 8P —e.PM (e > 0, e 4= 1),

where8

is a given point and

Mis the foot of the perpendicular from

P to a given line d (15.71). Choose 8 for origin, and let the given line

then have equation x = k (k > 0). The equation of the locus referred

to this set of axes is therefore

On ' completing the square ' with the terms involving x, this becomes

i.e.

x 2 + y2 = e 2 (x —k) 2

,

x 2 (l - e 2) + 2e 2 kx + y

2 = e 2 k 2.

d

Fig. 169

e 2 k \ 2

1-e 2/

+e 4 F e 2 k 2

(l-e 2)

2 + l-e

(l-e 2)

:



620 THE ELLIPSE [17.1

Omitting dashes, the equation of the locus becomes

V-e*)x*+y* = ~.

17.12 Focus-directrix property of ^+p=In the equation x 2 y 2

we can assume that a > and b > 0, and also that a #= b (otherwise

would be the equation of the circle with centre and radius

Without loss of generality we can therefore always assume a > b >Equation (ii) will be the same as equation (i) if

9 e j r.9 e ft '

a and 6 = T^-On dividing these we get

1 - e 2 = ~5 , i.e. e 2 = — < 1;a 2 a 2

hence < e < 1.

Wo the new (dashed) axes, S has coordinates

and the equation a; = A; becomes

_ e 2 & & _ a

Hence (ii) is the locus of a point whose distance from 8(ae, 0)

e times its distance from the line x = aje, where b 2 = a 2 (l —e 2) and

< e < 1. Therefore (ii) is the equation of an ellipse whose eccentricity

is e, whose focus is S(ae, 0), and whose directrix is the line x = aje.

17.13 Focus-directrix property of -= —v= =a 1 b 2



17.14] THE ELLIPSE 621

(iii) is the equation of a hyperbola of eccentricity e, focus S(ae, 0), and

directrix x = a\e.

In equation (iii) we may assume a > and b > 0; but we must

allow a $ b, because (iii) is not symmetrical in x and y like(ii).

17.14 Second focus and directrix

For both curves the condition SP = e.PM is expressed by

(x-ae) 2 + y2 = e 2 (--xY

which shows that the distance of P(x,y) from the point S'( —ae,0)

is e times its distance from the line x = —aje: S'P = e.PM'.

Hence both curves have a second focus S'( —ae, 0) and a corre-

sponding directrix x = —aje.

17.15 Further definitions

From the equations, both curves are clearly symmetrical about the

£c-axis and about the «/-axis. The origin is the centre of each.

The ellipse meets Ox at the points A (a, 0) and A'( —a, 0), and meets

Oy at B(0, b) and B'(0, —b). AA', BB' are called the major and minor

axes of the ellipse; their lengths are 2a, 2b respectively. The points

A, A' are called the vertices of the ellipse.

The hyperbola cuts Ox at the points A(a, 0) and A'{ —a, 0); but it

does not cut Oy. AA' is called the transverse axis; it has length 2a.

The points A, A' are the vertices of the hyperbola.

For either curve, the chord through S or S' at right-angles to AA'

is called a latus rectum.

We now consider the ellipse in detail, and leave the hyperbola

until Ch. 18.

i.e. x 2 —2aex + a 2 e 2 + y2 = a 2 —2aex + e 2 x 2

.

By adding 4:aex to both sides, this becomes




so that —a < x < a for all points on the curve. Similarly —b^y^bfor all points on the curve. Hence the ellipse lies in the rectangle whosides are x = ±a, y = ±b.

When x = a, the equation shows that y%

\b2

= 0, which hasrepeated root y = 0. Hence the line x = a touches the ellipse at A {a,

it is the tangent at the vertex A. Similarly the line x = —a touchesthe ellipse at A'; and y — ±b touch it at B, B' respectively.

Since e < 1 , therefore ae < a and so S lies between O and A ; similarly

S' is between and A'. Also aje > a, so the directrices cut Ox at point

D, D' beyond A, A' respectively. We have

OS = OS' = ae,

OD=

OD'= - . (iv

B

B'

Fig. 170

17.17 Circle and ellipse

When b = a, the equation (ii) becomes that of the circle with centr

and radius a; but equation (i) cannot represent a circle for any choic

of e and k. Hence a circle cannot be defined by the focus-directrix

property, and therefore is not a 'conic' in the sense of the definition

in 15.71; in particular, a circle is not a special {or degenerate) ellipse.

We may consider the ellipse represented by (ii) when b -> a. Trelation b 2 = a 2 (l —e 2

) shows that e->0; hence by (iv), S andapproach O and the directrices recede along the a;-axis. The ellipse

t d b i l i f i




Thus the circle with centre and radius a is the limit of the ellipse;

and the equation of this circle is certainly the limit when b -> a of

the equation of the ellipse.

17.2 Other ways of obtaining an ellipse

17.21 Auxiliary circle

The circle on diameter AA' is called the auxiliary circle of the

ellipse. Its equation is 2 2 _ 2x -\-y —a .

Fig. 171

Let the ordinate NP of any point P on the ellipse be produced to

cut this circle at Q. Then P, Q are called corresponding points on the

ellipse and circle. Since 9 „

PN = y = -J{a?-x*)

i.e.

= ^(OQ*-ON*) = ^QN,

PN : QN = b:a.

Conversely, an ordinate QN to a circle of radius a is divided at P




Example

The elliptic trammel.

Through P draw the line parallel to QO to meet Ox, Oy at jB, R'; thPR' = Q0 = a. From the similar triangles RPN, OQN,

PR _ PNQO ~ QN'

and hence PR = b.

If a rod PRR' has pegs on its under -side at R, R', and these pegs are madeslide in perpendicular grooves AOA', BOB', the point P will trace an ellips

whose major and minor semi-axes are PR', PR respectively.

17.22 Focal distances

The lengths SP, S'P are the focal distances of the point P on t

ellipse. Since SP = e.PM and S'P = e.PM' (fig. 170),

SP + S'P = e{PM + PM') = e.DD' = e(^j = 2a.

Hence the sum of the focal distances of a point on the ellipse is 2a (t

bifocal property).

Conversely, the locus of a point P such that SP + S'P —constant

an ellipse with foci S, S'.

Proof. Choose the mid-point of SS' for origin, and take Ox alon

S'S; then let S be (c, 0) and S' be ( —c, 0). Supposing that

SP + S'P = 2a,

any point P(x, y) on the locus satisfies

V{(* - c) 2 + y2

} + <J{(x + c) 2 + y2

} = 2a,

i.e. <J{(x + c) 2 + y 2} = 2a- J{{x - c) 2 + y

2}.

Squaring, (x + c) 2 + y2 = (x - c) 2 + y

2 + 4a 2 - 4a *J{(x - cf + y2

},

i.e. 4a J{(x - c) 2 + y2

} = 4a 2 - 4ca.

Again squaring,

a\x 2 - 2cx + c 2 + y2

} = a 4 - 2a 2 cx + c 2 x 2,

h ^jx 2 + 2 = 2 - 2




h 2 = a 2 -c 2. If e is its eccentricity, then since a 2 -b 2 = a 2 e 2 by 17.12,

we have c = ae. Hence S, S' are the points ( ± ae, 0), i.e. the foci.

Example

Mechanical description of the ellipse:'pin-construction' '.

Fix pins S, S' at distance 2ae apart. Tie the ends of a string of length 2a at

S and S'. A pencil point P moving so that the string S'PS is kept taut will trace

the ellipse with foci S, S', eccentricity e, and major axis 2a.

17.23 Orthogonal projection of a circle

Given two planes a, a' which intersect in a line I, consider per-

pendiculars dropped from points of a onto a'. If the perpendicular

from P in a has foot P' in a', then P' is called the orthogonal projection

of P on a'. If P lies on a locus 2 in a, then the corresponding locus

of P' in a' is called the orthogonal projection of S on a'.

Fig.172

We now obtain formulae relating the coordinates of P and P'.

Choose the common line I for a;-axis in both planes, and take any point

on it for origin. Draw lines Oy, Oy' in a, a' at right-angles to Ox.

Wo these axes, suppose P{x,y) in a projects into P'(x',y') in a'. If

6 is the angle between the planes a, a', i.e. between Oy and Oy', then

a;' = x and y' = y cos 0.



626

Choosing 6 so that cos 6

THE ELLIPSE [17

= bja (b < a), this equation becomes

x' 2 y' 2,h— = 1

a 2 ^ b 2

Hence the ellipse — = ]

a 2 o 2

is the orthogonal projection of a circle of radius a onto a plane makiangle cos -1

{bja) with the plane of the circle.

If we rotate the plane xOy about Ox until it coincides with pla

x'Oy', we obtain fig. 171 of the ellipse and its auxiliary circle.

17.24 General properties of orthogonal projection

( 1 ) Effect on angles. A line projects into a line, since Ix + my + n = becomlx' + m sec 6y' + n — 0. This line and its projection cut Ox at the same poiviz. {—nil, 0), and make angles <j>, <j>' with Ox, where

tan 6 = —— and tan 6' = —-.m msec

tan 0' = tan cos 6.

Hence parallel lines project into parallel lines, but in general angles are changeHowever, a line perpendicular or parallel to Ox projects into another such

(2) A tangent to a curve projects into the tangent to the projected curve at

corresponding point. The tangent to y = f(x) at (x lf y^) is

y-Vi =f,(î)'(x-x 1 ),

and projects into (y' —y'J sec = f'(x'j) . (x' —x'j),

i.e. y' - y[ = /'(a£) cos 6 . (x' - x[).

This is the tangent at (x[, y[) to the curve y' = f(x') cos 6, which is the projectionof y =f(x).

By (1), a normal does not in general project into a normal.

(3) Effect on lengths. If PQ in a has length r and makes angle <j> with Ox,its projection P'Q' on a' make angle 0' with Ox. Through O draw OP x equalparallel to PQ; then P 1 is (rcos0, rsm<j>) and so its projection P'

t is

(r cos 0, r sin (j> cos 0)

.




Hence, in general, lengths are changed by orthogonal projection; they are con-

served only on lines parallel to Ox.

Formula (i) also shows that the ratio of lengths on parallel l ines (or on the

same line) is unaltered; in particular, mid-points project into mid-points. On

intersecting lines the ratio of lengths is altered.

(4) Effect on areas. Since

orthogonal projection reduces areas in the ratio cos 6:1.

(5) By means of these results and that in 17.23, properties of the ellipse can

be deduced from certain properties of the circle by orthogonal projection.

In general only central properties can be obtained in this way; we cannot expect

to prove focal properties thus, since foci are foreign to the circle (see 17.17).

Example

PN is the ordinate of P; the tangent at P meets the major axis at T. Prove

ON.OT = OAKFirst sketch the corresponding figure for the circle. In it, OP _L PT, and by

geometry ON.OT = OP 2 = OA2, which can be written ON .OA = OA-.OT

and involves the ratios of lengths on the same line.

If the figure is projected orthogonally onto a plane through the line OT, this

relation will hold for the new figure, and is the required property of an ellipse;

for since PN JL OT, the ordinate PN to the circle will project into an ordinate

to the ellipse.

N.B. —For the circle it is true that ON.OT = OP 2, but this cannot be

generalised into a property of an ellipse because the lengths concerned donot he along the same or parallel lines.

P

T

Fig. 173




angle of P). Then x = ON = acos0; and since NQ = asin^, we hav

y = {b\a)NQ = bsm<j). Hence the coordinates of each point on t

ellipse x 2la

2 + y2 /b 2 = 1 can be expressed in the form (acos^, ftsin^).

As increases from to 2n, Q describes the auxiliary circle anddescribes the ellipse. Two values of

<f>which differ by an integral

multiple of 2n give the same point of the ellipse.

Conversely, for each the point (a cos <}>, b sin<fi)

lies on the ellipse,

as is clear by substituting in the equation. The point whose coordinates

are (a cos 0, 6 sin(f>)

is referred to as the point<f>.

17.32 The point t

By putting t = tan \<$ in the coordinates of the point 0,

1 —t2 2t

z = acos0 = aj-^, 2/ = 6sin0 = 6j-j-^.

Hence for each t, the point/ 1-t 2 2bt_\

\a

i+t 2 >l+t 2)

lies on the ellipse x 2 \a 2 + y2 \b 2 = 1 ; we refer to it as the point t.

Each value oft determines just one point of the ellipse ; but the point

A '( —a, 0) is not given by any value of t, although it can be obtained

as the limit when t->oo.

This algebraic representation can also be obtained directly; f

sincey

2

b 2

t y\° l-xla .

therefore 1^/5—57* -'

Hen*, (l-|):g:( 1+ ?)

and so 1 = (1 -f 2) : 2* : (1 + Z

2).

a b

Example



17.32J THE ELLIPSE 629

Since this is quartic in t, a circle and ellipse can cut in at most four points; the

actual number of intersections will be 4, 2, or (some of which may coincide),

since a quartic has 4, 2 or no roots.

Suppose the above circle cuts the ellipse at the points t lf t 2 , t 3 , t t ; then these

numbers are the roots of (ii). Since the coefficients of J8

and t are equal, therefore

Now by 14.23, (viii),

-i^î^. <Ui »

which is zero by the above condition. Hence S$0 X = nn for some integer n(positive, negative, or zero). Thus */ the points X , a , 3 , 4 of the ellipse are

concyclic, then 20 x = 2mr.

This necessary condition for concyclic points is also sufficient: see Ex. 17 (a),

no. 11.

Exercise 17(a)

' The ellipse'' means the locus x 2 /a*+y 2 /b* = 1.

1 Find the length of a latus rectum of the ellipse.

2 With the notation of 17.15, P is any point on the ellipse and the perpen-

dicular from P to AA' has foot N; prove that PN Z :A'N.NA = OB*: OAKFind a similar relation involving N', the foot of the perpendicular from Pto BB'.

3 If

Pis the point (x,

y)on the ellipse, prove SP = a —ex and S'P = a + ex.

4 Use the bifocal property to prove that BS = BS' = a.

Given an ellipse and its major and minor axes AA' ', BB', construct the foci

S, S' geometrically.

5 A rod AB moves with its ends on two fixed perpendicular lines; P is fixed

in the rod, and BP = a, PA = 6. Find the locus of P.

6 Prove that the gradient of the chord 6<j> is —(b/a) cot \(d + <f>).

7 Prove that the gradient of the chord < < a is —b(l —* 1 * 2 )/a(« 1 + * 2 ).

8 Show that the locus of the mid-point of SP is an ellipse whose centre is

the mid-point of OS. State the lengths of the semi-axes.

9 Prove that the mid-point of the chord 6$ has coordinates

(a cos £(0 + 0) cos $(0-0), b sin £(0 + 0) cos $(0-0)).

10 Find the locus of the mid-point of the chord 00 when (i) + = 2a;

(ii) — = 2a, where a is constant.

1 1 Prove that the condition E0 X = 2nn is sufficient for the points V 2 , 8 , 4

of the ellipse to be concyclic. [The circle through<f> v 2 , 3 cuts the ellipse again

at 04 ; prove 0£ —4 is a multiple of 2n, so that 0^ and 4 give the same point.

Cf. 16.12, ex. (ii).]



630 THE ELLIPSE [1

* 1 5 (i) Find the point where the circle of curvature at ^ meets the ellipse aga(ii) If the points

<fi lt <fi 2 , t are concyclic, and the circles of curvaturethese points cut the ellipse again at <j>' v <f>' ,

<j>'3 , <fi' v prove these points are

concyclic.

*16 With coordinate axes as in 17.23, show that the orthogonal projectionthe parabola y

2 = 4nax is another parabola.

*17 Prove that the tangent to the circle x 2 + y2 = a 2 at the point (a cos 0, asin0

has equation #cos0 + 2/sin0 = a. By projection deduce the equation oftangent to the ellipse at the point

<f>.

*18 Mis the mid-point of a chord HK of an ellipse whose centre is O; OMmethe ellipse at P; N is the mid-point oiPH; and ON, HK meet at Q. Prove tOH bisects the chord which passes through P and Q. [In the circle figure,

the orthocentre of triangle OPH.~\

*19 The chord PQ has mid-point M; the tangents at P, Q meet at T. Prothat O, M, T are collinear. If this line meets the ellipse at R between andprove OM.OT = OB 2

.

*20 Deduce the area of the ellipse with semi-axes a, b from that of a circle

radius a.

Further examples using orthogonal projection appear in Ex. 17 (d).

17.4 Chord and tangent

17.41 Chord P,P 2

The line joining P1 and P2 has equation

Since P r and P2 lie on the ellipse,

^ + and -2 + p-L

By subtracting and then rearranging,

y\-y\ = x\-x\b 2 a 2 '

so that ~——= —5 — 1.x 2 -x x a 2

y 2 + y t

The equation of the chord PjP 2 therefore becomes

b 2 x 2 + x t*« T *1/ \




Alternatively, consider the equation

{x-x x ){x-x %) (y-y 1 ){y-y 2 ) , a/2

,

a 2 b 2. a 2 b 2 '

When simplified it is linear in x and y, and therefore represents a line.

It is satisfied by the points (x v y x ), (x 2 , y 2 ) on the ellipse, and so must

be the chord PxP 2 .

17.42 Chord 6<£

The chord joining (a cos <j>, b sin <p) and (a cos 0, b sin 0) has equation

, . , &(sin0-sin0) ,y-bam.6 = -j „ ^ (a: - a cos g>)y Y a(cos0-cos0) v Y

_ 26cos£(0 + 0)sin£(0-0)- - 2arinlf/9-nA) sin^-^) ( ^

(# —acos0),

2a sin %{6 + <f>)sin £(0 - <p)

6cos|(0 + 0)a sin £(0 + 0)

i.e.

-cos + <j>) + 1 sin \{d + <j>) = cos $ cos £(0 + 0) + sin sin £(0 + <j>)

= cos {0- £(0 + 0)},

i.e. -cos£(0 + 0)+|sin£(0 + 0) = cos£(0-0). (ii)

17.43 Chord f t f a

The line ux + vy + 1 = cuts the ellipse at points t for which

1 —t2 2t

i.e. (l-ua)t 2 + 2vbt + (l+ua) = 0.

This line will be the chord t x t 2 if the above quadratic has roots t x

and t 2 , and then1+ua _ 2vb



632 THE ELLIPSE [17.

The required chord is therefore

U-¥2)^ + (*i + * 2 )f= 1+M 2 -

Alternatively, the chord t x t 2 cuts the ellipse at the points given

t = t v t = t 2 which are the roots of the quadratic (t —t x ) (t —t 2 ) = 0,

*2

-(*i + y* + M2 = 0-

Consider the equation constructed from (iv) by substituting

*2 :*:lfrom (i) in 17.32:

(i-g-(* 1+ y| + M2 (i +

3= o.

It is linear in x and y, and so represents a line. It cuts the ellipse

points t given by (iv), i.e. at t = t x ,t 2 , and therefore is the chord t x

VJM Tangent at P t

This can be obtained by making and y 2-> y x in equation

giving2?h.2m = _ 9

a2

^ 62

a2

1

^'

since Px lies on the ellipse. The tangent at Px is therefore

which can be written down by the usual 'rule of alternate suffixes'.

Alternatively, equation (v) can be found directly by calculus.

17.45 Tangent at 4>

Letting 6 ->(J)

in equation (ii), we obtain

x v-cos ^+^sin^ = 1. (a r

b

This could also be obtained directly by calculus. Cf. Ex. 17 (a), no.

17 46 Tangent




Alternatively, this can be found by using the theory of quadratic

equations, or by calculus, or as a particular case of the result in

Ex. 16 (e), no. 28.

17.47 Examplesx 2 y 2

(i) Condition for y = mz+c to touch —+— = 1.

The line cuts the ellipse at points for which

x % (mx + c)* _a 2

+6 2 ~ h

i.e. (a«m 2 + 6 2) x* + 2a 2 mcx + a 2 (c 2 - 6 2

) = 0.

The line will touch the ellipse if and only if this quadratic in x has equal

roots, i.e. = o2(a2m2 + ft2) (c2 _ ft2))

i.e. c a = a%i 2 + 6 2.

Hence for all values of m the lines

y = mx± V(a 2m2 + 6 2)

touch the ellipse.

(ii) Locus of the intersection of perpendicular tangents : director circle.

If a tangent found in ex. (i) passes through the point (# >2/o)> then

y —mx ± V(a 2 m2 + 6 2),

i.e. {x* - a 2) m2 - 2x y m+ (y\ - 6 2

) = 0.

Since this is quadratic in m, in general two tangents can be drawn from a given

point (x , y ) to the ellipse. The roots mlf m2 are the gradients of such tangents.

These tangents will be perpendicular if ml m2 = —1, i^ if

» 2 -a 2

i.e. if (x , y ) lies on the locus

x*+y 2 = a 2 + 6 2.

This is a circle concentric with the ellipse, and having radius A/(a 2 + 6 2) = AB

(fig. 170). It is the locus of points from which perpendicular tangents can be

drawn to the ellipse, and is called the director circle of the ellipse (by analogy

with the parabola, for which the corresponding locus is the directrix: see

Ex. 16 (6) no. 11 (ii), (iii)).



634 THE ELLIPSE

Exercise 17(b)

1 Use equation (i) in 17.41 to prove that the mid-points of all chords of ggradient m lie on the line y = —b 2 x\a 2 m.

2 Find the condition for the chord 6<f> to pass through the centre ofellipse.

3 Prove that 6<}> is a focal chord if and only if ± e cos £(0 + 0) = cosVerify that this is equivalent to tan yd tan £0 = (e —l)/(e + 1) or (e + l)/(e

4 Prove that the chord PP' of the ellipse and the chord joining the co

sponding points Q, Q' of the auxiliary circle meet on the major axis. Deda property of tangents at corresponding points P, Q.

5 If 6 + = 2a, prove that the chord d^> is parallel to the tangent at a,

conversely.

6 For a system of parallel chords prove that the sum of the eccentric anof the extremities is constant.

7 Find the envelope of the chords joining points of the ellipse wheccentric angles differ by a constant. [Let the angles be

<f>—a,

<f> + a>;

chord is (x/a) cos <j> + (y/b) sin = cos a, which is a tangent to the elli

x 2 /a 2 + y2 /b 2 = cos 2 a.]

8 Prove that lx + my + n = touches x 2 /a 2 + y2 /b 2 = 1 if and only

a 2l

2 + b 2m2 = n 2. [Method of 17.47, ex. (i); or compare with 17.45, equation

9 If m is the gradient of a common tangent to the circle x 2 + y2 = c 2 and

ellipse x 2 /a 2 +y 2 /b 2 = 1, prove m2 — (c 2 — 2 )/(a 2 — 2 ). Deduce that four commtangents exist only if b < c < a.

10 Prove that tangents from ( —2, —3) to 4x 2 + 9y 2 = 36 are perpendicular.

1 1 PN is the ordinate of a point P on the ellipse, and the tangent at P mthe maj or axis at T. Prove ON .OT = a 2

, and obtain the corresponding properfor the minor axis. (Cf. 17.24 (5), ex.)

A12 The tangent at P meets the directrix at Z. Prove PSZ is a right-angle.

1 3 (i) If Q is the point of contact of the other tangent from Z in no. 12, prgeometrically that PSQ is a straight line.

(ii) Deduce that tangents at the extremities of a focal chord meet oncorresponding directrix.

14 Prove that the focal radii SP, S'P are equally inclined to the tangent a

(Cf. 8.14, ex. (iv); and also Ex. 17 (c), no. 3.)

15 If, p, p' are the lengths of the perpendiculars from the foci S, S' to

tangent, prove that pp' = b 2.

16 Using the similar triangles SPY, S'PY', prove p/r = p'/r', where Yare the feet of the perpendiculars in no. 15 and r —SP, r' = S'P.

17 From nos. 15, 16 and the bifocal property r + r' = 2a deduce that the (



THE ELLIPSE 635

19 The chord PQ subtends a right-angle at the vertex A(a, 0). Prove that PQpasses through a fixed point on the major axis. [Use equation (iii) in 17.43.]

20 Prove that the chord joining the points given by the roots of ut 2 + vt + w =has equation (w —u) x/a + vyjb + (w + u) = 0.

21 Prove that the tangents at the points t x , t 2 meet at (x, y), where

1 —x/a _ 1+xja _ yjb

[Use equation (vii) in 17.46, and theory of quadratics.]

22 Write down the equation of the chord of contact from ( 1 , 2 ) to \x 2 + $y2 = 1

and deduce the equations of the tangents from this point.

23 The chord PQ of the ellipse x z /a 2 + y2 /b 2 = 1 touches the circle x 2 + y

2 = c 2.

Prove that the tangents at P, Q meet at a point T on the ellipse

x 2 y 2 _ 1

o*+

6* ~c 2

'

[The chord of contact from T(x lt y x ) to the ellipse touches the circle; use no. 8.]

24 Prove that the tangents at the extremities of a variable chord through P x

meet on the line xxja? + yyjb 2 = 1. What follows by taking P x at a focus?

*25 Given the points Pv P 2 , prove that the point dividing Px P2 in the ratio

k : I will be on x 2 /a 2 + y 2 /b 2 = 1 if

(îHHîHHM- 1 )''- -

(Joachim8thaVs ratio quadratic for the ellipse).

*26 By taking P1 on the ellipse, deduce the equation of the tangent at Pv[See Ex. 16(6), no. 26.]

27 Obtain from no. 25 the equation of the pair of tangents from P v [See

15.64, ex. (ii).] Deduce the equation of the director circle of the ellipse by using

the condition for these lines to be perpendicular (15.52 (2)).

*28 A variable chord through P t cuts the ellipse at A and B, and P a is chosen

so that P x and P 2 divide AB (one internally and the other externally) in the

same ratio. Prove that P 2 lies on the line xxjat + yyjb 2 = 1. [Method of

15.65(2).]

*29 (i) Denning the polar of P x wo a;2 /a 2 + y

2 \b 2 = 1 to be the line

a 2+

b 2 ~ L '



636 THE ELLIPSE [17

17.5 Normal

17.51 Equation of the normal

The reader should verify that the normals to the ellipse at Plt

and t have respective equations

x-x 1 ==y- yi

x x \a 2 yjb 2 '

ax sec <j> —by cosec = a 2 —b 2,

2atx - b( 1 - t2

) y = 2(a 2 - b 2) . (i

1

+1

17.52 Conormal points

The normal at t passes through the point (h, k) if

2ath -b{l-t 2 )h = 2(a 2 - b 2)

* (1 ~ ^ ,

i.e. bkt* + 2(ah + a 2 - b 2) t z + 2(ah -a 2 + b 2 )t-bk = 0. (i

Since this equation is quartic in t, at most four normals can be drawfrom a given point (h, k) to the ellipse; and (apart from coincidences)

there will be either 4, 2, or no such normals.

Suppose that the normals at t v t % ,t 3 ,

i 4 meet at (h, k); then these

numbers must be the roots of (iv). Since the term in t 2 is absent, anthe constant term is minus the coefficient of £

4, we have

Y>t x t 2 = and t^t^ = —1.

These necessary conditions for the points t v t 2 ,t z ,

£ 4 of the ellipse

be conormal (i.e. such that the normals at them are concurrent) ca

also be shown to be sufficient. This is reasonable, because two indepen-

dent conditions are required in order that four lines shall pass through

the same point.




Exercise 17(c)

1 The normal at P meets the major axis at G, and PN is the ordinate at P.

Prove that 00 = e 2. ON. What is the corresponding result for the minor axis?

2 Prove that SG = e.SP.

3 Use no. 2 to prove SG/S'G = SP/S'P, and hence that PG bisects SPS'.

Deduce that the tangent at P is equally inclined to SP, S'P. (Cf. Ex. 17 (6),

no. 14.)

4 Find the locus of the mid -point of PG.

5 Prove that 3x —y—l = is a normal to 2x 2 + 3y 2 = 14, and find the

coordinates of its foot.

6 If to + my + n = is a normal to x 2 /a 2 + 2 /b 2 — 1, prove that

a 2

+b 2 _ (a 2 -6 2

)2

7 PN, PN' are the perpendiculars to the axes from a point P of the ellipse

x 2 /a 2 + y2 /b 2 = 1. Prove that NN' is always normal to a fixed concentric ellipse,

and give the equation of this curve.

*8 (i) If the perpendicular from the vertex A to the normal at the point <j>

meets the ellipse again at the point <p', use Ex. 17 (6), no. 5 to prove <j)' = 20.

(ii) Deduce that the perpendiculars from a vertex to four concurrent

normals meet the ellipse again in four concyclic points.

*9 Prove that the circle through any three of four conormal points cuts the

ellipse again at the point diametrically opposite to the fourth. [If <f>' t arediametrically opposite, then ft = 4 + 7r by Ex. 17 (6), no. 2. Since

S0 X = (2w + 1) n, therefore $ x + <j> 2 + 3 + ^ = (2n + 2) n.~]

10 Prove that the feet of normals drawn from P x to x 2 ja 2 + y2 /b 2 = 1 lie on the

curve (a 2 —b^xy + b^x —a 2 x x y = 0. Verify that this locus passes through P x

and the centre of the ellipse. [Method of 16.32, ex. (iii) (6).]

*11 Prove that the curves a 2x x y = (a 2 —b 2 )xy + b 2y x x and b 2 x 2 + a 2

y2 = a 2 6 2

intersect at points for which y satisfies

b 2 (a 2 x x y)2 = 6 2

a:2 {(a 2 -6 2

)2/ + 6Vi}2 = a 2 (b 2 -y 2 ){(a 2 -b 2

)y + b 2y x }\

and that to each root of this quartic in y corresponds a single x given by{(a 2 —b 2 )y + b 2

y x } x = a 2 x x y. Deduce that in general four normals can be drawnfrom P x to the ellipse.


17.61 The line through P1 (« 1 , y^) in direction 6 (15.26) has parametric

equations Q . Q^ x = x x + r cos a, y = y x + r sin a.




This quadratic in r gives the (signed) distances from Px of the point

on the ellipse which he on the line through P\ in direction 6, andcalled the distance quadratic for the ellipse.

Example*

Newton's theorem.

The product of the roots of (i) is

(x\ y\ \ // cos 2fl sin 2 0\

K +// \~a?~

+~b*~)'

For chords Pj^QR, P x Q'R f drawn in direc-

tions 6, d' it follows that

P 1 Q.P 1 RP 1 Q'.P 1 R' P t

_ / cos 2 6>' sin 2 0'\ //cos 2 6 sin 2 0\ Fig. 174

which is independent of and Hence if chords P X QR, P X Q'R' are drawthrough P x in fixed directions d, d', the ratio P X Q .P X R:P X Q' .P X R' is

dependent of P x .

In particular, when P x is at the centre the ratio becomes OK2: GIT 2

, whethe radii OK, OK' lie in directions 6, 6'. Hence Newton's theorem:

P X Q.P X R _ OK*P X

Q' ,P X R' ~ 0K' %'

It is a generalisation of the 'product property ' of chords of a circle, and reduceto this property when b = a. Also see Ex. 17 (d), no. 16.

By letting R -> Q and R' -+ Q' we see that the ratio of the tangents fromis equal to the ratio of the parallel radii.

17.62 Chord having mid-point Pt

If the chord through P1 in direction 6 is bisected at Pv then troots of the distance quadratic will be equal and opposite, so that

b^cosd + a^sind = 0.

This determines the direction cos 6 : sin of the chord which is bisected

at P, . Its equationx ~ x i = y-Vx




17.63 Diameters

If the chord (ii) above has gradient m, then m = - b^xâhf^ Hence

the mid-points of all chords of gradient m satisfy

• 6 2

^ a 2mCf. Ex. 17 (b), no. 1. According to the definition in 16.41, this locus is

the diameterf bisecting all chords of given gradient m. It is a straight

line which passes through the centre of the ellipse, and is thus a

'diameter' in the usual sense.

Examples

(i) Tangents at the extremities of a diameter are parallel to the chords which

the diameter bisects (the ordinates to the diameter).

Let P 1 be a point where the ellipse is met by the diameter which bisects all

chords of gradient m; then y 1 = —bhcâhn.

The tangent at P x (17.44) has gradient —ft^/a 2 ^, i-e- ™- The result follows.

(ii) Tangents at the extremities of any chord meet on the diameter which bisects

that chord.

Let the chord PQ have gradient m, andlet the tangents at

Pand

Qmeet at

0»i>2/i)- Since PQ is the chord of contact from (x^y^), its equation is

xxJaP+yyxlb 2 = 1, and hence its gradient is m = —b^xja^. Therefore {x lf y x )

lies on y = —b^xjahn, which is the diameter bisecting PQ.

(iii) The sum of the squared reciprocals of two perpendicular semi-diameters is

constant.

Choosing O for pole and Ox for initial line, the ellipse x 2 /a 2 +y s /b 2 = 1 has

polar equation/cos 2 sin 2 0\ ,

(^- + -H = 1 -

If P on the ellipse has polar coordinates (r lt X ), then the coordinates of anextremity Q of the perpendicular semi-diameter are (r 2 , X + \ir), where fromthe equation,

1 ^ cos 2^7\~~ a*

+6 2

1 cos^ + ^tt) sin^ + iTT) sin 2 ^ cos 2 ând -r = 1— = — 1 ——.

r| a 2 6 2 a 2 6 2

1111Addmg, _ + _ = _+_






Choose OP, OD for axes of x and y, and then let Q be the point (x, y). Since

PV.VP' = {OP-OV)(OP + OV) = a\-x\

where a x = OP, the above equation becomes (on writing b x = OD)

a\ —x 2 a\

a\+

b\•

This has the same form as the standard equation of the ellipse, which corre-

sponds to the choice of the perpendicular conjugate diameters OA, OB for

coordinate axes.

Exercise 11(d)

1 Write down the equation of the diameter of 2x 2 + Sy 2 = 1 which bisects

the chords parallel to 2x —%+ 3 = 0.

2 Write down the gradient of the chords of 3a; 2 + %2 = 2 which are bisected

by the diameter y = Zx.

In the following, OP and OD are conjugate semi-diameters whose other extremities

are P', D'.

3 If P is the point in fig. 176, show that the eccentric angles of D, P', D',

Q, Q' can be taken as<fi + %7T,

<J) + tt, ^> —\tt, <j> + a, $ —a respectively. Writeout in full the coordinates of P, D, P', D'

4 If P(a cos <j>, b sin 0) and D(— asin^, ftcos^) are extremities of equal

conjugate diameters, prove that tan a = 1. Deduce that the equations of

the equi-conjugate diameters are y = ± bxja, and show that they lie along the

diagonals of the rectangle which circumscribes the ellipse. State the length

of OP.

5 Prove that OP 2 + OD2 = a 2 + b\



642 THE ELLIPSE

8 Prove PS. PS' = OD2. [Use Ex. 17 (a), no. 3.]

9 The two chords joining any point of the ellipse to the extremitiesdiameter are called supplemental. Prove that supplemental chords are parall

to conjugate diameters.

10 (i) Prove that the lines px 2 + 2rxy + qy 2 = are conjugate diametersx 2 /a 2 + y

2 /b 2 = 1 if and only if pa 2 + qb 2 - 0.

(ii) Find the condition for Ix + my + n = to be the join of extremitiestwo conjugate diameters of x 2 /a 2 + y

2 /b 2 = 1. [Use 15.54.]

(iii) Prove that the chord joining the extremities of conjugate diametersan ellipse touches another ellipse. [Use Ex. 17 (6), no. 8.]

