Co-ordinate geometry

Co-ordinate Geometry is one of the most scoring and easy to prepare Unit  in the entire preparation of IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations. It is of no doubt that along with Calculus, Co-ordinate Geometry is the most scoring of all topics in Mathematics.

The importance of Co-ordinate Geometry lies in the fact that almost all the students who aspire to get high All India rank in IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations give a good emphasis on Co-ordinate Geometry.

Co-ordinate Geometry is important not only from the point of view of Mathematics but also from the point of view of Physics and Physical Chemistry as it helps in problems pertaining to graphs, especially Cartesian Co-ordinates.

Co-ordinate Geometry can be further divided into many parts but The Straight Line and The Circle holds high importance not only because more questions appear from these topics but also from the point of view that these two chapters are prerequisite to Conic Sections.

Conic Sections can be further divided into Parabola, Ellipse and Hyperbola. It is well known that it is really easy to score in Conic Sections as questions which come every year has few fixed pattern and hence can be easily dealt with.

Since Co-ordinate Geometry is not new to the students as they do study some of its basics in earlier classes. Hence, it is easy to pick up and a good take-off is not difficult. This gives an early advantage to the students not only in Mathematics but also in Physics and Physical Chemistry.

There are few students who find Co-ordinate Geometry a bit difficult but to these students we can advise more and more practice.

Wishing you "All the Best" for the preparation of Co-ordinate Geometry with askIITians.

Topics Covered under Coordinate Geometry are:-

1.Straight lines2.Circle3.Parabola4.Ellipse5.Hyperbola6.3D Geometry

Straight LinesStraight Lines is a fundamental and an important topic under Co-ordinate Geometry.

Several methods have been developed by mathematicians to uniquely locate the position of the points in space. The easiest and most widely used one is the Cartesian coordinate system, which is based on mutually perpendicular axes. In this chapter you will learn about this system: locating the points in this system and finding equations of line passing through these points.

If we say a line passes through two given points, we are imposing two conditions on this line. The conditions can be imposed in several other manners also e.g. we may say that a line passes through one point and is perpendicular to another line. These two conditions are sufficient to uniquely known our line. In this chapter you will also learn various manners of imposing conditions and finding the equation of line under those conditions.

Having known the line uniquely in space we shall try to a lots timings with it e.g. divide it in a given ratio, find distances of other points or lines from it, find the angles, which this line make with some straight line.

“Straight Lines” is one of the easiest and important chapters of Co-ordinate Geometry in the Mathematics syllabus of IIT JEE, AIEEE and other engineering examinations. These laws are not new to any aspirant as he is using the principles even in day to day life.

The chapter is important not only because it fetches 2-3 questions in most of the engineering examination but also because it is prerequisite to the other chapters of Co-ordinate Geometry.

Topics Covered: 

Representation of points in a plane Distance between two points Centroid Incentre and Circum Centre Area of a triangle Standard equations of the Straight Line Different Forms of line Examples based on straight line Angle between two straight lines Equation of Locus Examples on Angle Between two straight lines Length of the Perpendicular from a Point on a Line Distance between two parallel lines Examples on distance between two parallel lines Family of lines Concurrency of Straight Lines Position of two points with respect to a given line Angle Bisectors Pair of Straight Lines Angle between pair of lines Combined equations of the angle bisectors of the lines

Joint Equation of Pair of Lines

Introduction Coordinate Geometry is the unification of algebra and geometry in which algebra is used in the study of geometrical relations and geometrical figures are represented by means of equations. The most popular coordinate system is the rectangular Cartesian system. Coordinates of a point are the real variables associated in an order to describe its location in space. Here we consider the space to be two-dimensional. Through a point O, referred to as the origin, we take two mutually perpendicular lines XOX’ and YOY’ and call them x and y axes respectively. The position of a point is completely determined with reference to these axes of means of an ordered pair of real numbers (x, y) called the coordinates of P where |x| and |y| are the distances of the point P from the y-axis and the x-axis respectively. X is called the x-coordinate or the abscissa of P and y is called the y-coordinate or the ordinate of P. 

Representation of points in a plane We are familiar with the representation of real numbers on a line, which we call a real line. In this representation we fix a point O (called origin) and represent a real number by a point A on this line such that its distance OA (see figure given below) is equal to the value of real number. In the left side of O we represent negative real numbers and in the right side of O we represent positive real numbers. Thus, not only the magnitude of OA but the direction of the line OA is also considered for representation. 

               →        →Hence OA = –AO Similarly ordered pairs are represented in a plane. To represent an ordered pair (a, b) we take two reference lines which are mutually perpendicular. The ordered pair (a, b) represents in such a plane, by a point P(a, b) such that (see figure given below) OA = a and OB = b. 

 This system is called Cartesian co-ordinate system. Since elements of an ordered pair are not inter changeable (i.e., (a, b) ≠ (b, a) unless a = b) so they are represented in particular order, the first element ‘a’ is represented on horizontal line called abscissa and the second element ‘b’ on a vertical line called ordinate. Like the real number notation the positive side of the x-axis is the right side of O and positive side of O and positive side of y-axis is upper side of O.  

 So, the two lines divide the region in 4 parts. These are called quadrants. These quadrants are characterized as                 I       quadrant    x > 0, y > 0                 II      quadrant    x < 0, y > 0                 III     quadrant    x < 0, y < 0                 IV     quadrant    x > 0, y < 0 Here the point ‘O’ represents x = 0 and y = 0, hence order pair becomes (0, 0). There is a second type of representation called the polar co-ordinate system. In this system a reference is fixed to a line (Called the initial line), and a point called the origin in the system. Any point P is represented by ordered pair (r, θ).         Such that 

        OP = r; The distance of point from origin.         and ∠POX = θ The angular displacement of line OP from fixed line i.e. the initial line, (in the anticlockwise direction)         Clearly ‘a’ = r cos θ and ‘b’ = r sin θ (see figure given below)

 

 

Distance between two points 

 

The distance between two points P(x1, y1) and Q(x2, y2) is (see the figure given below).

 

 Length PQ = √(x2 – x1)2 + (y2 – y1)2

 Proof:

Let P(x1, y1) and Q(x2, y2) be the two points and let the distance between them be d. Draw PA, QR parallel to y-axis and PR parallel to x-axis. 

 Angle QRP = 90o                   ⇒ d2 = PR2 + RQ2 

                ⇒ d2 = (x2 – x1)2 + (y2 – y1)2

                ⇒ d = √(x2 – x1)2 + (y2 – y1)2. 

Section Formula  

Let us say we want to know the co-ordinates of point which divides a line segment between two points A(x1, y1) and B(x2, y2) in the ratio m : n. 

The coordinates of such a point are given by 

(nx1 + mx2/m+n, ny1 + my2/m+n) (for internal division) 

 Note: 

This is called section formula. 

Let P divide the line segment AB in the ratio m : n. If P is inside AB then it is called internal division; if it is outside AB then it is called external division. 

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n. 

 Proof: 

        Consider ? ABB’  

        Since BB’||PQ and AP:PB = m:n (see figure given below)

 

 

                AQ/AB' = PQ/BB' = m/m+n (= AP/AB) 

               ⇒ x – x1/x2 – x1 = m/m+n 

               ⇒ x = nx1 + mx2/m+n and y = ny1 + my2/m+n 

If P is outside AB (less assume it is at P’) 

We have 

                x – x1/x2 – x1 = m/m+n 

               ⇒ x = nx1 + mx2/m+n and y = ny1 + my2/m+n 

Similarly if P is at P” then 

                 x = –mx2+m+n/n–m, y = – my2+ny1/m+n

 Note:  

m:n can be written as m/n or λ:1. So any point on line joining A and B will be P(λx2+x1/λ+1.λy2+y1/λ+1). It is useful to assume λ:1 because it involves only one variable. 

  illustration: 

Find the ratio in which line segment A(2, –1) and B(5, 2) is divided by x-axis. 

