SOLVED EXAMPLES Example – 1arbindsingh.com/wp-content/uploads/2012/04/Ellipse-5.pdf · to the tangent drawn at a point P on the ellipse. If F 1 and F 2 are the two foci of the ellipse,

LOCUSLOCUSLOCUSLOCUSLOCUS 58

Mathematics / Ellipse

Let F1 and F2 be the foci of an ellipse with eccentricity e. For any point P on the ellipse, prove that

1 2 2 1tan 1tan2 2 1PF F PF F e

e

Assume the ellipse to be 2 2

2 2 1,x ya b

and let P be the point ( cos , sin ).a b

y

xF2

P a b( cos , sin )

F1

21

F ae1 = (- , 0)

F ae2 = ( , 0)

Using the sine rule in 1 2 ,PF F we have

2 1 1 2

1 2 1 2sin sin sin( ( ))PF PF F F

2 1 1 2

1 2 1 2sin sin sin( )PF PF F F

1 2 1 2

2 2sin sin sin( )

a ae

1 2 will always equal the length of the major axis

PF PF

1 2

1 2

cos2

cos2

e

1 2 1 2

1 2 1 2

cos cos12 21cos cos

2 2

ee

1 2 1tan tan2 2 1

ee

This is the desired result.

SOLVED EXAMPLES

Example – 1



Let d be the perpendicular distance from the centre of the ellipse 2 2

2 2 1x ya b

to the tangent drawn at a point P on

the ellipse. If F1 and F2 are the two foci of the ellipse, prove that

22 2

1 2 2( ) 4 1 bPF PF ad

Let P be the point ( cos , sin )a b whereas F1 and F2 are given by ( , 0).ae

y

xF1F2

P d

By definition, the focal distance of any point on an ellipse is e times the distance of that point from thecorresponding directrix. Thus,

1 cos aPF e ae

cosae a

cosa ae

2 cos aPF e ae

cosa ae2 2 2 2

1 2( ) 4 cosPF PF a e ...(1)

Now, the equation of the tangent at P is

cos sin 0bx ay ab

Example – 2



The distance of (0,0) from this tangent is d. Thus,

2 2 2 2sin cosabd

a b

2 22 2

2 2sin cosb bd a

2 22

2 21 1 cosb bd a

2 2cose

22 2 2 2

24 1 4 cosba a ed ...(2)

From (1) and (2), we see that the equality stated in the question does indeed hold.

Find the radius of the largest circle with centre (1, 0) that can be inscribed inside the ellipse 2 2

1.16 4x y

The following diagram shows the largest such circle. Observe it carefully :y

x

R1

C(1,0)

R2

O

Note that the largest possible circle lying completely inside the ellipse must touch it, say, at the pointsR1 and R2, as shown. At these points, it will be possible to draw common tangents to the circle and theellipse.

Let point R1 be (4cos , 2sin ). The equation of the tangent at R1 is

cos sin 14 2

x y

cos 2 sin 4x y

Example – 3



If CR1 is perpendicular to this tangent ( which must happen if this tangent is to be common to both theellipse and the circle), we have

2sin 0 cos 14cos 1 2sin

1cos3

Thus, R1 is 4 4 2, .3 3

The largest possible radius is therefore

22

max 14 4 21 03 3

r CR

113

A tangent is drawn to the ellipse 2 2

1 2 2: 1x yEa b

which cuts the ellipse 2 2

2 2 2: 1x yEc d

at the points A and B.

Tangents to this second ellipse at A and B intersect at right angles. Prove that2 2

2 2 1a bc d

Let the point of intersection of the two tangents be ( , ).P h ky

x

E1

P h,k( )

E2

B

A

Note that since AB is the chord of contact for the tangents drawn from P to E2, we have the equationof AB as

( , ) 0T h k

2 2 1hx kyc d

2 2

2

d h dy xc k k

Example – 4



If AB is to touch the inner ellipse E1, the condition of tangency must be satisfied :

4 4 22 2

2 4 2

d d ha bk c k

4 4 2 4 2 2 4 2c d a d h b c k ...(1)Since PA and PB intersect at right angles, P must lie on the director circle of the ellipse E2. Thus,

2 2 2 2h k c d ...(2)(1) and (2) can be considered a system of equations in the variables h2 and k2 :

2 4 2 2 4 2 4 4( ) ( )a d h b c k c d h2 + k2 = c2 + d2

If these relations are to hold for variable h and k, they must in fact be identical. Thus, these variablescan now easily be eliminated to obtain :

2 4 2 4

4 4 2 2 4 4 2 2

1 1;a d b cc d c d c d c d

2 2 2 2

2 2 2 2 2 2;a c b dc c d d c d

2 2

2 2 1a bc d

Let ABC be an equilateral triangle inscribed in the circle 2 2 2.x y a Perpendiculars from A, B, C to the major

axis of the ellipse 2 2

2 2 1x ya b

(where a > b) meet the ellipse respectively at P, Q, R so that P, Q and R lie on the

same side of the major axis of A, B and C respectively. Prove that the normals to the ellipse at P, Q and R areconcurrent.