11 If a tangent to the ellipse cuts the director circle at U, V, prove thatO V lie along conjugate diameters.

Nos. 12-15 are easily solved by orthogonal projection.

*12 QR is a diameter, and P is any point on the ellipse. The tangent at Q mePR at T. Prove that the tangent at P bisects TQ.

*13 OP, OD are conjugate semi-diameters; the tangent at P meets theof the ellipse at Q and R. Prove PQ.PR = OD2

.

*14 PQ is a diameter and HK is any chord; PH, QK cut at X; PK, QHat Y. Prove that XY is parallel to the diameter conjugate to PQ.* 1 5 By projecting a circle and a circumscribing square, prove that the tangentat the extremities of conjugate diameters PP', DD' form a parallelogramconstant area 4ab. (Cf. no. 6 (ii).)

*16 Prove Newton's theorem (17.61, ex.) by projection. [When writtenthe form

^ p^, p ^Rt

OK' OK ~ OK' ' OK'

it involves ratios of lengths on parallel lines.]

* 1 7 If parallel chords P x QR, P 2 Q'R' are drawn through the fixed points P x

prove that the ratio P^.P^-.P^Q' .P 2 R' is independent of the direction.

[Use the distance quadratic]

*18 Prove that the mid -points of variable chords through P x he on

xx l|

yy 1 _ x 2 y 2

a 2 b 2 ~ a 2 b 2

'

Show that this is an ellipse whose axes are proportional to those of the origina

ellipse and whose centre is (%x lf ^y ± ).

*19 Use the distance quadratic to obtain the equation of (i) the tangent

Pi [two roots r = 0]; (ii) the pair of tangents from P x [equal roots r],

*20 Obtain the equation of the chord having mid-point P t from the ra

quadratic in Ex. 1 7 (6), no. 25. [If the chord is AP X P 2 , the roots k : I are the rati

P X P 8 : P a P 2 and P X A : AP Z , and hence are I = 0, kjl = —£.]



THE ELLIPSE 643

3 Prove that the tangents at the ends of a latus rectum are concurrent with

the major axis and corresponding directrix.

*4 P, Q are the points 6, $ on the ellipse x 2 /a 2 +y 2 /b 2 = 1, and the tangents

at P, Q meet at T. Prove that triangle TPQ has area

[The corresponding area for the circle is 2A0PT —A0PQ; use projection.]

5 Y is the foot of the perpendicular from S to the tangent at P. Prove OPand SY meet on the directrix corresponding to S.

6 Circles with centres S, S' pass through a point P on the ellipse. Prove that

the common tangents to these circles touch the auxiliary circle, and that their

points of contact with this circle lie on the common chord of the original circles.

[Use pure geometry and the bifocal property.]

7 S is a given point inside a given circle of centre O and radius a; Y is avariable point on the circumference, and p is the line through Y perpendicular

(i) Choosing OS for #-axis, let p meet Ox, Oy at T, T'. Explain why tri-

angles TYO, TST' are similar, and prove OY.TT' = OT.ST'.(ii) If S is (c, 0) and p has equation Ix + my + n = 0, prove that the equation

in (i) reduces to a 2 l2 + b^m 2 = n 2

, where 6 2 = a 2 —c 2.

(iii) Deduce that p touches an ellipse having S for focus and the given circle

for auxiliary circle.

*8 Give the result corresponding to no. 7 (iii) when S lies outside the given

circle. What happens when S lies on the circle?

9 Find the common tangents to x 2 +y 2 = 25 and a;2 /169 + y

2 /16 = 1.

10 The centre of an ellipse coincides with the vertex of a parabola, and they

have a focus in common. Prove that the common tangents meet the commonaxis at its intersection with the other directrix.

1 1 (i) Prove that two tangents can be drawn from P t to x 2 /a 2 + y2 /b 2 = 1 if

and only if x 2fa 2 + y

2 /b 2 - 1 > 0. [Use 17.47, ex. (i).]

(ii) From the ratio quadratic (Ex. 17 (6), no. 25) prove that if

then P x and P a lie on opposite sides of the ellipse. (Accordingly, the set of points

P x for which x\ja 2 + y\jb 2 —1 > is the outside of the ellipse.)

12 Using the contact condition c a = aha 2 + b 2 for the line y = mx + c and the

ellipse x 2 /a 2 + y2 /b 2 = 1, show that the equation of the pair of tangents from

P(* 1 ,y 1 ) can be written (y t x— x y)2 = a^y —y^ + b^x —Xj) 2

.

Let these tangents cut Ox at E, F. Find the locus of P (i) if PE, PF are per-

pendicular; (ii) if EF has fixed mid -point (k, 0).

13 TP, TP' are tangents to the ellipse. Find the equations of the line-pairs

ab sin 8\(<f>

—6) sec J(0 —6).

to S Y.



644 THE ELLIPSE

16 Prove that points whose chords of contact are normals to the ellipse

x 2 /a 2 + y2 /b 2 = 1 lie on the curve a«/x 2 + b 6

/y2 = (a 2 -b 2

)2

.

17 If the normal at P meets the major and minor axes at Q, Q', and if OFthe perpendicular from the centre O to this normal, prove that PF.PO =and PF.PG' = a 2 .

1 8 If the normal at an extremity of a latus rectum passes through an extremityof the minor axis, prove that e 4 + e 2 —1 = 0.

19 P and Q are variable points on x 2 /a 2 + y 2 /b 2 = 1 such that the mid -poinof PQ lies on x 2 /a 2 + y

2 /b 2 = k 2. Prove that the tangents at P, Q meet at a poin

T on x 2 /a 2 + y2 /b 2 = 1/k 2

. [The chord of contact from 2 7(ar 1 ,y 1 ) is the same li

as the chord PQ having mid-point (x 2 , y 2 ), where x 2 /a 2 + y2 /b 2 = k 2

.]

20 The tangents to x 2 /a 2 + y2 /b 2 - 1 at P and Q meet at T{h, k). Obtain t

equation of the line-pair OP, OQ. Find the locus of T when the diameters OOQ are {i) perpendicular; (ii) conjugate. Show that these two loci meet at fopoints which are the vertices of a rectangle.

21 OP, OD are conjugate semi-diameters of the ellipse. The circles witdiameters OP, OD meet again at Q. Prove that Q lies on the curve

2(x 2 + y2

)2 = a 2 x 2 + b 2

y 2.

22 Lines are drawn through O perpendicular to the tangents from P 1

x 2 /a 2 + y2 /b 2 = 1. If these lines are conjugate diameters of the ellipse, prov

that P x lies on the curve a 2 x 2 + b 2y

2 = a* + b*. [Use the equation of the pair

tangents in no. 12.]

Show that eachof

the following equations representsan

ellipse,and find

the centre,

semi-axes, eccentricity, foci, and directrices.

23 &x + 2)2 + \(y-l) 2 = 1. 24 8x 2 + 9y 2 - lQx-Qy = 63.

25 3(x-y+l) 2 + 4:(x + y-l) 2 = 12.



645

18

THE HYPERBOLA

18. 1 The hyperbola ^ - p = 1 ; asymptotes

18.11 Form of the curve

We now resume the discussion, begun in 17.13-17.15, of the

hyperbola ^ 2 oi2

7i -£ = i- »

x 2 y 2

From the equation, ^ = 1 4-p > 1,

so that \x\ $s a for all points on the curve; i.e. no part of the curve lies

between the lines x — ±a. Further, since

y 2 =6 2 a 2 '

we see that when x ~> + cc or x -> —oo, then also y -> ± oo. The curve

is therefore unbounded.

When x — a, the equation gives y2 /b 2 = which has the repeated

root y = 0. Hence the line x = a touches the hyperbola at A {a, 0): it

is the tangent at the vertex A . Similarly, the line x = —a is the tangent

at the vertex A'{ —a,0).

Since e > 1, we have ae > a and aje < a. Hence S lies on OAproduced beyond A ; and if the directix x = aje cuts Ox at D, then Dlies between O and A. Similarly S' lies on OA' produced beyond A',

and D' lies between and -4'.

18.12 Asymptote: general definition

We P b h f di



646 THE HYPERBOLA [18.

ellipse x 2 /a 2 + y2 /b* = 1 because for every point P1 on this curve

have ^ a and \y x \^ b.

(2) An asymptote of a curve is a line such that the distance o

point P1 of the curve from this line tends to zero when Pt-> oo alo

the curve.

In this definition, 'distance' may mean the usual 'perpendicular

distance' or the 'oblique distance' measured. parallel to some fix

direction, e.g. the y-axis. The informal use of the term 'asymptote'

in 1.41, Remark (a), is consistent with our definition.

18.13 Asymptotes of the hyperbola

If P1 lies on the curve, then x{ \a % —y\\b % = 1 and so

M=7(M)-The perpendicular distance of P1 from the line xja —yjb = is

which by (ii) is equal to

When Px-> oo along the curve, this expression tends to zero. Hen

the line xja —yjb = is an asymptote to the hyperbola (i). Similarly

the line x/a + y/b = Qis an asymptote.

The equation of the asymptotes, regarded as a line-pair, is therefore

a 2 6 2

Alternatively, the equation (i) can be written

6 / a a \*y = ±

aX

\1 -^) '

If y = mx + c is an asymptote to the part

b I a 2 \*y = +-x[l —- ,

th h X



18.14] THE HYPERBOLA 647

and this will tend to zero when x -» oo only if c = and bja —m = 0. Hencethe line y = bx/a is an asymptote. Similarly, y = —bxja is an asymptote to

the other part ^ /a2 \ i

y = —aX

\

l-x*)

'

Since

for any point on the hyperbola, therefore y2 < y\\ hence the, hyperbola

lies entirely within that double angle between the lines y = ± bx/a which

contains the x-axis.

The angle between the asymptotes is 2tan _1(6/a). Figures 177,

178, 179 illustrate the cases a < b, a = b, a > b respectively.

When b = a, the asymptotes are perpendicular and the curve is

called a rectangular hyperbola. The equation (i) becomes x 2 —y2 = a 2

;

and the eccentricity, given by b 2 = a 2 (e 2 —1), is e = <J2.

Fig. 177 Fig. 178 Fig. 179

18.14 The bifocal property: SP-S'P= ±2a

If P lies on the branch which

contains the focus 8, then

S'P-SP = e.PM'-e.PM

= e.MM'

=•(¥)

M'

y

M



648 THE HYPERBOLA [1

Conversely, the locus of a point P such that S'P —SP = 2a is

branch of a hyperbola having foci 8, S' and transverse axis of length

viz. that branch which contains S.

Proof. The algebra of 1 7 .22 shows that, with thesame

choiceof

ax

a 2 a 2 —c 2

From triangle P8S' we have PS' < PS + 88', i.e.

2a = PS'-PS <SS' = 2c,

so that a < c. Hence we can write b 2 = c 2 —a 2, and the equation

comes that of the standard hyperbola (i). Since b 2 = a 2 (e 2 —1),

have c2

= a2

+ b2

= a2

e2

, and so c = ae. Therefore 8, 8' are the point( + ae, 0), i.e. the foci. Finally, since SP < S'P, P lies on the brancwhich contains S.

Example

Mechanical construction of a hyperbola.

A rod APB is movable about the fixed end A, and a string BPG passinthrough a small ring P which slides along the rod is tied to the other end Bto a fixed point G. If the string is kept taut,

prove that in general P moves on one branch C '

'

of a hyperbola with foci A, G.Let I be the length of the string. Then

BP+PG = l. Since

AP = AB-BP = AB-(l-PC)= (AB-l)+PC,

therefore AP —PC is constant, viz. AB —I.

If AB =# I, the locus of P is that branch of

a hyperbola with foci A. G which encloses G. Fig. 181

If AB = I, then AP - PG, and P lies on theperpendicular bisector of AG.

18.2 Properties analogous to those of the ellipse ^+p= 1

Since the equations of the ellipse and hyperbola differ only in

sign of 6 2, it follows that many properties of the hyperbola can

written down from the corresponding results for the ellipse by replacing



18.3J THE HYPERBOLA 649

(ii) Tangent at Px : = L

(ui) Normals Pl :

^+^= 0.

(iv) The contact condition for y = mx + c is c 2 = a 2™2 —6 2 provided

\m\ < bja. The condition remains significant when m = ± bja, for then

c = 0; but the lines y = ± foc/a do not cut the hyperbola anywhere

(see 18.13).

(v) If —bja < m < b\a, the lines

y = mx ± *J(a2 m2 —b 2

)

touch the hyperbola.

(vi) Director circle. The locus of the intersections of perpendicular

tangents to the hyperbola is the circle

x 2 + y2 = a 2 —b 2

provided a > b. If a < 6, there are no perpendicular tangents.

(vii) Chord of contact from Px :

a 2 b 2

(viii) Pair of tangents from Px :

x l_yl_A 3L_t_A = (xx i_yyi_iY

a 2 b 2J [a 2 b 2 ) [a 2 b 2

J'

(ix) Chord having mid-point Px :

xx i Wi _ x l yl

a 2 b 2 a 2 b 2 '

(x) Diameters. The mid-points of all chords of gradient m lie on

y = Wxjahm. The properties of diameters stated and proved for the

ellipse in 17.63, exs. (i), (ii) remain true for the hyperbola.

(xi) Conjugate diameters, y = mx and y = m'x are conjugate if

and only if^ 2

mm' = -5.a 2




Analogously to the treatment in 17.32, put t = thf^; then

1 + t 2, 2t

x = a - y = b- -

1 —T 1 —T

If t is now unrestricted, these equations represent the whole of

curve except the point A'( —afi

0); but this can be obtained by lettin

t -> oo. (If t still denotes th \$, it can take only values betwe- 1 and + 1.)

18.32 The point t

The general representation above can be found directly. For si

the equation can be written

IHH(Hf xja-l y/b

therefore ._ = ~- —- = t, say.y/b x/a + 1

J

Hence

(H:

:

fi

+1

H: < :1 '

i.e. -:|:1 = (l + t*):2t:(l-t*).a b

Each value of t except t = + 1 gives just one point of the curve;

each point of the curve except A'( —a, 0) corresponds to just

value of t. We refer to the point

/ 1+t 2 2bt \

\ a i-t*'l-t 2)

as the point t.

18.33 The point <j>

The equation (i) is also satisfied by x = a sec $,y = b tan <j> for all

We refer to (a sec (j>, b tan 0) as the point

If we put t = tan \<f>, we obtain the point t in 18.32.



18.4] THE HYPERBOLA

we may put x_ y ^ xyla b tb

From these we find

x

These equations represent the whole of

the curve; t = +l gives A(a, 0), and pt = —1 gives A'( —a, 0). When t increases

from —oo to +oo, the parts traced are:

P'A't < -1 A'Q'-1 < t PA< t < 1

Fig. 182

AQ1 < t

651

18.4 Chord, tangent, and normal

To each of the preceding representations corresponds a standard

form of equation for chord, tangent, and normal. By methods already

illustrated in Ch. 17, the following results can be obtained.

Chord 6$: -cos£(0-0)-|sin£(0 + 0) = cos %($ + </>).

kh- (i+M 2 )|-(*i+yf = -m 2 .

Tangent at 0: -sec0— ^tan^ = 1.

t: (l+t 2 )--2t%= l-t 2.

a o

Normal at $ : ax cos <j) + by cot<f>

= a 2 + b 2.

1-t 2 't: 2atx + b{l + t*)y = 2(a 2 + 6 2

)

The reader should verify these; also see nos. 13-15 of the following

Exercise, which contains examples similar to those already given for

the ellipse together with some properties of the asymptotes of the



652 THE HYPERBOLAof P. [Let u = velocity of sound, v = velocity component of the shot parallel

to AB; then

AP AB PB , An nn (AB \\-n = 1 , so that AP—PB =u[ n I = constant.]

u v u \ v )

3 Find the locus of the centre of a circle which touches externally two givecircles whose centres are A, B.

4 (i) State the order in which the hyperbola is traced by the point

/ 1 + t 2 2bt \

i-t 2)

as t increases from —oo to + oo.

(ii) What is the condition for the points t lt t 2 to (a) lie on the same branch;

(6) be diametrically opposite?

5 P, P' are the points <p, — on the ellipse x2

/a2

+ y2

/b2

= 1 whose verticesare A, A'. If AP meets A'P' at Q, prove that the locus of Q is the hyperbolax 2 /a 2 -y 2 /b 2 = 1.

6 If the points 6, <j) are the extremities of a focal chord of x 2 / 2 — 2 /b 2 =prove that

1 _ e l + etan \d tan \<p = or .

7 Prove that Ix + my + n — touches x 2 /a 2 —y 2 jb 2 = 1 if and only

aH 2 -b 2m2 = n 2.

8 (i) The tangent at any point P of the hyperbola meets a directrix at

Prove that PZ subtends a right-angle at the corresponding focus.

(ii) Deduce that tangents at the extremities of a focal chord meet on t

corresponding directrix.

9 Y is the foot of the perpendicular from S to the tangent at any point Pthe hyperbola. Prove that Y lies on the circle x 2 + y

2 = a 2 (the auxiliary circle).

10 Prove that the foot F of the perpendicular from a focus to an asymptotelies on the auxiliary circle and on the corresponding directrix.

[OF = OS cos FOS = ... = a; also OS.OD = a 2 = OF 2, so ODF is a right-

angle.]

1 1 The line joining the focus S to the point P on a hyperbola is parallel toasymptote. Prove that this asymptote, the directrix, and the tangent at

are concurrent.

12 The point P on the hyperbola is such that the tangent at P, the latu

rectum through S, and an asymptote are concurrent. Prove that SP is parallel

to the other asymptote.

With the representation in 18.34, obtain the equation of

13 the chord t x t 2 . 14 the tangent at 15 the normal at t.




19 Show that the feet of the normals from P x to x %la % —y i \b % = 1 lie on thecurve a 2 y(x x —x) + b 2 x(y 1 —y) = 0.

20 If the points t lt t 2 ,t 3 , £ 4 are concyclic, prove S(f 1 + * a < 8 * 4 ) = 0.

18.5 Asymptotes: further properties

18.51 Lines parallel to an asymptote meet the hyperbola only once.

Any line parallel to the asymptote x/a + y/b = has equation

x y-+f = k.a b

It meets the hyperbola at points t for which

l+t 2 2t

l-. t2 + l-t 2 ~ '

i.e. (l + t)2 = k{l-t 2

),

i.e. * = -l or 1+* = jfc(l-J).

Since t = —1 gives no point of the curve, there is a unique inter-

section given by t = (k— l)/(k + l). A similar result holds for lines

parallel to x\a —y\b = 0.

18.52 The equation of the tangent at Px tends to the equation of anasymptote when Px

-> oo along the curve.

The equation xx x \a 2 —yy x \b 2 = 1 of the tangent at Px can be written

a b2

x x x x'

where - —=1 i e —=a 2 b 2

' a 2 x\'

The last equation shows that when Px-> oo, then y x /x x bja or —b\a.

The hmit of the equation of the tangent at Px is thus either

# V x y—£ = or +J = 0



654 THE HYPERBOLA [18.5

according to the definition in 18.12, but not according to that abovbecause the equation of the tangent may have no limiting form whPx

-> oo.

18.53 The family x 2 \a 2 —y2 \b 2 = k of hyperbolas has the same asymptotes

for all k.

For the equation can be written in standard form as

y2

=ka 2 kb 2 '

and the asymptotes have equation

^_ll_ ie t-t-oka 2 kb 2 ' a 2 b 2 ~ '

which is independent of k.

Example

In the quadratic which gives the meets of x 2 /a 2 —y2 /b 2 = k with the

Ix + my + n = 0, the coefficients of x 2 and x are independent of k. Hencesum of the roots is independent of k. Taking k = 1, 0, it follows that the a;-c

ordinate of the mid -point of any chord of the hyperbola x 2 /a 2 —y

2 jb % = 1 is

same as the as-coordinate of the mid -point of the same chord of the asymptotesx 2 /a z —y

2 /b 2 = 0. The same is true of the ^-coordinates since Ix + my + ngives y uniquely in terms of x. Consequently, */ the line meets the hyperbolaP, P' and the asymptotes at Q, Q', then PP' and QQ' have the same mid-point

Since PM = MP' and QM - MQ', therefore PQ = P'Q'. In particular,when P' P this becomes PQ = PQ'; i.e. the part intercepted on a tangentthe asymptotes is bisected at the point of contact.

18.6 The conjugate hyperbola

18.61 Definitions

In 17.15 we defined the transverse axis to be the intercept AA' maby the hyperbola on Ox. Although the curve does not cut Oy, it

now convenient to consider the points -6(0,6) and B'(0, —b) on th

axis of symmetry and (analogously to the ellipse) to call BB' t

conjugate axis of the hyperbola.

Two hyperbolas are said to be conjugate when the transverse a




IS

b 21. (ii)

For, with Oy for #-axis and Ox' for y-axis, the equation of the con-

jugate hyperbola is x 2 jb 2 - y 2 ja 2 = 1 . On rotating these axes clockwisethrough a right-angle to positions Ox, Oy, the equation becomes

y2 \b 2 -x 2 \a 2 = l,i.e. (ii).

The two hyperbolas have the same asymptotes, viz.

x 2 y 2

a2

b2

°* (iii)

Remarks

(a) Equation (iii) difiFers from (i) by the same constant that (ii)

differs from (iii). If we transform these equations by a change of axes

(15.73), this relationship will be preserved.

(/?) The equation of any hyperbola whose asymptotes are

a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 =

fax + brf + cj (a 2 x + b z y + c 2 ) = A,s

where A is constant; for this equation differs only by the constant Afrom the equation of the asymptotes, viz.

{a 1 x + b 1 y + c x ){a 2 x-\-b %y + c z ) = 0.

By writing —A for A, we obtain the equation of the conjugate

hyperbola.

I



656 THE HYPERBOLA [18.6

The argument which leads to property (xi) in 18.2 can be applied to

andgives

the same conclusionindependently

ofk:

y = mx and y—

m'xconjugate diameters of each of the hyperbolas (iv) if and only if mm' = b 2

The result stated concerns the cases k = + 1, —1.

(ii) Intersections of the hyperbolas with conjugate diameters.

The diameter y = mx cuts hyperbola (i) if and only if \m\ < bja. The c

jugate diameter y = b 2 x/a 2 m cuts (i) if and only if \b 2 /a 2 m\ < b/a, i.e. \m\ >Hence of two conjugate diameters, only one meets the hyperbola.

The other meets the conjugate hyperbola. For if y = mx cuts (i), then |m| b/a, so that y = b 2 x/a 2 m cuts hyperbola (ii).

(iii) Extremities of conjugate diameters.

The points of intersection are called the extremities of the diameters, despit

the fact that only one pair can lie on each curve.

If P(a sec <j>, b tan<fi)

is an extremity of one diameter, then m = b tan (j>/a sec

Hence the conjugate diameter has gradient m' = 6 a /a 2m = 6 sec <j> la tan <j>,

its equation is y = m'x. It cuts the conjugate hyperbola where

i.e. where x = ±atan0 and hence y = ±6 sec<fi.

Thus P(a sec <j>, 6tan0) a

D(a tan 0, b sec (j>) are extremities 6f conjugate semi-diameters.

1 If e, e' are the eccentricities of a hyperbola and its conjugate, prol/e a +l/e' 2 = 1.

OP, OD are conjugate semi-diameters of the hyperbola x 2 /a 2 —y2 /b 2 = 1. Pro

the following.

2 OP 2 ~OD 2 = a 2 -b\3 The mid-point of PD lies on an asymptote.

4 PD is parallel to an asymptote. 5 The area of triangle OPD is %db.

6 Prove that the tangents at the extremities of a diameter are parallel

the conjugate diameter.

7 Prove that the tangents at the extremities of two conjugate diametersintersect on the asymptotes.

8 Deduce from nos. 6-7 that tangents at P, D and their diametrically

opposite points P', D' form a parallelogram of constant area 4a6, whose vertices

li th t t

Exercise 18(6)




12 Find the equation of the hyperbola whose asymptotes are x —2y + 1 =and dx-y + 2 = and which passes through the point (0,1). [18.61,Remark (yff).]

*13 Given that the equation 2x % —5xy —By 2 + x + lly —8 = represents a

hyperbola, find the equations of the asymptotes and the equation of the con-jugate hyperbola. [Choose A so that 2a; 2 - 5xy - 3y* + x + 1 ly A = representsa line -pair.]

18.7 Asymptotes as (oblique) coordinate axes

18.71 xy = c 2

The product of the perpendiculars from a point Px on

Fig. 184

to the asymptotes x/a + yfb = 0, xja —yjb = is

<H

+y± 3h_yx *±_y\

a b a b _a? 6 2

\a i+ b 2) \a*

+b*) a 2 6 2

1

~ ± 1

because x\/a 2 —yf/b2 = 1.




i.e. Px satisfies, , . a?b 2

x y sin- 5o) = —

Since o) = 2 tan -1 (b/a),

.2 _ / 2tan^> \

2

_ / 2b I a \ 2 _ 4a 2 6 2Sm &>_\l + tan 2 |o>/ ~[l + b 2

la2

) (a 2 + 6 2)

2 '

and the equation becomesx'y' = J(a 2 + 6 2

).

Omitting dashes, this can be written

xy = c 2,

where c 2 = |(a 2 + 6 2), and is the standard equation of the hyperbola

referred to its asymptotes as (oblique) coordinate axes.

Example

Any equation of the form xy = ax + by + c represents a hyperbola whose asymtotes are parallel to Ox, Oy.

For the equation can be written (x —b) (y —a) = ab + c; i.e. x'y' = abafter a change of origin.

Thus the locus in ex. (iii) (6) of 16.32 is a rectangular hyperbola, knownthe hyperbola ofApollonius of (h, Jc) wo the parabola y

2 = 4ax. See also Ex. 17

no. 10, and Ex. 18 (a), no. 19.

18.72 Parametric representation

The equation xy = c 2 can be written

= *, say,

so that x = ct and y = c/t. For each value of t except t = the po

(ct, c/t), referred to as the point t, lies on xy —c 2; and to each point

the curve corresponds just one value of t. The parametric equations

can also be written ±l) , ±x:y:c = t2 :l:t.

The methods illustrated in 16.22 (2), (3) and 16.24 (1) do not appe

to 'gradient', and can be used in the present case of oblique axes





660 THE HYPERBOLA3 {(t)} The chord P x P s is c 2 x + x x x 2 y = c 2 (x x + x t )

.

4 {(o) The tangent at P x is xy x + x x y = 2c 2. Could this be written down fro

xy —c 2 by the 'rule of alternate suffixes'?

5 {o)} The chord of contact of tangents from P x is xy x +x x y = 2c 2.

*6 {to} Joachimsthal's ratio quadratic is

(x t y s -c 2 )k 2 + (xy x + x x y-2c 2 )kl+(x L y x -c 2 )l 2 = 0.

*7 {(o) The pair of tangents from P x is 4(^2/ —c 2 )(xy —c 2) = (xy x +x x y —2c

8 The normal at P L i s a ja^ —yy x = x\ —y\.

9 The distance quadratic is

r 2 sin 6 cos 6 + (x x sin + y x cos 0) r + (a? ^ —c 2) = 0.

10 The chord having mid-point P x is xy x + x x y = 2x l y x . [Use no. 9 or no. 3

1 1 The diameter bisecting all chords of gradient m is y = —mx.

12 y = mx, 2/ = m'a; are conjugate diameters if and only if m+ m' — 0.

13 {&>} The conjugate hyperbola is xy = —c 2. [Begin as in 18.71.]

14 Conjugate semi-diameters are equal in length and make complementaryangles with the transverse axis; and the asymptotes bisect the angles betweenthem.

1 5 {(i)} Prove that lines drawn from a variable point P on a hyperbola to atwo fixed points E, F on the curve intercept a constant length on either

asymptote.

16 {(t)} If the tangents at P, Q meet at T, prove that OT bisects PQ.

*17 {(o} A quadrilateral circumscribes a hyperbola. Prove that the li

joining the mid-points of its diagonals passes through the centre of thyperbola.

18 Prove that no two tangents to a rectangular hyperbola can be perpen-dicular.

19 If P, Q, R are points on a rectangular hyperbola such that PQ subtendsright-angle at R, prove that the tangent at R is perpendicular to PQ.

20 Concyclic points.

(i) Show that the circle x 2 + y2 + 2gx + 2fy+d = cuts the rectangular

hyperbola xy = c a at points t for which

c 2 * 4 + 2gct z + dt 2 + 2fct + c 2 = 0.

(ii) If the points t x , t 2 , t s ,t t are concyclic, prove <i< a * 8 $ 4 = 1.

(iii) By considering the fourth intersection of the hyperbola and the circle

through the points t x ,t 2 , t a , prove the converse of (ii).

21 Use no. 20 (ii) to prove that the common chords of a circle and rectangular

hyperbola are equally inclined in pairs to either axis of the hyperbola.

22 If circles touch a rectangular hyperbola at a fixed point, prove that thei

common chords not through this point lie in a fixed direction.



THE HYPERBOLA 661

respectively, prove that the mid-points of the chords AA', BB', CC, DD' lie

on another hyperbola having the same asymptotes as the given one.

25 Prove that the circumcircle of triangle t x 1 2 1 3 cuts the rectangular hyperbolaagain at the point diametrically opposite to the orthocentre of this triangle.

[Use 18.73, ex.]

26 If a circle and rectangular hyperbola meet at A, B, C, D, prove that theorthocentres of triangles BCD, CDA, DAB, ABC are concyclic. Prove alsothat the tangents to the hyperbola at G, D meet at a point on the diameterwhich is perpendicular to AB.

27 (i) Prove that the normal at t to the rectangular hyperbola meets thecurve again at the point —l/* 8

.

*(ii) Verify from no. 23 (i) that this is diametrically opposite to the pointwhere the circle of curvature at t meets the curve again.

28 The normal at P meets the rectangular hyperbola again at Q. Prove thatthe mid-point of PQ lies on 4x 3 y 3 + c 2 (x 2 -y 2

)2 = 0.

29 Conormal points.

(i) If the normal to the rectangular hyperbola xy = c 2 at the point t passesthrough (h, k), prove c* 4 —ht 3 + ht —c = 0.

(ii) If the normals at t lt t t , t a , t 4 are concurrent, prove that t l t 2 t 3 t l = —1 and= 0.

30 Prove that four conormal points are also orthocentric, but that fourdistinct conormal points can never be concyclic.

3 1 The axes being rectangular, prove that the equation y = (ox + b)j(cx + d),

c =}= 0, represents a rectangular hyperbolawhose asymptotes are x = —djc,y = a/c. What are the coordinates of its centre? [Use 18.71, ex.]

32 A variable line passes through a fixed point and meets two given inter-

secting lines at P, Q. Prove that the locus of a point dividing PQ in a given ratiois a hyperbola whose asymptotes are parallel to the given lines. [Use obliqueaxes.]

33 A variable chord of xy = c a passes through the fixed point (p,q). Provethat the locus of its mid-point is another hyperbola, and give the equations ofits asymptotes.

Miscellaneous Exercise 18 (J)

1 If o) is the angle between the asymptotes of a hyperbola, prove that its

eccentricity is sec %o>.

2 PN is the ordinate of a point P on x 2 /a 2 - y2 jb 2 = 1 ; NT is a tangent from

N to the auxiliary circle x 2 + y2 = a 2

. If P is the point ^, prove NOT =<f>

andNP:NT = b:a.

3 With the notation in 17.23 and cos# = b/a, what is the orthogonal pro-jection of the rectangular hyperbola x 2 —y

2 = a 2 1 Show that the rectangular



662 THE HYPERBOLA6 B is a fixed point in the plane of a given circle of centre A, and P

variable point on the circumference; BP meets the circle again at Q, andparallel to AQ through B meets AP at R. Prove that the locus of R is an ellip

or a branch of a hyperbola according as B is inside or outside the circle.

7 The tangent at P to x 2 /a 2 —y 2 /b 2 = 1 meets the asymptotes at Q,VQ, VR are parallel to the axes of the hyperbola. Prove that V lies on onethe rectangular hyperbolas xy = ± ab.

8 The tangents at P, Q to a hyperbola meet an asymptote at H, K. Prothat PQ passes through the mid -point of HK. [Use xy = c 2

, oblique axes.]

9 The tangents at two points P, P' on a hyperbola meet one asymptoteQ, Q', and the other at R, R'. Prove QR', Q'R are parallel.

10 For the hyperbola x = %a(t + -1), y = \b(t —£

_1), find the equation of

chord joining the points t given by ut 2 + vt + w = 0.

1 1 Prove that tangents toy

2 = kx at (kt\, kt^ and (kt\, kt2

) meet atpoint (kt^, mh + tj).

The tangent to y2 = kx at any point P cuts the hyperbola x 2 —ty

2 = kU and V. IfQ,R are the points of contact of the other tangents from U, Vthe parabola, prove that the chord QR touches the circle x 2 + y

2 = k 2.

12 Prove that the chord of contact from any point (x lf y x ) on the hyperbolax 2 /a 2 —y

2 /b 2 = 1 to the ellipse x 2 /a 2 + y2 /b 2 = 1 touches the hyperbola, and g

the coordinates of the point of contact.

13 A variable chord of x 2 /a 2 —y2 /b 2 = 1 touches the circle x 2 + y

2 = c 2. Pro

that the mid-point of this chord lies on

14 The normal at a variable point P on the rectangular hyperbola xy —meets the asymptotes at Q, R. Prove that the mid-point of QR lies on the cur4x 3

y3 + c\x 2 —y

2)

2 = 0. Explain why this is the same locus as in Ex. 18 (c), no.

15 Prove that the tangent and normal at P on a hyperbola bisect the interior

and exterior angles (respectively) between SP, S'P. [Either prove SO = e

and use pure geometry (Ex. 17(c), no. 3); or use r' —r= ±2a and calculu

(8.14, ex. (iv)).]

16 If F, Y' are the feet of the perpendiculars from S, S' to the tangent a

to the hyperbola, prove that SY.S'Y' = b 2 and that triangles SPY, S'PY'similar. Hence prove that p/r = (b 2 /p)/S'P, where p = SY andr = SP. Dedufrom the bifocal property that the (p,r) equation of the hyperbola wo the focusas pole is b 2

/p2 = 1 ± 2a/r, where + corresponds to the branch enclosing S.

17 If an ellipse and a hyperbola have the same foci S, S' (i.e. are confocal),

prove that they cut orthogonally at each common point P. [The tangents a

are respectively the external and internal bisectors of angle SPS'.]

18 (i) Prove that the equation x 2 /(a 2 + X)+ y 2 j(b 2 + X) = 1 (a > b) represents

an ellipse a hyperbola or nothing according as A > —6 2 —a 2 < A < —b





664

19

THE GENERAL CONIC; s = ks'

19.1 The locus 5 =

19.11 Scheme of procedure

It has now been shown that the general equation of the second degr

s = ax 2 + 2hxy + by 2 + 2gx + 2fy + c =

represents either a conic (in the wide sense of 15.72), a circle, a p

of parallel lines, or nothing; cf. the last paragraph in 15.74. In

chapter it will be convenient to refer to the locus s = as a 'conic',

even when it is a circle or a parallel line-pair.