Solution:  

Let x-axis intersect line at point P(xp, 0) such that AP/BP = λ/1⇒ yP = 0 = λy2+1.ya/λ+1 = 2λ+(–1)/λ+1 Þ λ = 1/2⇒ AP/BO = 1/2

 Illustration: 

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are 

(x1 tan A + x2 tan B + x3 tan C/tan A + tan B + tan C, y1 tan A + y2 tan B + y3 tan C/tan A + tan B + tan C), 

 Solution: 

In triangle A(x1, y1), B(x2, y2) and C(x3, y3), draw AD perpendicular to BC. Our effort now should be to find the co-ordinates of the point D.

 

 To do that, we need to find BC/CD. (figure is given above) 

                tan B = AD/BD and tan C = AD/CD 

            ⇒ BD/DC = tan C/tan B Now we apply section formulae. 

        xD = x2 tan B + x3 tan C/tan B + tan C                                        …… (i)

            yD = y2 tan B + y3 tan C/tan B + tan C                                          …… (ii)

 We know that orthocenter will lie on AD. We need to find this point and its co-ordinates. 

We should select a point H1 on AD and take the ratio AH1/H1D in such a manner so that xH1 and yH1 calculated form (i) should be symmetric in x1, x2, x3, tan A, tan B and tan C. Think before you proceed 

Let    AH1/H1D = tan B + tan C/tan A 

                         ⇒ xH1 = x1 tan A + x2 tan B + x3 tan C/tan A + tan B + tan C

        and   ⇒ yH1 = y1 tan A + y2 tan B + y3 tan C/tan A + tan B + tan C 

Since the result is symmetric, this point H1 will lie on other altitude as well i.e. the altitudes are concurrent

                         ⇒ xH = xH1 and yH = yH1 

 Illustration: 

Prove analytically that in a right angled triangle the midpoint of the hypotenuse is equidistant from the three angular points. 

 Solution: 

While proving a problem analytically take most convenient co-ordinates of known points. 

In the present case triangle is assumed as AOB with coordinates as shown in figure given below, C is midpoint of AB.

 

 So co-ordinates of C will be (a/2, b/2) 

Now AB = √a2 + b2

 CA = CB = AB/2 (C is mid point of AB) 

        = √a2 + b2 

and, we know that the distance between two points C and O is given by 

CO = √(a/2 – 0)2 + (b/2 – 0)2 = √a2 + b2/2 

Hence CA = CB = CO 

Coordinates of the point P dividing the join of two points A(x1, y1) and

B(x2, y2) internally in the given ratio λ1 : λ2 i.e., AP/BP = λ1/λ2 areP(λ2x1+λ1x2/λ2+λ1, λ2y1+λ1y2/λ2+λ1). 

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) externally in the ratio λ1 : λ2 i.e., Ao/BP = λ1/λ2 areP(λ2x1+λ1x2/λ2–λ1, λ2y1+λ1y2/λ2–λ1).

Centroid of Triangle

The centroid of a triangle is the point of concurrency of the medians. The centroid G of the triangle ABC, divides the median AD, in the ratio of 2 : 1. Illustration: Find the centroid of the triangle the coordinates of whose vertices are given by A(x1, y1), B(x2, y2) and C(x3, y3) respectively. 

 

 Solution:

              AG/AD = 2/1  Since D is the midpoint of BC, coordinates of D are (x2+x3/2, y2+y3/2)  Using the section formula, the coordinates of G are              (2(x2+x3/2)+1.x1/2+1, 2(y2+y3/2)+1.y1/2+1)          ⇒ Coordinates of G are (x1+x2+x3/3, y1+y2+y3/3).  Incentre of Triangle The incentre ‘I’ of a triangle is the point of concurrency of the bisectors of the angles of the triangle. Illustration: Find the incentre of the triangle the coordinates of whose vertices are given by A(x1, y1), B(x2, y2), C(x3, y3).

 Solution: By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA). If the lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b. Coordinates of D are (bx2+cx3/b+c, by2+cy3/b+c) IB bisects ÐB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.                                                     –  –

Let the coordinates of I be (x, y).        –                                                –

Then x = ax1+bx2+cx3/a+b+c, y = ay1+by2+cy3/a+b+c.  Circum Centre of Triangle  This the point of concurrency of the perpendicular bisectors of the sides of the triangle. This is also the centre of the circle, passing through the vertices of the given triangle.  Orthocentre of Triangle

 This is the point of concurrency of the altitudes of the triangle.  Excentre Excentre of a triangle is the point of concurrency of bisectors of two exterior and third interior angle. Hence there are three excentres I1, I2 and I3 opposite to three vertices of a triangle.  If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle ABC,

 coordinates of centre of ex-circle opposite to vertex A are given as              I1(x, y) = (–ax1+bx2+cx3/a+b+c/–a+b+c, –ay1+by2+cy3/–a+b+c). Similarly co-ordinates of centre of I2(x, y) and I3(x, y) are              I2(x, y) = (ax1–bx2+cx3/a–b+c, ay1–by2+cy3/a–b+c) 

             I3(x, y) = (ax1+bx2–cx3/a+b–c, ay1+by2–cy3/a+b–c)

Area of a triangle

Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is 

|1/2[x1(y2 – y3) + x2(y3 + y1) + x3(y1 – y2)]| = .

 It follows that the three points (x1, y1), (x2, y2) and (x3, y3) will be collinear

if   = 0. Area of a polygon of n sides First of all we plot the points and see their actual order. Let A1(x1, y1), A(x2, y2), …, An(xn, yn) be the vertices of the polygon in anticlockwise order. Then area of the polygon =

. Illustration: Calculate area of a triangle shown in figures given below. 

Solution:

 Using the just derived formula Area of a triangle            1/2 [3(2 – 4) + (1) (4 – 6) + 5 (6 – 2)]            = 1/2 [– 6 – 2 + 20] = 6

 Similarly, area of the triangle shown in the figure given above. Area of a ?ABC is            = 1/2 [3(4 – 2) + (5) (2 – 6) + (1) (6 – 4)]            =1/2 [6 – 20 + 2] = – 6 Caution: Thus we observe that the area of a triangle is positive vertices are taken in the anticlockwise direction and negative when the vertices are taken the clockwise direction. 

 Note: 

Area of a triangle can also be expressed as = 1/2             = 1/2 [x1 y2 – y1 x2 + x2 y3 – y2 x3 + x3 y1 – y3 x1] This form is important. It can be used to find area of a quadrilateral, pentagon, hexagon and polygons.  Important: If three points P1, P2 and P3 are collinear then the determinant at must vanish i.e. the area of triangle formed must be zero.  Note: If the vertices are in clockwise order then take modulus. Illustration: Prove that the area of the triangle with vertices at (p – 4, p + 5), (p + 3, p – 2) and (p, p) remains constant as p varies.  Solution: The area of the triangle is            

 

which remains constant for all values of p.

Straight Line

Any equation of first degree of the form Ax + By + C = 0, where A, B, C are constants always represents a straight line (at least one out of A and B is non zero). Slope If θ is the angle at which a straight line is inclined to the positive direction of the x-axis, then m = tanθ, (0 < θ < 180o) is the slope of the line.

 Standard equations of the Straight Line Slope Intercept From:         y = mx + c, where          m = slope of the line         c = y intercept  

   Intercept Form:      x/a + y/b = 1           x intercept = a        y intercept = b 

   Slope point form  (a)    One point on the straight line  (b)    The gradient of the straight line i.e., the slope m of the line

Equation:  

        y – y1 = m(x – x1), where (x1, y1) is a point on the straight line. Illustration:

             

Pause:Equation of line in figure (ii) is x = 3, because x-co-ordinate of each point on the line is 3.

Equation of line in figure (iv) is y = 2, because y-co-ordinate of each point on the line is 2. Although every line satisfied the above given basic definition, a line can be represented in many forms, some of which are given hereunder.  Two points form:Let there be two points A(x1, y1) and B(x2, y2) in a co-ordinate plane. If any point P(x, y) lies on the line joining A and b then m = tan θ = y–y1/x–x2 = y2 – y1/x2 – x1,  (see figure given below).

  y – y1 = y2 – y1/x2 – x1 (x – x1) which is the equation of the given line. Equation of line can also be written as

 y – y2 = y2 – y1/x2 – x1 (x – x2) or   = 0.