The following figure graphically portrays the situation described in the question :

y

xWe need to show thatthe normals at , and

are concurrentP Q

R

A

C

BQ O

R

P

Example – 5



All the information given about ABC being equilateral and all can be reduced to this single piece ofsignificant information : the polar angles of A, B and C, and hence, the eccentric angles of P, Q and R,

will be evenly spaced at 2 ,3 by virtue of ABC being equilateral.

Thus, we can assume the eccentric angels of P, Q and R to be 2 2, , .3 3

Now, the equation of a normal to 2 2

2 2 1x ya b

at eccentric angle is given by

2 2sec cos .ax by ec a b

2 2( ) sin 2sin cos .2

a bax by

Thus, the normals at P, Q and R are respectively given by

2 2( ) sin 2: sin cos2P

a bN ax by

2 22 2 4: sin cos sin 23 3 2 3Q

a bN ax by

2 22 2 ( ) 4: sin cos sin 23 3 2 3R

a bN ax by

Let us evaluate , the determinant of the coefficients of these three equations :

2 2

sin cos sin 2( ) 2 2 4sin cos sin 2

2 3 3 32 2 4sin cos sin 23 3 3

ab a b

Using the row operation 1 1 2 3,R R R R the first row reduces to zero, which means that

0

Thus, the normals at P, Q and R must be concurrent.



An ellipse slides between two lines at right angles to one another. Show that the locus of its centre is a circle.

It would be easiest to assume the two lines to be the coordinate axis, and an ellipse of fixed dimensionssliding between these two lines as shown below :

y

x

The ellipse has fixed dimensions, say, Major axis = 2Minor axis = 2Assume the centre to be ( , )

ab

Sh k

O

Q

P

S h,k( )

From the view-point of the ellipse, since the tangents to it at P and Q intersect at right angles at O, the

point O must lie on the director circle of the ellipse. Since the radius of the director circle is 2 2 ,a bwe must have

2 2OS a b

2 2 2OS a b

2 2 2 2h k a b

This must be the locus of the centre S! It can be written in x – y form as

2 2 2 2x y a b

It is evident that this is a circle centred at the origin and of radius 2 2 .a b

Find the locus of the point P such that tangents drawn from it to the ellipse 2 2

2 2 1x ya b

meet the coordinates axes

in concyclic points.

Example – 6

Example – 7



y

x

P h, k( )

C

A

B

The tangents from ( , ) to the ellipse meet the coordinate axes in , , and (not shown) which are concyclic.

P h k

A B C D

The pair of tangents PA and PC has the joint equation2

1 : ( , ) ( , ) ( , )J T h k S x y S h k22 2 2 2

1 2 2 2 2 2 2: 1 1 1 0x y h k hx kyJa b a b a b

The coordinate axes has the joint equation

2 : 0J xy

We can treat J1 and J2 as two curves, which intersect in four different points A, B, C, D. Any seconddegree curve through these four points can be written in terms of a parameter as

1 2 0J J

We now simply find that for which this represents a circle, since A, B, C, D are given to beconcyclic.

22 2 2 2

2 2 2 2 2 21 1 1 0x y h k hx ky xya b a b a b

This represents a circle if

2 2

2 2 2 2 2 2

1 1k ha b a a b b

2 2 2 2h k a b ...(1)

2 2

2hka b ...(2)

(1) itself gives the locus of P(h, k) as2 2 2 2x y a b



Through any arbitrary fixed point ( )P on the ellipse 2 2

2 2 1,x ya b

chords at right angles are drawn, such that the

line joining the extremities of these chords meets the normal through P at the point Q. Prove that Q is fixed for allsuch chords.

y

x

Q

P( )

BA

We can assume the eccentric angles of A and B as 1 and 2.