We begin the chapter by applying to the locus s = the genera

methods for finding chords, tangents, etc. already illustrated

dividually for the parabola, ellipse, and hyperbola. The work will th

be a summary and unification of ' conies ' , and will contain as particularcases many results already obtained for the special standard forms

equation of these curves. Had the general case been treated fir

considerable repetition would have been avoided; but dealing w

the special forms separately increases the appreciation of the method

themselves. Some of our work will apply unmodified when 5

represents a line-pair. The reader should consider whether the result

remain significant in this degenerate case.

The second part of the chapter is independent of the first, andconcerned with a principle which permeates the whole of coordinate

geometry. The student who is pressed for time may turn at once

19.5 etseq.

19.12 Notation

To simplify the writing we introduce the following notation.

W





666 THE GENERAL CONIC; S = ks' [

by Px , for s xx = because Px lies on s = 0; and it is satisfied bsince s 22 = 0. Hence it is the equation of the chord PX P2 .

Letting P2 Px along the conic, we obtain 2s x = s xx as the limit

the above equation, i.e. 2s ± = 0. Hence the equation of the tangents = at Px is s

x= 0.

19.2 JoachimsthaPs ratio equation

19.21 The ratio quadratic for s =

The point dividing PX P2 in the ratio k : I has coordinates

/lx x + kx 2 ly x + ky 2 \

\ l + Jc'

l + k )'

It will lie on s = if and only if

a(lx x + kx 2 )2 + 2h{lx x + kx 2 ) (Z^ + ky 2 ) + b{ly x + ky 2 )

2

+ 2g(lx 1 + kx 2 ) (l + k) + 2f{ly x + ky 2 ) (l + k) + c(l + kf =

i.e. {ax\ + 2hx 2 y 2 + by\ + 2gx 2 + 2fy 2 + c) k 2

+ 2{ax x x 2 + h{x x y 2 + x 2 y x ) + by x y 2 + g{x x + x 2 ) +f(y x + y 2 ) + c}

+ (axl + 2hx x y x + by\ + 2gx x + 2fy x + c) Z2

=i.e. s 22 fe

2 + 2s 12 fe/ + s 11 /2 = 0.

This quadratic in k:l is Joachim-

sthaVs ratio equation for 5 = 0. Its

roots are the values of Px A : ^4P 2 for

the points of intersection A of the

line P]P 2 and the conic. Since the

quadratic will have either two, one,or no roots, f a line meets a conic in Kg- 186

two, one, or no points.

The above work applies even if s = represents a line-pair.

We now make a sequence of deductions from the ratio equation.

19.22 Sides of a conic

When and have opposite signs the product of the roots





668 THE GENERAL CONIC; S = ks' [19.25

19.25 Chord of contact from P,

Method 1. Let the tangents from Px (supposed outside the conic)

touch 8 = at P2 ,P3 . The tangent at P2 nas equation s

2= 0; and sinc

it passes through Pv we have s 12 = 0. This shows that P2 lies on t

line 5 X = 0. Similarly s 13 = 0, so that P3 also lies on s x = 0. Hence t

equation of P2 P3 is s t= 0.

Method 2. The points of contact of tangents from Px satisfy bot

s = and s xl s = s\. Hence they also satisfy s x = 0. Since s x is linear,

st

= is the equation of the chord of contact of tangents from Pv

19.26 Examples; polar of P, wo s =

(1) Tangents at the extremities of a variable chord through P x meet on s x = 0.

Let the tangents at the extremities A, B of such a chord meet at P 2 . Ththe chord of contact from P 2 is AB, whose equation is therefore s 2 = 0. Sincthis line passes through P x , s 12 = 0; and this condition shows that P 2 lies

the line s ± = 0.

(2) Let a chord through P x cut the conic at A and B. The point P a such that

and P 8 divide AB in the same ratio {one internally and the other externally) l

on the line s 1 — 0.

From the hypothesis it follows thatA

andB

dividePiP 2

hithe same

rati

(one internally and one externally). Hence the ratio quadratic must haveroots k : I equal and opposite; this is so if and only if s 12 = 0, which shows thP 2 lies on * t = 0. P 2 is called the harmonic conjugate of P x wo 8=0.

(3) The diagrams in 15.65, with the circle replaced by any conic, show thafor some positions ofP 1 the locus of P 2 in (1), (2) above will be only part of tline « x = 0. We unify these results by making the following definition.

The whole line s t = is called the polar of P t wo s = 0.

Remarks

(a) If P x lies outside s, the polar coincides with the chord of contact from P(/?) If P x lies on s, the polar coincides with the tangent at P x .

(y) In full, the polar of P x is

* = (ax x + hy 1 + g)x + (hx x + by 1 +f)y + (gx x +fy x + c) = 0.

It does not exist if ax k + hy± + g = and hx x + by x +/ = 0; in general there is

unique point P x satisfying these conditions.

(3) The argument shows (without modification) that the polar of P x woline-pair s = is s x = 0. Since

= h ) b f)



19.27] THE GENERAL CONIC; S = ks' 669

(£) Reciprocal property. If the polar of P x passes through P a , then the polar

of P a passes through P x . (This is valid for a line-pair provided P a is not the

vertex.) For the polar of P x wo s = is = 0, and this passes through P a if

s 12 = 0; this shows that P x lies on s 2 = 0, which is the polar of P a wo s = 0.

Infig.

187, P X Qis

thepolar of

P 2.

Fig. 188

19.27 Chord whose mid-point is P,

Let the extremities of the chord be A and Pa

(fig. 188) . Then s22

= 0,

and the ratio quadratic has one root 1 = 0. The other root, given by

2k8 li + U11 = 0,

corresponds to the point A ; and since A divides P1 P2 externally in the

ratio 1 : 2, this root must be k\ls= —\. Hence s 12 = «s n , which shows

that P2 li es on * ne locus s x = s n . Since this is linear, and is also satisfied

by Pl5 it is the required chord. Thus the chord ofs = wJiose mid-point

is Px is st

= s

u.

Example

Show that the mid-points of all chorda of s = which pass through P x lie onthe curve s — s x .

Let P 2 be the mid-point of such a chord, which will have equation * a = « aa .

Since the chord passes through P x , « ia = s 2t . This shows that P a lies on the

locus s x = s which, being of second degree, is also a conic; clearly it passes

through P r

Fig. 187



670 THE general conic; s = ks' [19.2

Its equation is also s x = s xx , which can be written

ax x (x - x x ) + h{x x (y - y x ) + y x (x - x x )} + by x (y - y x )

+ g(x-x 1 )+f(y-y x ) = 0.

Hence ax x + h(mx x + y x ) + bmy x + g +fm = 0,

which shows that Px lies on the line

(ax + hy + g) + m(hx + by+f) = 0.

By definition (16.41) this is the equation of the diameter bisecting

all chords of gradient m.

19.29 Conjugate diametersThe diameter bisecting all chords of gradient m has gradient

a + hmh + om

so that a + h(m + m') + bmm' = 0.

The symmetry in m, m' of this relation shows that the diameter

gradient m bisects all chords of gradient m'. Two such diameters

are conjugate.


The line through Px in direction 6 has parametric equations

x —x x + r cos 6, y = y x + r sin 6.

It meets s = at points for which r is given by

a(x x + r cos d)2

+ 2h(x x + r cos 6) (y x + r sin 6) + b(y x + r sin d)2

+ 2g(x x + r cos 6) + 2f(y x + r sin d) + c = 0,

i.e. r 2 (a cos 2 6 + 2h cos 6 sin 6 + b sin 26)

+ 2r{(ax x + hy x + g) cos 6 + (hx x + by x +/) sin 6} + s xx = 0.

This quadratic in r gives the signed distances from Px of the points

of the conic s = which he on the line (i) through Px in direction 0.




which gives the direction cos# :sin# of such a chord. If (x,y) is any point onthis chord, then from equation (i),

cos 6 sm.6'

and so (ax x + hy x + g)(x-x x ) + (hx 1 + by x +/) (y - y x ) = 0.

This is therefore the equation of the chord of s = which has mid -point P x ;

it can be arranged in the form s x = s u (cf. 19.27).

(ii) Segments of a chord.

If the line (i) cuts the conic at Q, R, then by considering the product of the

roots of (ii) we have

P t Q.P x R = r x r 2 =a cos2 d + 2h e g

.

n e + b gin2.

(a) For chords QB, Q'B' through P x in given directions 6, 6', it follows that

P X Q.P X R _ a cos 2 6' + 2h cos B' sin d' + b sin a 6 '

P X Q'.P X R' ~ acos 2 + 2/&cos0sin0 + &sin a '

which is independent of the coordinates of P x . Hence the ratio of the product of

the segments of two chords drawn in given directions through the same point is

constant for all points.

(6) For chords P X QR, P Z Q'R' drawn through P lt P 2 in the same direction 6,

it also follows that „ ^ „ DP X Q.P X R _

P%Q 'P%R *22

which is independent of 6. Hence the ratio of the products of the segments of

two parallel chords drawn through given points P x , P 2 is constant for all

directions.

19.4 Tangent and normal as coordinate axes

Sometimes it is convenient (as for example in 8.42(1)) to choose

the tangent and normal at a particularyj

point for axes of coordinates; this point

thus becomes the origin.

If s = passes through 0, then c = 0.

If «s = touches y = at (0, 0), the equation

ax % + 2gx + c = must have both roots zero,

so that also g = 0. The equation of the conic

is therefore o x

ax 2 + 2hxy + by 2 + f = 0 Fi 8- 189



672 THE GENERAL CONIC; S = ks' [19

Choosing axes as above, let PQ have equation Ix + my = 1. (This formequation for PQ is legitimate since the hypothesis that

l PQ subtends a righ

angle at O ' implies that PQ does not pass through 0.)

The line-pair OP, OQ has equation

ax 2 + 2hxy + by 2 + 2fy(lx + my) = 0,

by 15.54. Since OP JL OQ, then from 15.52 (2)

a + (b + 2fm) = 0.

If a + b 4= 0, then m = —(a + 6)/2/+ 0. Hence Ix + my = 1 passes throug

(0, —2//(a + 6)), which is a fixed point on Oy, the normal at O.

If a + b = 0, then the perpendicularity condition becomes fm = 0. If / =the conic s = is an orthogonal line -pair —excluded by hypothesis; in suchcase every 'chord' subtends a right-angle at O. If m = 0, the equation of

is Ix = 1, which is a line parallel to Oy.

Exercise 19(a)

Write out in full the expressions for &xx , s Xi when s denotes

1 ^ + ^-1- 2 y2 -4ax. 3 xy-cK

*4 (lx + my + n)(l'x+m'y + n'); and if u — Ix + my + n, u' = I'x + m'y+n'show that s ia = ^{u-û^ + uû

1

^.

5 Interpret the equation s lx + 2s 12 + s 2iS = 0.

6 Obtain the equation of the tangent to s = at P x (i) by Calculus; (ii) frs« n = si by taking P x on s = 0.

7 Prove that the normal to s = at P x has equation

(2/ - Vx) (oî + %i + g)~(x- x x ) (hx x + by x +/)

8 Find the condition for the lines px 2 + 2rxy + qyz = to be parallel

conjugate diameters of * = 0.

9 Prove that the line-pair joining O to the meets of x 2 + y2 = r 2

ax 2 + 2hxy + by 2 = 1 is

a-^jx 2 + 2hxy+(b-^jy 2 = 0.

Show that these will be conjugate diameters of the conic if

1 _ 2{ab-h 2)

r 2 a + b

Hence obtain the equation of the equi-conjugate diameters of

ax 2 + 2hxy + by 2 = 1.

10 Writing u = qx + hy + g v ~ hx + by+f w = gx +fy + c, verify that



19.5] THE GENERAL CONIC; S = ksr

673

12 (i) Prove that the equation of the pair of tangents from P x is

{u x (x - xj) + v x {y - y x )}2 = {a(x -xtf + 2h(x - Xl )(y- yi ) + b(y- y^} s u .

(ii) Using no. 10, show that this is equivalent to (s l—s 11 )

2 = (s —2s 1 +8 11 ) $ nand hence to s\ = 8S n .

1 3 A line through O cuts 8 = at P and Q. A point R is chosen on this line sothat OR is (i) the arithmetic mean; (ii) the geometric meanj (hi) the harmonicmean, of OP, OQ. Find the locus of R in each case.

14 K is the point ^ on the ellipse x 2 /a z + yz /b 2 — 1. Write down the equation

of the chord PQ for which KP, KQ are parallel to the axes, and find where it

meets the normal at K. Deduce from the example in 19.4 that chords whichsubtend a right-angle at the fixed point <j> are concurrent at

/a 2 -6 2J a 2 -6 2

. AI ~z — a cos p, ——o sin <p |

.

15 If equation (i) in 19.4 represents lines OP, OQ equally inclined to Oy, use15.52 (3) to prove h +fl = 0. Deduce that if OP, OQ are equally inclined to thenormal to * = at O, then PQ passes through a fixed point on the tangent atO, or is parallel to this tangent.

19.5 Number of conditions which a conic can satisfy

If k is any non-zero constant, the equations

s = ax 2 + 2hxy + by 2 + 2gx + 2fy + c =and Ics = Jcax 2 + 2khxy + Jcby 2 + 2hgx + 2kfy + he =

represent the same locus; for if P1 satisfies one, then it also satisfies

the other. Since a, b, c, f, g, h are not all zero, we can always choose hso that one coefficient in ks is 1. The remaining five coefficients (i.e.

the five ratios a:b:c:f:g :h) can be chosen so as to satisfy five con-

ditions, but in general no more. The equations which express these

conditions may have more than one set of solutions but, since theyare polynomial equations, the number of sets will be finite if theconditions are independent.

It follows that a conic can be chosen to satisfy five independentconditions, and that the number of such conies is finite.

Examples



674 THE GENERAL CONIC; S = ks' [1

Alternatively, a, b, c, f, g, h can be eliminated from the above equationtogether with s = 0, giving the required equation in determinant form.

An easier way of finding the equation of a conic through five given poiwill be illustrated in 19.64, ex. (i).

(ii) If three of the five points in ex. (i) lie on a line I, then I must be part oconic, which therefore consists of two lines I, V.

For no line can meet a non-degenerate conic in three points (see the footnoon p. 666); and the only line-pair through these five points is I and the lin

which joins the other two.If four of the points are collinear on I, the conic consists of I and any lin

through the fifth point. If all five points lie on I, the conic consists of I andline in the plane. In these two cases of ex. (i) the conic is not unique.

19.6 The equation s = ks'

19.61 Number of possible intersections of two conies

The equations of the two given conies can be arranged in the for

s = ax 2 + 2(hy + g)x + (by 2 + 2fy + c) = 0,

s' s a'x 2 + 2(h'y + g')x + (by + 2f'y + c') = 0.

The coordinates of any common point of these must satisfy

equation in y obtained by elimination of x. Since this equation

in general be quartic in y, there are 4, 2 or no possible values o

(apart from possible coincidences). By eliminating x 2 only, wethat to each value of y corresponds at most one value of x, given

2{(a'h -ah')y + (a'g - ag')} x

+ {(a'b - ab') y2 + 2{a'f- af) y + {a'c - ac')} = 0

Hence, ignoring coincidences, the number of possible intersections

two conies is 4, 2 or 0.

19.62 s = ks'

This equation has already been discussed when s = 0, s' = repr

sent lines (15.41) or circles (Ex. 15 (d), no. 16).

(1) If s = 0, s' = are conies and k is constant, then s = ks' is

the equation of a conic, which passes through the meets {if any) ofs and

For if P satisfies both 5 = and s' = 0, then it also satisfies s =




viz. k = s n ls' u (assuming that P± is not on s' = 0: see the Remarkbelow). For this k, the conies <r and s = ks' have five points in common,and hence coincide (19.5, ex. (i)).

Exceptions to the last statement may arise when three of the fivepoints are collinear. Since Px is an arbitrary point of a, Px need notbe collinear with any two of A, B, C, D. If A, B,C are collinear, thens, s' must be of the forms txfi, ay where a = is the line ABC and

fi = 0, y = are lines through D (19.5, ex. (ii)). Every conic throughA,B,C,D then consists of a == and a line = Icy through D. Hence

cr = a(fi-ky) = s-Jcs',

and the theorem is still valid.

Remark. The only conic through the meets A, B, C, D of s and s'

which does not have an equation of the form s = ks' is s' itself. This

exception can be avoided by using As = AV, where A and A' are

independent constants; but usually the equation s = ks' is adequate.

For different k or A : A', the equation s = ks' or As = AV represents

a family or system or pencil of conies through the intersections (if

any) of s and s'.

19.63 Degenerate cases

The results in 19.62 apply when one or both of s, s' degenerate, andthese cases are particularly useful.

(i) s — kecfi. This represents a conic through the meets (if any)

of the distinct lines a = 0, fi = with s = (fig. 190).



676 THE GENERAL CONIC; S = ks' [19.6

(iii) s = JcotT, where r is a tangent to s. When ft-» t in (i), we hav

a conic meeting s at its intersections (if any) with a = 0, and touching

it at the contact of t = with s (fig. 192).

(iv) s = kaft, where a, ft meet on s. If one meet of a and s coincideswith one meet of ft and s in (i), we obtain a conic meeting s at a poin

on a and at a point on /?, and touching 8 at the meet of a and /? (fig. 1 93

Fig. 196 Fig. 197

(v) s = foxT, where a passes through the contact of r. If one meet

d i i id ith th t t f we obtain a conic meeting



19.64] THE general conic; s=ks' 677

(viii) a/? = ky 2. If S -> y in (vii), we have a conic touching the lines

<x, /? at their meets with y (fig. 197).

(ix) a 2 = ky 2 represents the two lines a = ± ^/fcy, which pass

through the meet of a, y if this exists.In the next two cases s' represents a single line.

(x) s = kcc is a conic through the meets (if any) of s and a.

(xi) s = &t is a conic touching s at its contact with t.

19.64 Examples

The 's = ks principle has already been used for conies in Ex. 15 (/),

no. 15; Ex. 16 (b), no. 1; and in the alternative method in 17.41.

(i) Find the equation of the conic through (3, 0), (0, —2), (5, 0), (0, 1), (15, 6).

The lines joining the first four points in pairs are as follows.

(3,0), (0,-2): |-|-1 = 0; (5,0), (0,1): | + j,_l = 0;

(0,-2), (0,1): x = 0; (3,0), (5,0): y = 0.

Hence any conic through the first four points is (by 19.63, (vii))

This passes through (15, 6) if 1.8 = k. 15.6, i.e. k = The required conic is

therefore ^ _ 3y _ 6){x + 5y _ 5) = 8xy ^

(ii) Find the equation of the circumcircle of the triangle formed by the lines

ax 2 + 2hxy + by 2 = and Ix + my + n = 0.

Let the tangent at O to the required circle be px + qy — 0. Then by 1 9- 63, (iii)

its equation will beax 2 + 2hxy + by 2 — (Ix + my + n)(px + qy)

provided we choose p and q so that

a—pl —b —qm and 2h = mp + Iq.

If we arrange the above three equations in terms of p and q, we have

(Zoj + my + n) xp + (Ix + my + n)yq— (ax 2 + 2hxy + by 2) = 0,

Ip —mq —(a —b) = 0,

and mp + lq —2h = 0



678 THE GENERAL CONIC; S = ks' [19.

(iii) Goncyclic points on the ellipse

xl + yl _ ia 2 b 2

The chords0i0 2

and3 4

are a =0,

/? = 0, where

a = - cos £(0 X + 2 ) + - sin £(0 X + 2 ) - cos ^(fa -2 ),

= -cos£(0 3 + 4 ) + -sin£(0 s + 4 )-eos|(0 3 -0 4 ).

The points l5 2 , 3 , 4 will be concyclic if and only if one conic of the syste

a 2 o 2

isa

circle.This requires that the

coefficientsofx

2

and y2

shall be equal, andcoefficient of xy zero. In general it is not possible for the single numbersatisfy these two conditions; but the coefficient of xy is zero if

cos £(0 X + 2 ) sin £(0 3 + 0J + cos J(0 3 + 4 ) sin + 2 ) = 0,

i.e. sin + 2 + 3 + 4 ) = 0,

i.e. <j) x + 02 + 03 + 04 = 2rwr

for some integer «,

The coefficients of x 2 and ?/2 are then equal if

^-

^= k c os

i(0x + 2) cos

|(0 3 + 4) -

^sin

£(0 X + 2) sin

|(0 3 + 4 )

J

= &j-^ cos 2£( 0 + 2 ) + ^ sin 2

J( 0 + 2

)

J cos nn ,

and here the coefficient of k is non-zero.

Hence if S0 X = 2nn, k can be chosen uniquely to satisfy the second condition.

Thus the relation S0 X = 2nn is necessary and sufficient for the points <j> x , 2 , 3

of the ellipse to be concyclic. Compare the example in 17.32.

*(iv) Normals at the extremities of the chords Ix + my = 1, 1'x + m'y = 1 of

ellipse 9 9

a 2 b 2

The conic x 2 v 2

- + f--l = k{lx + my-l){l'x + m'y-l)a 2 b 2

passes through the extremities of the chords, by 19.63, (i). If the normalsthese extremities are concurrent, say at (x^y^), then their feet lie on

(a 2 —b 2 )xy + b 2 y 1 x —a 2 x 1 y =

by Ex 17 (c) no 10 Consequently this conic must belong to the above system




*(v) A conic passes through four given points A, B, G, D on s = and also

through the meet P x of tangents at A, B. Prove that it passes through the meet P 2

of tangents at C, D.

Since the equations of the chords of contact AB, CD are respectively s x = 0,

« 2 = 0, the conic is 8 = A» x * a by 19.63, (i). This passes through P lt so s u = ks n s l2 ;

hence k = l/s 12 , and the conic has equation s 12 s = Clearly this is satisfied

byP 2 .

*(vi) Pascal's theorem. If 1, 2, 3, 4, 5, 6 denote six points

on a conic, then the meets of the lines

12, 45; 23, 56; 34, 61

are collinear.

For suitable k and I, the line-pair (12) (56) has equation

8 = k{25) (16), and the line-pair (23) (45) has equation

s = Z(25)(34). The points X(12,45) and F(23,56) lie onboth pairs, but do not lie on s = or on (25). Hence theylie on the line &(16) = i(34), which passes throughZ(34, 61).

The argument holds even when * = is a line-pair provided that the sets

1, 3, 5 and 2, 4, 6 lie on different lines; the result is then known as Pappus's

theorem.

19.65 Contact of two conies

Although in general two distinct conies intersect in at most four

points A,B,C, D, there may be coincidences of the following types:

AAGD 2-point contact at A (figs. 192, 193),

AACC double contact along AC (fig. 191),

AAAD 3-point contact at A (fig. 194),

AAAA 4-point contact at A (fig. 195).

Since a circle can be determined to pass through only three general

points, we may expect that in general a circle cannot have more than

3-point contact with a conic at By regarding such a circle as the

limit when P A along the conic of a circle touching the conic at Aand cutting it at P, we see from 8.42 (2) that the circle having 3-point

contact at A is the circle of curvature of the conic at A.

A conic having 3-point contact with s = at A is s = Tear, where



680 THE GENERAL CONIC; S = ks' [

Example

Find the circle of curvature of y2 = 4ax at the point t.

The tangent at the point t is t = x —ty + at 2 = 0, and has gradient 1/t.

chord through {at 2, 2at) with gradient —ljt is

y —2at = —-(x —at 2),

i.e. a = x + ty —Sat 2 = 0.

The required circle is

y2 - 4ax = k(x + ty- Zat 2

) (x-ty + at 2),

where k must be chosen so that the coefficients of x 2 and y2 are equal

xy-term is already absent). This gives k = —kt 2 —1, i.e. k = —1/(1 + t2

that the circle is x 2 + y2 _ 2a(3i 2 + 2) x + 4 a% - 3a 2 < 4 = 0.

The centre of curvature at t is therefore (2a + 3a£ 2, —2at s

); and the radius

curvature is found to be 2a(l + £ 2 )*. These results can of course be obtained

the methods of Ch. 8.

Remark. Iif(x) and g(x) are polynomials, then Remark (a) in

shows that when the curves y = f(x), y = g(x) have mth-order conta

at x = a, they have (m + l)-point contact in the sense of the prese

section, and conversely. For general curves the conceptsare

equivalent.

19.7 Equations of a type more general than s= ks'

Iff = 0, g = represent any two curves, their intersections (if any) sat

every equation which can be deduced algebraically from / = and g = 0.

new equation need not be of the form/ = kg; and even when it is, k may no

constant. The problem in 15.54 is of this type.

Sometimes the equations are combined in such a way that the deduction

do not hold for all the common points.

Examples

(i) Circle through the feet of conormal points on a parabola.

The points in question satisfy the equations

y2 = 4ax and xy + (2a —h)y —2ak = 0,

where (h, k) is the point of concurrence of the normals (see 16.32, ex. (iii)




(ii) Find the equation of conies through the meets, other than the origin, of

s = ax 2 + 2hxy + by 2 + 2gx + 2fy =

and s' = a'x 2 + 2h'xy + b'y 2 + 2g'x + 2f'y = 0,

where fg' *f'g.The given equations can be written

x(ax + 2hy + 2g) = -y(by + 2f),

x(a'x + 2h ,

y + 2g') = -y(b'y + 2f).

Coordinates other than (0, 0) which satisfy both these equations also satisfy

<r = (ax + 2hy + 2g) {b'y + 2f) - (a'x + 2h'y + 2g') (by + 2f) = 0.

This is therefore one conic through the meets of*, s', and does not pass through O.

Clearly<r + ks + k's' = (i)

also passes through the same points for all k, k'.

If 8, s' meet in three points other than O, then k, k' can be chosen to make theconic (i) pass through two other general points. Hence any conic through thethree points has an equation of this form. The system (i) is called a net of conies.

Exercise 19(6)

Find the equation of the conic through

1 (0,0), (0,1), (3,0), (2,1), (2,-3).

2 (2,3), (-1,1), (4,0), (3, -2), (5,3).

3 Assuming that ax 2 + 2hxy + by 2 = c and a'x 2 + 2h'xy + b'y 2 = 1 intersect

in four points, find the condition for these points to be concyclic.

4 Find the common chords of x 2 + y2 = 25 and x 2 + xy + y

2 — 36.

5 Two chords are drawn through a focus of an ellipse, and a conic is drawnthrough their extremities and the centre of the ellipse. Prove that this conic

cuts the major axis in another fixed point.

6 (i) Prove that, for any c,

(l+t 2

)(y2 -4ax)

+(x-ty

+at 2 )(x

+ty

+c) =

is the equation of a circle which touches y2 = Aax at the point t.

(ii) PSQ is a focal chord of a parabola. Circles are drawn through thefocus S to touch the parabola at P, Q respectively. Prove that these circles

cut orthogonally.

7 A circle has double contact with ax 2 + by 2 + c = 0. Prove that the chord ofcontact is parallel to either x = or y = 0. [Let the chord be Ix + my + n = 0.]

8 Find the equation of the circle touching the ellipse x 2 /a 2 + y2 /b 2 = 1 at each

end of a latus rectum.



682 THE GENERAL CONIC; S = ks r

1 1 Write down the equation of the tangent at the point of x 2 ja 2 + y2 jb

and find the equation of the chord through which has the same inclination

Ox as this tangent. Hence show that the circle of curvature at is

and find its centre and radius.

12 Find the circle having 3 -point contact with xy = c a at the point t,

deduce the centre and radius of curvature at this point.

13 Prove that the circle of curvature of ax 2 + 2hxy + by 2 = 2y at

a{x 2 + y2

) = 2y.

*14 Use example (iv) in 19.64 to prove thatif

normals at the ends of the ch

0i0 2 and 3 4 of x 2 /a 2 + y2 /b 2 — 1 are concurrent, then

COS £(0 X + 2 ) COS £(0 3 + 4 ) + COS J(0j -2 ) °OS £(0 3 - 4 ) =

and sin i(0 x + 2 ) sin £(0 3 + 4 ) + cos £(0 X - 2 ) cos £(0 3- 4 ) = 0.

By subtraction, deduce that E0 t = (2n + 1) ir is a necessary (but not sufficien

condition for the points 1? 2 , 03> 04 to be conormal. (Cf. 17.52, ex.)

*15 (i) Write down the equation of a conic circumscribing the quadrilateral

whose opposite sides are the lines a,/?; y, 8.

(ii) If p is the length of the perpendicular from P to the line a = 0, p

that p = constant x a .

(iii) If a conic circumscribes a quadrilateral, prove that the ratio of

product of perpendiculars from any point P of the conic onto two oppo

sides, to the product of perpendiculars from P onto the other two sides

constant (a result due to Pappus).

*16 Show that the feet of normals from (h,Jc) to y2 = 4ax lie on the parab

x 2 + (2a-h)x = iky. [See 19.7, ex. (i).]

*17 Explain why the conic H$x —t 2 y + at$ (x —t z y + at\) = passes throuthe vertices of the triangle formed by the tangents to y

2 = 4ax at t x , t 2 ,£ 3 .

Find the ratios fc x : k 2 : k 3 for which this conic is a circle, and verify that

circle passes through the focus. State this result as a geometrical theorem.

Miscellaneous Exercise 19(c)

1 A pair of tangents to ax 2 + by 2 = 1 intercepts a constant length 2c onProve that the point from which the tangents are drawn lies on the curve

by 2 (ax 2 + by 2 ~ 1) = ac 2 (by 2 —l) 2.

2 Prove that chords of y2 = 4aa? which subtend a right-angle at (at\,

cos0 —-sin0 —cos 20J

=b



THE GENERAL CONIC; S - ks' 683

4 If ax 2 + 2hxy + by 2 + 2gx + 2fy + c = represents a line -pair, prove that

the pair (other than the coordinate axes) passing through the meets of these

with the axes has equation ax 2 + 2(2fg/c —h)xy + by 2 + 2gx + 2fy + c = 0.

5 The lines 6a; 2 + ay 2 = (a + 6 4= 0) meet the conic s = in four points

A, B, G, D. If the diagonals not through O of the quadrilateral ABGD areperpendicular, prove that

a —b h g

h b-a f = 0.

g f o

6 ABGD is cyclic; AB, DC produced meet at E; BG, AD produced meetat F. Prove that the bisectors of angles GEB, CFD are perpendicular. [With Afor origin, let the pair AB, DC bet = px 2 + 2rxy + qy

2 = 0, and BG, AD be * = 0.

Express the condition for t = ks to be a circle, and use 15.52 (3).]

7 Prove that the points of intersection of x 2 —2xy + By 2 + 9x —16y + 24 =and 5a; 2 + 6xy —y

2 —43a; + 60y —62 = are concyclic. [Eliminate xy.]

8 Find the equation of the rectangular hyperbola through the meets of theellipse x 2 /a 2 + y

2 jb 2 = 1 and the line-pair y2 = m2 x 2

. Show that when m = b/a,

this rectangular hyperbola cuts the ellipse orthogonally.

9 A circle through the origin touches the rectangular hyperbola xy = c 2 andmeets it again at P, Q. Prove that the foot of the perpendicular from the origin

to PQ lies on 4acy —c 2.

*10 Obtain the general equation of conies through (3,2), ( —3,2), (2, —3),

( —2, —3). Show that this family contains a parabola, and find its equation.[Use Ex. 16(e), no. 26 (i).]

*11 The normal to y2 = 4oa; at the point P(t) meets the parabola again at

Q(d), and the normal at Q meets the curve again at R. Prove that the equationof the parabola which passes through P, B and touches the given one at Q canbe written

X{y 2 - lax) + (tx + y- at 3 - 2at) (6x + y- ad 3 - 2a6) = 0,

where A = (d —t) 2 /4dt. Prove that its axis is parallel to SN, where S is the focus

of y2 = 4ax and N is the point where the normal at P toy 2 = 4ax cuts Oy. [In

thelast

partuse

6 = —t—2/t.]

*12 Prove that the tangent at P 1 to the conic obtained in ex. (v) of 19.64 is theline joining P x to the meet of AB and CD. [Use Ex 19 (a), no. 4.]



684

20

POLAR EQUATION OF A CONIC

20.1 The straight line

Polar coordinates and equations have been used in various par

of the book, particularly in calculus applications. The purpose of t

chapter is to obtain and use the polar equation of a conic to pro

geometrical properties, especially focal ones.

Webegin with sho

sections on the straight line and circle, in which we assemble

material required.


Given two points A(r v X ) and B(r 2 ,d 2 ), the length of AB can

found by applying the cosine rule to triangle AOB (fig. 199):

AB2

= r\ + r\ - 2r t r 2 cos {6 1 - 2 ).

Fig. 199 Fig. 200

20.12 Line joining two points

Let P(r, 6) be any point on AB. If P lies between A and B, th

(fig. 200)^ AQB = ^ AQp + ApaBj



20.13] POLAR EQUATION OF A CONIC 685

Accordingly, since this equation is satisfied by any point (r, 6) of

the line AB, it is the equation of the line.

Remark. The line joining O to the point (p, a) has equation = a.

This fact, already noticed in 1.63 (c), is obvious from first principles.

20.13 Line in 'perpendicular form'

Let the perpendicular from O to the given line have length p andmake angle a with Ox. If P(r, 6) is any point on the line, then fromtriangle ONP we have

p = rcos(0 —a). (ii)

This (cf. 15.28) is the equation of the line in 'perpendicular form'.

Fig. 201

Remark. The foot N of the perpendicular from has polar co-

ordinates (p,oc).

20.14 General equation of a line

When converted to polar coordinates, the general linear equation

Ax + By = C becomesÂeosd + Bsvcid = —. (iii)

r

Example

Any line perpendicular to (iii) has an equation of the form

CAcoa(d + $7r)+Bsin(d + in) = —



686 POLAR EQUATION OF A CONIC [2

20.2 The circle

20.21 Polar equation

Let P(r, 6) be any point on the circle with centre C(p,a) and radius(fig. 202). From the distance formula (20.11),

a 2 = CP 2 = p2 + r 2 -2prcos(d-oc).

Hence the circle has equation

r 2 —2pr cos (6 —a) —a 2 — 2.

P(r, 6)

Fig. 202 Fig. 203

We notice the following particular cases.

(a) If C lies on Ox, then a = 0; the equation is

r 2 —2pr cos 6 = a 2 — 2.

(b) If O lies on the circle, then p = a; the equation becomes

r 2 —2arcos(6 —a) = 0,

so that r — 2a cos (d —a) represents a circle through O.

(c) If C lies on Ox and O lies on the circle, then a = and p =the equation is then r —2a cos 6, which is easily written down fr

fig. 203.

20.22 Chord P t P 2 oir = 2a cos 6 ; tangent at P t




and hencef 26 1 —ex.— —(20 2—a),

i.e. a = X + 6 %.

From (i) we now havep

— 2a cos 61

cos2

> and the equation of the

chord becomes ID n a . n ar cos (0 —

1 —d 2 ) —2a cos X cos u % . (u)

By letting 2 -> 0 we obtain the equation of the tangent at Pv viz.

r cos (0 - 2^) = 2a cos 2 6 X . (iii)

20.23 Examples

(i) Simeon's line.% From any point O on the circumcircle of triangle ABG,perpendiculars are drawn to the sides. Prove that their feet are collinear.

Choose O for pole, and the diameter through O for initial line. The equationof the circumcircle is then r — la cos 6.