Different Forms of line

Illustration: The ends of a rod of length l move on two mutually perpendicular lines. Find the locus of the point on the rod, which divides it in the ratio 2 : 1. Solution:

Suppose the two perpendicular lines are x = 0 and y = 0 and let the end of the rod lie at the point (0, a) and (b, 0).

 The point P has coordinates given by

            h = b+2.0/2+1, k = 2.a+1.0/2+1 ⇒ a = 3k/2, b = 3h. Also l2 = a2 + b2.         ⇒ l2 = (3k/2)2 + (3h)2. 

Thus the required locus is x2 + y2/4 = l2/9, which represents an ellipse. Similarly, take the ratio AP : BP as 2 : 1 and proceed. We get the result as y2 + x2/4 = l2/9.  Illustration: Find the locus of the point of intersection of the liens xcosα + ysinα = a and xsinα – ycosα = b. where α is a variable.

Solution: Let P(h, k) be the point of intersection of the given lines. Then hcosα + ksinα = a.                                         … (1)         hsinα – kcosα = b.                                         … (2) Here a is a variable. So we have to eliminate a. Squaring and adding (1) and (2), We get, (hcosα + ksinα)2 + (hsinα – kcosα)2 = a2 + b2

         ⇒ h2 + k2 = a2 + b2. Hence locus of (h, k) is x2 + y2 = a2 + b2. Two intercept form: If intercepts of a line on x and y-axis are known then equation of the line can also be found in two-intercept form. Intercepts are OA and OB on x and y-axis respectively, where A(a, 0)

and B(0, b) are two points through which line is passing. Treating it as a special case of two-point form, one can write a unique equation of the line as y–0/x–a = 0–b/a–0, where P(x, y) is any point on the line (figure given below) 

 ⇒ y/b = –x/a + 1. ⇒ x/a + y/b = 1. This is intercept from of the equation of a straight line. Parametric form:

Consider line PQ with points Q(x1, y1). Then Co-ordinates of any points P(x, y) are x = x1 + r cos (see figure given below) 

            y = y1 + r sin θ Equation of the line is obtained as follows:-           ⇒ x–x1/cos θ = y–y1/sin θ = r This is parametric form of the equation of a straight line. Note:

        That tan θ = m = slope of line.  Normal form: Consider line l as shown in figure given below 

           ON ⊥ l and |ON| = p We have in ?ONA           OA = p/cos α A and B are intercept points of line l. So intercepts on x and y-axes are p/cos α and p/sin α respectively. So equation of line l will be (both intercept form)           x cos α/p + y sin α/p = 1        ⇒ x cos α + y sin α = p This is the equation of a straight line in normal form, where p is perpendicular distance of the line from origin. Caution:

        tan a ≠ m (slope of line) Note: P is always measured away from the origin and is always positive in value, αa is a positive angle less than 360o measured from the positive direction OX of the x-axis to the normal from the origin to the line.  General form: The general form of the equation of a straight line is         Ax + By+ C = 0 

This equation can be reduced to any of the above forms with some rearrangements.

Examples based on straight line

Illustration:  

Reduce 3x – 4y + 5 = 0 to all other forms.  Solution: (a)    Slope intercept form           y = 3/4 x + 5/4         where, m = slope = 3/4         c = 5/4 (y – intercept) (b)    Intercept form         3x – 4y = –5         ⇒ 3x/–5 + 4y/5 = 1         x/–5/3 + y/5/4 = 1 (intercept form)         x – intercept = –5/3         y – intercept = 5/4  (c)    point – slope form         Let x = 1, then y = 3/4 + 5/4 = 2         y – 2 = 3/4 (x – 1)  (d)    parametric form          x–1/cos θ = r, where tan θ = 3/4  

         ⇒ x–1/4/5 = y–2/3/5 = r 

(e)    Normal form 

 

        3x – 4y = – 5         –3x + 4y = 5        Dividing by ((–3)2 + 4)½           ⇒ –3/5 x + 4/5 y = 5/5 = 1         ⇒ x cos α + y sin α = p, p > 0         Where cos α = –3/5, sin α = 4/5, p = 1  Illustration: The straight line drawn through the point P(0, 3) and making an angle of 30o with positive x-axis, meets the line x + y = 6 at Q. Find the length PQ. 

Solution:

Method 1. Equation of the line through the point P is        x–0/cos 30 = y – 3/sin 30 = r         ⇒ xQ = √3/r r, yQ = 3 + r/2, r = distance PQPoint Q lies on x + y = 6           ⇒ r√3/2 + (3 + r/2) = 6        ⇒ r = 6/√3+1 

Method 2. Equation of the line through the point P is y = 1/√3 x + 3[because here m = tan 30o = –1/√3, c = 3] Solving this line, with x + y = 6, we get         xQ = 3√3/√3+1, yQ = 3√3+6/√3+1         distance PQ = √(xQ – xP)2 + (yQ – yP)2

                          = 6/√3+1

Illustration: Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15o. 

Solution:  Here, we are given p = 4 and ω = 15o.Now cos 15o = √3+1/2√2and sin 15o = √3–1/2√2By the normal form, the equation of the line is

x cos 15o + y sin 15o = 4 or √3+1/2√2 x + √3–1/2√2 y = 4or (√3 + 1) x + (√3 – 1)y = 8√2. This is the required equation. 

Illustration: Given a line 2x – 3y + 5 = 0. Write various forms of the line.

Solution: Slope intercept form:        y = 2x/3 + 5/3, C = 5/3

        and m = – (coeffcient of x/coefficient of y) = 2/3

Intercept Form:         x/(–5/3) + y/(5/3) = 1, a = –5/2; b = + 5/3. 

Normal Form:         sin α = 3/√13, cos α = –2√13         ⇒ –2/√13 ⇒ p = 5/√13.

Illustration: Find the equation to the straight line which passes through the point (–5, 4) and is such that the position of it between the axes is divided by the give point in the ratio 1 : 2.  Solution:  Let the required straight line be (x/a) + (y/b) = 1. Using the given conditions, P (2a+1.0/2+1, 2.0+1.b/2+1) is the point which divides (a, 0) and (0, b) internally in the ratio 1 : 2.   But P is (–5/4)Hence –5 = 2a/3, 4 = b/3 ⇒ a = –15/2, b = 12.  Hence the required equation is x/(–15/2) + y/2 < θ < π.

⇒ tan θ = –√8 = slope of line.We know that the equation of the straight line passing through the point (x1, y1) having slope m is y – y1 = m(x – x1).Therefore the equation of the required line is y – 2 = –√8 (x – 1)⇒ √8 x + y – √8 – 2 = 0.

Illustration:Find the equation of the line joining the points (–1, 3) and (4, –2).

Solution:Equation of the line passing through the points (x1, y1) and (x2, y2) is        y – y1 = y1–y2/x1–x2 (x – x1)Hence equation of the required line will be        y – 3 = 3+2/–1–4 (x + 1) ∝ x + y – 2 = 0.

Illustration: Represent the straight-line y = x + 2 in the parametric form.

Solution: Slope of the given line is = 1 = tan π/4.Equation of the straight line can be written as y – 2 = x.or y–2/1/√2 = x/1/√2= r.Any point on the line is (r/√2, 2 + r/√2).The point (x, y) is at a distance r from the point (0, 2).

Illustration: A line joining two points A(2, 0) and B(3, 1) is rotated about A in the anticlockwise direction through an angle of 15o. Find the equation of the line in the new position. If B goes to C, what will be the coordinates of C, in the new position?

Solution: Slope of BAB(m) = 1 ⇒ m = tanθ = 1 ⇒ θ = 45o.                         = tan (60o) (because angle between AB and AC = 15o).Also AB = AC = √2 and A is (2, 0). 

 Hence equation of the line AC is        x–2/cos60o = y–0/sin 60o

          or x–2/1/2 = y/√3/2 = r = √2         ⇒ C is (2 + √2.1/2, 0 + √2.√3/2) i.e. C is (2 + 1/√2, √3/√2).

Angle between two straight lines

Angle bisector of two lines (the line which bisects the angle between the two lines) is the locus of a point which is equidistant (having equal perpendicular distance) from the two lines.   