The normal at P has the equation :

2 2: sec cosec PQ ax by a b ...(1)

The chord AB has the equation

1 2 1 2 1 2: cos sin cos2 2 2

x yABa b ...(2)

Also, since ,PA PB we have

1 2

1 1

slope of chord slope of chord

cos cos2 2 1

sin sin2 2

PA PB

b b

a a

2 21 2 1 2sin sin cos cos 02 2 2 2

a b

Using trignometric formulae, this expression can be rearranged to

2 21 2 1 2

2 2cos cos2 2

a ba b ...(3)

Example – 8



From (2) and (3), we have

2 21 2 1 2 1 2

2 2: cos sin cos2 2 2

x y a bABa b a b ...(4)

The point Q can now be obtained as the intersection of the lines represented by (1) and (4). Let uswrite them as a system and solve for Q using the Cramer’s rule:

2 2: ( sec ) ( cosec ) ( ) 0PQ a x b y b a

2 2

2 2

( ): ( cos ) ( sin ) cos( ) 0ab a bAB b x a ya b

where 1 2

2 has been substituted for convenience.

We now have

2 2 2 2 2 2

2 2 2 22 2 2 2( ) ( )cosec cos( ) sin ( ) ( )cos sec cos( )

x yab a b a b a ba b a b b a

a b a b

2 2

1sec sin cosec cosa b

2 2 22 2

2 2

2 2

( ) cos( ) sin sin ( )

sin ( sec sin cosec cos )

ab a b a b aa bx

a b

2 2

2 2

( ) cosa a ba b

...(5)

and

2 2 22 2

2 2

2 2

( )( )cos cos cos( )

cos ( sec sin cosec cos )

a b a bb b aa by

a b

2 2

2 2

( )sinb b aa b

...(6)

Thus, the point Q, whose x and y coordinates are given by (5) and (6) respectively, can be seen to be

independent of or 1 2 .2 Q is the therefore fixed for such pairs of chords PA and PB and

depends only on the eccentric angle of P.



Consider three points on the ellipse 2 2

1 22 2 1, ( ), ( )x y P Qa b

and 3( ).R What is the area of ?PQR When

is this area maximum ?

The three points have the coordinates

1 1 2 2 3 3( cos , sin ); ( cos , sin ); ( cos , sin )P a b Q a b R a b

The area of this triangle, by the determinant formula, is

1 1

2 2

3 3

cos sin 11 cos sin 12

cos sin 1

a ba ba b

...(1)

1 2 3 1 3 2 2 3 2 3cos (sin sin ) sin (cos cos ) cos sin sin cos2

ab

2 1 1 3 3 2sin( ) sin( ) sin( )2

ab

1 2 1 22 1 3sin( ) 2sin cos

2 2 2ab

1 2 1 2 1 23sin cos cos

2 2 2ab

2 3 3 11 22 sin sin sin2 2 2

ab

This is the area of the triangle PQR.

To find its maximum value, we use a rather indirect route. Suppose we had to calculate the area ofa triangle inscribed in the circle 2 2 2x y a with the same polar angles as P, Q, R. The only differencebetween and will be that in the determinant expression for in (1), we will have all ‘a’ insteadof ‘b’ in the terms of the second column.

This means that and will always be in a constant ratio :

ba

Thus, the maximum for will be achieved in the same configuration as the one in which the maximumof be achieved !

Example – 9



Since the area of a triangle inscribed in a circle has the maximum value when that triangle is equilateral(this should be intuitively obvious but can also be easily proved), and hence will be maximumwhen

1 2 2 3 3 123

Thus, the three eccentric angels must be equally spaced apart at 2 .3

Prove that the circle on any focal distance as diameter touches the auxiliary circle of the ellipse.

Let ( )P be an arbitrary point on the ellipse 2 2

2 2 1x ya b

and let F1 be one of its foci.

y

xF1

C

P

O

Auxiliary circle

The radius of the auxiliary is a. The circle on PF1 as diameter will touch the auxiliary circle (internally)if :

(radius of this circle)OC a

C, being the mid-point of PF1, has the coordinates

cos sin,2 2

a ae bC

Example – 10



Thus,

2 2cos sin2 2

a ae bOC

2 2 2 2 2 2 21 cos sin 2 cos2

a b a e a e

2 2 2 2cos (1 ) sin 2 cos2a e e e

2 21 cos 2 cos2a e e

(1 cos )2a e

Also, the radius of the inner circle is

2 2

1cos sin

2 2a ae bCF ae

2 2 2 2cos (1 )sin 2cos2a e e

(1 cos )2a e

This gives

1OC CF a

which proves the stated assertion.



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT[ LEVEL - I ]

If the normals at the points 1 2, and 3 on the ellipse 2 2

2 2 1x ya b

are concurrent, prove that

1 1

2 2

3 3

sec cosec 1sec cosec 1 0sec cosec 1

The tangent and normal at a point P on the ellipse 2 2

2 2 1x ya b

meet the minor axis at A and B respectively.