Let A, B, G correspond to the values = a,/?,y. Then the chord BC has

equation2^ cos ^ cqs y = r COjg ^Q_p_y^

and hence by the Remark in 20.13, the foot of the perpendicular from O to

BG has polar coordinates (2a cos ft cosy, ft + y). Similarly, the other feet are

(2acosy cosa, y+a), (2acosa cos/ff, oc+fi).

These three points clearly lie on the line whose polar equation is

2a cos a cos ft cosy = rcos(0 —a— ft— y). (iv)

This is known as the Simeon line of O wo triangle ABC.

(ii) Extension of ex. (i). If D is another point on the circumcircle of triangle

ABC, then A, B, G, D can be selected three at a time in 4 ways, and hence there

are 4 Simson lines wo O corresponding to the 4 possible triangles.

By the Remark in 20.13, the foot of the perpendicular from O to (iv) is

(2a cos a cos ft cosy, oc+ft+y). Similarly, if D corresponds to the value 6 = 8,

the feet of the perpendiculars from O to the other Simson lines are

(2acosft cosy cos#, ft+y + S), (2a cosy cos S cosa, y + 8+a),

(2a cos 8 cos a cos ft, 8 a +ft).

Clearly these four points lie on the line

2acosa cos ft cosy cos# = rcos(0 —a— ft— y —8).

This result is likewise capable of extension.




Exercise 20(a)

1 Sketch the figures for the eases in 20.12 when P lies on (a) AB produced(6) BA produced. Verify for each that the equation (i) still holds.

2 Show that the area of the triangle whose vertices are A^rÔ^), 2?(r 2 ,C(r 3 ,0 z ) is £{r 2 r 3 sin (0,-0,)+ ,^ sin (0 x

-0t ) + r x r 2 sin(0 2 —#1)}. Deduce

condition for A, B, G to be collinear.

3 Prove that the perpendicular from (r lf X ) to the line r cos (0 —a) = plength ± {p - r x cos (d x

—a)}.

4 A variable line through a fixed point meets three given lines at poinP lt P 2 , P 3 . On this line is taken a point P such that

OP ~ OP* OP* OP 3

'

Prove that the locus of P is another straight line.

5 (i) Write down the polar equation of (a) the circle whose centre is (pand which touches the initial line; (6) the circles of radius a which touchinitial line at the pole.

(ii) Explain why the circles r = acos(0 —a), r = 6sin(0 —a) cut orthgonally.

6 Find the centre of the circle r = a cos + b sin 0.

7 If OPQ is a chord of the circle r 2 —2pr cos (8 —a) +p 2 — 2 = 0, prove tOP.OQ = p 2 -a 2

. Deduce the 'product property' that, for chords OPQ, Othrough O, OP.OQ = OR. OS.

8 Let ON be the perpendicular from to the tangent at A^dj) oncircle r = 2a cos 6, and P(r, 6) be any point on this tangent. By expressingin two ways as r cos (6 —26 x ) and r x cos lf obtain equation (hi) of 20.22.

*9 Prove that the normal to r = 2a cos 6 at 6 = a has equation

rsin (2a —6)= asm 2a.

10 Find the equation of the chord joining the points of the circle

r = 2a cos {6 —a)

for which = 6 lt 2 . Deduce the equation of the tangent at V1 1 Two circles meet at O, and a line through meets them again at P,

respectively. Find the locus of the mid-point of PQ.

12 Find the condition for the line IJr = a cos 6 + b sin 6 to touch the cir

r = 2c cos 0.

*13 Prove the Simson line property (20.23, ex. (i)) by pure geometry.




In figs. 204 and 205, P lies on the same side of the directrix as S.

Fig. 206, where P and S are on opposite sides, will arise when P is onthat branch of a hyperbola which is remote from S. We have

r = SP = e.PM = e.ND

(e(SD-SN)in 204

e{SD + 8N)in 205

e(SN-SD)

in 206

(e(SD-rcoad)

in 204, 205

e(rcoad-SD)in 206

l —ercoad

in 204, 205,

ercoad —l

I in 206,

p M

A\S N D X N S D

M M— r

1

i

s D N x

Fig. 204 Fig. 205

since I = SL = e.SD. Hence

Fig. 206

in figs. 204, 205 - = 1 + e cos 6;T

I- = —1 + ecosd.r

in fig. 206

The last equation represents the branch of a hyperbola remote from S.

M ~^jP^ \(J>r > -

D N ^

Fig. 207

If we replace r by — and 6 by 6 —n ( if we use the t ti f




as for figs. 204 and 205. Hence the equation Ijr = l + ecos#

represent the branch of a hyperbola remote from S if r is allowe

to take negative values.

It follows that, if negative values of r are allowed, then every prop(i.e. non-degenerate) conic is completely represented by Ijr = 1 + e co

(a) If the line DS (in the sense from D towards S) is taken

initial line, this is equivalent to replacing by 71 —6 in the abodiscussion. The equation becomes Ijr = 1 —ecosd; it will completely

represent all proper conies if r is allowed to take negative values.

(/?) The equation Ijr = —1 + e cos 6, obtained from fig. 206,completely represents all proper conies (with the same proviso abo

negative values of r). Indeed, the two equations Ijr = —1 + ecos#

Ijr = 1 + ecosd represent the same conic, but any chosen point of

conic will of course correspond to different pairs of values of r,

the two representations. This fact is used in 20.33, ex. (hi).

(y) The equation of a conic whose major axis is inclined at angle

to the initial fine is Ijr = 1 + e cos (6 —a). For, wo its axis JSD as initi

line, the equation is Ijr — 1 + e cos 6; rotation of the initial line cloc

wise through angle a gives the new equation.

20.32 Tracing of the curve l/r = l+e cos 8

(1) When e = 1, the conic is a parabola whose equation can

Remarks

written

r-= l + cos0 = 2 cos 8 40.




(2) When e < 1, the conic is an ellipse which is completely repre-

sented by _—n < 6 ^ tt and r > 0.

(3) When e > 1, the conic is a hyperbola. The branch enclosing Sis given by , n .—7r + sec -1 e < 6 < n —sec -1 e and r > 0;

and the branch remote from $ by

n —sec -1e < < n + sec -1

e and r < 0.

The reader should check the above statements.

Fig. 210

20.33 Examples

(i) Eqitations of the directrices,

(a) From the focus-directrix definition, SL = e.SD, so that SD = l/e.

If (r, 6) is any point on the directrix corresponding to 8, then

rcostf = SD = Z/e.

Hence this directrix has equationI- = e cos 0.r

*(b) For central conies we obtain the equation of the directrix remote from

S as follows.

Jl—- -

6' D D'



692 POLAR EQUATION OF A CONIC [20.3

and since l/a = 1 —e 2 (see Ex. 20(6), no. 1), this becomes

A 11 2 \ Zl + e 2

—rcos0 = -i- - —II = .

e\l-e 2/ el-e 2

Hence the equation is r cos 6 = -Hee 2 -l

If e > 1 (fig. 212),

rcostf = SD' = SD+DD'I 2a= -+ —e e

11 2a\s;(

1+ t)

e\ e 2 -l/ ee 2 -l'

since l/a = e 2 —1. The equation is thus the same for each case.

*(ii) Equations of the asymptotes of a hyperbola.

The asymptote whose cartesian equation is y = bx/a is inclined to Oxangle A, where tan A = b/a and hence cos A = aQ(a 2 + b 2

) = 1/e. Thereforeperpendicular from S to this asymptote makes angle Zn— sin -1 (1/e) with O

its length is6 6

Fig. 213

and so its equation is(20.13)

6 = rcos —2n + sin -1.

Since 6 = l/*J(e2 —1), this can be written

rcos | + sin -1 - | = —r~ —rr*\ e) V(e 2 -1)

Similarly j the asymptote y = —6sc/a has polar equation

6 = r cos Id —sin -1 - i e r cos id —sin -1 - 1 = -j~ —




If we subtract these equations, we obtain

I V—e cos 6 = e' cos (0 —a),r r

which is the equation of a line. Since it is satisfied by the points (r, 6) whichsatisfy the equations of both conies, it represents a common chord. Clearly it is

also satisfied by the points (r, 6) which satisfy both the equations

I V- —e cos 6 = 0, e' cos (6 —a) = 0,r r

and these represent the directrices corresponding to S, by ex. (i). This commonchord therefore passes through the meet of these directrices.

The first conic is also given by

I- = —1 + e cos 6r

(see Remark (fi) in 20.31). Proceeding similarly, another common chord is

I V—ecos#H e'cos(0 —a) = 0,r r

which also passes through the meet of the same directrices.

20.34 Chord and tangent

Let P,Q

be the points of the conic Ijr = 1+e cos 6 corresponding

to 6 = a, 6 = /?. Since by 20.14 the polar equation of any line can be

written ^.4 cos 0+ 2? sin = —

r

let the required chord have equation

r

Then ^?cosa+

gsina = 1

+ecosa

and pcosfi + qainft = 1 + ecos/?;

i.e. {p —e)cosa + 2sina = 1, (p —e)cos/#+gsin/? = 1.

Solving these,

sina —sin/? 1# m mp-e = -7— : = cosA(a + tf)secA(a-/?)r sm(a-yff) 2V r/ 2V r '



694 POLAR EQUATION OF A CONIC [20.3

6 + sec i(a - /?) cos [d - .

i.e. I- — e cosr

For an alternative method, see Ex. 20 (b), no. 7.

By letting J3 -> a we find that the tangent at the point 6 = a is

- = e cos # + cos (d —a).

For a calculus method, see Ex. 20 (b), no. 24.

Examples

(i) The tangents at the points P, Q corresponding to 6 = a,fi meet at the poT(r, 6), where 6 = %(ot+j3).

For the coordinates of T satisfy

e cos 6 + cos (0 —a) = - = e cos 6 + cos (d —fi),

r

i.e. cos {6 —a) = cos (0

from which 6 = £(a + /?).

Remark. The general solution is d —a = 2nn±(d—fi). The sign + is

admissible since P, Q are assumed to be distinct; the sign — gives

6 = nir + Ha+fi).

Since —n < a < n and —7r < /? < 7T, also —at < £(a + ^n, and so thereno loss of generality by taking n — because any point in the plane canspecified by an angular coordinate which lies between + n.

(ii) With the notation of ex. (i), ST bisects PSQ.If P, Q are not on different branches of a hyperbola (fig. 214),

PST = Ha+fi)-a = Hfi -a)

and TSQ = fi-â+p) = £(/?-a).A

Hence ST bisects PSQ (internally).

If P, Q are on different branches of a hyperbola, let Q be on the branch remofrom S. Then /? is the angle xSQ', and -|(a + /?) is angle xST', as shown in fig.

PST' = i(a+yff)-a = £(/?-a)

and T'SQ' = fi-\{a+P) = ^ - a),

Aso that ST bisects PSQ externally.




Tangents at A and B meet where 6 = \{<x + ft), by ex. (i), and hence

- = cos £(a ft) + cos \{ft —a) = 2 cos \ct cos \ft,r

i.e. at the point (£Zsec £a sec \ft, \{a +/?)). Similarly the other meets are

(£Z sec \ft sec £y , + y )) , (|Z sec \y sec £a, £(y + a))

These three points satisfy the equation

(

a + ft 4. y\^ 2

/'

which by 20.21 (6) represents a circle of radius R = \l sec \a sec £/? sec £y, centre

(R, \(ct + ft + y )), and passing through the pole S.

Fig. 214

(iv) If a variable chord of a conic subtends a constant angle at the focus, then

tangents at the extremities of the chord meet on a fixed conic, and the chord itself

touches another fixed conic.

b d d d l



696 POLAR EQUATION OF A CONIC [20.

This is the equation of the tangent at a to the conic

I cos B n= 1 + ecos/f cos0,T

whose eccentricity is e cos /? and latus rectum 21 cos {$.

The tangents at the extremities of the chord are

- = ecos0+*cos(0 — - = ecos0 + cos(0 —a —B),r r

and these meet where (by ex. (i)) = a and l/r = e cos a + cos/?. This pointon the conic

^- = e cos + cos B,r

IseoB an.e. — = 1 + e sec B cos 0,

r

which has eccentricity esec B and latus rectum 2Zsec /?.

The conies (i), (ii) have the same focus and directrix as the given one.

Exercise 20(6)

1 If a, 6 are the semi-axes of l/r = 1 + e cos 6 (e 4= 1)» prove that when e

a = Z/(l-e 2) and 6 = Z/V(l-e 2

).

Give thecorresponding results when e >

1.

2 What locus is represented by a = r sin 2 £0? Sketch this curve.

3 What is represented by kjr = a + 6 cos + c sin 0?

4 If P(r 1 , a) is an extremity of a focal chord of l/r = 1 + e cos 0» show tthe other extremity Q has coordinates of the form (r 2 , n+ a). Prove that

J_ J__2SP +

SQ ~ Vi.e. that /&aZ/ i7ie Zaiws reciwm is the harmonic mean of the segments of any fo

chord.

5 Prove that the sum of the reciprocals of the lengths of two perpendicular

focal chords of a conic is constant. Express this constant in terms of e and

6 Prove that mutually perpendicular focal chords of a rectangular hyperbolaare equal in length.

7 Obtain the equation of the chord joining the points of l/r = 1 + e c

which correspond to 6 = a,fi from the result in 20.12.

8 Find the equation of the chord of l/r = l + cos(0 —y) which joins

points 6 = a, /?. Deduce that the tangent at tx is l/r = e cos (6 —y) + cos (0 —



POLAR EQUATION OF A CONIC 697

12 A chord oiljr = 1 + e cos subtends an angle 2a at the focus. If the bisector

of this angle meets the chord at M, prove that Mlies on the conic

Zcosa , a= 1 + ecosa.cosc/.

r

13 P is any point on a conic with focus S. A line through S making a given

angle with SP meets the tangent at P in T. Prove that T lies on a conic which

has the same focus 8 and corresponding directrix as the given conic.

14 Two conies have a common focus, about which one of them rotates. Prove

that the common chord touches a conic which has the same focus as the given

conies, and whose eccentricity is the ratio of the eccentricity of the fixed conic

to that of the rotating one.

15 An ellipse of eccentricity e and a parabola intersect at the ends L, L' of

a common latus rectum, and 8 is their common focus. If a common tangent

touches them at P, Q respectively, prove that SP, SQ are each inclined to LL'at an angle sin -1 {£(l + e)}. Explain geometrically how the two cases arise.

16. P, Q are the points of the conic l/r=l + e cos 6 corresponding to 6 = ct, fi.

If the tangents at P, Q meet at T, prove that T has coordinates

1 1

H*+P)).\cos £(a -

fi) + e cos £(a fi)

'

17. If P, Q are points on a parabola with focus S, and the tangents at

P, Q meet at T, prove that SP.SQ = ST*.

18 (i) P, Q are the points = a, fi of the parabola ljr=l+ cos 0. Prove thatthe tangents at P, Q contain an angle £ |

a /?|

(ii) Tangents to a parabola with latus rectum 21 contain the constant

angle<f>.

Prove that they meet at a point on a hyperbola having the samefocus as the parabola, and give the latus rectum and eccentricity. [Use no. 16.]

* 19 Prove that the chord of contact of tangents from (r x , 6 t ) to l/r = 1 + e cos d

has equation .^ v „ >.

/ ecosfljj ecos#j = cos^ —^).

[If its extremities are given by 6 = a, ft, the chord is

cos4(a —fi) ^ —ecos#^ =cos^0 —

Use no. 16.]

20 Show that the tangents at the points 6 = a,/i of l/r = 1 +ecos# are per-

pendicular if and only if e 2 + (cos<x + cos/?) e + cos(a —ft) = 0. [Change the

equation of the tangent into cartesian coordinates.]

*21 Prove that the meet of perpendicular tangents to l/r = 1 + ecos0 lies onthe locus 21(1 jr —ecostf) = r(l —e 2

). If e 4= 1, show that this locus is a circle

concentric with the given conic (the director circle) What is the locus when



698 POLAR EQUATION OF A CONIC*23 Show that the equation of any line perpendicular to the tangent at

l/r = 1+e cos 6 has an equation of the form

kl—— e cos (6 + in) + cos (6 + \n —a).T

[See 20.14, ex.] If this line passes through the point a, prove that

k = —esina/(l + ecosa).

Deduce that the normal at a has equation

esina I . n= esin0 + sm(0 —a).1 + e cos a r

*24 Prove that the tangent at A(p, a) to the curve r =f(6) has equation

1 1 fdr\ 1-cos(0-a)- —

( —) sin(0-a) =

[If P(r, 6) is any point on the tangent at A, apply the sine rule to triangle Oand use cot $5 = drjrdd (8.14).] Deduce the equation of the tangent to

l/r — 1 + e cos at 6 — <x.

*25 Obtain similarly the equation of the normal at (p, a) to r = /(#), andit to obtain the result of no. 23.

Miscellaneous Exercise 20(c)

1 A, B are the points (r^dj), (r 2 ,6 2 ), and M(r,6) is the mid-point of

Prove that

, „ » ~ „ _ , „ ?\ sin + To sinr = i + r\ + 2r x r 2 cos X - 2 } and tan 6 = -± ^1 2 ricos^ + racosâ

2 A variable line meets n given lines through at P lf P 2 , n respectively.n

If S ( 1 jOP r ) is constant, prove that it passes through a fixed point.r=l

3 Find the equation of the common chord and of the line of centres ofcircles r = 2a cos (6 —a), r = 2b cos(d-fl).

4 Obtain the polar equation of the circle whose diameter is the join

(a, a) and (&,/?).

5 Prove that the chord of contact from (r lt dj) to the circle r —2a cos d

cos 6 cos Qr _ cos (6 — X )

r x r a

[Use the cartesian equation of the circle and chord.]



POLAR EQUATION OF A CONIC 699

8 (i) If the tangent to the parabola a = r cos 2 \d at the point P given by6 = a meets the initial line at T, prove ST = a sec 2 \a. —SP.

(ii) Find the equation of the circle SPT.

9 Find the value of a for which the tangent at a to Ijr = 1 + e cos 6 is parallel

to the initial line. Hence find the locus of the ends of the minor axis of thisellipse as e increases from to 1.

10 Prove that the line l/r = a cos 6 + b sin 6 touches the conic l/r = 1 + e cos 6

if and only if (a - e) 2 + 6 2 = 1.

1 1 Two parabolas have a common focus and their axes perpendicular. Provethat the directrix of either passes through the point of contact with the other

of their common tangent.

12 Find the locus of the foot of the perpendicular from the pole to the line

l/r = e cos 6 + cos (6 —a) when a varies. Interpret the result geometrically.

13 The perpendicular from the focus S to the directrix meets the conic at A.Prove that */ the foot of the perpendicular from S to any tangent lies on the

tangent at A, then the conic is a parabola. [Do not prove the converse result.]

14 A variable chord PQ of the ellipse Z/r = 1 + e cos 6 with focus S is parallel

to the major axis. If its extremities are the points 6 = 2a, 2fi, prove that

cos (a + /?) + ecos (a —/?) = 0.

If the internal bisector of PSQ meets PQ at K, prove that

QTr 2SP.SQ aSK = spTsQ

coaiPSQ '

and deduce that K lies on a parabola whose vertex is S.

*15 PQ is a focal chord of a conic, and its mid-point is M. The normals at

P, Q meet at N. Prove that MNis parallel to the (major) axis. [Use no. 7 (i)

and Ex. 20 (6), no. 23 to prove that r sin 6 is the same for Mand N.]



700

21

COORDINATE GEOMETRY IN SPACE:THE PLANE AND LINE

21.1 Coordinates in space

21.11 Rectangular cartesian coordinates

The method of locating points in a plane by coordinates x, y referre

to two intersecting lines Ox, Oy (15.11) can be extended to points

space. Clearly three axes of reference will be required, and for s

plicity we shall choose them to be mutually perpendicular.

Take any two intersecting perpendicular lines; let them meet a

Fig. 216 Fig. 217

On each select a positive sense, indicated by an arrow, and la

these directions Ox, Oy. Through draw a line perpendicular to b

Ox and Oy (and therefore to the plane xOy), and choose the positiv

sense along it to be the direction of advance of a right-handed scr

turned in the sense from Ox towards Oy, label this direction Oz.

Then we have constructed a right-handed system of rectangular a

Ox, Oy, Oz.f The planes xOy, yOz, zOx determined by these axes

l t ll di l d ll d th di l



21.12] THE PLANE AND LINE 701

edges through be OA, OB, 00 along Ox, Oy, Oz respectively. Thedistances OA, OB, OC can be specified in magnitude and sense bysigned numbers x, y, z, and the ordered triplet of numbers (x, y, z) are

the rectangular cartesian coordinatesof P wo

theaxes Ox, Oy, Oz. In

particular, observe that L has coordinates (0, y, z), M(x, 0, z), N(x, y, 0)

and that A is (x, 0, 0), J3(0, y, 0), C(0, 0, z).

Conversely, given three signed numbers x, y, z, the points A, B, Oon the axes can be determined, and the box then completed. Thecorresponding point P is the extremity of the diagonal through O.

It follows that to each point P in space there corresponds a unique set

of coordinates x, y, z wo the axes Ox, Oy, Oz; and to each ordered set of

numbers x, y, z corresponds a unique point P in space.The three coordinate planes divide space into eight regions called

octants; to each corresponds a distribution of + and — signs in the

coordinates (see Ex. 21 (a), no. 1), the number of such possible sign-

distributions being 2 s = 8. In fig. 217 all coordinates of P are positive.

Finally, observe that A in fig. 217 is the point in which Ox is metby the plane through P drawn parallel to the plane yOz. Hence PAis perpendicular to Ox, so that A is the orthogonal projection of P

on Ox. Similarly, B and C are the projections of P on Oy, Oz respec-tively. The coordinates of a point P are therefore the (signed) lengths

of the orthogonal projections of OP on the axes.

Notation. In the remainder of this book Px is the point whosecoordinates are (x v y v Zj), and so on.

21.12 Other coordinate systems

(1) Cylindrical polar coordinates. We may replace the coordinates (x,y) in

the plane xOy by polar coordinates (/?, <j>), where p = ON and is the angle

z

z



702 THE PLANE AND LINE [2

xON, measured positively in the sense from Ox towards Oy. P is then det

mined by the triplet of numbers (p, <j), z), called the cylindrical coordinates o

(2) Spherical polar coordinates. Further, in the plane zONP the coordinates

p and z can be replaced by polar coordinates (r, 6), where r = OP and 6 is

angle zOP, measured positively from Oz towards ON. P is then determined

(r, 6, 0), the spherical coordinates of P.

4*

Fig. 220

If the sphere in fig. 220 represents the Earth and xOy, xOz are the planes

the equator and Greenwich meridian, then 6 is the colatitude and <p is

longitude of P.

21.2 Fundamental formulae


Given two points Px and P2 , let the perpendiculars from them to

plane xOy be PXNX , P2 N2. The coordinates of

Nx are (#1,^,0), and of N2 are (x 2 ,y 2 ,0).

Regarded as points of the plane xOy, their

cartesiancoordinates wo axes Ox, Oy are

( x i> Vi)> (*2> y*)> so tn at NX N2 is given by

NX N\ = {x x -x 2 Y + {y x -y 2 )\

Through P2 draw P2 R parallel to N2 NX ;

since P2 B lies in the plane of PXNX and P2 N2 ,

it meets PXNX ,say at R. From the right-angled

triangle P x P2 B,

N,

Fig. 221




21.22 Section formulae

Given two points Pv P2 , let P{x, y, z) be the point dividing PXP2 in

the ratio k:l; then PXP\PP 2 = k\l.

Draw PX NX , P2 N2 , PN perpendicular to plane xOy. Through Pdraw PR parallel to NX N2 to meet P2 N2 at R; and through i\ drawPX Q parallel to N± N2 to meet PN at Q.

Fig. 222 Fig. 223

From thesimilar trianglesPj PQ,PP 2 R we have PQ\P 2 R = P^/PPj.(i) If the division is internal (fig. 222), it follows that

z —z, k

z 2 -z I*

from which z = ^ +^

Za.

l + k

Similarly, by dropping perpendiculars from P lf P2 , P to plane yOz,

we find that x = (Z# 1 + &a; 2 )/(Z + &); and by perpendiculars to planezOx, that y = {ly x + ky 2 )f{l + k).

Alternatively, since N divides îVg internally in the ratio k : I andhas the same x- and ^-coordinates as P, the last two results follow

from the section formulae in the plane xOy (15.14).

(ii) If the division is external (fig. 223) we have

z —z x k








sense as PQ (fig. 224). We call OA the unit ray corresponding to

(sensed) line PQ.The direction of OA, and therefore that of PQ, will be determined

by the coordinates of A. Since these coordinates are the orthogonalprojections of OA on the axes Ox, Oy, Oz, hence A is the po

(cos a, cos/?, cosy), where a = xOA, /? = yOA and y = zOA. It

customary to write

I = cosa, m = cos /?, n = cosy,

and call {I, m, n} the direction cosines of PQ (and therefore of anyparallel to and in the same sense as PQ).

Since OA has unit length, the distance formula shows that

Z2 + m2 + rc

2 =l.

Fig. 224 Fig. 225

Remarks

(a) If the line is described in the sense from Q towards P, then

corresponding unit ray through is OA', where A' is the point

(cos (77 —a), cos(7T— /?), cos (it —y)),

i.e. (— cosa, —cos /?, —cosy), i.e. ( —I, —m, —n).

(/?) A line parallel to the plane yOz has a = \tt, so that 1

Similarly, a line parallel to zOx has m = 0, and one parallel to

has n = 0.

(y) In particular, the coordinate axes described in the positive

senses Ox Oy Oz have direction cosines




The angle between the {sensed) lines PQ, BS (intersecting or not) is

denned to be the angle AOB for which < AOB ^ it (fig. 225).

Let AOB = 6; then from triangle AOB, in which OA = 1 = OB,

we have by the cosine rule that

AB2 = 2-2cos0.

Using the distance formula for AB, we also have

AB2 = (l- I')2 + (m- m'f + {n- n') 2

= Z2 + m2 + n 2 + V2 + m' 2 + n' 2 - 21V - 2mm' - 2nn'

= 2 - 2{IV + mm' + nn'),

since by (i), Z2 + m2 + n 2 = 1 and similarly I'

2 + m' 2 + n' 2 = 1 . Comparing

the two expressions for AB2, we find

cos = U'+ mm' + nn'. (ii)

In particular, the directions {l,m,n}, {l',mf,n'} are perpendicular

if and only if IV + mm' + nn' = 0. They are parallel (in the same sense)

if and only if I = V, m = m' , n = n'\ and 'antiparaUel' if and only if

I =—V, m

=—m', n

=—n'.

21.33 Direction ratios

It is often convenient to specify the direction of a line by a set of

three numbers which are proportional to the direction cosines I, m, n.

Any such three numbers are called direction ratios of the line. For

example, the line in the plane xOy which bisects the angle xOy has

a = \n, f$ = \-n, y = \n, so that its direction cosines are {1/^2, 1/^2, 0};

a convenient set of direction ratios for this bisector would be 1 : 1 : 0.

In general, suppose I : m : n = p : q : r;f then I = Xp, m = Xq,n = Xr

where A is determined from equation (i) by

\ 2(p

2 + q2 + r 2

) = 1.

Hence when the direction ratios p:q:r are given, the direction cosines

can be written down as



708 THE PLANE AND LINE [21.33

p, q, r specifying a direction are meant to be direction cosines

direction ratios. We remark that direction ratios do not distinguish

the two senses of a line.

Examples

(i) If 6 is the acute angle between two lines having direction ratios p:q:r,p' : q' : r', then by equation (ii)

a ± {pp ' + qq' + rr ')

COS U ——: : .

V(2>8 + ?

2 + r*) V(P'2 + ?' 2 + r'*)

'

where the sign is chosen so that the expression is positive.

In particular, the lines will be perpendicular if and only if

pp' + qq' + rr' =(the 'perpendicularity condition').

They will be parallel or ' antiparallel ' if and only if p : q : r = p' : q' : r'.

(ii) If P is (x,y,z), the direction cosines of OP are {x/OP,y/OP,z/OP};cos a = x/OP, etc. A convenient set of direction ratios for OP is x : y : z.

(iii) The projections of P x P a on Ox, Oy, Oz are x 2 —cc l9 y 2 —yi, z % —z x . Tha set of direction ratios for P 1 P i is x 2 —x1 :y 2 —y l :z 2

—z 1 .

(iv) The lines OP, OQ have direction ratios I: mm, V : m' : n'. Find direction

ratios for the line OB which is perpendicular to both OP and OQ.Let OR have direction ratios p : q : r. Then by the perpendicularity condition

'* lp+mq+nr = and Vp + m'q + n'r = 0.

Solving these for p : q : r (see 1 1.43 (2)), we have

p:qir = mn'—m'n:nl'—n'l:lm' —l'm.

Exercise 21(6)

1 For .4(3, 2,5),

B( - 1, 6,4),

C(5, - 3,7),

D( - 3,5, 5),

proveAB

||

CD.

2 For A{2, -4, 3), B(3, 2, - 1), C( -4, - 3, 0),D(6, -2, 4), prove AB JL C

3 Prove that the points ( - 5, 2, 7), ( - 9, 3, 6), (3, 0, 9) are collinear.

4 Prove that (2, 3, 0), (1, 5, 2), (3, 7, 1), (4, 5, —1) are the vertices of a square.

5 Find the direction ratios of the lines through O which make equal angles

with the coordinate axes.

6 Calculate the angle between any two diagonals of a cube.

7 Find the angles between the diagonals of a rectangular box whose edge




10 If M is the foot of the perpendicular from N to OP in no. 9, find thecoordinates of M.* 1 1 If the directions p± : q 1 : r lt p a : q 2 : r 2 , p 3 : q 3 : r a are parallel to the same plane,

prove that'

Pi ?i *i

p % q % r 2

Pa ?s r s

= 0.

[They are perpendicular to the normal to the plane.]

*12 Prove that the converse of no. 11 is true. [Use Theorem II of 11.43.]

*13 If Ox, Oy, Oz and Ox', Oy', Oz' are two systems5f, Z? of rectangular axes with the common origin O, ^let Ox', Oy', Oz' have direction cosines (^.w^Wj),

y(l 2 ,m 2 ,n 2 ), (l 3 ,m 3 ,n 3 ) wo S?.

(i) Prove that

andZ* + m*+n* = l\+m\ + n\ = Zf+m» + n§ = 1

i a i s +m a m3 +w 2 W3 = l 3 l 1 +m3 ml +n 3 n 1

(ii) Verify that l 2 , l 3 ), (m^ m2 , m3 ), (n ly n 2 , n 3 ) are

the direction cosines of Ox, Oy, Oz wo £f , and hencethat

+ + ll

= mf +m|+m| = n*+n§+n* = 1

Fig. 226

and m1 n 1 +m 2 n i +m 3 n 3 = n^+n^+n^ = l 1 ml + l 2 m2 + l 3 m3 = 0.

14 Rotation of axes.

(i) If P has coordinates (x, y, z) wo £7 and (x' t y', z') wo Sf , prove that

x' = l l x + m1 y+n 1 z, y' = l 2 x+m 2 y+n 2 z, z' = l 3 x+m 3 y+n z z.

[x' —projection of OP on Ox'; use no. 9.]

(ii) Also prove that

x = l 1 x' + l a y' + l 3 z', y = ml x'+m 2 y' + m3 z', z = n x x' + n 2 y' + n 3 z'.

[Symmetry; or multiply x', y', z' by l lt l %,l 3 in (i), add, and use no. 13 (ii).]

*15 Explain geometrically why x 2 +y 2 +z* = x' 2 +y' 2 +z' 2, and verify this

algebraically.

21 4 The plane




P(x, y, z)

N(pl, pm,

21.41 Equation of a plane in 'perpendicular form'

Let the perpendicular from to the given plane have foot N.ON has length p and direction cosines {I, m, n}, then N is the poin

(pl,pm, pn).

If P(x, y, z) is any point of the plane,

then PN±0N. Hence the direction

given by x—pl:y—pm:z—pn is per-

pendicular to that given by {I, m, n},

and so

(x —pi) l + (y —pm) m+ (z —pn) n = 0,

i.e. Ix + my + nz —p(l 2 + m2 + n 2 ) = p.

The equation

Ix + my + nz = p,

which is satisfied by the coordinates of any point of the plane,

consequently the equation of the plane. We observe that it is linear

in x, y, z.

Fig. 227

21.42 General linear equation Ax+By + Cz + D = Q

(1) Let Pj and P2 be two distinct points which satisfy the abov

equation; then

Ax x + By x + Cz 1 + D = 0, Ax 2 +By 2 + Cz z +D = 0.

Multiply the first of these by + h), the second by kj(l + h), and add

W6get

H^)+B

^)+c

^)+D '

'

which shows that the point dividing PX P2 in the ratio k : I also satisfies

Ax + By + Cz + D = 0. (

Since k : I is arbitrary, this means that all points of the line P^satisfy (ii).

By definition a plane is a locus in space (or surface) such that th

line joining any two points of it lies entirely on the locus; henc




This shows that the direction A : B : C is perpendicular to the direction

x 1 —x 2 :y 1—y 2 :z 1

—z 2 , i.e. to PX P^ for any line PjP 2 in the plane.

Hence the direction A : B : C is perpendicular to the plane.

(3) At least one of the coefficients in (ii) is non-zero. On dividing byit we obtain an equation containing three arbitrary coefficients. This

confirms the geometrical fact that a plane is determined by three

independent conditions.

21.43 Conditions for Ax + By + Cz + D = 0, A'x +B'y + C'z + D' = Q

to represent the same plane

The planes are parallel if and only if their normals are parallel, i.e.

A' B' C' . .

A=B =C

=k (say) '

The equation of the second plane can therefore be written

k{Ax + By + Cz) + D' = 0.

These parallel planes will be the same plane if and only if they have

a point in common, say (x , y , z ); and then

Ax + By + Cz + D = 0, k{Ax + By + Cz ) + D' = 0,

so that Jc{-D) + D' = 0, i.e. k = D'/D.

Hence the planes are identical if and only if

^1/ _ IT _ (7 _ £>'

A ~ B ~ C ~ D'

i.e. if corresponding coefficients are proportional; in general the latter

will not be equal.

21.44 Plane having normal I: mm and passing through P,

By 21.41 the plane with normal in direction l:m:n has an equation

of the form ,,J lx+my + nz = p,



712 THE PLANE AND LINE [21.

This relation holds for any point (x, y, z) of the required plane,

therefore is the equation of the plane.

Alternatively, consider the equation (iii). It is linear in x, y, z

therefore representssome plane (21.42(1)), whose normal has dire

tion l:m:n (21.42(2)). It is satisfied by {x x ,y x ,z x ). Hence it is

required equation.

21.45 Plane P,P 2 P 8

Method 1. Suppose this plane has normal I : m : n; then by 21.44

equation is , . . . ,l(x - x x ) + m(y - yd + n(z - z x ) = 0.

The points P2 , P3 must satisfy this, so

l(x 2 - x x ) + m(y 2- y x ) + n(z 2

- z x ) =

and l(x 3- x x ) + m(y s - y x ) + n(z 3

- z x ) = 0.

Ehmination of I, m, n from this homogeneous system of thr

equations givesx ~ x x y-Vx z ~ z i

x% —xx y%~ y\ z% ~ i\

x z~ x i Vz~ Vl Z Z ~ Z l

=

as the equation of the plane P1 P2 P3 .

Method 2. Let the required equation be

Ax + By + Cz+D = 0.