Say, we have two lines           L1 : A1x + B1y + C1 = 0           L2 : A2x + B2y + C2 = 0 If point R(p, q) lies on the bisector, then length of perpendicular from P to be both lines should be equal. 

        i.e.  generalizing for any point (x, y) the equation of the angle bisector is obtained as:        A1x+B1y+C1/√A1

2 + B12 = + A2x+B2y+C2/√A2

2 + B22

 

Note:

1. This equation gives two bisectors: one-acute angle bisector and the other obtuse bisector. 

2. Rule to decide the particular bisector 

To determine a bisector which lies in the same relative position with respect to the lines as a given point S(x3, y3) does, make the signs of the expressions A2x3 + B1y3+ C1 and A2x3 + B2y3 + C2 identical. (say positive) then A1x+B1y+C1/√A1

2 + B12 = +A2x+B2y+C2/√A2

2 + B22 gives the bisector

towards this point. If the signs are different multiply one of the equations with ‘–1’ throughout, so that positive sign is obtained. Then above equation with changed equations of lines will given the required bisector. 

3. If (x3, y3) ≡ (0, 0) and A2A1 + B2B1 > 0 then the bisector towards the origin is the obtuse angle bisector. 

4. Alternative method: to determine whether the bisector is of the obtuse or acute angle, determines both the bisectors and calculate angle between one of them and the initial line. The bisector for which |tan θ| > 1 is the obtuse angle bisector.

 Locus 

If a point moves according to some fixed rule, its co-ordinates will always satisfy some algebraic relation corresponding to the fixed rule. The resulting path (a curve) of the moving point is called the locus of the point. The locus i.e. the curve now contains all the points satisfying the specified condition and no point outside the curve satisfies the condition.When a point moves in a plane under certain geometrical conditions, the point traces out a path. This path of the moving point is called its locus.  

Equation of Locus The equation to the locus is the relation which exists between the coordinates of all the point on the path, and which holds for no other points except those lying on the path.  Procedure for finding the equation of the locus of a point (i) If we are finding the equation of the locus of a point P, assign coordinates (h, k) to P. (ii) Express the given conditions as equations in terms of the known quantities and unknown parameters.  (iii) Eliminate the parameters, so that the eliminant contains only h, k and known quantities.  (iv) Replace h by x, and k by y, in the eliminant. The resulting equation is the equation of the locus of p. The problem of determining the equation of locus of points every pair of which has constant slope. (see figure given below) 

Slope is the tangent (i.e. tan q) of the angle made by a line with the positive x-axis (remember positive) taken in anticlockwise direction from x-axis to the line. For any two points P(x1, y1) and Q(x2, y2). 

Examples on Angle Between two straight lines.Illustration: Draw the lines 3x + 4y – 12 = 0 and 5x + 12y + 13 = 0. Find the equation of the bisector of the angle containing the origin. Also find the acute angle bisector and obtuse angle bisector. 

  Solution:Let us make the expression on the left-hand side of the given equations of the same sign – or + ve. After substituting x = 0 and y = 0. L.H.S. of (i) is 3.0 + 4.0 – 12 = – 12 = – ve R.H.S. of (ii) is 5.0 + 12.0 + 13 = 13 = + ve         So, multiply equation (i) by (–1), we get                 – 3x – 4y + 12 = 0                                              …… (1) Equation of the bisector of the angle containing origin is given by +ve sign i.e. –3x – 4y+12/5 = + 5x+12y+13/13 ⇒ 64x + 112y – 91 = 0                                                …… (3)         Again, the given lines are                 – 3x – 4y + 12 = 0                              …… (1)                 5x –+ 12y + 13 = 0                             …… (2) To find out whether this is an acute angle bisector or obtuse angle bisector, let us find the sign of a1 a2 + b1 b2 from equation (1) and equation (2). a1 a2 + b1 b2

 = (–3) (5) + (–4) (12) = – 15 – 48 = – 63 = – ve 

        the bisector containing the origin is the acute angle bisector. Now, For obtuse angle bisector, we take –ve origin.         i.e. –3x – 4y+12/5 = + 5x+12y+13/13         i.e. 14x – 8y – 221 = 0                                                …… (4) Well, to confirm all this, let us find angle between one of the lines and one of the bisectors i.e. 5x + 12y + 13 = 0                                              …… (2) 64x + 112y – 91 = 0                                           …… (3)         Slope of line (2) is m2 = –5/12         Slope of line (3) is m3 = –64/112         Let q be the angle between these two lines 

                ⇒ tan θ =   < 1         ∴ 64x + 112y – 91 = 0 is an acute angle bisector. 

If θ is the angle between two lines, then tanθ =                                    

 where m1 and m2 are the slopes of the two lines.         (i)     If the two lines are perpendicular to each other then m1m2 = –1. Any line perpendicular to ax + by + c = 0 is of the form bx – ay + k = 0.          (ii)    If the two lines are parallel or are coincident, then m1 = m2. Any line parallel to ax + by + c=0 is of the form ax – ay + k=0. Let there be two-lines l1 and l2 with slopes m1 and m2 respectively. So tan α = m1, tan β = m2 Angle between them is either α – β or π – (α – β) depending on the side on considers 

 Now, tan (a – b) = tan α – tan β/1+tan α tan β         ⇒ tan (θ) = m1+m2/1+m1m2                    (α – β = θ say) Since lines can be taken in any order and tan(– θ) = – tan θ. So only the magnitude of θ can be obtained. Further tan (π – θ) = – tan θ. Since magnitude also includes the other angle i.e. Supplementary angle. So θ is given by 

        tan θacute =   

Important:         1.     If lines are parallel                 tan θ = 0 ⇒ m1 = m2

         2.     If lines are perpendicular                 tan θ = tan (π/2) = ∝                 1 + m1 m2 = 0 ⇒ m1 m2 = – 1 3.     Equation of a line parallel to y = mx + c is y = mx + k, i.e. Equation of a line parallel to ax + by + c = 0 is ax + by + k = 0 4.     Equation of a line perpendicular to y = mx + c is y = 1/m  x + k i.e. Equation of a line perpendicular to ax + by + c = 0 is bx – ay + k = 0 5.     Lines          a1x + b1y + c1 = 0                               …… (i)                         a2x + b2y + c2 = 0                               …… (ii)         represents         (i)     intersecting lines if a1/a2 ≠ b1/b2

 (ii)    parallel lines if a1/a2 = b1/b2

 

(iii)    Coincident lines if a1/a2 = b1/b2 = c1/c2

 Length of the Perpendicular from a Point on a Line The distance of a point from a line is the length of the perpendicular drawn from the point on the line. Given the equation of the line are different forms, the length of the perpendicular can be obtained in different forms. First form: The normal equation helps us in finding the distance of a point from a straight line. Suppose we have to find the distance of the point P(x1, y1) from the line l1 whose equation is x cos α + y sin α = p. Let l2 be the line through P parallel to the line l1. Let d be the distance of P from l1. Then, the normal from O to l2 is of length p + d. Hence the equation of l2 is x cos α + y sin α = p + d. Since P(x1, y1) lies on it. 

        ∴ x1 cos α + y1 sin α = p + d         ∴ d = x1 cos α + y1 sin α – p. Note: 1.     Rule to find the perpendicular distance of a given point from a given line in normal form.         In the left side of the equation (right side being zero), substitute the coordinates of the point. The result gives the perpendicular distance. 