Prove that the circle with AB as diameter passes through P as well as the two foci of the ellipse

Let AB be a focal chord of the ellipse 2 2

2 2 1.x ya b

The tangent at A and the normal at B intersect in P. Find

the locus of P.

A line intersects the ellipse 2 2

2 2 1x ya b

at A and B and the parabola 2 4 ( )y d x a at C and D. AB

subtends a right angle at the centre of the ellipse. Tangents to the parabola at C and D intersect in E. Provethat the locus of E is

2 2 2 22 2

1 14 4 ( 2 )y d d x aa b

Prove that the common chords of an ellipse and a circle are equally inclined to the axes of the ellipse.

Prove that the minimum length of the intercept made by the axes on the tangents to the ellipse 2 2

2 2 1x ya b

is equal to a + b.

If d is the length of the perpendicular from the focus F of 2 2

2 2 1x ya b

upon a tangent at any point P on the

ellipse, prove that 2

2

21 b ad FP

With a given point and line as focus and directrix, a series of ellipses are described. Prove that the locus ofthe extremities of their minor axis is a parabola.

The tangent at any point P on the ellipse 2 2

2 2 1x ya b

meets its auxiliary circle in A and B. AB subtends a

right angle at the centre of the ellipse. Let e be the eccentricity of the ellipse and be the eccentric angle ofP. Show that

2

1sin 1e

Show that the common tangent to the ellipses2 2

2 2

2x y xa b c

and 2 2

2 2

2 0x y xa b c

subtends a right angle at the origin.



[ LEVEL - II ]

If a variable point P on an ellipse is joined to the two foci F1 and F2, show that the incentre of 1 2PF F lieson another ellipse.

Two tangents to a given ellipse intersect at right angles. Prove that the sum of the squares of the chordswhich the auxiliary circle intercepts on these tangents, is a constant.

From any point P on the ellipse 2 2 2 2 2 2 21 : ( ) ,E a x b y a b tangents are drawn to the ellipse

2 2 2 2 2 22 : ,E b x a y a b which touch E2 at A and B. Prove that the orthocentre of PAB lies on E2.

Normals to the ellipse 2 2

2 2 1x ya b

at the points ( , ), 1, 2, 3,i ix y i are concurrent. Show that

1 1 1 1

2 2 2 2

3 3 3 3

0x y x yx y x yx y x y

A point P moves so that circle with PQ (where ( , 0))Q a as diameter touches the circle 2 2 24x y ainternally. Prove that the locus of P is an ellipse.

Normals at the points 1 2 3, , and 4 on the ellipse 2 2

2 2 1x ya b

are concurrent. Prove that

1 2 3 4 1 2 3 4(cos cos cos cos ) (sec sec sec sec ) 4

Tangents are drawn from any point on 2 2

116 9x y

to the circle 2 2 1.x y Find the locus of the

mid-point of the chord of contact.

A parallelogram circumscribes the ellipse 2 2

2 2 1x ya b

and two of its opposite vertices lie on the straight

lines .x c Find the locus of the other two vertices of the parallelogram.

The base of a variable triangle is of a fixed length d and the sum of its side, s, is also fixed. Prove that thelocus of the incentre of the triangle is an ellipse.

Let ( )P be a point on the ellipse 2 2

2 2 1.x ya b

A parabola is drawn having its focus at P and passing

through the foci of the given ellipse. Show that two such parabolas can be drawn. Prove also that theirdirectrices will be inclined at an angle 2 .



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT

( ANSWERS )

[ LEVEL - I ]

2 2 2

2 2 2 2 1(2 )

x b ya a b

[ LEVEL - II ]

2 22 2 2( )

16 9x y x y

2 2 2 2 2 2 2

2 2 2 2 2 2 21 1a y x y a x yb c a b c a b



ANSWERS

TRY YOURSELF - I

abe2 2

19 25x y

3 1, 2 2 25 9 54 36 0x y y

32

e

TRY YOURSELF - II

2 2

2 2 2x ya b

2 22

2 2 secx ya b

2 2 2x y a 2x ya b

TRY YOURSELF - III

2 2

4 4 2

1x ya b c

TRY YOURSELF - IV

11 1,cos5 5 4 3 12x y

Documents

SOLVED EXAMPLES Example – 1arbindsingh.com/wp-content/uploads/2012/04/Ellipse-5.pdf · to the tangent drawn at a point P on the ellipse. If F 1 and F 2 are the two foci of the ellipse,