This is satisfied by P x , P 2 and P s , so

Ax x + By x + Cz x + Dz=0,

Axz + Byz + Czz + D = 0,

Ax 3 + By z + Cz z + D = 0.

Elimination of A, B, G, D from this homogeneous system of four equationsgives

x y z 1

x z 1



21.46J THE PLANE AND LINE 713

21.46 Intercept form: plane making intercepts a, b, c on the

coordinate axes

The required plane passes through the points (a, 0,0), (0,6,0),

(0, 0, c). Any plane through (a, 0, 0) has an equation of the form

l(x —a) + my + nz = 0.

This is satisfied by (0, b, 0), (0, 0, c) if

—al + bm = 0, —al + cn = 0.

Hence al = bm = cn,

so that l\m\n = -\\\-.a b c

and the required equation is

x y z-+ +- = i.a o c

21.47 Angle between two planes

The angle between two intersecting planes (a dihedral angle) is bydefinition the acute angle between the perpendiculars drawn in each to

the common line, and is clearly equal to that between their normals.

Given two planes

Ax+By + Cz + D = 0, A'x + B'y + C'z+D' = 0,

their normals have directions A : B : C, A' : B' : C. The angle 6 between

the planes is therefore given by (see 21.33, ex. (i))

±(AA' + BB' + CC)cose/ =

V (A* + B 2 + C2

) y/(A' 2

+ B'* + C2

)

In particular, the planes are perpendicular if and only if

AA' + BB' + CC = 0.

They are parallel if and only if their normals are parallel, i.e.

A:B:C = A':B':C'.




4 Obtain the equation of the plane through (2, 2, - 1), (7, 0, 6), (3, 4, 2).

5 Find the equations of the two planes through (0, 4, —3) and (6, —4,other than the plane through O, which make intercepts on the axes whose sis zero.

6 If ^1(3, 0, 0), B(2, 3, 0), 0(1, 1, 1), find (i) the angle between planes OOAB; (ii) the angle between line BC and plane OAB.

7 At three points O, X, Y on a horizontal plane, where OX = c = OYXOY = \tt, vertical shafts are sunk to meet a seam of coal at depths p,respectively. Assuming that the coal face is plane, find the angle betweenplane and the horizontal, and the distances from O of the points whereplane meets OX and OY.

8 If A(a, 0,0), B(0,6,0), (7(0, 0,c), P(a,b,c), prove that the angle betweeOP and plane ABC is sin 1 {Bq/p), where p 2 = a 2 + 6 2 + c 2 andg~ 2 = a~ 2 + &~ 2 +Also prove that the distance between the feet of the perpendiculars from

P to ABO is <J{p* - 9g 2), and find the angle between planes PBO, PGA.

9 Find the equation of the plane bisecting P X P Z a * right-angles.

*10 Show that the two detenriinantal equations in 21.45 are equivalent.

1 1 Prove that the plane through P 1 and P 2 which is parallel to the direction

p : q : r has equation

Xo X-i

y-yx z —z 1

= 0.

12 Prove that the plane through P 1 whichis

parallel to each of the directionsp : q : r, p' : q' : r' has equation

x-x x y-y x z-z x

p q r

p' q' r'

= 0.

13 Sides of a plane. Find the value of k : I if the point dividing P-lP 2 in

ratio k : I lies in the plane Ax + By + Cz + D = 0. Deduce that P x , P a are onsame or opposite sides of this plane according as Ax 1 + By 1 + Cz 1 +Ax 2 + By z + Cz 2 + D have the same or opposite signs.

21.5 The line

A straight line in space can be determined by (a) the meet of t

planes ; (6) the join of two points ; (c) a single point together with a

of direction ratios.

We begin most conveniently with case (c).




This can be expressed in the formf

x-x 1 ^ y-y 1 = z-z 1 ^.

I m n

These equations are satisfied by any point of the line, and are therefore

the equations of the line.

Notice that two equations are necessary to determine a line in space:

a single linear equation represents a plane. The two equations (i) are

in fact the equations of two planes, each of which contains the line;

G ' g x-x x = y-y x x-x x = z-z x

I m ' I n

represent two such planes.

We refer to (i) as the symmetrical equations of a line.

Example

Equation of the line P X P 2 (case (6) above).

Direction ratios of the line P XP % are x x —x % : y x—

1/ 2 : z x —z 2 . Hence the line

has equations

x-x x y —y x z-z x

2/1 —2/2

21.52 Parametric equations of a line

Putting each of the ratios in (i) above equal to A, we have

x = x x + M, y = y 1 + \m, z = z x + Xn. (ii)

These are the parametric equations of the line through Px in direction

l:m:n, A being the parameter (cf. 15.26$). They are particularly

useful when dealing with intersections of the line with a line, plane,

or other surface.

t If the line is parallel to a coordinate plane, say to yOz, then w = 0. The equation

x—x 1 :y—y 1 :z—z 1 = l:m:0 (i)'




Examples

(i) Find the length of the perpendicular from A(3, 7, —2) to the line

x-10

_ y+8 _z

+5

3 ~ ~~~T'and find the coordinates of its foot N.

The parametric equations of the line, obtained by putting the ratios in

equations equal to A, are

x = 3A+10, y = -4A-8, z = A-5.

Calling this general point P, the direction ratios of AP are

3A + 7:-4A-15:A-3.

APwill

beperpendicular to the given line (whose direction ratios

3:-4: 1) if and only if

3(3A + 7)-4(-4A-15) + l(A-3) = 0,

from which A = —3. Hence the foot N of the perpendicular from A to the

is obtained by putting A = —3 in the coordinates of the general point P;

gives N(l, 4,-8).It now follows that

AN2 = 2 2 + 3 2 + 6 2 = 49, .\ AN = 7.

(ii) Find the image of the point P(3, - 5, 2) in the plane 2x + ly - 3a = 27.

By definition, the image P' of P in the plane is such that PP' is perpendicular

to the plane and is bisected by it.

Since the normal to the plane has direction ratios 2 : 7 : —3, the equation

the normal through P is6 x-Z _ y + 5 _ z-2~2~~

7 ~ -3'

or parametrically „» « oi.oJ # = 2A + 3, y = 7A-5, z = - 3A + 2.

This point will be P', the image of P, provided that the mid-point of Plies on the plane, i.e.

2(A + 3) + 7(|A-5)-3(-|A + 2) = 27,

from which A = 2. Hence P' is the point (7, 9, - 4).

(iii) Find the image of the line

x-2 _ y- 1 _ z + 3__ __5

in the plane x —2y + 3z = 5.

We find (a) the point Q where the line and plane meet; (6) the image P'h f




(6) The method of ex. (ii) shows that the image of P in the given plane is

P'(4, -3, 3).

(c) The line P'Q has direction ratios 1 : 6 : - 1, so its equations can be written

x-5

_

y-S _ g-2

__ ____

_

(iv) Condition for two linen to be coplanar.

Suppose that the lines

x —a y —b z —c x —a' y —b'_z —c'

I m n ' V m' n'

are coplanar. Then either they are parallel, or they intersect.

If they are parallel, then I : m : n = V : m' : n'.

If the lines intersect, then the common point has coordinates

(a + Al, b + Am, c + An) for some A,

and also (a' + pi', b' + pm', c' + fin') for some fi.

Hence there exist values of A and p for which

a-a' + Al-/il' = (U

6 —6' + Am pm' = 0, r (iii)

c —c'+An—pn' = 0J

Elimination of A,

p(11.43, Corollary I (6)) gives the condition

a —a' I I'

b —b' m m' =0.

c —c' n n'

The condition is also satisfied when the lines are parallel, for in this case the

second and third columns of the deterrninant are proportional.

This necessary condition for two lines to be coplanar is also sufficient. Forwhen it is satisfied, then by 11.43, Theorem II, there exist numbers a, /?, ynot all zero such that

C£ (a-a')+ fil+yV = 0,

a(b-b')+fim + ym' = 0,

and a(c c') +fin+ yn' — 0.

If a = 0, then 1 1 m : n = V : m' : n', so that the lines are parallel, and hencecoplanar.

If a 4= 0, the above equations can be written in the form (iii), which showthat, for some pair of values of A and p, the point (a + Al, b + Am, c + An) of the

first line coincides with the point (a' + pi', V + pm', c' + pn') of the second. Thed




which intersect, we require to express the equations of their commlinef in the symmetrical form.

The planes will not be parallel if and only if not all of

be' —b'c, ca' —c'a, ab' —a'b

are zero. When this is so, let the common line have direction ratio

l:m:n. Then since this line lies in each plane, it is perpendicular

the normal to each plane, i.e. to the directions a\b:c,a' :b' :c' '. Hen

al + bm + cn — 0, a'l + b'm + c'n = 0,

from which (11.43(2))

I m nbe' —b'c ca' —c'a ab' —a'b

'

We now require the coordinates of one point on the line of inte

section. This can be chosen in infinitely many ways; e.g. we may f

the point where the line meets any one of the coordinate planes.

Example

Find symmetrical equations for the line of intersection of the planes

2x-3y+z = 6, 5x + 4y-3z = l.

If the required line has direction ratios I : m : n, then

2l-3m + n = 0, 5Z + 4m-3n = 0,

and hence l:m:n = 5: 11 :23.

Let us find the point where the line cuts the plane x — 0. The y- andcoordinates are given by _ , . „ _

hence y = —5, z = —9, and a point on the line is (0, —5, —9). Symmetrical

equations of the line are therefore

x y+5 z+95

=~TT

=_23

_*

21.54 Distance of a point from a plane

Given a point Px and a plane ax + by + cz + d = 0, let N be the f

of the perpendicular from P1 to this plane. We require to find

P




or parametrically

x —x x + Aa, y = y x + A6, z = z x + Ac.

These expressions will be the coordinates of N if they satisfy the

equation of the plane, i.e. if

a{x x + Aa) + b(y x + A6) + c{z x + Ac) + d = 0,

from which A = _ ^+^+2Y d-

With this value of A we have, by the distance formula,

PXN2 = (Aa) 2 + (A6) 2 + (Ac) 2 = A2 (a 2 + 6 2 + c 2)

Hence PXN = ±

a 2 + 6 2 + c 2

ax x + by x + cz x + d

V(a 2 + 6 2 + c 2)

9

where the sign is chosen so that the expression is positive.

Example

Planes bisecting the angles between two given planes.

JfP(x, y, z) is any point on a plane which bisects an angle between the planes

ax+by + cz + d = 0, a'x + b'y + c'z + d' = 0,

then the perpendiculars PN, PN' to these two planes are equal. Hence

ax + by + cz+d _ a'x + b'y + c'z+d'

ât + tf + c*)~ ±

V(«'2 + &' a + c /2

)

'

and these are the equations of the two bisector planes.

21.55 Areas and volumes

(1) Area of triangle P X P 2 P 3 . Let a, ft, y be the angles between the plane

P X P 2 P 8 and the coordinate planes yOz, zOx, xOy. Then cos a, cob ft, cosy are

the direction cosines of the normal to this plane, and hence

cos 2 a + cos 2ft cos 2 y = 1 . (iv)




(0, y 3 , z 3 ) of space. Regarded as points of the plane yOz referred to axes Oy,they have coordinates (y^Zj), (y 2 ,z 2 ), iy 3 ,z 3 ). Hence by 15.16(2),

2/i «i 1

Ax = ± iy

2 z 2 1

2/s z 8 1

Similarly A a = + x 2 z 2

^3 2 3

A3 = ±

2/i

2/s

On substituting these expressions in (v), we obtain a formula for A in termof the coordinates of the vertices of the triangle.

(2) Volume of tetrahedron P P 1 P 2 P 3 . The volume is given by

i x (area of one face) x (perpendicular to that face from the remaining vertex).

If V is the required volume, A the area of triangle P X P 2 P 3 , and h the perpendicular from P to the plane P]P 8 Pg, then

V - \bJi.

The plane P X P %P 3 has equation (see 21.45, Method 2)

x y z 1

2/i 1= 0.

»i

y 2z

2i

XZ 2/3 2 3 1

If we expand this determinant by the first row, we obtain (11.7)

2/1 «1 1 x t *1 1 2/1 1 x x 2/1 2 1

2/2 1 -y x 2 1 + 2 2/2 1 x 2 2/2 2 2 = 0.

2/3 z 3 1 x 3 23 1 ^3 2/3 1 x 3 2/3 2 3

The coefficients of x, y, z are numerically equal to the expressions 2A X , 22A 3 in (1).

The perpendicular distance h of P from the plane P 1 P 2 P 8 is obtained (21.54) by writing x

, y , z Q for x, y, z in the left-hand side of this equation, athen dividing by the square root of the sum of the squares of the coefficients

of x, y, 2:

h= + 2/i «i

Xi 2/2 2 2

»3 2/8 2 3

Using (vi) and (v) we now have

Xa 2/0

-r2V(A» + A» + A«).




Example*

With the notation of Ex. 21 (b), no. 13, let I, J, K be the points which have

coordinates (1, 0, 0), (0, 1, 0), (0, 0, 1) wo Ox'y'z'. Then triangle OJK has area J,

and tetrahedron OIJK has volume \.

Referred to axes Oxyz, the points are I(l v m^rtj), J(J 2 ,m 2 ,n a ), K{l 9 ,m 3 ,n a ),

and so

volume OIJK - ±\

1

h ml 1

h ra 2 n 2 1

h ma n a 1

Hence = ±1.

Exercise 21(d)

1 Find the point where the line

x— 1 _ y + 3 _ z —2 ~ 5

~

cuts the plane Zx —2y+z = 8. Also find the distance between (1, —3, 4) andthis point.

2 Find the points where the line through (a, b, c) in direction l:m:n cuts

the coordinate planes.

3 O is the centroid of the triangle whose vertices are the points where the

plane Ix+my+nz = p cuts the coordinate axes, and ia +m a + n a = 1. The

perpendicular at O to this plane meets the coordinate planes at A, B, C. Prove

1 1 13QA + QB + QG~V

4 Find the perpendicular distance of (5, 4, 1) from the line

x-6 _ y+15 _ 2-145 ~ 1 ~ 8

5 Obtain the equations of the line through (1, 2, 3) which meets the line

a; + l _ y-2 _ g + 4~~

2 1 ~ ~Y~

and is parallel to the plane x + 5y+z = 7. Give the point of intersection.



722 THE PLANE AND LINE

8 Obtain the equation of the plane which contains the parallel lines

x—1 y+2 _z— 4 x —4 y —3

2=

-3 4 ' 2=

-3

9 Prove that the lines

a-2 _ y-3 _ z + 4 a;-3_2/+l__________ ______

are coplanar. Find their common point, and the equation of the planetaining them.

10 Obtain the condition in ex. (iv) of 21.52 by using Ex. 21 (c), no. 12.

11 Prove that the line 3x + 2y+z = 4, x + y —2z = 1 is perpendicular to

line 2x-y-z = 16, Ix+lOy-Sz = 2.

12 Find the equation of the plane through O which is perpendicular to

common line of the planes x + y + z = 2, 3x —y + 2z = 1.

13 Q is the point (f , f , §), and a line I through Q has direction ratios —4:1:The line OQ meets the line V at P, where V has equations 3% —3y + 4x + 2y + 2z — 12. Prove that OPQ is perpendicular to both I and V.

If R is a point on V and JS is a point on I such that PR = QS, proveOR2 —OS 2 is constant, and state its value.

14 A plane makes angle 60° with the line x = y = z and 45° with thex = = y —z. Find the angle which it makes with the plane x = 0.

15 Find the equations of the two lines through O which meet the line

at angles of 60°.

16 Find the equations of the lines which meet the line

x + 5 _ y + 6 _ z + 7

2 ~ 2 Fat 60° and which lie in the plane x + y + 2z+l = 0.

17 Calculate the distance of the point (3, —1, 2) from the plane

5x - 6y -30z = 23.

18 Find the equations of the planes which bisect the angles betweenplanes 2x-3y + 6z = 1, 4x + 3y~12z + 2 = 0.

*19 Find the area of the triangle formed by the lines in which the plaIx + my + nz = p cuts the coordinate planes (Imn 4= 0).

2+14

'

g-1~^2



21.6J THE PLANE AND LINE 723

21.6 Planes in space

21.61 Planes through a common line

Consider the equation a + fox' = 0, where k is constant and

a = ax + by + cz + d, <x' = a'x + b'y + c'z + a .

It is linear in x, y, z, and therefore represents some plane. It is satisfied

by the coordinates of any point P which satisfies both a = and

a' = 0, i.e. by any point on the common fine of these planes.

Hence a + ka' = represents a plane through the line of intersection

of the planes a — 0, a' = 0, provided that such a line exists. If a = 0,

a' = are parallel, then clearly a + fox' = is a plane parallel to both.Conversely, every plane through the common line of a = 0, a' =

can be represented by an equation a + fox' = for some value of k

(except the plane a' = itself f). For if P1 is a point (not on the com-

mon line) of such a plane, then <x x + ka' x = gives a unique value of k,

viz. —cl x \ol' x , and the required equation is aa^ = a'a vA case of a + fox' — arose in 21.54, example.

Examples(i) Find the equations of the orthogonal projection of the line

x—1 y+l z —3

2=

-1 ~ ~~4~

on the plane x + 2y+z = 8.

Two planes through the given line are

x—1 y+l , x—1 z —3— and —2-1 2 4

i.e. a;-|-2y+l = and2x-z+l

= 0.

Hence any plane through the line has an equation of the form

(2x-z+l) + k(x + 2y+l) = 0,

i.e. (2 + k)x+2ky-z + (l + k) = 0.

The orthogonal projection of the line is the meet of the plane x + 2y + z = 8

with the plane through the line and perpendicular to this given plane. Wetherefore choose k so that the directions 1:2:1 and 2 + k : 2k : —1 are per-

pendicular:(2 + ^ + 4^-1 = 0,




(ii) If the direction ratios of a vertical line are 1 : : —2, find direction ratio

of a line of greatest slope of the plane 2x —y —3z = 4.

Let the vertical through the origin cut the given plane at P, and let

be the line of greatest slope through P. Then the plane OPQ must be p

pendicular to the given plane.The equations of OP aren x y z

1= 0~~^2'

which are equivalent to y — 0, 2x + z = 0.

Hence any plane through OP has an equation of the form

2x+z + ky = 0.

This will be the plane OPQ if k is chosen so that the direction 2 : k : 1 is p

pendicular to 2 : —1 : —3 (the normal to the given plane)

4-fc-3 = 0, i.e. k = 1.

Hence the plane OPQ is 2x + y + z = 0, and this equation together w

2x —y —3« = 4 determines the line of greatest slope through P. Its direction

I '.mm, where n , , „ , OJ A2Z + m+ n = and 2l —m—3n = 0;

hence Z : m : n = 1 : —4 : 2.

(hi) Show that the planes Zx + y + Zz = 4, a; —2/-5« = 2, and 6x + 4y+ 15z

pass through a common line, and find direction ratios for this line.

Any plane through the line of intersection of the first two planes has

equation of the form

3x+y + Sz-4: + k(x-y-5z-2) = 0,

i.e. (k + 3)x + (l-k)y + (3-5k)z = 2k + 4.

This will represent the same plane as 6x + 4y + 15z = 7 if and only if (21.43)

k + S _ 1-k _ 3- 5k _ 2k + 4~6~~ 4 ~ 15 7

The first of these equations gives k = —f ; and this value is found to satisfy

the second and third. Hence the three given planes possess a common line, whodirection ratios l:m:n are given by any two of the equations

3Z+m + 3n = 0, l-m-5n — 0, 6Z + 4m+ 15n = 0.

Thus l:m:n = -2:18:-4 = l:-9:2.



21.62] THE PLANE AND LINE

If (x,y,z) satisfies all three equations, then as in 11.41 we have

Ax = A(1>, Ay = A< 2

>, Az = A< 3>,

where

725

(i)

A =

A(2 > =

a 2

a a

«i

a 2

a.

—d„

do Co

A< x >

A< 3) =

-d x &i

—d % 62 c 2

—d a 63

a x b x

a 2 6 2—d z

a 3 6 3 —d z

If A =|= 0, the equations possess a unique solution (x, y, z), i.e. the planescc L , a 2 , a 8 fcaue a imigtte common point.

If A = 0, then by Theorem II of 11.43 there exist numbers I, m, n not all

zero such that 7 ,c^f + o 1 m+ c 1 w = 0,

a 2 Z + &2 ra + c 2 n = 0,

a s l + b a m+ c 3 n = 0.

Hence the line with direction ratios I: mm is perpendicular to each of thedirections a x : b x : c x , o 2 : 2 : c 2 , a 3 : 6 S : c 3 . Since these are the directions of thenormals to <x x , a 2 , a 8 , these planes must all be parallel to some line.

Conversely, if the planes are all parallel to a line having direction ratios

l'.m:n, then the above three homogeneous equations in three unknownsI, m, n hold, and so

A=

by11.43, Theorem I.

Hence a lf a 2 , a 3 are parallel to some line if and only if A = 0. There are variouspossibilities; to illustrate them, let a plane perpendicular to this line cut

a 2 , a s in lines Ax , A2 , A3 .

(i) All of a l9 a 2 , a 3 are parallel (fig. 228). This is so if and only if

a 1 :b 1 xc 1 = a 2 : &2 : c 2 = a 3 : 3 : c 3 .

(ii) Two of a x , a 2 , a 3 are parallel, and the third intersects them (fig. 229). Thishappens if and only if just two of the ratio-sets a x : b x : c lt a 2 : &2 : c 2 , a 3 : 3 : c 3 areequal.

Fig. 228

In both of these cases we certainly have A = since at least two columns

.




Since (iii) (a) and (iii) (6) are mutually exclusive, (iii) (a) occurs when andwhen A = but at least one of A(1)

, A(2), A(3) is non-zero; and the planes hav

common line (case (iii) (&)) if and only if A = A(1) = A(2) = A(3) = 0.

The algebra in 11.42(1) corresponds to the case (iii) (6), but includes

possibility that two of the planes may coincide; that of 11.42 (2) correspondsthe possibility of all the planes coinciding.

Fig. 230 Fig. 231

Exercise 21 (e)

1 P is the point ( —2, —3, —2), and I is the line

x-2 _ y-1 _ 2 + 3~~ 3~ ~ ~T~ ~ ~2~*

Find the equation of the plane determined by P and Z.

2 Find the equation of the plane through the line

x— 1 _ y _ 2+2

and parallel to the line 2x + 5y + 3z = 8, x —y —5z = 3.

3 Find the plane throughy —Z 2 —5

~3~=

and perpendicular to the plane 2x + ly —Sz = 4.

4 Find direction ratios for the projection of

x— 1 _ y _ z + 2

on the plane 2x + y —3z = 9.

5 Find the equations of the projection of the line

Bx-y + 2z=l, x + 2y-z = 2

on the plane 3x + 2y+z = 0. Verify that ( —1, 1, 1) lies on the projection,

obtain its equations in symmetrical form.

6 Taking Oz to be vertical, find direction ratios for a line of greatest slope

of the plane through (0, 0, 0), (3, 5, - 2), (4, 1, 1).




9 Find the equations of the three planes which pass through the line of

intersection of two of the planes

x+y+z + 3 = 0, 2x + y + 2z + 5 = 0, x + Sy + 2z + 6 =and are perpendicular to the third.

Prove that the planes so obtained have a common line, and that the plane

through perpendicular to this line is Ix + 5y —2z = 0.

10 Show that the planes x + 2y —z = 0, Zx —±y + z = 3, and <ix + 3y —2z = 24

form a triangular prism, and that the lengths of the sides of a normal cross-

section are in the ratio ^13 : ^/58 : 5^3. [Use the sine rule.]

21.7 Skew lines

21.71 Geometrical introduction

Two lines in space may (i) intersect, (ii) be parallel, or (iii) neitherintersect nor be parallel. In cases (i) and (ii) the lines are coplanar;

in (iii) the lines are said to be skew.

Fig. 232

Two skew lines have infinitely many common transversals; for any

point on one can be joined to any point on the other. Cf. the example

below. We now prove that there is exactly one transversal which cuts

both lines at right-angles, and that the distance between the inter-

sections is the shortest distance between the two lines.

Let AB, CD be the given skew lines. Through any point X on ABdraw XY parallel to CD. Then the plane AXY contains AB, and is

parallel to CD; for if CD met plane AXY, say at E, we could draw

a line through E parallel to XY, i.e. parallel to CD; and this is im-




We prove that PQ is the unique common perpendicular of AB aCD. For since PQ is perpendicular to plane AXY, hence PQperpendicular to AB and to PF. Also, since CD

\\PF, PQ is p

pendicular to CD. The uniqueness follows from the steps ofconstruction for PQ.

If H, K are any two points of AB, CD respectively, then HK > PTo prove this, let the perpendicular from K to plane AXY hafoot N; then N lies on PF, and KN±NH, so that HK > KN (

hypotenuse being the longest side of triangle KNH). Since bo

PQ and NK are perpendicular to PF and coplanar, hence PQ \\NK

and since QK \\PN, PQKN is a rectangle, so that PQ = KN. Th

HK > PQ.It follows that PQ is the shortest distance between the two sk

lines. It is equal to the perpendicular distance from any point KCD to the plane AX Y.

Example

Prove that through a given point there is a unique common transversal ofskew lines I, V.

Let the given point be P, not on either line I, V . If a line through P meetsit must lie in the plane (P, I). If a line through P meets V, it must lie in the pla(P, V). Hence a line through P meeting both I and V would have to lie in bthese planes. Now these planes certainly intersect, because P lies in bohence they have a unique common line which passes through P, and this

the common transversal of I, V through P.

21.72 Length of the common perpendicular

Let the lines AB, CD in 21.71 have equations

x —a y —b z —c x —a' y —b' z —c'

I m n * V m' n'

Then the plane AXY, which is parallel to CD and contains AB, has

equation of the form

p{x-a) + q(y-b) + r{z-c) = 0,

where pi + qm + rn =

and pi' + qm' + rn' =




as the equation of plane AXY. Its normal therefore has direction

ra * 10Smn' —m'n : nl' —n'l : lm' —I'm.

The length of PQ is equal to the perpendicular distance of any pointK of CD from this plane. Since (a', b', c') is a point on CD, we have

PQ =

a' —a b'-b c'-c

± I m n

V m' n'

J{{mn' - m'n) 2 + (nl' - n'l) 2 + (lm' - I'm) 2 }

where the sign is to be chosen so that the result is positive.

21.73 Equations of the common perpendicular

The line PQ is the line of intersection of the planes QAB, PCD.The plane QAB has an equation of the form

f{x-a) + g(y-b) + h(z-c) = 0,

where fl+gm + hn =

and f{mn' - m'n) + g(nl' - n'l) + h(lm' —I'm) = 0,

since its normal is perpendicular to both AB and PQ. Ehmination of

/, g, h givesx —a y —b z —c

I m n =

mn' —m'n nl' —n'l lm' —I'm

as the equation of QAB. Similarly, plane PCD has equation

x—a' y—b' z—c'

V m' n' =0.

mn' —m'n nl' —n'l lm' —I'm

These are the two equations of the line PQ.




Since P lies on AB, its coordinates are of the form

(a + XI, b + Xm, c + Xn).

Similarly, Q on CD has coordinates (a' +/il',b' +/im',c' +/in'Direction ratios for PQ are therefore

a —a' + XI —/il' : b —b' + Xm—/im' :c —c' + Xn—/in'.

As PQ is perpendicular to AB (whose direction ratios are l:m:n)to CD (l':m':n') } hence

l(a - a') + m(b - b') + n(c - c')

+ X(l 2 + m2 + n 2) - /i(W + mm' + nn') =

and l'( a -a') + m'{b-V) + n'{c-c')

+X(ll' + mm' + nn')-/i(l' 2 + m' 2 + n' 2) = 0.

From these A, /i can be calculated, and then the coordinates o

and Q found.

Example

Find the shortest distance between the lines

x _ y _ z x —2__y—\_z + 2

2~^3~1 ~~3~ ~~2~'

and find equations for the line along which it lies.

Any point P on the first line is (2A, —3A, A), and any point Q on the second(2 + Z/i, 1 —5/i, —2 + 2/jl). Hence PQ has direction ratios

2A-3/t-2:-3A + 5/t-l:A-2/t + 2.

If is perpendicular to the first line,

2(2A-3/t-2)-3(-3A + 5/t-l) + (A-2/t + 2) = 0,

i.e. 14A-23/t+l = 0.

If PQ is perpendicular to the second line,

3(2A-3/t-2)-5(-3A + 5/t-l) + 2(A-2/t+2) = 0,

i.e. 23A- 38/« + 3 = 0.

On solving we find A = /i = Hence P is —31, ^), andB




21.75 Standard form for the equations of two skew lines

Calculations for proving properties of skew lines can be simplified

by a suitable choice of coordinate axes as follows.If AB, CD are the skew lines, let their common perpendicular PQ

be chosen for z-axis. To introduce as much symmetry as possible,

choose at the mid-point of PQ, and through O draw lines OB', OD'parallel to AB, CD respectively. For Ox, Oy choose (again with a

view to symmetry) the bisectors of the angles between OB', OD', andlabel them in such a way that Oxyz is a right-handed system.

Fig. 233

From the construction, the plane xOy is parallel to both AB andCD, and hence is perpendicular to PQ, i.e. to Oz; also xOy is a right-

angle. The coordinate axes are thus mutually perpendicular.

Let angle B'OD' be 20; by definition this is the angle between ABand CD (21.32). Then OB' makes angles \ir-0, 0, \tt with Ox, Oy, Oz,

and therefore has direction cosines {sin#, cos#, 0}; these are the

direction cosines of the parallel line AB. Also OD' makes angles

\n + 0, 0, \tt with the axes, and its direction cosines are

{— sin#, cos#, 0},

which are also those of CD.If PQ = 2c, then P is (0, 0, -c) and Q is (0, 0, c). The equations of

AB CD h f




Example

A line is drawn to meet the lines (i) and to make a constant angle with Oz. P

that the locus of the mid-point of the intercept is an ellipse.

The point P on the first line having ^-coordinate A has y = mX, z =and so is P(A, mA, —c). Similarly, a point Q on the second line is Q{/i, —m/i

If M(x, y, z) is the mid-point of PQ, then

x = J(A+/t), y - £m( A -//,), z = 0.

Direction ratios for PQ are A - (i : m( A +/*):- 2c. If PQ makes angle aOz, which has direction cosines {0, 0, 1}, then

-2cCOS QL — — - - #

V{(A-/*) 2 + m2 (A+/t) 2 + 4c 2}

On squaring and rearranging, and vising the above equations toelimina

A and/*, vI _£

J+ m2 (2a;) 2 + 4c a = 4c 2 sec 2 a,

k 2 /m 2fc

2 m2

where k = c tan a. The fact that the z-coordinate of Mis zero shows that Min the plane xOy ; and the above equation in (x, y) represents an ellipse in

plane.

Exercise 21(/)1 Obtain direction ratios of the line through O which meets each of the

x-l _ y-2 _ z-3 x + 2 _ y-3 _ z-4 ~ 2~~~~ 3~ __ 4~'

1 ~ 3 ~ 2'

[Method of 21.71, ex.]

2 Find the equations of the line through ( —6, —4, —6) which meetsof the lines ,

-= =- x 2 = V ~ = z 12^3' X2

Z

Also find the points of intersection.

3 Prove that there are two lines which meet Ox at right-angles andmeet the lines _ , „ , , . ~ n

y —1 z— 13 a;+l_^ + 2_z —a ._ 4== __ ___________

and find the points where they meet Ox. [Use the parametric equations of

lines.]

4 Prove geometrically that three given skew lines have infinitely mcommon transversals, no two of which intersect.



THE PLANE AND LINE 733

x + 1 y-5 z-4 , a; + 4 y 2-19 , , .

7 ——= ——= ——and ——= - = — [use the formula in 21.721—o 6 1 43 —2

and find equations of the line along which it lies.

8 A fixed line APB passes through A{2, 2, 2) and B( - 1, - 1, - 3), and avariable line CQD passes through (7(2, 3, 1) in such a way that the commonperpendicular PQ has direction ratios : 5 : —3. Find the equations of the locusof Q and the length of PQ.

9 From the result of 21.72 deduce the condition for the lines

x —a_y —b_z —c x —a' _ y —b' z —c'

I m n ' V m' n'

to intersect.

Two skew lines I, I' are met by their common perpendicular at

Aon I and

Bon I';

P, Q are variable points on I, V respectively, and M is the mid-point of PQ(new. 10-13).

10 If AQ 1_ BP and I, V are not perpendicular, prove that Mlies on a hyper-bola whose asymptotes are parallel to I, V. [Choose axes as in 21.75.]

11 If AP 2 + BQ 2 is constant, prove that the locus of Mis an ellipse.

12 If PQ = AP —BQ and 2a is the angle between I, V , prove that (with axesas in 21.75) PQ makes an angle \tt —a with Ox, and that the meet of withthe plane x = lies on the circle whose diameter is AB.

13 Prove that AP.BQ is constant when (a) the lines AQ, BP are perpen-dicular, or (6) the planes APQ, BQP are perpendicular.

14 Prove that the coordinates of any point equidistant from the lines

y—mx = = z + c, y + mx = = z —c satisfy the equation mxy = (l+m 2 )cz.

Miscellaneous Exercise 21 (g)

1 If AB 2 + CD 2 = AC2 + BD\ prove BC _|_ AD.2 Two edges AB, CD of the tetrahedron ABCD are perpendicular. Prove

that the distance between the mid-points of AC, BD is equal to the distance

between the mid-points of AD, BC.If also the edges AC, BD are perpendicular, prove that the third pair BC,

AD are perpendicular.

3 Find the ratio in which N, the foot of the perpendicular from O, dividesthe join of A(4t, 6, 0) and 5(1, 2, —1); and prove that N does not lie betweenA and B.

4 P moves with uniform speed in a straight line from A(0, 0, 12) to B(3, 4, 0)in 13 seconds. Find the coordinates of P after it has been moving for t seconds.For what value of t is P nearest to 01

C CP



734 THE PLANE AND LINE

6 Find the equations of the line through which intersects and is

pendicular to the line x + 2y + 3z + 4 = 0, 2x + Zy + 4z + 5 = 0, and find

point of intersection.

7 Find the angle between the line 6x + Ay —5z = 4, x —5y + 2z = 12

2 -1 1'

Prove that these lines intersect, and obtain the equation of their plane.

8 Prove that any point equidistant from the lines through O which h

direction cosines {I, m, n}, {V, m',n'} lies in one of the planes

(l + l')x + (m + m')y + (n + n')z = 0, (l-l')x + (m-m')y + (n-n')z = 0

Hence find the equations of the lines which bisect the angles betweengiven lines.

9 Find the coordinates of the point P on the intersection of the plan

2x + 2y —z = 0, x + y —3z = which has least distance from the line I joini

the points (-2,3, 2), ( —5, 5, —3). Find the point Q on I which is nearest to

10 The plane 4x + ly + 4z + 81 = is rotated through a right-angle about

line of intersection with the plane 5x + Zy + lOz = 25. Find the equation of

plane in its new position. Also find the distance between the feet of the

pendiculars from O to the plane in its two positions.

1 1 Obtain the equation of the plane through the origin and the line

x-1 _ y-2 _2-3

~2~ ~1 -2

'

Find the equations of the line meeting Ox at right-angles whose projection

this plane coincides with the given line.