 2.     Complete distance formula. If the point P and the origin O, instead of lying on the opposite sides of l as in figure given above, lie on the same side of line l1 it may be proved by proceeding exactly in the same manner that 

  

                 d = –(x1 cos α + y1 sin α – p)         Hence, the complete distance formula is                 d = + (x1 cos α + y1 sin α – p)   

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point P(x1, y1) is d. Draw a perpendicular PM from the point P to the line L. If the line meets the x and y-axes at the points Q and R respectively, then coordinates of the points are Q(–C/A, 0) and R(0, – C/B). Thus the area of the triangle PQR is given by area(?PQR) = 1/2 PM × QR, which gives PM = 2 are (PQR)/QR                 … (1) also, area (?PQR) = 1/2 |x1 (0 + C/B) + (–C/A) (–C/B – y1) + 0(y1 – 0)| = 1/2 |x1 C/B + y1 C/A + C2/AB| 

 Or 2 area (?PQR) = |C/AB| |Ax1 + By1 + C| and QR = √(0 + C/A)2 + (C/B – 0)2 = |C/AB| √A2 + B2

 Substituting the values of area (?PQR) and QR in (1), we get         PM = |Ax1+By1+C|/√A2+B2.         Or d = |Ax1+By1+C|/√A2+B2. Thus the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |Ax1+By1+C|/√A2+B2.

Distance between two parallel lines We know that slopes of two parallel lines are equal. Therefore, two parallel lines can be taken in the form         y = mx + c1                                                       … (1)         and y = mx + c2                                                 … (2) Line (1) will intersect x-axis at the point A (–c1/m, 0) as shown in figure. 

   

Distance between two lines is equal to the length of the perpendicular from point A to line (2). Therefore, distance between the lines (1) and 92) is         |(–m)(–c1/m)+(–c2)|/√1+m2 or d = |c1–c2|/√1+m2. Thus the distance d between two parallel liens y = mx + c1 and y + mx + c2 is given by d = |C1–C2|/√A2+B2.  

Illustration:         Find the distance of the point (3, –5) from the line 3x – 4y – 26 = 0. Solution:         Given line is 3x – 4y – 26 = 0.                       …… (1)         Comparing (1) with general equation of line Ax + By + C = 0, we get         A = 3, B = –4 and C = –26. Given point is (x1, y1) = (3, –5). The distance of the given point from given line is d = |Ax1+By1+C|/√A2+B2 = 3/5.  

Illustration: Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0.  Solution:         Here A = 3, B = –4, C1 = 7 and C2 = 5. 

        Therefore, the required distance is d = |7–5|/√32+(–4)2 = 2/5.  

Illustration: The coordinates of the vertices A, B, C of a triangle are (6, 3), (–3, 5) and (4,–2) respectively and P is any point (x, y). Show that the ratio of the areas of the triangles PBC and ABC is |x + y – 2| : 7.  Solution: Equation of the line BC is x + y – 2 = 0. Let PG and AD be perpendiculars from p and A on BC. 

         Ratio of areas of triangles PBC and ABC is 1/2.BC.PG/1/2.BC.AD = PG/AD. But PG is the length of the perpendicular form P(x, y) on x + y – 2 = 0 and AD is the length of the perpendicular from A(6, 3) on x + y – 2 = 0. ⇒ Ratio of the areas of ?PBC and ?ABC. ⇒ |x+y–2/√2|/|6+3–2/√2| = |x+y–2/7|. 

Illustration:        Find the distance from the line 3x – 4y + 35 = 0 of the point (0, 0).

Solution:        Writing the given equation in the normal form, we get                 –3/5 x + 4/5 y – 7 = 0         Substituting x = 0, y = 0 in it, d = –3/5 (0) + 4/5 (0) – 7 = –7         Changing sign the required distance = 7. 

Second form: 

To find the perpendicular distance of the point (x1, y1) from the line ax + by + c = 0. Let us convert the given equation in the normal form.         a/√a2+b2 x + b/√a2+b2 y + c/√a2+b2 = 0 ∴ The perpendicular Distance of (x1, y1) is         d = a/√a2+b2 x1 + b/√a2+b2 y1 + c/√a2+b2

         d = ax1+by1+c/√a2+b2

 This formula can also be obtained independently as under: Let PM be the perpendicular form P on AB. Then coordinates of A and B are (–c/a, 0) and (0, –c/b) respectively, AB = √c2/a2+c2/b2

 = c/ab √(a2+b2)         Area of ?PAB (Recall from page M4 – M – 4)                 = 1/2 [x1 (–c/b – 0)+0(0 – y1) – c/a (y1 + c/b)]                 = –c/2ab (ax1 + by1 + c)                                           …… (1) 

         Also, area of ?PAB = ½ PM.AB                 = 1/2 PM.c/ab √(a2+b2)                                              …… (2)         From (1) and (2), we have                 1/2 PM.c/ab √(a2+b2) (ax1 + by1 + c) 

                ⇒ PM = ax1+by1+c/√a2+b2

 Neglecting the negative sign, as the length of a segment is always positive, we have 

PM = |ax1+by1+c/√a2+b2|

Note: 1.     Actually                        d = + ax1+by1+c/√a2+b2

 2.     To find the distance of a point from the given line, in the left side of the equation (right side being zero) substitute the co-ordinates of the point, and divide the result by √(coefficient of x)2 + (coefficient of y)2

 3.     When to use complete perpendicular Distance formula? The complete perpendicular Distance formula is used when the length of the perpendicular to the line is given. 

Illustration: Find the distance of the point P(–2, 3) from the line AB which is x – y = 5.

Solution:The equation of the line is         x – y – 5 = 0 [Making right side zero (note this step)] ∴ Perpendicular Distance of the point (2, 3)         = (–2)–3–5/√(1)2+(–1)2 = –10/√2         = –5√2 ∴ Changing the sign, perpendicular Distance in magnitude = 5√2. 

Enquiry:    We can now find the distance of a point form a line but how can we determine as to which side of the line does the point lie?         From the figure, we observe that                 ax0 + by0 + c = 0     (? point (x0, y0) lies on the line)        …… (1)

         Consider, ax1 + by1 + c                 = (ax0 + c) + by1                [x0 = x1 = x2]                 = b(y1 – y0)                 = –ve         Consider, ax2 + by2 + c                 = (ax0 + c) + by2

                 = b(y2 – y0)                 = +ve Thus we observe that the point is on one side of the line, if put in the expression of line is gives one sign, while the point is on the other side of the line, if put in the expression of line it gives opposite sign. 

Illustration: Final the condition so that the points (x1, y1) and (x2, y2) lie on the same side, of the line ax + by + c =

Solution:Since, (ax1 + by1 + c) and (ax2 + by2 + c) Should be of the same sign. ∴ Their product should be positive i.e.         (ax1+by1+c)(ax2+by2+c) > 0, which is the required condition.

Family of lines The general equation of the family of lines through the point of intersection of two given lines is L + λL’ = 0, where L = 0 and L’ = 0 are the two given lines, and λ is a parameter.  Illustration: A variable line through the point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 meets the coordinate axes in A and B. Show that the locus of the midpoint of AB is the curve 2xy(a + b) = ab(x + y).  Solution:         Let (h, k) be the midpoint of the variable line AB.         The equation of the variable line AB is                (bx + ay – ab) + λ(ax + by – ab) = 0 

 

         Coordinates of A are (ab(1+λ)b+λa).         Coordinate of B are (0, ab(1+λ)b+λa).         Mid point of AB is (ab(1+λ)2(b+λa), ab(1+λ)2(a+λb))                 ⇒ h = ab(1+λ)2(b+λa); k = ab(1+λ)2(a+λb) ⇒ 1/2h = b+λa/ab(1+λ); 1/2k = a+λb/ab(1+λ)                 ⇒ 1/2h + 1/2k = a+b/ab ⇒ (h + k)ab = 2hk (a + b). 

        Hence the locus of the midpoint of AB is (x + y) ab = 2xy (a + b).  To find the equation to the straight lines which pass through a given point (x1, y1) and make equal angles with the given straight line y = m1x + c. If m is the slope of the required line and α is the angle which this line makes with the given line, then tana = + m1–m/1+m1m. 