12 Write down the equation of the plane n AB which bisects the line ABright-angles, where A is (a,b,c) and B is (d,e,f). Prove that, for any poi

A, B, C, the planes n AB , n BC ,tt ca have in general a common l ine A. If .4(1, 0

B(0, 1, 1), C(2, 2, 0), show that the distance of from A is -&V 22 -

13 Prove that the planes vy —/iz = p, ?te —vx = q, fix-Ay = r possess

common line if and only if Xp +/iq + vr = 0, and show that this line lies inplane px + qy + rz = 0.

14 A line meets the planes a = 0, of = at P, P', and meets the pla

a ± Aa' = at Q, Q'. Prove that Q, Of divide PP' internally and externally

the same ratio.

1 5 Prove that the two lines through (a, a, a) which lie in the plane x + y + z

and are inclined at 30° to the plane x = have equations

x —a y —ct z —a x —a_ y —a _ z —a~~2~ =

T+V6=

1 V5' ~2~ ~ 1 V5 1 + V5'



THE PLANE AND LINE 735

17 Verify that the line x = p, cy —mpz cuts each of the three lines

y = = z, y —mx = = z —c, y+mx = Q = z + c.

As p varies, prove that all such lines lie on the surface whose equation is

cy —mxz.18 If r is any transversal of the three lines

y—mx = = z —c, y = = z, y + mx = = z + c,

and a is any transversal of the three lines

y —mz = = x—c, y = = x, y + mz = = x+c,

prove that r and s intersect.

19 Points P, P' are taken on the lines y = x,z = \; y ——x, z ——1 respec-

tively, such that OP = 30P'. Prove that the meet of PP' with the plane z =lies on the curve 2a; 2 —5xy + 2y a + 1 = 0, z = 0.

20 The common perpendicular to two skew lines y—mx = = z + c,

y + mx = = z —c meets them at .4, P respectively. Points P on the first andQ on the second line are such that AP = BQ. Prove that PQ lies on one of the

surfaces mxz + cy = 0, yz + cmx = 0.

21 Discuss the intersections of the line

x-x x _ y-y x _ z-z x

I m n

and the plane ax + by + cz + d = (a) algebraically, (6) geometrically, showingthat:

(i) if al + bm + cn 4= 0, there is a unique point of intersection;

(ii) if al + bm + cn = 0, the line and plane are parallel unless also

(iii) ax x + by x + cz x + d = 0, in which case the line lies in the plane.



736

22

THE SPHERE; SPHERICALTRIGONOMETRY

22.1 Coordinate geometry of the sphere

22.11 Equation of a sphere

The distance formula shows that the sphere of centre C{a, b, c)

radius r (denned as the locus of points P in space such that CPhas equation

{x . a f + { y- 0? + {z - C )2 = rK

When expanded, this takes the form

x 2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0.

Conversely, (ii) can be written in the form (i) by separately c

pleting the square for the terms involving x, y, and z:

(x + u)2

+ (y + v)2

+(z

+ w)2 —

u2

+ v2

+ w2

—d.

If u 2 + v 2 + w2 > d, (ii) is therefore the equation of the sphere havi

centre { —u, —v, —w) and radius ^(u 2 + v 2 + w2 —d).

If u 2 + v 2 + w2 = d, (ii) represents the single point ( —u, —v, —

We may call this a point-sphere.

When u 2 + v 2 + w2 < d, the equation does not represent any locus.

The slightly more general equation

cx2

+ cy2

+ cz2

+ 2ux + 2vy + 2wz + d = 0,

where c 4= 0, can be written in the 'normalised' form (ii) on dividin

by c, and hence in general this also represents a sphere. We therefore

take (ii) as the standard general equation of a sphere, and observ

that it involves four arbitrary coefficients u, v, w, d. Consequently

a sphere can be determined by four independent conditions.



22.13] sphere; spherical trigonometry 737

(i) If a plane and sphere intersect, they do so in all points of a

circle (whose centre is the foot of the perpendicular from the centre Gof the sphere to this plane).

If the plane already passes through C, the circle cut on the sphere

is called a great circle of the sphere; otherwise it is a small circle.

(ii) If two spheres intersect, they do so in all points of a circle

(whose plane is perpendicular to the line of centres and whose centre

lies on this line).

(iii) If a line touches a sphere at P, it is perpendicular to the radius

GP. Conversely, all lines through P which are perpendicular to CPwill touch the sphere at P. All such lines therefore he in a plane through

Pwhich has

PGfor normal.

(iv) If a plane touches a sphere at P, it is perpendicular to the

radius GP. Conversely, a plane through P and perpendicular to CPtouches the sphere at P.

It follows that just one plane can touch a sphere at a given point P,

and that all tangent lines at P he in this tangent plane.

Example

Sphere on diameter P XP VIf P(x, y, z) is any point on the required sphere, then the plane PP x P a cuts

the sphere in a great circle having diameter PiP a , hence (by 'angle in asemi-circle') PP 1 J_ PP a . Since PP lt PP a have direction ratios

x-x 1 :y-y 1 :z-z 1 , x-x 2 :y-y 2 :z-z it

hence by the perpendicularity condition,

{x - xj) (x -<r a ) + (y-yx) (y-y 2 ) + (e-z 1 ) (z-z a ) = 0.

This relation, satisfied by any point P of the sphere on diameter PiP a , is the

required equation.

22.13 Tangent plane at P x

We employ some of the results in 22.12. Algebraical treatments

using the ratio quadratic or the 'distance quadratic' are indicated

in Ex. 22 (a), nos. 26-30, 23-25; these are applicable to surfaces more

general than the sphere, whereas the present is not.

For simplicity we first discuss the sphere of radius r and centre the



738 sphere; spherical trigonometry [22.

Since Px lies on the sphere, we have xf + yf + zf = r 2, and the abo

equation becomes, 9^ xx 1 + yy 1 + zz 1 = r 2

.

By 22.12, (iv) this represents the tangent plane at Px

.

If P1 is a point on the general sphere

x 2 + y2 + z 2 + 2ux + 2vy + 2wz + d = 0,

whose centre is C( —u, —v, —w), then direction ratios for CP 1 are

Xj^ + uiyj^ + v.z-L + w.

The plane through Px and perpendicular to this direction has equation

(xû) {x-xj + fa + v) (y-yj + fa + w) {z-z x ) = 0,

i.e. (x 1 + u)x+(y 1 + v)y+(z 1 + w)z = x\ + y\ + z\ + ux x + vy x + wz x .

Since Px lies on the sphere,

xl + yl + zl + 2ux 1 + 2vy 1 + 2wz 1 + d = 0.

By adding the last two equations,

(x 1 + u)x + (y 1 + v)y + (z 1 + w)z + ux 1 + vy 1 + wz 1 + d = 0,

which is the equation of the tangent plane at Px to the general sphere

Since (iv) can be arranged as

xx 1 + yy 1 + zz 1 + u(x + x 1 ) + v(y + y 1 ) + w(z + z 1 ) + d = 0,

we see that the 'rule of alternate suffixes' (15.63, Remark) is

valid as a mnemonic for writing down the equation of a tangent pla

to a sphere.

22.14 Examples

(i) Condition for the plane Ix + my + nz = pto touch the sphere x* + y2 +z 2 =

The plane and sphere will touch if and only if (22.12, (iv)) the perpendiculardistance of the centre from the plane is equal to the radius, i.e.

±p/<J(l2 + m2 + n 2

) = r,

i 1 = 2 2 + m2 + n 2




The sphere has centre (3, - 3, - l)andradius J{( - 3)2 + 3 2 + l 2 -( -2)} =J21.

Any plane through the given line is (21.61)

(x-2z + 4) + ky = 0,

and this will touch the sphere if and only if (as in ex. (i))

3 -3fe + 2 + 4 =-V{l 2 + ^ 2 + (-2) 3

}V '

i.e. (9-3A;) 2 = 21(5 + F),

from which 2k 2 + 9k + 4 = and so k = —4 or —£. The required tangent planes

are therefore x-fy-2z + 4: = 0, 2x-y-4z + % = 0.

(iii) Plane of contact from P x to the sphere x 2 + y2 + z 2 = r 2

.

Let a line through P x touch the sphere at P 2 . Then the tangent plane at P 2 is

xx z + yy 2 + zz 2 = r 2

and (since this contains the tangent line P 2 Pi) passes through P x . Hence

x x x 2 + y x y 2 + z x z z = r 2,

which shows that the coordinates of P 2 satisfy the linear equation

x x x+y x y+z x z = r 2. (v)

Therefore the points of contact of all tangent lines from P x lie in the plane (v),

which is called the plane of contact from P x . Its normal has direction ratios

x \ ' Vi •' z i» so * nat tne plane is perpendicular to OP x .

This plane cuts the sphere in a circle, and hence the points of contact oftangent lines fromP x lie on this circle. The tangent lines themselves lie on a right

circular cone with vertex P x , called the tangent cone from P x .

(iv) Condition for the line

x-x x _ y-y x _ z-z x

I m n

to touch the sphere x 2 + y2 + z 2 = r 2

; tangent cone from P x .

The point (x 1 + M, y x + Am, z x + Aw) of the line also lies on the sphere if and° my lf

(x x + AZ)2

+ (t/x + Am)2

+ ( Zl + Aw)2

= r\i.e. if A satisfies

A2(Z

2 + m2 + w2) + 2A(Z* 1 + my x + nz x ) + (x 2 + y\ + z\ - r 2

) = 0. ( vi)

In general this quadraticf gives two values of A (corifirming algebraically

that a line can meet a sphere in at most two points); but the line and sphere

will touch when there is a repeated root, i.e.

(lx x + my x + nz x )2 = {I 2 + m2 + 2

) (x 2 + y\ + z\ - r 2).

When this condition is satisfied, the line lies on the tangent cone from P x .



740 sphere; spherical trigonometry [22.14

(v) Orthogonal spheres.

The angle of intersection of two spheres at a common point P is denned to

the angle between the tangent planes at P. If C, C are the centres, then t

angle is equal to that between the radii CP, CP because these are normal

the respective tangent planes. When the angle is a right-angle, the spheresorthogonal; in this case triangle GPC is right-angled at P, and so

CO2 = CP 2 + CP 2.

If the spheres are 9 a . a o , o . o j nx 2 + y2 + z 2 + 2ux + 2vy + 2wz + a = 0,

x 2 + y2 + z 2 + 2u'x + 2v'y + 2w'z + d' = 0,

then C(— u, —v, —w), C(—u',—v',—w'), and the orthogonality condition

becomes

(u -u') 2 + (v- v') 2 + (w- w') 2 = (u 2 + v 2 + 2 - d) + (u' 2 + v' 2 + w' 2 - d'),

which reduces to 2mm' + 2vv' + 2ww' = d + d'. (v

Conversely, if this condition is satisfied, then by adding the expression

u 2 +v 2 + w2 +u' 2 +v' 2 + w' 2 to both sides we obtain the previous equation,

which is equivalent to G'C 2 = CP 2 + CP 2. The converse of the theorem

Pythagoras then shows that CP JL CP, so that the spheres are orthogonal.

Exercise 22(a)

1 A point moves so that the square of its distance from ( —1, 2, 1) is equalits distance from the plane

2x—Sy

+6z — 8. Show that it lies on a sphere, a

state the centre.

2 A, B are fixed points, and P varies so that AP :PB = A : 1, where Aconstant. Find the locus of P. [Choose axes so that A is (a, 0, 0), B( —a, 0, 0)

Find the equation of the sphere through

3 (0,1, 3), (1,2, 4), (3, 0,2), (2, 3,1).

4 (1, 2,1 ), (0, 3, 1), ( —1, 1, 2) and having its centre in the plane 3w + 2z =5 (1, 0, 0), (0, 1, 0), (0, 0, 1) which has its radius as small as possible.

6 and the points where the plane 2x + 3y + z = 6 cuts the coordinate

axes, and find its diameter.

7 Find the equation of the sphere which circumscribes the tetrahedron

whose faces lie in the planes x = 0, y = 0, z = 0, x + 2y + Zz = 4.

8 A sphere of radius r passes through O. Show that the ends of the diameterparallel to Oz lie one on each of the spheres x 2 + y

2 + z 2 ± 2rz = 0.

9 Spheres are drawn through (2,0,0) and (8,0,0) to touch Oy andProve that there are four such spheres, and give their equations.

10 Find the centres of the two spheres which touch the plane 3x + 4z =at (5 4 8) and which touch the sphere x 2 + y

2 + z 2 = 1



sphere; spherical trigonometry 741

12 Find the centres and radii of the two spheres whose centres lie in the octantfor which all of x, y, z are positive, and which touch the planes x = 0, y = 0,

2 = 0, 2x + 2y + z = 4.

13 A sphere is inscribed in the tetrahedron whose faces lie in the planes

x = 0, y = 0, z = 0, 2x + 6y + Bz = 14. Find its equation.14 Three spheres have centres (0, 0, 0), (3a, 0, 0), (0, 4a, 0) and radii a, 2a, 3a

respectively. Two planes, including an acute angle 0, are such that every oneof the spheres touches both planes. Prove that cos = -£g.

15 Spheres 8 lt 8 2 , * 3 have centres (0, 0, 0), (3, 0, 0), (0, 30, 0) and radii 1, 1, 19respectively. Find the equations of all common tangent planes n of the spheressuch that and s s lie on opposite sides of n, and s 2 , 3 lie on the same side. Showthat two such planes exist, and that the acute angle between them satisfies

9cos0 = 7. [Use Ex. 21 (c), no. 13.]

16 Find the tangent planes to the sphere x 2 + y2 +z 2 —2x + fy —6z+ 10 =

which pass through the line

x + S _ y+1 _ z-514 -3 ~ ~T~'

17 Obtain the result of 22.14. ex. (i) by comparing the equations

lx+my + nz=p and xx 1 + yy 1 + zz l = r 2.

Hence obtain the point of contact when the condition is satisfied.

18 Find the condition for the plane Ix + my + nz = p to touch the sphere

x 2 + y2 + 2 a + 2ux + 2vy + 2wz + d = 0.

19 Find the equation of the sphere having centre (5, —2, 3) and touching the

6 2 ~ -3*

Also find the equation of the tangent plane which contains this tangent line.

20 Prove that a tangent line from the point P x outside the sphere

8 = x 2 + y2 + z 2 + 2v,x + 2vy + 2ivz + d -

has length ^8 lt , where 8 n = x\ + y2 + z\ + 2ux^ + 2vy x + 2toz 1 + d.

21 Show that the points from which equal tangents can be drawn to the threespheres

x z + y * + z 2 = 1, x 2 + y2 + z 2 + 2x-2y + 2z- 1 = 0,

and x 2 + y2 + z 2 -x + 4:y-6z-2 =

he on the line = = .

2 5 3

Also find the point of this line from which the length of the tangents is least.

22 (i) Prove that the points from which tangents to the spheres

s = x 2 + y2 +z 2 + 2ux + 2vy + 2ivz + d = 0,




other than P x in common with the sphere (i.e. will be a tangent line at P x

and only if lx x +my x + nz x = 0. (This confirms that the tangent line is pe

pendicular to OP x : see 22.12, (iii).)

*24 If P s (x 2 , 2 , 2 ) is any point on the tangent line at P x in direction l:m:n,

prove that (x 2 —x x ) x x + (y 2 —y x ) y x + (z 2 —z x ) z x = 0. [Eliminate I :

m: n from t

result of no. 23 and the equations {x 2 —x x )jl = (y 2 —y x )/m = (z 2 —z x )fn.~\

Deduce that all tangent lines at P x lie in the plane xx 1 + yy 1 +zz 1 —r 2 (t

tangent plane at P x ).

*25 Use the 'distance quadratic ' (vi) to show that the mid-points of all chordin direction I : m : n lie in the plane he + my + nz = (which is clearly normalthis direction).

*26 The ratio quadratic. Show that the point dividing P X P% hi the ratio klies on the sphere s = if and only if

s 22 k 2 + 2s X2 kl + s xx l2 - 0.

(Here s may denote x 2 + y2 +z z —r 2 or the general expression in no. 20; t

notation of 19.12 is used, extended to three variables in the obvious way.)

*27 Taking P x on s = 0, deduce from no. 26 that P 2 lies on a tangent li

at P x if and only if s X2 = 0. Hence show that the tangent plane at P x is s x =*28 By expressing the condition for equal roots, deduce from no. 26 that ttangent cone from P x has equation ss xx —s 2

.

*29 Prove that the plane of contact of tangents from P x has equation s x = 0.

*30 A line through the fixed point P x meets s at A, B, and P 2 is chosen so th

P x , P 2 divide AB internally and externally in the same ratio. When the li

through P x varies, prove that P 2 moves in the plane s x = 0.

22.2 s = ks'

22.21 The general principle

If s = 0, s' = represent any two loci in space or in a plane, the

s = ks' represents some locus which passes through their commopoints (if any). For, points whose coordinates satisfy 5 = and s' —

will also satisfy s = ks'.

In this section we consider the equation s = ks' when k is constant

(i.e. independent of x, y, z) and one of s = 0, s' = represents a sphere,

while the other represents either a plane or a sphere.

22.22 Spheres through a given circle

(1) If s = x 2 + y2 + z 2 + 2ux + 2vy + 2ivz + d =




the straight line, more than one equation is needed; a circle cannot be

specified more simply than by one linear and one second-degree

equation, although the latter may not be the equation of a sphere:

e.g.see ex.

(iii)

below, whereit

represents acylinder.

The equation s —ks' = (k constant) will represent a sphere, a

single point, or nothing, since it has the form appropriate for the

general equation of a sphere (22.11). If s, s' intersect, then the locus

is a sphere through the circle determined by s = 0, s' = 0. If s, s'

touch, then s = ks' represents a sphere through their common point,

i.e. touching s at its contact with s'. In both cases the constant k can

be chosen to make the sphere satisfy one further condition.

(2) If s = x 2 + y2 + z 2 + 2ux + 2vy + 2wz + d =

and s' & x 2 + y2 + z 2 + 2u'x + 2v'y + 2w'z + d' =

both represent spheres, then s = ks' (k constant) will represent a

sphere, a single point, or nothing, provided that k =|= 1; if k = 1, it

represents a plane.

If s, s' intersect, then they do so in a circle, and s = ks' represents

a sphere through this circle unless k = 1, in which case the locus

becomes s —s' = and is the plane containing the common circle

(cf. Ex. 22 (a), no. 22 (ii)). The two second-degree equations 5 = 0,

s' = could be taken as equations of the circle of intersection, but

this circle can be represented more simply as in (1) by the linear

equation s —s' = and one second-degree equation, say s = 0.

If s, s' touch, then s = ks' represents a sphere (k #= 1) or a plane

(k = 1) through their single common point, i.e. touching both s and

s' at this point.

Examples

(i) Find the equation of the sphere having the circle

x 2 + y2 + z 2 = 16, 2x-3y + 6z = l

as a great circle, and give its centre and radius.

Any sphere through the given circle has an equation of the form

x* + y* + z 2 -16 + k(2x-Sy + 6z-l) = 0.



744 sphere; spherical trigonometry [22

(ii) Find the equation of the sphere which touches the sphere

x i + y2 + z i + x + 3y-6z-B =

at (1, 2, 3) and passes through (2, 0, 1).

The point (1,2, 3) certainly lies on the given sphere. The tangent plane(1, 2, 3) has equation f

a; + 22/ + 3z + i(a:+l) + |(2/ + 2)-3(2 + 3)-3 = ;

i.e. Zx + ly-11 = 0.

Any sphere touching the given sphere at (1, 2, 3) has an equation of the fo

x 2 + y2 + z 2 + x + 3y-Qz-B + k(3x + 7y-n) = 0.

It passes through (2, 0, 1) if 4 + 1 + 2 - 6 - 3 + k(6 - 17) = 0, i.e. k = Henthe required equation is

1 l(sc 2 + y2 + 2 2

) + 5x +19y- 662+I = 0.

(iii) Find the equation of the sphere through the circle

x 2 + y2 -4x + 2y = 1, 2 =

which is orthogonal to the sphere x 2 + y2 + z 2 + Sx —5y + 62 + 8 = 0.

The equation x 2 + y 2 —4x + 2y = 1 does not represent a sphere but a rigcircular cylinder: see 22.3, ex. (i).

The given circle is that in which the plane z = cuts the sphere

x 2 + y2 + z 2 -4x + 2y-l = 0.

Any sphere| through this circle can be written

(o;2 + t/

2 + 2 2 -4a; + 2t/-l) + ifc2 = 0.

It will be orthogonal to the given sphere if (22.14, ex. (v))

3(-2)-5.1 + 3& = 8-1,

i.e. if k = 6. The required sphere therefore has equation

a;2 + 2/

2 + z 2 -4a; + 22/ + 62-l = 0.

22.3 Surfaces in general

An equation fix, y,z) = explicitly involving some or all of x, y, z representsa locus in space called a surface (cf. 9.23). For example, when the functionlinear in x, y and z, the equation represents a plane surface or plane; whenthe type considered in 22.11, it represents a spherical surface or sphere. Aother special classes of equations will be noticed here because of their gmetrical significance.

Examples

= Th i h l O h d




varies on %', the line PQ generates a right cylinder whose cross-section by the

plane xOy is Thus f(x, y) = represents a cylinder with generators parallel

to Oz.

Thus in ex. (hi) of 22. 22 the equation x 2 + yz —4x + 2y = 1 represents a cylinder

whose section by the xy -plane is the circle with centre (2, —1, 0) and radius ^/6.

(ii) Homogeneous equation f(x,y,z) = 0. If f(x,y, z) is homogeneous of degree nin (x,y,z), then (1.52(4))

f(Xx, Xy, Xz) = Xnf(x,y,z)

for all A and x, y, z for which the functions are denned. If P(x, y, z) is a pointon the surface f(x, y, z) = other than O, then (Ax, Xy, Xz) also lies on this

surface for all X; i.e. every point of the line OP lies on the surface, which is

therefore a cone with vertex O.

(iii) Surface of revolution. Let the curve/(a;, y) = in the xy-jA&ne be rotatedabout Ox, thereby generating a surface of revolution with axis Ox. Let Q(x , y , 0)

be any point on the curve, and P(x, y, z) be its position in space at some stage

of the rotation; then P lies on a circle whose plane is perpendicular to Ox andpasses through Q, and whose centre C lies on Ox. Thus

y 9 = CQ = CP = J(y» + z*), x = OG = x.

Since Q lies on the curve, we have f(x , y ) = 0, and hence

f(x,^y 2 + z 2)) = 0,

which is the equation of the surface of revolution. Cf. the Remark in 7.5.

If two surfaces intersect, they do so usually in points of a curve (although

they may possess isolated points in common). In general this curve does not lie

entirely in one plane (as was the case for the intersection of two spheres), andconsequently is called a twisted curve or a skew curve or a space curve. It is

determined by two equations, viz. those of the surfaces of which it is the inter-

section. Any other pair of surfaces through the curve would equally well serveto detemiine it.

Exercise 22(b)

1 Write down the centre and radius of the circle of intersection of the spherex 2 + y

2 + z 2 - 16 and the plane 3x + 2y-z = 0.

2 If a sphere of radius r is cut by a plane distant c from the centre (c < r),

prove geometrically that the circle of intersection has radius *J(r2 —c 2

). Whathappens when c = r?

3 Show that the meet of the sphere x 2 + y2 + z 2 —2x —4y— 6z— 11 = and

the plane 2x —y + 2z— 15 = is a circle of centre (3, 1, 5), and find its radius.



746 sphere; spherical trigonometry6 Find the circumcentre of the triangle whose vertices are (2, 0, 4). (2, 4,

(0,2,4).

7 Prove that the spheres

(x - a) 2 + (y - b) 2 + (z - c) 2 = r 2

and (x —a —l)(x —a) + (y —b —m)(y —b) + (z —c —n)(z —c) = r 2

cut along a great circle of the former.

8 Find the equations of the two spheres through the circle

x 2 + y2 + z 2 -2x-4y = 0, x + 2y + 3z = 8,

which touch the plane 4tx + By = 25.

9 Find the equations of the spheres through the circle

x 2 + y2 + z 2 + 2x + 2y = 0, x + y + z + 4 = 0,

which intersect the plane x+y = in circles of radius 3.

10 Find the equation of the sphere through the circle x 2 + y2 + 2x = 0, z

and the point (1, 2, 1).

1 1 Spheres are drawn through the circle x 2 + y2 = a 2

, z = 0. Prove thatlocus of the ends of their diameters parallel to Ox is the rectangular hyperbolax 2 -z 2 - a 2

, y = 0.

12 Show that every sphere through the circle x 2 + z 2 = a 2 , y = is orthogonalto every sphere through the circle x*+ y

% + 4ax + a 2 = 0, z = 0. Find the tspheres (one from each of the above systems) which cut the sphere

3(£c a + y2 + 2

) + 2a(2x + 2y-z) = a 2

in the same great circle.

13 Find the equation of that sphere through the intersection of the sphere

s = x 2 + y2 + z 2 + 4:x-2y + 2z + 5 = 0, s' = x 2 + y

2 + z 2 + 2x-8y + 2z + 9 =

which has the smallest radius, and state this radius. [The system of sphereis best written s + k(s —*') = 0.]

14 Prove that the circles

x 2 + y2 + z 2 -9x + 4y + 5z - 1 - 0, 7%-2y + z = 4,

x 2 + y2 + z 2 + 6x-10y + Qz-l = 0, 4%-6y=l

lie on the same sphere, and find its equation.

15 Show h th h 2 + 2 + 2 = 2 + 2 + 2 6 8 24 + 120



22.4J sphere; spherical trigonometry 747

pass through a fixed circle. [Choose for origin, and the common tangent plane

at for ay-plane.]

* 1 8 Interpret the following

(i) = 0; (ii) f(x,z) = 0; (iii) f(y,<J(x*+z*)) = 0; (iv) f{z,J(a? + y*)) = 0.

*19 Write down the equations of the projection on the plane xOy of the circle

in no. 1.

*20 Show that the right circular cone with vertex O, semi-vertical angle a,

and axis Oz has equation x 2 + y2 = z 2 tan 2 a.

22.4 The spherical triangle

22.41 Some definitions and simple properties

Consider a sphere of centre O and radius r. In 22.12, (i) we defined

a great circle of the sphere to be the section made by a plane through 0,

and a small circle to be any other plane section. Thus although all

great circles have radius r, a small circle can have any radius between

r and zero.

The ends of a diameter of the sphere are called antipodal points.

There is a unique great circle through two non-antipodal points

P, Q; for there is a unique plane through O, P, Q, and this cuts the

sphere in a unique great circle. Consequently there are just two great

circle arcs joining P, Q, viz. the minor and major arcs of this great

circle. The minor arc is called the arc PQ; its length is the spherical

distance between P and Q.

The axis of a circle on the sphere is that diameter of the sphere

which is perpendicular to the plane of the circle. Its extremities are

poles of the circle, and the nearer one is called the pole.

A circle on the sphere which passes through the poles of a given

circle is called a secondary to that circle. Hence every secondary is a

great circle; and a given circle has infinitely many secondaries. Twogreat circles have a unique commonsecondary: it lies in the plane deter-

mined by their axes.

Defining the angle between two curves

at a common point to be the angle be-




angle B'AC between the arcs is the angle BOC between t

planes. Hence

the angle between two great circles

= the angle between their planes

== the angle at subtended by the arc intercepted on their

common secondary.

A lune is the figure on a sphere formed by two great semi-circles.

The angle between the semi-circles is called the angle of the

Assuming that areas of lunes on the same sphere are proportional

their angles, then for a lune of angle 6,

area of lune _area of sphere 2n

since the whole sphere can be regarded as a lune of angle 2n. S

the sphere has area 4fly 2,

area of lune = 2r 2 6.

22.42 Sides and angles

The figure on a sphere formed by three minor arcs of three great

circles is called a spherical triangle. Its sides a, b, c are the three

lengths, and the angles between the arcs b, c; c, a; a, b are its an

A, B, C.

There are two different ways of joining two non-antipodal po

by an arc of a great circle. Hence if no two of the points f A,B,Cantipodal, there are eight different figures having these for vertice

Only one such figure is a spherical triangle, viz. that formed by

minor arcs BC, CA, AB.Taking the sphere to have unit radius, J the length of an arc of a g

circle is the radian measure of the angle subtended by it at the centre

Hence the sides a, b, c are the angles at subtended by the

BC, CA, AB; and since these are minor arcs, each of a, b, c is less tha

The side a of the spherical triangle ABC is BOG, a plane angle; and

angle A is that between the planes OBA, OCA (a dihedral angle). Theref

i l l i




since in a trihedral angle the sum of two plane angles is greater than the third;t

the sum of the sides of a spherical triangle is less than 2n, (in)

(i.e. less than the circumference of a great circle) since the sum of the plane

angles of a trihedral angle is less than four right-angles, f Cf. Ex. 22(d), no. 20.

A

Fig. 235

Each angle of a spherical triangle is less than n. For if possible suppose

B^tt\ let arc AB be produced to meet arc CA again at A'. ThenA, A' are antipodal, and arc CA ^ &tqAA' = n, i.e. b ^ n, contra-

dicting the definition of 'spherical triangle'.

However, a spherical triangle may have one, two, or three of its

angles obtuse; and one, two, or three may be right-angles. Also see

22.43 (3).

22.43 Polar triangle; supplemental relations

(1) If A' is that pole of BG which is on the same side of the plane OBG as Aand B', G' are defined similarly by cyclic interchange of the letters, thenA'B'G' is a spherical trianglej called the polar triangle of ABC. We now provethe following reciprocal property.

A

Fig 236




As A' and A are on the same side of plane OBC, and the distance of A' fany point on BG is \tt, hence AA' < \tt. Thus, since the distance of A frompoint on B'C is \n, A and A' must be on the same side of the plane OB'C.

Therefore A is that pole of B'G' which is on the same side of plane OB'C

A'; and B, G have the corresponding properties obtained by cyclic interchange.Hence ABC is the polar triangle of A' B'C.

(2) Sides, angles of the polar triangle are respectively the supplements ofcorresponding angles, sides of the original triangle.

Proof. Let G'A' cut AB at X. and BG at Y. Since CA' is secondary to

and BG, therefore XY = B. Also CX = \<n = YA'. As YA' = XY+ XA',CX + XY+ XA' = Ti, so that CA' = n-XY, i.e. V = n-B. We infer

supplemental relations

a' = tt-A, b' = n-B, c' = 7T-C.

Since ABC is the polar triangle of A' B'C, we also have

a = tt-A', b = n-B', c = n-C.

( 3 ) The sum of the angles of a spherical triangle lies between 2 and 6 right-angles.

Proof. By applying (iii) to the polar triangle A'B'C,

< a' + b' + c' < 2n.

From (iv) this becomes < 3n —(A+B + G) < 2n,

i.e. 7T < A + B + C < 3n.

The expression E = A+B + C —n is called the spherical exc

of the spherical triangle ABC ; it is the amount by which the sumthe angles of the spherical triangle exceeds the sum of those of

plane triangle. By (vi), < E < 2n.

22.44 Area

Given the spherical triangle ABC on a sphere of centre

radius r, produce the sides AC, BC to cut the great circle AB aga

at A', B' . By considering areas on the hemisphere shown (fig. 23

and using (i),* K' ABC+ s x = lune ABA'C =2Ar\

ABC+ s % = lune BAB'C = 2Br*,

ABC C 2




22.5 Triangle formulae

We continue to assume that the sphere has unit radius until 22.54.

22.51 Cosine rule

cos a = cos b cos c + sin b sin c cos A, (viii)

and two similar formulae which can be obtained from this by cyclic

interchange of the letters.

c

Fig. 237 Fig. 238

The cosine rule is important because all the remaining triangle

formulae can be deduced from it. In calculation it can be used to find

(a) a side of the triangle when the other two sides and the included

angle are known; (6) an angle when the three sides are known; this

is exactly the situation in the plane case.

First proof, using projection (fig. 238).

Let N be the projection of C on the plane AOB, and P, Q be the

projections of N on OA, OB respectively.

Since OP is perpendicular to CN and NP, therefore OP is per-

pendicular to the plane CNP, and hence OP j_ PC because PC is aline in this plane. Similarly OQ _L QC.

P j ti g OB




and since CPN is the angle between the planes OAB, OAC, viz. A

PN = PC coa A = OCsin&.cosA

Substituting, we get formula (viii).

Second proof, using coordinates (fig. 239).

Choose axes at the centre O of the sphere, with Oz along OA, and let the ot

vertices B, C have cartesian coordinates (x^y^Zj), (x 2 ,y%, %) and spherica

polar coordinates (1, d x , <f> x ), (1, d 2 , 2 ): see 21.12 (2) and Ex. 21 (a), no. 15.

Since OB, OC have direction cosines {x x , y x , z x ), {x %, y 2 , z 2 },

cos a = x x x 2 + y x y 2 + z x z 2

= sin 6 X cos<f> t . sin 6 Z cos 2 + 8U1 î sul 0i • si 11 ^2 sul 02 + cos ^1 • cos

= COS d x COS2 +

SU1 #1 sul #2 COS(01

—02)

= cose cos 6 + sin c sin 6 cos .4

because 6 X = c, d 2 = b, and <j) 1 —<fi 2 = ±A.

Fig. 239

Example

Median of a spherical triangle.

The arc of the great circle joining a vertex A to the mid-point A' ofopposite side BC is called a median of the triangle ABC (fig. 240).

Since angles CA'A, AA'B are supplementary,

cos + cost's = 0.

By applying the cosine rule to each of triangles AA'G, ABA', this becomes

cos 6 —cos AA' cos cose —cos AA' cos \a

sin .4.4 ' sinâ

d AA'

+sin AA' sin \a

= 0,




Geometrical proof.

From fig. 238 we have

NG = PCainA = OCsin6 sin 4,

and NC = QCsinB = OC sin a sin Bsince CQN is the angle between the planes OBA, OBG, viz. B. Hence

sin 6 sin A = sina sinjB,

o , . , sina sin bfrom which - - = - —

sini sin£

Similarly, by projecting fromA

onto plane OBC,we

canprove

sin b _ sine

sinJ5 sinC

Deduction from the cosine rule.

sin.2 A — 1 —cos 2 A

_ (cos a —cos b cos c) 2

sin 2 6 sin 2 c

_ ( 1 —cos 2 b) ( 1 —cos 2 c) —(cos a —cos b cos c) 2

sin 2 b sin 2 c

_ 1 —cos 2 a —cos 2 6 —cos 2 c + 2 cos a cos b cos c

sin 2 b sin 2 c

sin^l _ +a/(1 —Scos 2 a + 2cosacos6 cose)

sina sina sin6 sine '

and this expression is symmetrical in a, b, c. Hence it is also equal to

sin^B/sinfe and to sin C/sinc. The sine rule can therefore be written in

the 'completed form'

sin^L _ sini? _ sinff _ +V(1 -Scos 2 q + 2cosa cos6 cose)

sina sin b sine-

sin a sin b sine 'sine

where the positive sign of the square root is taken because all the



754 sphere; spherical trigonometry [22.5

Examples

(i) If A' is the mid-point of BG, fi = BAA' and y = A' AG, prove

sinCcot/? = cot A + -

and cot y —cot /? =

sin .4 sinB

sin (6 + c) sin (6 —c)

sinfe sine sin .4

By applying the sine rule to triangles ABA', AA'G in fig. 240,

sin/5 sinâ siny

sinB sinÛ.' sinC'

sin /? sin G = sin (A —/?) sin B,

sin/?(eos.4 sinB + sinC) = cos/? sin.4 sin B,

sin Gand so cot /? = cot A + -

Similarly , coty = cot A +

.'. coty —cot ft =

sin A sinJ5

sin BsmA sinC

sin 2 B- sin 2 Gsax A sinSsinC

sin 2 b —sin 2 c

sin .4 sin 6 sine

sin (6 + c) sin (6 —c)

(ii) Prove that

siab sine sin .4

sin (B + G) cos b + cos c

sin^l 1 + cosa

by the sine rule,

after reduction.

sin(B + C) sinB „ sinC JX ^:—-—= cos G H - cos is, and by using both rules,

sini sin 4 smisin 6 cos c —cos a cos b sin c cos 6 —cos c cos a

= -. : r-r +-. . :

sin a sin a sin o sina sine sin a

sin b sin c (cos c —cos a cos 6) + sin b sin c (cos 6 —cos c cos a)

sin 2 a sin 6 sine

_ sin 6 sin c(cos 6 + cos c) ( 1 —cos a)

( 1 —cos 2 a) sin 6 sin c

cos 6 + cos c

1 + cos a



22.54] sphere; spherical trigonometrygives —cos A = cos B cos G —sin B sin G cos a,

i.e. cos A = —cos i? cos (7 + sin B sin C cos a.

Similarly the sine rule in its completed form gives

sin a _ sin b sin c + V( 1 —^ cos2 ^ —2 cos ^1 cos 2? cos C)sinA sinB sinC sin A sin S sin G

22.54 Triangle on the general sphere

When the sphere has radius r #= 1, the sides of the spherical triangle

ABC are no longer equal to the angles which they subtend at the

centre. If we continue to denote these angles by a, b, c, and call the

sides a, /?, y, then a Bya = -, b = -, c — -.

r r r

All the preceding formulae remain valid because their proofs used the

angles at the centre and not the actual sides of the triangle.

Example

By using the expansions of coax, sin* (Ex. 6(6), nos. 21, 20), the cosine

formula (viii) can be written

l-ia 2 + ... = (l-^6 2 + ...)(l-ic 2 + ...) + (6-...)(c-...)cos^,

where the dots denote terms of order 3 or 4.

On introducing the sides a, /?, y and simplifying, this becomes

a a = ^2 + 72-2^7003^+0

When r -» 00, the spherical triangle ABC tends to the plane triangle ABCwith sides a, /?. 7. and in the limit we obtain the cosine rule of plane trigono-

metry. The sine rule can be treated similarly.

Exercise 22(c)

1 Write out the formulae for cos 6, cos c corresponding to formula (viii).

2 If G = \n, a = \n, b = \n, find c and the area of the triangle.

3 A = \n, b = c = \n, and D is the mid-point of side AG. If BDG = a,

show that cos a = 1/^3 and that the areas of the triangles ABD. ABC are in

the ratio (3n —8a) : tt.

4 If B, G are points on the equator whose longitudes differ by 90°, and A is

a point in latitude A on the meridian through G, prove that in the triangle

755

(xi)



756 sphere; spherical trigonometryTaking the radius of the Earth to be 3960 miles, find (correct to three sign

ficant figures) the area included by the path of the ship, the meridian throughB, and the equator.

6 Two points A, B on the Earth's surface have latitude A, and their differ

ence in longitude is 8 radians. If r is the radius of the Earth, show that (i) thedistance apart along the parallel of latitude joining them is r8 cos A; (ii) the

distance apart along the great circle joining them is 2r sin -1 (cos A sin %8). [F(ii) use half the isosceles spherical triangle ABC, where C is the north pole.]

7 In no. 6 (ii) prove that the greatest latitude L reached on the great circ

course is given by tanZr = tan A sec %8. [If D is the point of latitude L,spherical triangle ACD is right-angled at D.]

8 If D is any point on BC, prove that

cos AD sin a = cos b sin BD + cos c sin DC.

[Method of 22.51, ex.]

9 The internal and external bisectors of angle A meet BC at D, D'. Prove

sin i?D sine sinPD'

sin DC sin 6 saxD'C'

10 The arc of the great circle through the vertex A which meets the oppositeside BC at right-angles is called the altitude from A. If a, /?, y are the altitudes

from A, B,C, prove that

sina sina = sin/? sin6 = siny sine = 2w,

where n = £ <J( 1 —S cos 2 a + 2 cos a cos b cos c).

*n -o xi. x sin(6 + c) cosP + cosC*11 Prove that ; - = —

.

sin a 1 —cos A*12 Verify as in 22.54 that the sine rule of plane trigonometry is the limit

the spherical case.

Miscellaneous Exercise 22{d)

1 A rectangular box has edges of lengths a, b, c, and P moves so that the sof the squares of its distances from the six faces is equal to m2 d 2

, where d is

length of a diagonal of the box and m is constant. Prove that the locus of Pa sphere if 2m 2 > 1.

2 Through a point P three mutually perpendicular lines are drawn:passes through a fixed point C on Oz, while the others meet Ox, Oy respectively.

Prove that the locus of P is a sphere with centre C.

3 Find the coordinates of the point of the sphere x 2 + y2 + z 2 —4x + 2y —4

which is nearest to the plane x + 4y + z = 19, and calculate its distance frothis plane.



sphere; spherical trigonometry 757

6 If the plane ax + by + cz + d = touches the sphere

(x-l) 2 + y2 + (z-l) 2 =l,

prove that a 2 + b 2 + c 2 = (a + c + d) 2.

If this plane also passes through (0, 0, 2), show that it cuts the plane z = in

a line which touches the parabola y2 = 4x, z = 0.

7 Show that there are two spheres through (0, 0, 0), (2a, 0, 0), (0, 26, 0) whichtouch the line .

x —a _ y —o _ z —c

I m n

and that if I2 + ma + wa = 1 the distance between their centres is

2—{c 2 -w 2 (a 2 + 6 2 + c 2)}*.

8 Prove that the centre of a sphere which touches each of the lines

y—mx = = z —c, y+mx = = z + c

lies on the surface mxy ( 1+ m2) cz = 0.

9 A line A passes through O and touches each of the spheres

x 2 + y 2 + z 2 + 2ax+p = 0, x 2 + y 2 + z 2 + 2by + q = 0.

Show that the angle between A and Oz is given by sin 2<f>

= p/a 2 + q/b 2. Also

find the distance between the contacts of A with the two spheres.

10 Find the condition for xjl = y/m = z/n to touch the sphere

x 2 +y 2 +z 2 + 2ux + 2vy + 2wz + d = 0,

and find the point of contact. Hence show that the tangent cone from O hasequation d(x 2 + y 2 + z 2

) = (ux + vy + wz) 2.

11 Two skew lines AP, A'P' are met by their common perpendicular at

A, A', and O is the mid-point of AA'; 2a, is the angle between the lines, andAA' = 2c. If AP. A 'P' has either of the values c 2 sec 2 a, — 2 cosec 2 a, prove thatPP' touches the sphere with centre O and radius c.

12 Prove that there are eight spheres of radius 5 which are orthogonal to thesphere x 2 + y

2 + z 2 = 16, touch Ox, and cut off a segment of length 2 from Oy.

13 Through the circle of intersection of x 2 + y2 + z 2 = 25 and x + 2y + 2z = 9,

two spheres s lf s 2 are drawn to touch the plane 4* + 3y = 30. Find the equationsof these spheres, and the coordinates of the point through which pass all thecommon tangent planes of * x and s 2 .

14 Prove that in general two spheres can be drawn to pass through a givencircle and to touch a given plane. If the circle lies in the plane z = and hasa given radius r, and if the plane is x cos 6 + z sin 6 = 0, show that when thedistance between the centres of the two spheres is constant and equal to 2c,



758 sphere; spherical trigonometry2a 2 /(u 2 + v 2 + w2

). Hence write down the coordinates of P. (This gives a rational

parametric representation of the sphere x 2 + y2 +z 2 = a 2

, the parametersbeing u, v.)

16 (i) From the cosine formulae for cose and cos b, eliminate cos 6, and th

substitute sin 6 = sinB sine/sin (7, to prove the 'four-parts formula'

cosa cos 2? = sina cote —sinB cotC.

(ii) Similarly prove

cosa cos C = sina cot 6 —sin O cot B.

17 If AD bisects angle A internally, prove cot^4D = (cot6 + cotc)/2cos^.[Apply no. 16 to triangles ABD, ADC; angles ADB, CDA are supplementary.]

18 If D is a point on BO, BAD — /?, and DAG = y, prove that

cot AD saiA = cot6 sin /#+ cote sin y.

[Method of no. 17.]

19 If a spherical triangle has area one-quarter that of the whole sphere, provthat cos a + cos b + cos c = —1, and deduce that each median is the supplementof half the side which it bisects. [Use 22.52, ex. (ii); and 22.51, ex.]

20 (i) Prove (ii) of 22.42 by using the cosine formula. [Since A<cosa > cos 6 cose —sin 6 sine = cos(6 + c), which implies a n, then 6 + c> a since a < n. ]

(ii) Use the result of 22.44 to infer that A + B + C —n> 0. By writing

down the supplemental inequality (cf. 22.53), deduce (iii) of 22.42.



(25)

ANSWERS TO VOLUME II

Exercise 10(a), p. 369

1 (a + l) 2 (2a-l). 2 (a-2)(3a+l)(2a + 3).

3 (a-l)(a + 2)(a 2 +a + 2). 4 -1,2,3.

5 a = -l, b = -2. 6 a = -2, 6 = 3. 7 7,1.

8 |(9a-l). 9 3a 3 + a 2 + lx+ 1. 10 a 3 + a 2 + 3a + 4.

11 1, 3, 6. 0. 12 1, -6, 0, -2.

13 -1, 25, ±(a 2 -3a-5). 14 a = 3, 6 = - 1.

15 a = 4, b = -2, c = 1. 17 Put a; = - (b + c), etc.

19 A ( x - b )( x - c )(x-d)B

(x-a)(x-c)(x-d)

(a-b)(a-c)(a-d) (b-a) (b-c) (b-d)

(x-a)(x-b)(x-d) (x-a)(x-b)(x-c)

(c-a)(c-b)(c-d) (d-a)(d-b){d-c)'21 (i) q = 3a*.

Exercise 10(6), p. 373

1 (x-2y + 4)(2x + y-l). 2 (x + y + 8z)(x-y-6z).

3 10, -7£. 5 3(y + z)(z +x)(x+ y).

6 -(y-z)(z-a)(a-2/)(a + 2/ + z). 7 (?/-z)(z-a)(a-y)(a + 2/ + z).

8 5(?/ + z)(z + a)(a + y)(a 2 + 2/2 + z a + 2/z + za + a2/).

9 Si/z(2/ 2 -z 2), £a(?/-z) 3

.

10 (i) a 2 + y2 + z 2

; (ii) a 2*/

2 + y2 z 2 + z 2 a 2

; (iii) bc(b - c) + ca(c - a) + ab{a - b) ;

(iv) a 2 6c + 6 2 ca + c 2 a6; (iv) 6c 2 + ca 2 + a6 2 + 6 2 c + c 2 a + a 2 6.

15 (ii) Babc(b-c)(c-a)(a-b);(iii) 5(6-c)(c-a)(a-6)(a 2 + 6 2 + c 2 -6c-ca-a6).

Exercise 10(c), p. 378

1 (i) 15; (ii) f ; (iii)-l; (iv) 19; (v) 13; (vi) (vii) 80; (viii) 343.

2 (i) 4a 2 -2a-3 = 0; (ii) 9a 2 -28a + 16 = 0; (iii) 12a 2 + 2a- 1 = 0.

3 b/a < 0 cja > 0



(26) ANSWERS9 (i) 2p 3 + 21r = 9pq; (ii) p 3 r = q

3.

10 x 3 + (3q-p 2 )x 2 + q(3q-p 2 )x + q3 -p 3 r = 0;

(i) a 2 —J3y = ft2 —yoc = y

2 —a/?; (ii) a 2 —/?y = —pa, etc.; p 3 r —qa

.

11 x* + 3px 3 +3p2

x2

+p 3 x + q = 0.

Exercise 10 (J), p. 382

1 y = x 2 + 2. 2 x 2 /a 2 + y 2 /b 2 = 1.

3 (a 2 -6 2)

2 = 2(6 4 -c 4). 6 6 4 c 4 + c 4 a 4 + a 4 6 4 =

7 (i) l + m+ n = 0; (ii) see 16.32.

8 (i) y2 -4ax = c 2

(a; + a) 2;

(ii) the locus of points from which the tangents to y2

= 4aa; includeangle tan -1

c.

9 y2 = a6 2

a; + a 2 6 2 (&-2). 10 a;2 + t/

2 = 2(a 2 + 6 2).

11 x = abc, y ——E6c, z — Ea. 12 1, 1, 3 in any order.

13 1, 2, —5 in any order. 14 1. —2, 3 in any order.

16 4p 8 + 27g 2 = 0. » 17 27p 4 = 256g 3.

18 (6r —eg) (ag —bp) 2 = (op —ar) 3.

19 If p(aj) = (a;2 + 2&»; + c) , - 1 g(a:), then

p'(x) = (r - 1 ) (x 2 + 2bx + c) r ~ 2 2(x + 6) g(x) + (x 2 + 2bx + o)' 1 g'(x)

.

Hence x 2 + 26a; + c is a factor of (x + b) g(x). If b 2 4= c, then x + 6 is not a factor

a;2 + 26a; + c, so a;

2 + 26a; + c is a factor of g(x), and the result follows. However,if 6 2 = c, then x 2 + 2bx + c = (a; + 6)

2, so that a; + 6 (but not necessarily (x + b

is a factor of g(x).

Exercise 10(<?), p. 388

1 a; + 3. 2 2a; + 7. 3 a;2 -2a; + 5.

4 (a;-2) 2 (a;+l)(a; + 3). 5 (z + 2)(a; 2 -a;+ l) 2 .

6 (a;-l) 3 (a;+l) 2 (a:-2). 7 -f twice.

8 2; i(-l±V6) twice. 9 --^(12a; + 35), -^(24a; 2 - 14a; + 43)

Miscellaneous Exercise 10(/), p. 388

2 No.

6 (i) (b-c)(c-a)(a-b)(a + b + c);



ANSWERS (27)

14 (i) Coeff. of x 2 zero; (ii) constant term zero; (iii) coeff. of x a equal to minusthe constant term; (iv) coeff. of x 2 equal to constant term.

17 2p\ 18 t/4 -52/ 2 + 6 = 0; ±V2-3, ±V3-3.

19 b2

x* + 2(ab - 2h2

) x2

+a2

= 0. 20 x2

ja + y2

/b +2

/c = 2.

24 (i) k = 3; x = 1 twice; (ii) k = —f; x = thrice.

25 (i) m = - 20, c = 7; m = 108, c = 135; (ii) m = 44, c = - 121.

27 p/g is constant.


1 2. 2 40. 3 82. 4 0.

5 Zi(y2-ya)+x< s (y a -y 1 )+x 9 (y 1 -y i ).

6 a 3 + 6 3 + c 8 -3a6c. 11 0. 12 -90.

13 0. 14 0. 15 18. 16 0. 17 0.

18 x(x-l) z (x*-l). 19 a 4 (a;-l) 2(a;

2 -l).

20 Sa 8 -3afec = (Sa)(Sa 2 -26c). 22 (i) and (ii) A. 23 9, ±V3.

24 0. 26 S cos 2 ^ + 2 cos J. cos B cos C = 1. 28 (ii) 3 8.

Exercise 11 (ft), p. 410

1 x = -|, = f,z = l. 2 4,3,-1. 3 2,-1,-2.

4 (*-^-<*>, etc .

(o —a) (c —a)

8 A = 1, x-.y.z = 1:-1:0; A = - 2, J2 : J2 : 1; A = 3, 1:1:- 2^2.

9 A = 3, a = £, y = |; A = 14, x = -f, y = f.

10 (i) x = -3,y = 3/(1 —A;), z = (l-4fc)/(l-fc);

(ii) a; = A, y = —A, z = —2 —A; (iii) inconsistent.

11 bc + ca + ab = 2abc+ 1.


1 (b-c)(c-a)(a-b). 2 (b-c)(c-a)(a-b)(a + b + c).

3 (b —c)(c-a)(a-b)(a + b + c). 4 (6-c) (c-a) (a-6) (bc + ca + ab).

5 (&-c)(c-a)(a-6). 6 -(6-c)(c-a)(a-6).

2 2 2 2 2 6 2



(28) ANSWERS

Exercise 11(d), p. 415

1 1 2 3 2 1 1 I

X X2 Xs 1 2 2x

2 6x 1 3 3x 2

3 logx X

1/x 1 2x

-2/x 3

[Deriveby

columns.

8 2(y-z)(z-x)(x-y).

9 56. [First interchange c x and c 2 , then r x and r 2 .]

10 -32. 11 -550. 13 x = -8, y = 14, z = -2, « = 4.

15 —(ic —a) (a; —6) (a; —c) (6 —c) (c —a) (a —b). With notation of Ex. 11(e),

no. 22, v^ = (b-c)(c-a)(a-b), V2 = (b-c)(c-a)(a-b)(a + b + c),

Vx = (b —c) (c —a) (a —6) (6c + ca + a6).

Miscellaneous Exercise 11(e), p. 416

1 The equation is linear in x, y, and is satisfied by (x x , y x ) and (x % , y 2 ).

4 (6-c)(c-a)(a-6). 5 4=(b-c)(c-a)(a-b)(a + b + c).

6 (6-c)(c-a)(a-6)(a6c+l).

7 (6-c)(c-a)(a-6)(a-d)(6-d)(c-d).

8 2abc(a + b + c) 3. 10 sin 2 0-2cos 2 + 2sin0 costf.

11 ap aq ap aq+

br bs+

br bs+

dr dsp cq dr ds cp cq

12 x — 1 twice, 2. [c 2 -> c 2 — x , remove cc— 1; c 3 -> c 8 —c 2 .]

13 x = 6, c, a 3 /6c.

15 Three distinct collinear points cannot be concyclic.

17 (i) Unique solution x = 3, y — 1, 2 = 0; (ii) a; = 3 —2A, y = 1 —A, z =(iii) inconsistent.

18 (i) A =(= 3; (ii) A = 3, /* 4= 10: (iii) A = 3, fi = 10.

19 A = -l, fc= 18, x= 6-4a-3/?, 2/ = a, z = /?; A = - 19, k = - 18, x =y-2aL,z = 3a + 2.

20 A= l,a:6:c = 0:-2:l; A = |,4:4:-3; A = |,4:4:5.

22 (ii) 1 1 1 1 1 1



ANSWERS (29)


1 81a^-216a; 3 + 216a; a -96a;+16. 2 x 5 + 5x 3 + 10x + 5/x* + 1/x*.

3 l + 3x + 2x 2 -2x*-Zx*-x*. 4 1 + 12x + 54x 2 + 100* 3 + I5x* + ....

5 60. 6 969. 7 -105x2- . 8 7920.

9 3 7 xl54». 10 -16x217. 11 (i) 1-2166; (ii) 0-9039.

12 7. 13 5th term = £f x 5 8 x 2 18.

14 Coefficient of the 6th term = 231 x 4 11 x 5 6.

15 n Cr + 2 CU + CU = «+ a Cr (r ^ 2).

18 »+ 1 Cr = (7r + -^.i + «- 2 CV_ 2 + . . . + »-'+ 1 C 1 + 1. 19 724.

Exercise 12(£), p. 425

3 n.2 - 1. 4 0. 8 0.

10 t 2 ™ 1 )* (l +a .)n+2_ (n + 2 ) a; -l{(n-1) } 2 ' (n+ l)( n + 2)

13 (i) 4 6; (ii) 2080, 2016.

14 (i) (l + *+a; a)

B = a x 2n + aJ x 2n - 1 + ...+a zn ,

(1— x + x 2)

n = ao-^x + a^x 2 —...+a 2u x Zn;

(ii) a + a 1 a;a + a 2 iB

4 + ... +a 2B a;*«; (iv)£(3 n -l); (v)(n + l)3».


1 (n+l) a (2ra + l)«. 2 |n(4w a -l).

3 (-l^-^w+l). 7 £n(n+l)(7w + 2).

8 a = 2, 6 = -3, c = -3, eZ = 0; %n(n+ 1) (n 2 + 3w- 1).

9 $n(w+l)(n-4). 10 ^n(n+l)(n +2)(8n+l).

11 f(2n+l)(2ra + 3)(2w + 5)(2n + 7)-|.3.5.7.

12 in(n+l)(n + 6)(n + 7). 13 Jsn(n+l)(n + 2)(3n + 5).

14 i- 1 1

3 3(3w+l)' 3 4\3.5 (2w + 3)(2n + 5)/' 60'

5 4w + 5 5 11 12n+ll 111

4~2(n+l)(n + 2); 4 120~ 8(2n + 3) (2n + 5)' 120'

3 1 13 17 1 2 17



(30) ANSWERS

8 £(a + 26).


Exercise 12(e), p. 442

4 d. 5 o. 6 d. 7 c. 8 c. 9 o.

10 d. 11 c. 12 d. 13 d. 14 c. 15 o.

Exercise 12(f), p. 446

1 0. 2 D.

4 c for all x ^ 0. 5 s$ a; 1, C; x > 1, D.

6 < a; < 1, c; x 3* 1- d. 7 d for all a; > 0.

8 < a: < 1, C; x ^ 1, D. 9 ^ a; < |, C; a; > |, d.

10 < a; < 1, C; a; 2* 1, D. 11 ^ a: < 1, C; x > 1, D.

12 {ra^-fn+lja^+l} (a:-l)- 2 1 if x * 1; £n(n+l) if x = ].

Exercise 12(g), p. 450

2 p > 1, C; p < 1, d. 4 7r/2a, 7r/2a+l/a 2. 5 e -1

.

Exercise 12(A), p. 459

1 o. 2 c. 3 c.

4 d. 5 c. 6 No. 2.

7 a.c. by comparison with 2(l/r 2). 9 |a;| < 1, O; \x\ > 1,

10 c for all x. 11 \x\ < 1. c; |a;| > 1, d.

12 |*| < 1, c; \x\ 1, d.

13 \x\ < 2, C; \x\ > 2, D; (x = 2, D; x = -2, C).

Exercise 12(0, P- 465

1 log 2. 2 i?r. 3 See 12.72(2).

4 £ 2*»- sin (ir?r) - , all x. 5 £ 2* cos (%rn) —, all x.

r=l r\ r =0 »



ANSWERS (31)

Exercise 120), P« 468

1 i(r+l)(r + 2)(-x) r; 1 -3x + 6x 2 - I0x* + \x\ < 1.

31 ' 3,5

;(2r ~ 1)

(-fa;)%r> 1; 1 -fa + ^ 2 - W*3 + -i W < hr

41 ' 3 - 5

-;,

(2r ~ 3)(-l)'- 1 ^ I>

r > 2; 2 + ^-^* + ^-...; |*| < 2.

5 (r+l)a; r + l; * + 2a: 2 + 3:« 3 + 4a 4 + \x\ < 1.

6 (S^ 1) (-»)', r ^ 1; i-|a; + fa;

a —1V3 +---; M < 2.

7 (r 2 + r+l)a; r; 1 + 3* + 7a: 2 + 13rc 3 + |ar| < 1.

8 672a; 6. 9 - 10(3aO- u

. 10 —12 5th term. 13 27th and 28th terms.

14 1st term. 15 33-6241. 16 9-8995.

17 l_| a; _| a; 2 + i| a;8 + .. < . 18 < £;

A{5 + (-2)'}.

19 |*| < 2; i{2-'-(-3)-'}. 20 |*| < 1; 2 1 '-r-4.

21 |a;| < 1; +1, -1, according as r = 3p, Sp + 1, 3p + 2 (p = 0, 1,2, ...).

( w + r _l)i . 1.3.6... (2r-l)22 W< l J^h' 23 EaGhlS

2.4.6...(2r) 'r>L

24 fV3. 25 #f 26 10^-11.

i ™ l + 4*-3* 2

27 1. 28 —(I-*) 3

Exercise 12 (A:), p. 472

3 r r-1 r 2 -4r+l1 -e 2

. 2 -. 3 (-1)' ~

+ (- )'4 —71—

111 2 2 2 2 3

5 W 1-Ii + 2l-3i + -' (») 1+ n +

2-+

3+ - ;



(32) ANSWERS

Exercise 12(/), p. 475

1 -£-*', a 2 log2-S -(-*)', -2 <a>< 2.

r=l ** r=l r

3 _2 5iliV-l<x<l. 4 jli-iy-i-V' + Wx', -%<x<%r=l r r=l r

00 2r— 1

5 *+£ -n :(-l) r * r> -1 1.

1 26 -ifr=3p, -ifr = 3p±l; -l<a;<l.

r r1 J

i Of IV7 -- - for a:

2 ' 1, - —~ —- for x 2r

; \x\ < 1.2r-l 2r

oo /_J)r-l8 2a + a:

2 -\%; 8 + |a; 4. 9 S

15 *log2.

18 log4-l.

1 rx r

16 -£log(l-a; 2).

19 log 2-*.

14 logf.

17 log 2.

20 ilog2.

1 / 1\ 1 + x21 -|a;--)log + J if a; 4= 0; if x = 0.

4\ ay 1— a;

» -H)23 logW-

log(l-a?)-i —if x 4=0; if 35 = 0.

24 a; 8 > 1. 25 Each is —2 log cos

5 1-4142136.

13 l + âr + fa;2

.

17 l + hp-&P*.

Exercise 12 (m), p. 481

8 l+x 2 + fa:8

. 12 f + f<r.

14 0037. 15 0-393.

18 19 loga.

Miscellaneous Exercise 12 («), p. 482

3 (2n) /(n ) 2.

5 (i) |(r-l)r; (ii) (r- l)r+ 1, r(r+ 1) - 1; (iii) r 8; (iv) m8

.

6 2-n+1 ; 2.

, 5 3

8 in.

12 2(n + 2) 2(n + 3)' 12*

9 $(w 2 + n + 2). 10 -7T.

$ 1



ANSWERS (33)

31 (l + 3a;+a: a )e .

33 (i) |(cha; + cosa;); (ii) i(ahx-ainx). 34 1-609.


1 5 + 11*. 2 -1 + 3*. 3 7 + 3*.

4 7 + 3*. 5 -9 + 19*. 6 41 + 0*.

7 tt+A*. 8 9 2i, -4, -8*.

10 -10. 11 cos (0 + 0) + * sin (0 + 0).

12 cos (0-0) + i sin (0-0).

14 X= 2+,y :- 1

„ , Y= 2y

{x-l)* + y*' {x-l)* + y2

'

15 o, + (±*=°E\\ 16 2 + *.

17 (a?3 -3a^/ 2

) + *(3a; 2 y-t/ 3); (a;

3 - 3a>*/2

) -i(3x 2 y-y 3).

26 {ad —be, bd], {ad, be}.

Exercise 13(6), p. 4981 2(cos^7r + *sin|7J-). 2 2(cos( —£7r)+*'sin( —£tt)).

3 V2 rcos|7r+*'sin|7r). 4 V2(

c °s( +*sin( -%nj).

5 2(cosf7r + *sin^7r). 6 cos \n + * sin \tt.

7 cos \tx + * sin

8 2 cos £0 (cos £0 + * sin \Q) or - 2 cos £0 {cos (\d + 7r) + *' sin (\Q + 7r)}.

9 2 cos i(a - /?) {cos £(a fi) + * sin £(a + /?)} or

- 2 cos £(a - /?) {cos £(a +/? + 2?r) + * sin £(a +^ + 27r)}.

10 (a;2

+2/2

)n

. 13 (w-l) 2 + « 2 = 9.

16 (£C + t/)(a? + W2/)(o; + w22/); (x + coy) (x + (o 2

y).


2 2z s + z 1 = 32 a . 3 (Zz x + A;z a )/(Z + A;)-

4 The region between the circles of centre ( —1,2) and radii 2, 3.

A l d



(34) ANSWERS


2 1 ± *, 1 ± V3. 3 - 2, 2 ± V3. 4 o),(o, (o\ w2, f.

5 ±V

2 ±V

3 ' <°, 0)2 ' 6 The root z = —b/a is real, so /?

7 (ii) The division process introduces coefficients of the same type a

divisor and dividend.

9 pr-4s. 11 -3. 12 +4, 5.

13 -4, 2, w, (oK 14 -4,-1,2,5.

15 2/8 + 3(a-/fc)y 2 + 3(fc 2 -2afc + &)y-fc 8 + 3afc 2 -3&fc + c; a.

16 = 1 or -3; x = -2, -4, 1±V15-

17 = 2; 2/4 + 22/

3 —9t/ 2 —16t/ —6 = 0.

18 (i) y3

-2qy 2 + (q*+pr)y + (r*-pqr) = 0;

(ii) ry a -(q + 3r)y 2 + (p + 2q + dr)y-(p + q+l) = 0.

20 q2 (y+l) 3 +p*(y + 2) = 0.


If. 2 |,-2. 3 None.

4 (-2,-1), (0,1). 5 (-1,0), (1,2). 6 (1,2).

7 Boots in (ax ,

a2 ),

(a3 ,

a4 ),

(a5 ,

a6

). 1 1 ( 1,2),

(

—6, —5).

12 (2,3). 14 One root. 15 8 < < 11.

16 - 47 < k < 98. •


1 2-0946. 2 0-2261. 3 -2-0945. 4 4-25.

5 1-52. 6 0-3399, 2-2618,-2-6018.

7 1-3569, 1-6920, -3-0489. 8 0-567. 9 0-057, 1-468.

10 1-3503. 11 4-4934. 12 1-87.

14 K+1/2K*-1I4JZ:*.

15 Newton's method fails; but put x = 2nn + h and expand.

Miscellaneous Exercise 13(g), p 525

2 s + ic. 4 r 2 +/) 2 —2r/>cos(0 —a).

7 (ii) i(ll-t), -i(3 + 5i).



ANSWERS (35)

12 a;2 +2/ a -(a/w)a; = 0, a;

a +2/2 + (a/v)t/ = (if u 4= 0, v 4= 0).

14 If P describes |z| = 1 counterclockwise, Q moves from —oo to + oo along Oy.

16 y = or (c-a)(a; 2 +y 8) + 2(d-6)a; + (ad-6c) = 0.

18 (i) 3 £ as** 8 ; (ii) 32a 3l >_ 2 a:

3 »- 2, Bâ^x 3 ^.

p=i p=l p=i

20 b 2 c-abd + d* = 0; or b = = d and a 2 5* 4c.

22 a;8 - 3a: 4 - 6a: 3 + 14a; 2 - lx+ 1 = 0. 24 * > p*.

25 6= l-|+i-...-l/w. 26 -f, -|,i2.27 1, i(3±V5), £(1±*V3). 28 - 1, i(l ±ijZ), i(- 1 + iVl5).

29 (n + i)7r + 4(-l)»- 1 /(2n+l)7r.


2 (2cos|0) 6 cis70. 5 32,-fTT-.

(8/**4 1 \

tt), r = 0, 1,2, 3.

8 cis(£0 + p7r), & = 0, 1,...,5; cis(£0 + for), k = 0, 1.

9 cos 30 + i sin 3(9 = cos 3 + 3i cos 2 sin0- 3cos0 sin 2 0-isin 3 0;

tan 30 = ( 3 tan - tan 3 0)/( 1 - 3 tan 2 0).

10 (i) cos0 —isin0; (ii) 2cos0; (iii) 2*sin0; (iv) 2cosn0; (v) 2*sinn0.

/2k+l \ /2k+l \11 cisl n], k = 0, 1, ...,n— 1. 12 —i cot ( — n) , k = 0, 1, ...,n —1.

V w / \ 2n J

1 / faA13 -|l+icot— ), k = 1,2, ...,n-l.

2 \ n/

14 itan —, where & = 0, 1, ...,n— 1 if n is odd, and the value k = in is

excluded if «, is even.

(2kn \—^-±ol\, k — 0, 1, ...,n- 1.

18 z 2 + z + 2 = 0. The + sign is correct since

sinf7r + sinf7r = sinfTT— sin^7r > and smfyr > 0.

20 (i) ]zj = constant; (ii) argz is a rational multiple of n.




(36) ANSWERS

7 -i(ffrsmlO0-isin80-£sm60 + 2sm40 + sm20-60) + c.

8 Th(w + fW3).

9 (i) 32c 8 - 48c* + 18c 2 - 1; (ii) 1- 18« 2 + 48* 4 -32s 6 .

10 7s-56s 3 + 112s B -64s 7. 11 1 - 24s 2 + 80s 4 - 64s 8

.

12 = 2nn±$iT. 13 = nn or ( l) n %n + nn.

16 (i) (a) (- l)* s n, (-l)* - 1 wcs n - 1

; (6) ( - l)k n -»tu» n - 1, ( - l)*<»-«««.

(ii) (a) (-l)* - 1 ™* - 1,

(-1)* * ; (6) ( - l)*< -««», ( - l)k»-»frf»-i.

17 See answer to Ex. 14(a), no. 9; tan -&n, tan -fen.

18 (i) £* = SWt?(ii

) (c —o)/( 1 —6 + d) ; S0 X is zero or an integral multiple of n.

\ / e»n— 5/.n—20 (v) (a) 2 n - 1 c n -n.2 - 3 c B - 2 +- 1 -T —-2 n - 6 c2

3

. I n 2 n 2 /n 2 -2 2 ^ n 2 fw 2 -S(6)

n(n —4)(w —5) „ „m _ / 2 - 7 c n - 6 + ...;

, f n 2„ n»(n -2») , n (n«-2« n 2 -4 2

„

v y

\ 2 4 6 J

, i,fw(w 2 -l 2

) ,n(n 2 -l 2 )(n 2 -3 2

)

\ 3 5


1 (a 2 - 2z cos \tt + 1) (a: 2 - 2a: cos |7T + 1) (a?2 - 2a; cos %n + 1).

2 (x 2 - 2 V2 a; cos -^tt + 2) (a;2 + 2 ^2a; cos J^7r + 2) (x 2 - 2 ^2* cos + 2)

x (a;2 + 2 V2a; cos &n + 2).

3 (a; 2 - 2a; cos tt + 1) (a:2 - 2x cos $n + 1) (x 2 - 2x cos |7T + 1).

4 x = 1, a = 2j3 + 7T/n. 5 a; = - 1, a = 2^.

6 » = -l, a = 2fi + n/n. 7 rtann^ = £ cot{^+- —-——r=0 I * w ;

8 Derive wo fi the result of 14.33, ex. (iv).

10 2 |- cosec na sin â + —̂j^ja; 2 - 2a;cos â + ^^ + lj.



ANSWERS (37)


1 8x z -6x-<j3 = 0. 2 16z 4 -20a; a + 5 = 0.

3 4r 2 + 2a;-l = 0; i(V5±l). 5 (i)

y

6 - 7y4 +

Hy2 - 1 = 0.

9 1.


1 —sin nff sec£/?cos{a + (n — 3 2 n cos n sinn0.

4 cosntf. 5 {2nx n (x-l) + (x+l)(l-x n )}/(x-l)K

6 jsinJ.-^ sin(w+l)^l + ^ sinwJ.|^l -2^cos4 +^7 cosec cos (a —6).