  (i)     The above expression for tanα, gives two values of m, say mA and mB.  (ii)    The required equations of the lines through the point (x1, y1) and making equal angles α with the given line are y – y1 = mA (x – x1), y – y1 = mB(x – x1).  Illustration: Find the equations to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).  Solution:         The problem can be restated as:         Find the equations to the straight lines passing through the given point (2, 2) and making equal angles of 45o with the given straight line 3x + 4y – 4 = 0         Slope of the line 3x + 4y – 4 = 0 is m1 = –3/4 

 

                 ⇒ tan 45o = + m1–m/1+m1m, i.e., 1 = + m+3/4/1–3/4m         So that mA = 1/7,           and   mB = –7.         Hence the required equations of the two lines are                 y – 2 = mA (x – 2) and y – 2 = mB(x – 2).                 ⇒ 7y – x – 12 = 0 and 7x + y = 16.  Illustration: The straight lines 3x + 4y = 5 and 4x – 3y = 15 intersect at the point A. On these lines points B and C are chosen so that AB = AC. Find the possible equations of the line BC passing through (1, 2).  Solution:         The two given straight lines are at right angles.         Since AB = AC, the triangle is an isosceles right angled triangle.         The required equation is of the form y – 2 = m(x – 1)              … (1)                 with tan 45o = + m+3/4/1–3m/4 = + m–4/3/1+4m/3                 ⇒ 1 = + m+3/4/1–3m/4 = + m–4/3/1+4m/3 ⇒ m = – 7, 1/7.         Substitute the value of m in (1). We get the required equations.  Illustration: Find the equation of the straight line passing through (–2, 7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and 4x + 3y = 3.

  Solution:         Distance AC between the two given parallel lines               = |c1–c2/√a2+b2| = 12–3/√16+9 = 9/5. 

 

         Let AB be the intercept of length 3 units.         ⇒ BC = 12/5. If θ is the angle between BC and AB, then tanθ = 9/12 = 3/4.         Slope of the parallel lines = –4/3 = m2. If m1 is the slope of the required line, then tanθ = m1–m/1+m1m ⇒ 3/4 = + m1–4/3/1+4/3m1

 i.e. m1 + 4/3 = 3/4 (1 – 4/3 m1) and m1 + 4/3 = 3/4 (1 – 4/3 m1). The slopes are (i) m1 = –7/24  (ii) m1 = ∞ (the line is parallel to the y-axis). The required equations of the lines are 7x + 24y + 182 = 0 and x + 2 = 0.  Alternative solution: Equation of the line, through P(–2, 7) and making angle θ with the x-axis, is x+2/cosθ = y+7/sin θ = r. If this line intersects the given lines at A and B, with AB = 3, the points A and B are A(–2 + r1 cos θ, – 7 + r1 sin θ) and B (–2 + (r1 + 3) cos θ, – 7 + (r1 + 3) sin θ). Since A and B lie on the lines 4x + 3y = 3 and 4x + 3y = 12, we have 

4r1 cos θ + 3r1 sin θ = 32 and 4r1 cos θ+ 3r1 sin θ + 12 cos θ + 9 sin θ = 41, so that 12 cos θ + 9 sin θ = 0 or 4 cos θ + 3 sin θ = 3. Solving this equation we find that θ = π/2 and tan θ = –7/24. Hence the required lines are x + 2 = 0 

and y + 7 = –7/24 (x + 2) i.e. 7x + 24y + 182 = 0.

 Concurrency of Straight Lines The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is 

        (i)      = 0.  (ii)    There exist 3 constants l, m, n (not all zero at the same time) such that IL1 + mL2 + nL3 = 0, where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines.  (iii)    The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines. Illustration:         Check if lines                 a1 x + b1 y + c1 = 0                                             …… (1)                 a2 x + b2 y + c2 = 0                                             …… (2)                 (2a1 – 3a2)x + (2b1 – 3b2)y + (2c1 – 3c2) = 0           …… (3)         are concurrent?Solution:        We can try to find α, ß and γ by observation as follow:                 L3 – 2L1 + 3L2 = 0 Enquiry:    many lines can pass through the intersection of two lines. Can we find a general equation of these lines? 

If L1 = 0 and L2 = 0 are two lines then equation of family of lines passing through their intersection is given by L1 + λ L2 = 0                                                       …… (A) Where ‘λ’ is any parameter. (Equation A is satisfied by the point of intersection of L1= 0 and L2 = 0) 

Note: To determine a particular line one more condition is required so as to determine or eliminate λ . 

Illustration:         If x (2q + p) + y(3q + p) = 0         (x + y – 1) + q/p (2x + 3y – 1) = 0, p ≠ 0

Solution:This equation represents the family of lines passing through the intersection of lines x + y – 1 = 0 and 2x + 3y – 1 = which is fixed point i.e. (2, –1).         If p = 0 then equation becomes                 q(2x + 3y – 1) = 0         this also represents a line which passes through fixed points (2, –1). Hence the given equation represents family of lines passing through a fixed point (2, –1) for variable p, q. 

Illustration: Find the equation of a line, through the intersection of 2x + 3y – 7 = 0 and x + 3y – 5 = 0 and having distance from origin as large as possible. 

 Point of intersection of two lines is A(2, 1) Now, with OA as radius and O itself as centre draw a circle.         There will be infinitely many lines through A and each except one of them produces a chord of circle and hence their distance from origin i.e. centre of circle is less than OA i.e. radius of circle.         But the exceptional one which infact is a tangent to circle at A will be at a distance OA from O.         Thus, tangent to circle at A will be the line through A and is farthest from origin. Now, OA ⊥ tangent at A. ∴ (slope of OA) × (slope of tangent at A) = –1 Or, 1–0/2–0 (slope of tangent at A) = –2 ∴ equation of required line is         (y – 1) = –2(x – 2) Or     2x + y – 5 = 0 

Illustration: Find the point of concurrency of the altitudes drawn from the vertices (at1t2, a(t1 + r2)), (at2t3, a2t2 + t3)) and (at3t1, a(t3 + t1)) respectively of a triangle ABC.  Solution: 

        Slope of AD = –t3.         Equation of AD is y – a(t1 + t2) = –t3(x + at1t2).                … (1)         Equation of CF is y – a(t3 + t1) = –t2(x – at3t1).                 … (2) 

 

         Subtracting (1) from (2), we get                 x = –a Þ y = a(t1 + t2 + t1t2t3).         Hence the point of concurrency of the altitudes is         (–a, a(t1 + t2 + t3 + t1t2t3)).

Position of two points with respect to a given line Let the line be ax + by + c = 0 and P(x1, y1), Q(x2, y2) be two points.  Case 1: If P(x1, y1) and Q(x2, y2) are on the opposite sides of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m1 : m2, where m1/m2 must be positive. Co-ordinates of R         are (m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2). Point R lies on the line ax + by + c = 0.         ⇒ m1/m2 = ax1+by1+c/ax2+by2+c > 0 

 

 So that ax1 + by1 + c and ax2 + by2 + c should have opposite signs.  Case 2: If ax1 + by1 + c and ax2 + by2 + c have the same signs then m1/m2 = –ve, so that the point R on the line ax + by + c = 0 will divide the line PQ externally in the ratio m1 : m2 and the points P(x1, y1) and Q(x2, y2) are on the same side of the line ax + by + c = 0.  Illustration: Find the range of θ in the interval (0, π) such that the points (3, 5) and (sinθ, cosθ) lie on the same side of the line x + y – 1 = 0.  Solution:         3 + 5 – 1 =7 > 0 ⇒ sinθ + cosθ – 1 > 0         ⇒ sin(π/4 + θ) > 1/√2         ⇒ π/4 < π/4 + θ < 3π/4         ⇒ 0 < θ < π/2.  Illustration:         Find a, if (α, α2) lies inside the triangle having sides along the lines                 2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.  Solution:         Let A, B, C be vertices of the triangle.

         A ≡ (–7, 5), B ≡ (5/4, 7/8),         C ≡ (1/3, 1/9).         Sign of A w.r.t. BC is –ve. 

 

 If p lies in-side the ¦ABC, then sign of P will be the same as sign of a w.r.t. the line BC ⇒     5α – 6α2 – 1 < 0.                         …… (1)         Similarly    2α + 3α2 – 1 > 0.                                …… (2)         And,          α + 2α2 – 3 < 0.                          …… (3)         Solving, (1), (2) and (3) for α and then taking intersection,         We get      α ? (1/2, 1) ∪ (–3/2, –1).  Illustration: The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, –2), find the equation of BC.  Solution:         From the figure,         E ≡ (x1+1/2, y1–2/2),         F ≡ (x2+1/2, y2–2/2).