1-1. 2 ie(l+*V3). 3 rcistf. * 4 2cosl.

* sin ten,

5 2e co80 cos(sin0). 6 2ie oa sin6a;. 7 S af. all a;.

r-i r

8 e* 008 6 sin (a; sin (9), all x, 6. 9 e- cos '^cos(a-sin/? cos/?), all a, /?.

10 e* sin (a; tan /?), all x. 11 cos (sin 0) sh (cos 0), all 0.

12 \e- Je 00825 cos (sin 20). 13 Line v = 0; line v = u

14(i)

w2

+ v2

=e 2x

;(ii)

wsini/ = vcoay;circle

and a diameter.

Exercise 14(g), p. 560

4 cha; cosy + isha; siny.

5 2(cos x ch y + i sin x sh y) /(cos 2x + ch 2t/).

6 (sh2a; + tsin2t/)/(ch2a: + cos2y).

8 (i) x + nni; (ii) a; + 0i or n7T + yi. 9 (n + n + \i log 2.

10 (2w7r ± ^7r) + *, both + or both — signs being taken.

12 (i) Hyperbola, (ii) ellipse, in the (w,v)-plane.

13 —+—+ -—-+..., all z. 14 sh(cos<9) sin(sin0), all 0.3 7 1 1

15 £{sh (cos 0) cos (sin 0) + sin (cos 0) ch (sin 0)}, all 0.

Miscellaneous Exercise 14(A), p. 561

4 i k = i \ i (



(38) ANSWERS11 8c 8 -9c 8 + 1 = 0; = 0, \n, |tt, 2n.

16 -i, -Kl+*cotir7r). r = ± 1, ±2, ±3. Put 2 = - sin 2 0.

17 z —acis|r7r, r = 0, 1, 5.

v = {(»— 1) sin a —y cos a}{x-l) 2 + y-

1

23 cos u = cos £0 —sin £0, chv = cos £0 + sin £0.

25 Denoting the vertices + a, + a + *a, —a+ia, —ahyA,B,G,D respec-

tively, then as z moves from O to A, w moves along the w-axis from to (a 2,

as z moves from A to B, w moves along the parabolic arc v 2 = 4a 2 (a 2 —u) from(a 2

, 0) to (0, 2a 2); from B to G, w moves along the parabolic arc v 2 = 4a 2 (a 2 +

from (0,2a 2

)through (

—2

, 0) to(0,

-2a 2); from G to D, w moves along a

v 2 = 4a 2 (a 2 —u) from (0, —2a 2) to (a 2

, 0) ; from D to O. w moves along the M-axi

from (a 2, 0) to 0.

27 \e-\ + \e eos 206 cos (sin 2a), all a. 28 ch (a; cos 0) sin (x sin 0), all x, 0.

CO3}

n p n

29 2 cosn0, where r = + ^(a 2 + b 2) and cos0:sin0: 1 = a:b:r.

31 z = e^^Ce^'+De '/ ), where /*a = n 2 + fc

2 and C, D are arbitrary complex constants.

33 x' = x cos 6 + y sin 0, ?/' = —a; sin + y cos 0.


6 a; + y = 2c cosec w. 7 # + = csec2

Jw.

8 x 2 + 2^ cos <y + ?/2 = 4c 2 cosec 2 o>. 9 Straight line.

1 1 h/x + kjy — 2. 12 Circle with centre 0.


1 x-2y = 0. 2 3x-5y + 9 = 0. 3 10x-9y = 5.



ANSWERS (39)


1 V2 - 4 bx 2 -2kcy + ay* = 0.

5 (i) tan- 1(2^/21); (ii) llaj a -26a^-llt/ a = 0.

7 pq(x* + y*) = (p* + q*)xy.

8 (i)47r, (-1,4); (iijtan- 1**, (-•&.-*); (iii) K ( - 2, 3)

(iv) tan- 1 ^,^), (1, -2).

9 (i) 3a; a -7«y + 2y a = 0; (ii) 2x* + xy- I5y* = 0; (iii) xy = 0;

(iv) 2« 2 + 4»2/-2/ 2 = 0.

10 3cc a + 40a*/ - lty 2 = 0. 12 2(flr - g') x + 2(/-/') j/ + (c - c') = 0.

Exercise 15 (d), p. 592

1 x 2 + y*-2px-2qy = 0.

4 (i) (* + a; a ) x + (</i + s/ a ) y = x^x % + y x y %- a 2

, xx x + yy x = a 8.

5 (aH/n t a*m/n). 6 c 2 = a 2 (l+m 2).

7 (i) x\ + y\ = 2a 2; (ii) x 2 + 2/

2 = 2a a.

8 (i) Transverse common tangents divide the line of centres internally in

the ratio of the radii.

(ii) 5x± 12y = 29, Ux±4:j3y = 61.

9 {(Ti+r^x-ia^ + a^ = {fa-a^-^+r^y*.13 x* + y*-3x-5y+l = 0. 14 a 2

+2/2 + 2a;-82/ + 5 = 0.

16 Straight line. 17 8 —s' = is a diameter of tr.

20 * 2+2/

2 +a;-42/-2 = 0. 21 x*+y 2 + x + y-2 - 0.

1 -2,0.

4 New equation is x

6 z a + 3t/ a = 1.


2 x 2 -xy-6y 2 = 1. 3 xy = —£a 2.

5 Jtan- 1 ^

Miscellaneous Exercise 15(/), p. 601



(40) ANSWERS19 (1 + 2gl + cl 2 )x 2 + 2(fl+gm + clm)xy + (l + 2fm + cm 2

)y2 - 0;

/ b(b-a)~2mh m(a-b)-2lh \ _\2{am 2 -2hlm + bl 2 )' 2(am 2 -2hlm + bl 2 ))''

™X+ V ~ '

20 (l-h)(x 2 + y2

) = h(x + y); (a + k) (x 2 + y2

) + k(x —y) = 0.

21 6(a; 3 + 2/3

) = Zaxy(x —y).

Exercise 16(a), p. 60S

1 2/(f 1 + * 2 ). 2 a(t+l/t) 2. 10 Focus A, directrix


5 y = x + a, x + 2y + 4a = 0. 10 The directrix.

11 Each is equivalent to t x t % = —1. 12 y2 = 4a(x + a).

15 2:1. 17 (i) Equally inclined to the axis.

19 xa* + ybi + (a 2 6 2 )* = 0.

20 $y = 2(* + |); 2y = 4a+l, 6y = 9x + 4.

25 The directrix.


3 y 2 = a(x-Ba). 4 s = -t-2jt. 6 (-2a, 0).

7 x —y = 3a, 3a; —t/ = 33a, 4x + y = 72a.

13 x + 4a -0,y 2 = 16a(a;-6a), y2 = 2a(#-4a).


1 y = 3x. 2 -f.


1 Mid-point of OG, where O is the point of contact and C is the centre;

perpendicular to OG at distance \OG from O on the side remote from G.

r 2 + vq2 x 2 +y 2 -(p + g)x —y+pq = 0.

T

/ _b 4ac-b 2 \ 14

\~2a'~~ 4a-

)'' H*8 *'

P



ANSWERS (41)


1 26 2 /a. 2 PN' 2: B'N' . N'B = OA2

: OB 2.

5 Ellipse with semi-axes a, 6. 8 %a, \b.

10 (i) ay = bxt&aa; (ii) x 2 /a 2 + y2 /b 2 = cos 2 a. 15 (i) -30.

17 (a/a) cos<fi + (y/b) sin = 1. 20 nab.


2 d-<j) = ±7T.

4 Tangents at corresponding points meet on the major axis.

11 ON'.OT' = bK 19 (a(a2 -6 2 )/(a 2

+6 2

),0).

22 \x + %y - 1; x+y-S= 0, a;-5t/ + 9 = 0.

24 Tangents at the extremities of a focal chord meet on the corresponding

directrix.

»l*4-i)$4-i)-@+*-i)->*+*-*+*.


1 b 2 .OQ' = -a 2 e 2 .PN.

4 1-— = -. 5 (1. 2).(2a 2 -6 2

)2 6 2 4

V'

1

7 a 2 x 2 + b 2y 2 = a*6*/(a 2 - 6 2

)2

. NN' is the normal at the point

(Aft cos (it —0), Xa sin (tt —0)), where A = ab/(a 2 —b 2).

Exercise17

(J),p.

641

1 5x + 3y = 0. 2 -\. 4 Ji&tf + b 2)}.

10 (ii) aW + btm 2 = 2n 2; (ii) x 2 /a 2 + y

2 /b 2 =


1 x 2 /a 2 + y2 lb 2 = 1.

8 A hyperbola having focus S and the given circle for auxiliary circle;



(42) ANSWERS20 b^h 2 - a 2

) x 2 + 2a 2 b 2 hkxy + a*(k z - b 2) y

2 = 0.

23 (-2,1); 2,V3; i;(

- 1, 1),(

-3,

1);x

= 2, x =- 6.

24 3, 2V2; f; (2,f), (0,*); x = 10, a = -8.

25 (0,1); 2, V3; £; (*,*),(-*,*); s-y-3 = 0, + 5 = 0.


2 A branch of a hyperbola with foci A, B: if .45 > m>, the branch enclosin

B; HAB < nv. the branch enclosing A. If AB = nv, the perpendicular bisect

of^B.3 That branch of a hyperbola with foci A, B which contains the centre

the smaller circle.

4 (i) Referring to fig. 182: - oo < t < - 1, A'Q'; - 1 < t < 0, PA; < t

AQ; 1< t < +oo : P'A'. (ii) (a) > 0; (6) < x « a = 1.

13 (1 + x/a + il-WJy/b = h + t z .

14 (l + * 2 Wa + (l-* 2 )y/6 = 2t.

15 (l-* 2 )aa;-(l+« 2 }fo/ = (a 2 + 6 2)

(1 -< 4)/2*.


12 3x 2 - 7*2/ + 2y 2 + 5x - 5y + 3 = 0.

13 2x + y-Z - 0, x-3i/ + 2 = 0; 2* 2 -5^-3i/ 2 + a; + ll2/-4 = 0.


4 Write the equation as xy + ?/£C = 2c 2.

31 (-d/c,a/c). 33 » = 4i>, y = i?-

Miscellaneous Exercise 18 (a*), p. 661

3 x 2 \a 2 -y 2 \b 2 = 1 (a> 6).

5 Branch of a hyperbola which contains the centre of the smaller circle,

both contacts are external or both internal. Ellipse if one contact is extern

and the other internal.

6 Foci A, B in either case. 10 (u + w) xja + (u-w)y/b + v = 0.

12 (x lt—

2/ x ). 14 See 18.53, example.

2 b



ANSWERS (43)

22 The case k = in no. 21 (ii): the same argument applies.

23 (0,0); 2, 3; ^13; (0, ±^13); y = ±4A/13; y = ±f».

24 (-1,-2); |,2; f; (|, -2), (-£, -2); a; = * = -if;

4#- 3y = 2, 4aj + 3y+10 = 0.

25 (0,1); ¥; ( + ¥,i); * = ±*; y=i±¥«-


1 -1 + ^-1. ^ + ^-1. 2 y 2 -4ax l ,y 1 y 2 -2a(x 1 + x 2 ).

a 2 b 2 a 2 o 2 1

3 x^-c 2, ^(x^ + x^-c 2

.

4 (Zajj + my l + n) (l'x l + m'y x + n') ,

i(lx + my + n) (l'x 2 + m'y 2 + n') + l(lx 2 + my 2 + n) (l'x t + m'y x + n').

5 8 = passes through the mid-point of P X P 2 .

8 pb - 2rh + ga = 0. 9 (ax 2 + 2hxy + by 2) (a + b) = 2(ab - h 2

) (x 2 + y2

).

11 (ax 1 +hy 1 +g){x-x 1 ) + (kx 1 + by 1 +f)(y-y 1 ) = 0.

13 (i) ax 2 + 2hxy + by 2 + gx+fy + c = 0; (ii) ax 2 + 2hxy + by 2 + c = 0;

(iii) gx+fy + c = 0.

14 (*/o) sin^

+ (y/6) cos<f>

= 0; at the point at the end of the question.

15 Either h * 0, or h = and Z = 0.


1 3aj 2 + 3a«/ + 2?/ 2 -9a;-22/ = 0.

2 165a: 2 - 294a«/ + 120?/ 2 - 273a; + 691*/+ 1572 = 0.

3 (a-b)h' = (a'-b')h.

4 lla; 2 -25a^+llt/ 2 = 0, (x + y)2 = 47, (a-y) 2 = 3.

/ a 2 -b 2 \ 2 /6 2 \ 8

• 8 * 2 + 2/2 + 2ae 8

ar = a 2 (l-e 2 -e 4). 9 la —I +y 2 = I —I .

11 - cos —7 sin ci = cos 20; j cos 3 0, —sin 30j;

a o \ a b j

(a 2 sin 2 + 6 2 cos 20)

lJ

% ah



(44) ANSWERS

Miscellaneous Exercise 19(c), p. 682

8 (a 2 m2 + b 2) (x 2 - y

2) = ( 1 - m2

) a 2 b 2.

10 25x 2 - (k + 1)y

2 - (k + 26) y + (Qk+ 169) = 0; y = x 2 + i£±.

12 ^12^1 — ~ ®*


2 r 2 r 3 sin (0 2 - 3 ) + r 3 r x sin (0 3 -X ) + sin (0 X - 2 ) = 0.

5 (i) (a) r 2 -2/orcos((9-a)+p 2 cos 2 a = 0; (6) r = ± 2asin0;

(ii) tangents at O are Oy, Ox.

6 ( V( a2 + 62)> tan- 1 (6/o)).

10 r cos (0 —X — a + a) = 2a cos (0j —a) cos (0 2 —a);

r 008(0-20 + a) = 2a cos 2 (0x-a).

11 Circle through 0. 12 6 2 c 2 + 2ac=l.


1 a = Z/(e 2 - 1), 6 = Z/V(e2 - 1). 2 Parabola 2a/r = 1 - cos 0.

3 Conic with I = ifc/a, e = (b 2 + c 2) */a. 5 (2 - e 2 )/2Z.

8 l\r — e cos (0 —y) + sec \{<x —/?) cos {0 —£(a + /?)}, which is therefore obtain

able by writing a —y,

/? —y, —y

for a, /?, in the standard equation.

9 (Z,£tt).

15 Concavities of the arcs LL' may be the same ( —) or opposite ( + ) in sen

18 (ii) 2Zsec0, sec^.

21 The directrix Ijr = cos of the parabola Ijr = 1+ cos 0.

25 ( 1 jp) cos (0 - a) + (d6/dr) a sin (0 - a) = 1 /r.

Miscellaneous Exercise 20(c), p. 698

3 (a sin a —6 sin /?) sin = (6 cos /? —a cos a) cos 0;

(a6/r)sin(a —/?) = asin(a —0) —6sin(/? —0)

4 r 2 -{acos(0-a) + &cos(0-/?)}/- + afecos(a-/?) = 0.

6 (2-e 2)/Z. 8 (ii) r = asec 8 Ja cos(0-^ Ja).

9 Ife < l,a = ±cos _1 (— e); ife ^ 1, there is no such a; r = Zcosec 2 0.

12 ( 1 —e 2) r 2 + 2eZr cos = Z

2; the auxiliary circle if e 4= 1 > the directrix if e

14 e2/ 2 + Z# = 0.



ANSWERS (45)

7 2V46, 2V38. 8 3:4,5:7,-2:1.

13 x = pcos^, y = psintfi; p = *J{x2 + y

2), cos^:sin0: 1 = x:y:p.

14 (i) On a cylinder with axis Oz, radius h; (ii) on a half -plane through Oz;

(iii) on a line parallel to Oz.

15 x = rsin# cos0, y = rsin# sin0, z = rooad;

r = J(x 2 + y2 + z 2

), cos <j> : sin ^ : 1 = x : y : *J(x2 + y

2),

cos : sin 0:1 = z : ^/(a; 2 + f/2

) : r.

16 (i) On a sphere with centre 0, radius k; (ii) on a cone with vertex O andsemi- vertical angle a; (iii) on a half -plane through Oz; (iv) generator of cone;

(v) great semi-circle of sphere; (vi) small circle of sphere.


5 1:1:1, 1:-1:1, 1:1:-1, -1:1:1. 6cos-H.

7 cos- 1{(6

2 + c 2 - a 2) /(a 2 + b 2 + c 2

)}, etc.

109 -7

10 (p2x piy p * z

\\0P 2 ' OP 2 ' OP 2 }'

15 Each is equal to OP 2.


1 x-4y + 5z = 0. 2 6x + 2y-10z = 31. 3 3x + 3y-z = 6.

4 5a? + 2y-3z = 17. 5 6x + 3«/- 2z = 18, 2a;- 3y- 6z = 6.

6 (i) cos 1(1/^14); (ii) sin-^l/Ve).

7 cos-^c/Vttp-g^ + fp-r^ + c 2}], cp/(p-q), cp/(p-r).

8 cos- 1 {(a- 2 + 6 2 - c- 2 )/(a~ 2 + 6~ 2 + c~ 2)}.

9 (aJi-âj + ^-^Jy-KZi-z^z = l{x\-x\ + y\-y\ + z\-z\).

1 3 - (A Xl + By x + CZi + D)/( Ax 2 + %2 + Cz 2 + Z>).

Exercise 21 (d), p. 721

1 (11,22,19); 5V38.

2 (0, 6 —om/i, c —ero/Z), (a —bl/m, 0, c —bnjm), (a —cZ/w, 6 —cm/n, 0).



(46) ANSWERS

_ x-3 y-2 2 + 3 rt . . , A , .

16 — = _ = j- (both upper signs, or both lower).\2 + 1 *J2+1 —

*J2

17 2. 18 2a; + 60y- 1622 + 27 = 0, 54a- lSy- 62+1 = 0.

19 ±-^— V(^ + w2 + ^ 2 )- 20 6 x volume of OP^P^.2lmn


1 12a;-lty + 42 = 1. 2 3x + ±y-2z = 7.

3 9x-3y-z+U = 0. 4 2:-l:l.

a;+ 1 v— 1 2—15 3a;-% + 72 + 4 = = 3a; + 2?/ + 2, —_ = 2-^- = _ '

6 7:-ll:10.

7 3a; + 2(3 ± a/3) t/ - 3(4 ± *J3) z = (both upper or both lower signs).

8 2x-y-z + 3 = = x + y + z-6; 1 : : - 1 or 1 : - 1 : 0.

9 x-y + z+1 = 0, 4x- 6^-2-3 = 0, lx-9y + 2z - 0.


1 10:3:-4. 2 ^ = *±* = ^; (4, 2, 6), ( - 1, - 1, 0).5 3 6

3 (3,0,0), (ff,0,0). 5 2V3. 6 5.

z-1 2/-2_2-3 x-2_ 2/ -^_2-ff_7 13; —- —- —. 8 —- —- —, Vi7-

9 See 21.52, ex. (iv).

Miscellaneous Exercise 21(g), p. 733

I3t 4t 156-12^ 144

18:5, 4 \13'13' 13~~ J' l3 5

V709325

x y z

2=

1=

i'6 —- —~ = ~l (t> -i, -*). 7 60°; * + 2/-2 = 0.

8 {I ±V) x + (m ±m') y + (w ±n')z = = S(mn' - m'w) x (all + or all-).

9 (0,0,0), (1, 1, -1). 10 »-42/+62 = 106; ^293.

11 lx-8y + 3z = 0; 13a; = 122, 20y + 23z = 0.

12 (,-^),a-i) + ( y - 6

f6

)(6-e) +(

S -^)(o-/) = 0.



ANSWERS (47)

3 2(x 2 +y 2 + z 2 )-lx-6y-9z+lB = 0.

4 x 2 + y2 + z 2 + 2x-2y + 2z-% - 0. 5 3(x 2 + t/

2 + z 2) - 2a; - 2t/ - 2z - 1 = 0.

6 6(z 2 + 2/2 + z 2 )-18a;-36i/-12z = 0; 7.

7 3(x 2 + y 2 + z 2 )-12x-6y-4z = 0. 9 (»- 5)2 + (t/ ± 4)

2 + (z ± 4) 2 = 41.

10 (2,4,4), (f, 4, ft). 11 (-1, -5,|), |V14.

12 (*,*,*), i; (2,2,2), 2.

13 81(x 2 + y 2 + z 2 )-126(x + y + z) + 98 = 0. 15 2x + 2y±z-3.

16 a; + 2y - 2z + 15 = 0, 4y + 3z - 1 1 = 0. 17 (Zr 2/p, mr 2

/p, rar 2/p).

18 {lu+mv+nw-p) 2 = (l 2 + m2 +n 2 )(u 2 + v i + w2 -d).

19 (a;-5) 2 + (2/ + 2)2 + (z-3) 2 = 49; 2x + 3y + 6z = U. 21 (A>A>-A)-

22 (ii) Tangents from a point on the common circle have equal lengths zero.


1 (0, 0, 0). 2 Plane touches circle.

3 4. 4 x 2 +y

2

+z 2 -8x-6y = 0.

5 3x + y-3z = 3, x 2 + y2 +z 2 -x-3y+z = 0; (A, tf, -*)•

6 (2,2,3).

8 a 2 + 2/2 + z 2 + 6z-16 = 0, 5(* 2 + 2/

2 + z 2 )-14a;-28y-12z + 32 = 0.

9 x 2 + y2 + z 2 -2z-8 = 0, a;

2 + 1/2 + z 2 + 2(te + 20y + 18z + 72 = 0.

10 x 2 + y2 + z 2 + 2x-8z = 0.

12 a;2 + 2/

2 + z 2 + 2a2/-a 2 = 0, a;2 + 2/

2 + z 2 + 4oa;-2az + a 2 = 0.

13 5(* 2 + 2/2 + z 2

) + 19a;-132/+10z + 27 = 0; 3/VlO.

14 * 2 + </2 + z 2 -2a; + 2</ + 6z-5 = 0. 15

16 a;2 + y

2 + z 2 + a; + 92/-4z = 0.

18 Cylinder with generators parallel to (i) Ox, (ii) Oy; surface of revolution

with axis (iii) Oy, (iv) Oz.

19 10« 2 +12a;i/ + 52/2 = 16.



(48) ANSWERS

Miscellaneous Exercise 22 (</), p. 756

2 x* + yz + z 2 -2cz = 0.

3 (2

+ JV2,-1

+ 2V2,£V2 )>

£(7

V2-6).

4 ^ = | = ^, cos- 1 . 5 (-alln,-am/n). 9 |Vp~ V«l-

a D O

10 (SwZ) 2 = dSZ 2; (AZ,Am,An) where A = —dj(Lul).

12 (a? ±4) 2 + ±4) 2 + (2 ±3) 2 = 25.

13 a;2 + 2/2 + 2 ;2_2 a ;_42/

_4 2 -7 = 0,

5(a;2 + 2/2 + z 2) + 22a; + 442/ + 44z-323 = 0; (3,6,6).

' 2a 2 w 2a 2 v a(u 2 + v 2 - a 2 )'

w2

+ v2

+ a2'

u2

+ v2

+ a2 '

W2

+ y2

+ a2



xxiii

INDEX TO VOLUME II

Numbers refer to pages.

* means 'Also see this entry in the Index of Vol. I\

Absolute convergence (a.c.), 452, 551

A.c, 452, 551Affix, 495d'Alembert's

ratio test, 443, 456theorem, 507

Alternate suffixes, rule of, 587Altitude of spherical triangle, 756

am, amp, 495Amplitude, 495Angle

dihedral, 713, 748of lune, 748trihedral, 748

Angle betweencircles, 591curves, 747line-pair, 580planes, 713sensed lines, 707spheres, 740

Antiparallel, 707Antipodal points, 747A.P., sines or cosines of angles in, 431,

548Apollonius

circle of, 504hyperbola of, 658

Approximations, 476formal, 479

Area

and orthogonal projection, 627of lune, 748of spherical triangle, 750of triangle, 568, 719

arg, 495Argand, 488Argand representation, 488, 499, 504

difference, 501loci, 500, 504product; 502quotient, 503

Auxiliary circle, 623, 652

Axes (of coordinates)

change of. 595left-handed, 700oblique, 565rectangular, in space, 700right-handed, 700rotation of, 596, 597, 709

Axisconjugate, 654major, 621minor, 621of circle on sphere, 747

of parabola, 603transverse, 621

Bifocal property, 624, 647Binomial

coefficients, 424, 466series, 463, 466

Binomial theorem, 419, 434general term, 420greatest term, 421

use trigonometrically, 534

c, 437C + iS method for

integration, 555series, 548

Cardan and complex numbers, 487Cardan's method, 390

Centre ofcircle, 586ellipse, 621hyperbola, 621line-pair, 584, 597sphere, 736

Centroidand coordinate axes. 601of tetrahedron, 704

Ceva's theorem, 574Change of



xxiv INDEX

Chord of

circle (polar equation), 686conic (polar equation), 693ellipse, 630, 631hyperbola, 648, 651, 658

parabola, 605, 606a = 0, 665

Chord of contact, 587, 610,

649to s = 0, 668

Circle

and ellipse, 622Cotes's properties, 542equations in space, 742general equation, 586great, 737, 747

of Apollonius, 504of curvature, 611, 679, *on given diameter, 586polar equation, 686small, 737, 747

Circles, family or system of, 593cis, 530Cofactor, 402

alien, 403true, 403

Colatitude, 702Collinear points, 570Comparison

of series and product, 543series, 440tests, 439

Complete quadrilateral, 577

Complex algebra

and real algebra, 492completeness of, 492

Complex numbers, 489and laws of algebra, 494conjugate, 497

'real', 496Compound angles, 537Concurrence of lines, 576, 577Concyclic points on

ellipse, 628, 678hyperbola, 660parabola, 604

Cone, 745Confocal conies, 662Conic, 594

general, 664

standard equations, 600system of, 675

Conjugate diameters, 640, 649, 655, 67extremities of, 640, 656

Conjugate

hyperbolas, 654of z, 497

633, surds, 511Conormal points on

ellipse, 636hyperbola, 661parabola, 613

Contactdouble, 611, 675of two conies, 679mth-order and (m+ 1) -point, 680, *

Contact condition for line withcircle, 587ellipse, 633, 634hyperbola, 649parabola, 609s = 0, 667sphere, 739

Contact condition for plane and sphere, 7Convergence, 436, 551

absolute, 452, 551

conditional (c.c), 454interval of, 456speed of, 446

Coordinate planes, 700Coordinates

cylindrical polar (p, 0, z), 701

plane cartesian and polar, 565, *rectangular cartesian in space, 701

spherical polar (r, 6, <j>), 702Coplanar lines, condition for, 717Coprime, 384Corresponding points on ellipse a

auxiliary circle, 623

Cosine rule, 751Cotes's properties, 542Cramer's rule, 404Cube roots of unity, 497Curvature, circle of, 611, 679Curve, skew, space or twisted, 745Cyclic

expression, 373interchange, 373order, 373

Cylinder, 745



INDEX XXV

Determinant (cont.)

elements of, 394expansion of, 394, 395factorisation of, 411

fourth-order, 414leading diagonal of, 395leading term in, 395minor in, 402multiplication of, 417notation c, r, 397row and column operations, 393rows and columns, 393second-order, 393symmetric. 403third-order, 394transpose of, 393

•triangular', 399Determinants and linear equations, 404Diameters, 615, 639, 649, 669

conjugate, 640, 649, 655.. 670equi-conjugate, 641extremities of, 640, 656ordinates to, 615

Difference method, 423Dihedral angle, 713, 748Direction

cosines, 706ratios, 707

Director circle, 633, 649, 697Directrices, polar equations of, 691

Directrix, 594Discriminant of 8, 583Distance formula

cartesian, 566, 702polar, 684

Distance of a point from aline, 575plane, 718

Distance quadratic, 617, 638, 670

Divergence, 437proper, 437

Divisionexternal, 567, 703internal, 566, 703long, 363successive, 384

Dominated, 518Double contact, 611, 675

e irrationality of, 470 *

pin construction, 625

(jj, r) equation wo centre, 641

(p, r) equation wo focus, 634Elliptic trammel, 624

Equatingcoefficients, 367, 509real and imaginary parts, 496

Equationsapproximate solution of, 521, *cubic, 375, 390homogeneous linear, 407

linear simultaneous, 391, 404quadratic, 374quartic, 377reciprocal, 527transformation of, 513

with rational coefficients, 511, 516with 'real' coefficients, 510

Error in «„=«, 476Euclid's algorithm, 384Euler's

constant y, 450, *exponential forms, 555

E volute, 613, * •

exp, 480, 552Expansion

in power series, 461formal, 464, 479of a determinant, 394, 395of circular functions of compound

angles, 537of circular functions of multiple angles,

535

Factorrepeated, 381simple, 381

Factorisation

in complex algebra, 507

in real algebra, 515of a determinant, 411of a polynomial, 366, 507, 515, 538of sinnfl, 542

Finite series, 422trigonometric, 431, 548

Focalchord, 604distances, 624, 647radii, 634

Focus 594



xx vi INDEX

Functions of a complex variable, 550circular, 557exponential, 552hyperbolic, 558

Fundamental theorem of algebra, 507

Gauss, 489Gauss's theorem, 507General conic s = 0, 664General equation of second degree in x, y,

581, 664and line-pair, 582, 584

General linear equation(s)

in x, y, 571in x, y, z, 710representing the same plane, 711

Geometrical progression (g.p.), 437, 551G. P., 437, 551Gradient of a line, 567von Graeffe's method of root-squaring,

523Great circle, 737, 747Gregory's series, 463, 476

•

Hamilton, 493Harmonic

conjugate, 668division, 590series S(l/r), 438

H. O.F., 383process, 384theorem, 385

Highest common factor (h.c.:f.)» 383Horner's method, 522Hyperbola, 594, 621, 645, 658

asymptotes, 646asymptotes as coordinate axes, 657mechanical construction, 648of Apollonius, 658

(p, r) equation wo focus, 662rectangular, 647, 659

i, 487i as an operator, 488Identity, 364

of form, values, 367Image, 716Imaginary

expression, 487number, 487

Integration by C+iS, 555Intersect, 736Interval of convergence, 456Irreducible

polynomial, 385rational function, 386

Isomorphism, 491

Joachimsthal's ratio quadratic, 58% 610

635for * = 0, 666for sphere, 742

Latus rectum, 603, 621

Legendre polynomial P n (x), 519Leibniz's

series for \ix, 476theorem on alternating series, 451theorem on nth derivative of uv, 434,

Limit of

a complex function, 550x n \n , 455

Limits, calculation of, 480Line (see 'straight line')

Line of intersection of two planes, 717Linear system of equations, 391, 404

homogeneous, 407inconsistent, 391, 406indeterminate, 391, 407non-homogeneous, 408solution in ratios, 409three equations, three unknowns, 39two equations, two unknowns, 391

two homogeneous equations, three

unknowns, 409Line-pairs, 579

angle between, 580angle-bisectors, 581

centre of, 584, 597

coincident, 580, 582general, 581intersecting, 584intersection of, 584, 597necessary condition for, 582parallel, 582through origin, 580, 584

vertex of, 584, 597

Loci in the complex plane, 500, 504

Long division, 363Longitude, 702



INDEX xxvii

Medians of

spherical triangle, 752tetrahedron, 704

Menelaus's theorem, 574Minor, 402

Modulus, 495, 499Modulus-argument form, 494

of expz, 554de Moivre's

property of the circle, 542theorem, 528

Multiple angles and powers, 535

Net of conies, 681

Newton's theorem, 638Normal to

conic (polar equation), 698ellipse, 636hyperbola, 649, 651parabola, 612, 614plane, 710rectangular hyperbola, 659

Notationc, r, 397P x , 565, 701s, 8 U « n , * 12 , 664.

u r , s n , s, Ew r , 422Number, *

complex, 489imaginary, 487of a point, 495purely imaginary, 496real, 488'real', 496

Number-pairs, 489

Oblique axes, 565Octants, 701Operator i, 488

Order of a root, 381Ordered

pairs, 489triplets, 493

Ordinates to a diameter, 615Orthocentre, 609Orthocentric points, 659Orthogonal

circles, 591projection, 625, 626spheres, 740

Pappus's theorem, 679Parabola, 594, 603

general equation, 618Parametric representation(s) of

ellipse, 627, 628

hyperbola, 649, 650, 658parabola, 603

Partial fractions, *theory, 386use of, 430, *

Pascal's

theorem, 679triangle, 421, 470

Pencil of conies, 675Period of

expz, 554

generalisedcircular functions,

557generalised hyperbolic functions, 558Perpendicular from a point to a

line, 575plane, 718

Perpendicularity condition for

directions in space, 707, 708lines in a plane, 571, 581planes, 713

n, calculation of, 476II-notation, 539Plane

as a surface, 710, 744distance of a point from, 718of contact, 739, 742

Plane, forms of equation, 709general equation, 710intercept form, 713

P^Pa, 712perpendicular form, 710through P x perpendicular to l:m:n,

711Planes

bisecting angles between two planes,719

incidence of three, 724through a common line, 723

Polar coordinates (see 'coordinates'), *Polar equation of

circle, 686conic, focus as pole, 688, 690line, 684, 685

Polar of a point, 591, 612, 635reciprocal property, 591, 669





INDEX xxix

Sphere (cont.)

on diameter P 1 P 2 , 737Spheres through a given circle, 742Spherical

coordinates (r, 6, 0), 702distance, 747excess, 750

Spherical triangle, 748. 755altitude, 756angles, 748. 755area, 750cosine rule, 751four-parts formula, 758median, 752sides, 748, 755sine rule, 752

Straight line, distance of a point from,575

Straight line, forms of equation, 570, 684general equation, 571, 685gradient form, 570intercept form, 571joining (r lf dj, (r 8 , 2 ), 684P X P 2 , 572, 715parametric form, 572, 573, 715perpendicular form, 573, 685symmetrical equations, 715through P t in direction l:m:n, 714through P 1 with gradient m, 570

Successive division process, 384Supplemental

chords. 642formulae, 754relations, 750

Sumof finite series, 422to infinity, 423, 436, 551to n terms, 422

Summation by diagonals, 553

Surds, conjugate, 511Surface of revolution, 745Surfaces, 744

Takingarguments, 496conjugates, 497moduli, 496

Tangency and repeated roots, 608Tangent

and normal as coordinate axes, 671

general curve, 618hyperbola, 649, 651, 658parabola, 607s = 0, 666, 667

Tartaglia, 390Theory of Numbers, 386Touch, 608, 736, *Transformation of equations, 513Translation of axes, 596, 705Transpose, 393Trial exponentials, 555, *Triangle

formulae, 751inequalities, 501, *inequality for infinite series, 454

Unit ray, 706

Vectors, 493, 496Vertex of

line-pair, 584, 597parabola, 603

Vertices of ellipse, hyperbola. 621Volume of tetrahedron, 720

wo, xxi

[a, 6], 489c, 397c r , 424 C r , 419e, 470. *{I, m, n}, 706l:m:n, 707 and footnote

0'«2>(°°), P( —°°)> 518P tk

(x), 519

Pi, etc., 565, 701r, 397s, a lf s lv s l2 , 664a, s n , 422u r , 422\z\, 495z, z*, 497X , 496

||, is parallel to

_|_, is perpendicular to

Documents

Pure Mathematics (Sixth Form) Vol.2