 Alt text : equations of the perpendicular bisectors of sides of triangle

       Since E and F lie on OE and OF respectively,                 x1 – y1 + 13 = 0                                  … (1)         and   x2 + 2y2 – 3 = 0                                  … (2)         Also, slope of AB = –1 and slope of AC is 2, so that                 x1 + y1 + 1 = 0.                                  … (3)         And   2x2 – y2 – 4 = 0                                  … (4)         Solving these equations, we get the co-ordinates of B and C as                 B ≡ (–7, 6) and C ≡ (11/5, 2/5) ⇒ Equation of BC is 14x + 23y – 40 = 0.  Illustration: Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’ + OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.  Solution:         Let A ≡ (a, 0), B (0, b), A’ ≡ (a’, 0), B’ ≡ (0, b’).         Equation of A’B is x/a' + y/b' = 1.                                          …. (1)         and equation of AB’ is x/a + y/b'  = 1.                                    …. (2)         Subtracting (1) from (2), we get, x (1/a – 1/a') + y(1/b' – 1/b) = 0.         ⇒ x(a'–a)/aa' + y(b–b')/bb' = 0.                             [Using a’ – a = b – b’]         ⇒ x/a(b–b'+a) + y/bb', 0 ⇒ b’ = a(a+b)y/ay–bx.                              ….. (3)         From (2) b’x + ay = (4) we get x + y = a + b

         which is the required locus. 

Angle Bisectors To find the equations of the bisectors of the angle between the lines         a1x + b1y+ c1 = 0 and a2x + b2y + c2 = 0. A bisector is the locus of a point, which moves such that the perpendiculars drawn from it to the two given lines, are equal. The equations of the bisectors are         a1x+b1y+c1/√a1

2+b12 = + a2x+b2y+c2/√a2

2+b22.

 

AP is the bisector of an acute angle if, Tan (∠PAN) = tan (θ/2) is such that |tan θ/2| < 1. AP is an obtuse angle bisector if, Tan (∠PAN) = tan (θ/2) is such that |tan θ/2| > 1.  Notes : • When both c1 and c2 are of the same sign, evaluate a1a2 + b1b2. If negative, then acute angle bisector is a1x+b1y+c1/√a1

2+b12 = + a2x+b2y+c2/√a2

2+b22.

  • When both c1 and c2 are of the same sign, the equation of the bisector of the angle which contains the origin is a1x+b1y+c1/√a1

2+b12 = + a2x+b2y+c2/√a2

2+b22.

  

• Bisectors of the angle containing the point (α, ß) is a1x+b1y+c1/√a1

2+b12 = +a2x+b2y+c2/√a2

2+b22 if a1α + b1ß + c1 and a2α + b2ß + c2 have the

same sign.  • Bisectors of the angle containing the point (α, ß) is a1x+b1y+c1/√a1

2+b12 = +a2x+b2y+c2/√a2

2+b22 if a1α + b1ß + c1 and a2α + b2ß + c2 have the

opposite sign.  Illustration: For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 , find the equation of the (i)     bisector of the obtuse angle between them, (ii)    bisector of the acute angle between them, (iii)    bisector of the angle which contains (1, 2)   Solution:         Equations of bisectors of the angles between the given lines are                 4x+3y–6/√42+32 = + 5x+12y+9/√52+122

 ⇒     9x – 7y – 41 = 0 and 7x + 9y – 3 = 0. If θ is the angle between the line 4x + 3y – 6 = 0 and the bisector 9x – 7y – 41 = 0, then tan θ =  > 1. Hence (i)     The bisector of the obtuse angle is 9x – 7y – 41 = 0. (ii)    The bisector of the acute angle is 7x + 9y – 3 = 0. For the point (1, 2) 4x + 3y – 6 = 4 × 1 + 3 × 2 – 6 > 0. 5x + 12y + 9 = 12 × 12 + 9 > 0. Hence equation of the bisector of the angle containing the point (1, 2) is 4x+3y–6/5 = 5x+12y+9/13 ⇒ 9x – 7y – 41 = 0.

Pair of Straight Lines 

The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously. Let a ≠ 0. Now, the above equation becomes         a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac on completing the square on the left side, we get,         a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac. i.e.    (ax + hy + g) = + √y2(h2–ab)+2y(gh–af)g2–ac We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is, (gh – af)2 = (h2 – ab) (g2 – ac) i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc         cancelling and diving by a, we have the required condition                 abc + 2fgh – af2 – af2 – bg2 – ch2 = 0 Illustration: What is the point of intersection of two straight lines given by general equation ax2+ 2hxy + by2 + 2gx + 2fy + c = 0?

Solution:The general solution is         ax2 + 2hxy + by2 + 2gx + 2fy + c = 0                   …… (1)         Let (α, ß) be the point of intersection we consider line paralleled transformation.         x = x’ + α,         y = y’ + ß From (1) we have a(x’ + α)2 + 2h(x’ + α) (y’ + ß) + b(y’ + ß)2 + 2g(x’ + α) + 2f(y’ + ß) + c = 0 ⇒     ax’2 + 2hx’y’ + by’2 + a α2 + 2hαß + bß2 + 2gα + 2fß + 2x’(a α + hß + g) + 2y’ + 2y’ (hα + bß + f) = 0 ⇒     ax’2 + 2hx’y’ + by’2 + 2x’(aα + hß + g) + 2g’ + 2y’ (hα + bß + f) = 0

 

         Which must be in the form                 ax'2 + 2hx’y’ + by’ = 0         This cannot be possible unless                 aα + hß + g = 0                 hα + bß + f = 0         Solving                 α/hf–bg = ß/hg–af = 1/ab–h2

                 α = hf–bg/ab–b2, ß = hg–af/ab–h2

 

Illustration: Represent lines y = 2x and y = 3x by a homogeneous equation of second degree 

Solution:(y – 2x) (y – 3x) = 0 Or 6x2 – 5xy + y2 = 0 

Illustration: Represent lines parallel to y = 2x and y = 3x by a second degree equation

Solution:        (y – 2x – c1) (y – 3x – c2) (where c1 and c2 are constants)         = 6x2 – 5xy + y2 + (3c1 + 2c2) x + (– c1 – c2) y + c1 c2 = 0 

Note: 1.     Homogeneous part is same as for the equation of above illustration. Therefore, the homogeneous part of a general second degree equation determines the slope of the lines i.e. lines parallel to ax2 + 2hxy + by2 + c = 0 and through the origin are represented by the equation ax2 + 2hxy + by2 = 0 2.     The equation ax2 + 2hxy + by2 + 2fy + c = 0 represents a pair of parallel straight lines if h/a = b/h = f/g or bg2 = af2

         The distance between them is given by 2√g2–ac/a(a+b) or √f2–bc/b(a+b)  

Illustration:         Does the second degree equation x2 + 3xy + 2y2 – x – 4y – 6 = 0 represents a pair of lines. If yes, find their point of intersection.

Solution:        We observe that                 a = 1, h = 3/2, b = 2, g = –1/2, f = 2, c = – 6         \ abc + 2fgh – af2 – bg2 – ch2 = – 12 + 3 – 4 – 1/2 + 27/2 = 0 Therefore the given second-degree equation represents a pair of lines, x2 + 3xy + 2y2 – x – 4y – 6 = (x + 2y + 2) (x + y – 3). 

Consider the equations formed by first two rows of  .         i.e. ax + hy + g = 0 and hx + by + f = 0         i.e. x + 3/2 y – 1/2 and 3/2 x + 2y – 2 = 0 Solving these, we get the required point of intersection.         i.e. 2x + 3y – 1 = 0              3x – 4y – 4 = 0 Solving the above equation, we get x = 8, y = –5. Note: 

        (2x + 3y – 1)(3x + 4y – 4) ≠ x2 + 3xy + 2y2 – x – 4y – 6.

 Angle between pair of lines represented by ax2 + 2hxy + by2 = 0

         Comparing the coefficients of x2, y2 and xy, we get                 b(y – m1x) (y – m2x) = ax2 + 2hxy + by2

                 m1 + m2 = –2h/b and                 m1 m2 = a/b                 tan θacute = |m2–m1/1+m1m2|                                   = |√(m1–m2)2 – 4m1m2/1+m1m2|                                   = |2√h2–ab/a+b| Caution:         1.     If two lines through the origin are represented by                         y = mx and y = m2x, we cannot write                 ⇒     (y – m1x) (y – m2x) ≡ ax2 + 2hxy + by2

 Because coefficient of y2 on left hand side is one on right hand side, it is b.         2.     The given equation represents real lines only when h2 – ab > 0 Note:         1.     If two lines are coincident then tan θ = 0 ⇒ h2 = ab         2.     If two lines are perpendicular then                         m1m2 = 1 ⇒ a + b = 0 i.e. x2 + 2hxy – y2 always represents pair of mutually perpendicular lines through origin. 3.     Two lines are equally inclined to axes but are not parallel. For such a case let us take a line l1 which is inclined at an angle θ, then l2 is inclined at (π – θ).                 tan (π – θ) = – tan θ.                 which is the condition for two lines inclined equally to axes. 4.     m1 = –h+√h2–ab/2 and m2 = –h–√h2–ab/b 

Illustration: What is the equation of the pair of lines through origin and perpendicular to ax2 + 2hxy + by2 = 0

Solution:Let ax2 + 2hxy + by2 = 0 represents the lines y = m1x (i) and y = m2x (ii) Lines perpendicular to the lines (i) and (ii) are y = –1/m1 x and y = –1/m2 x respectively and passing through origin i.e. m1y + x = 0 and m2y + x = 0 Their combined equation is given by         (m1y + x) (m2y + x) = 0          ⇒ m1m2 y2 + (m1 + m2) xy + x2 = 0          ⇒ a/2 y2 – 2h/b xy + x2 = 0 ⇒ bx2 – 2hxy + ay2 = 0 is the equation of the pair of lines perpendicular to pair of lines ax2 + 2hxy + by2 = 0 Note: 

Coefficient of x2 and y2 are interchanged and the sign of xy term is reversed.

Combined equations of the angle bisectors of the lines represented by ax2 + 2hxy +by2 = 0

        Let ax2 + 2hxy + by2 = 9 represent lines                 y – m1x = 0 and y – m2x = 0         Let P(α, ß) be any point on one of bisectors.                 ⇒ (ß–m1α)/√1+m1

2 = + ß–m2α/√1+m22

                 ⇒ (1 + m2

2) (ß – m1α)2 – (1 + m12) (ß – m2α)2 = 0

                 ⇒ Locus of P(α, ß) is                         x2 – y2 = 2hxy 1–m1m2/m1+m1

                 ⇒ x2–y2/a–b = xy/h; is required equation of angle bisectors   …… (1) Note: 

        1.     If a = b, then bisectors are x2 – y2 = 0 i.e. x – y = 0, x + y = 0 2.     If h = 0, the bisectors are xy = 0 i.e. x = 0, y = 0 3.     If in (i), coefficient of x2 + coefficient of 2 = 0, then the two bisectors are always perpendicular to each other 

Illustration: Prove that the angle between one of the lines given by ax2 + 2hxy + by2 = 0 and one of the lines given by ax2 + 2hxy+ by2 + K(x2 + y2) = 0 is equal to the angle between the other two lines of the system.

Solution:        Let L1 and L2 be one pair and L3 and L4 be the other pair of lines.         If the angle between L1 and L3 is equal to the angle between L2 and L4 then pair of bisectors of L1 and L2 would be same as that of L3 and L4. Pair of bisectors of L3 and L4 is              x2–y2/(a+k)–(b+k) = xy/h         ⇒ x2–y2/a–b = xy/h Which is the same as the bisector pair of L1 and L2. 6.     Angle bisectors of ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 (i) are given by (x–x0)2–(y–y0)2/a–b = (x–x0)(y–y0)/h where (x0, y0) is the point of intersection of (i) Enquiry:    If we shift the origin of the coordinate system how can the coordinates of a point be known in the new system? What will happen if we rotate the axis?         (i)     Linear Transformation:                 CP = x, DP = y; AP = x, BP = y

 If origin is shifted to (h, k) then the coordinates of a point P(x, y) in the new system are

 X = x – h Y = y – k You can check it by putting (h = 0, k = 0) that it gets reduced to the same original coordinate system.         (ii)    Rotation of Axes:

  If axes are rotated anticlockwise by angle q then the coordinates of a point P(x, y) changes to say P(x, y): OL = x, PL = y OM = X, PM = Y OP = R (say) cos (α + θ) = x/R                                                    …… (i) sin (α + θ) = Y/R                                                     …… (ii) cos α = x/R                                                           …… (ii) sin α = Y/R                                                            …… (iv) Eliminating α from (i), (ii), (iii) and (iv) we get X = x cos θ + y sin θ Y = –x sin θ + y cos θ It can be checked that if θ = 0; then coordinates remain unchanged. Illustration: Find the equation of the line 2x + y = 7 when co-ordinates system is shifted to the point (3, 1)Solution:        x = 3 + X and y = 1 + Y Equation of line becomes

         2(3 + X) + (1 + Y) = 7         or 2X + Y = 0 Note:         Slope of the line remains the same.  The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c=0 represents a

pair of straight lines if   = 0. ⇒ abc + 2fgh – af2 – bg2 – ch2 = 0 and h2 > ab. The homogeneous second degree equation ax2 + 2hxy + by2 = 0 represents a pair of straight lines through the origin. If lines through the origin whose joint equation is ax2 + 2hxy + by2 = 0, are y = m1x and y = m2x, then y2 – (m1 + m2)xy + m1m2x2 = 0 and y2 + 2h/b xy + a/b x2 = 0 are identical. If θ is the angle between the two lines, then tanθ = + √(m1+m2)2–4m1m2/1+m1m2 =   + 2√h2–ab/a+b. 

The lines are perpendicular if a + b = 0 and coincident if h2 = ab.

 Joint Equation of Pair of Lines Joining the Origin and the Points of Intersection of a Line and a Curve If the line lx + my + n = 0, (n ≠ 0) i.e. the line not passing through origin) cuts the curve ax2 + by2 + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e. 

  ax2 + 2hxy + by2 + (2gx + 2fy) (kx+my/–n) + c (lx+my/–n)2 = 0. is the equation of the lines OA and OB.  Illustration: Prove that the straight lines joining the origin to the points of intersection of the straight line hx + ky = 2hk and the curve (x – k)2 + (y – y)2 = c2 are at right angles if h2 + k2 = c2.  Solution: Making the equation of the curve homogeneous with the help of that of the line, we get                 x2 + y2 –2(kx + hy) (hx+ky/2hk) + (h2 + k2 – c2) (hx+ky/2hk) = 0         or 4h2k2x2 + 4h2k2y2 – 4h2x(hx+ky) – 4h2ky(hx + ky) + (h2 + k2 – c)(h2x2 + k2y2 + 2hxy) = 0. This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if Coefficient of x2 + coefficient of y2 = 0 i.e.         (h2 + k2) (h2 + k2 – c2) = 0 ⇒ h2 + k2 = c2       (since h2 + k2 ≠ 0). 

Circle Circle holds a high pedestal in the entire Syllabus of Co-ordinate Geometry in Mathematics. 

In this chapter, we discuss the algebraic equations representing a circle and the lines associated with it i.e. a tangent, a pair of tangents and a chord of contact. The concepts of family of circles and that of common tangents to two circles under various configurations are given due importance. 

Whole of the analysis is illustrated by appropriate examples, while giving the student ample scope to test him skills through assignments of different levels. 

“Circle” is one of the most scoring and hence an important

chapter of Co-ordinate Geometry in the Mathematics syllabus of IIT JEE, AIEEE and other engineering examinations. The preliminary knowledge of the concept of Straight Lines is prerequisite to study Circles. 

The chapter is not vast and it fetches 2-3 questions in most of the engineering examination. 

Index 

1. Basic Concepts2. Chord of a Circle3. Chord of Contact4. Common Tangents5. Family of Circle6. Power of a Point with Respect to a Circle7. Radical Axis8. Solved Examples

Circle is most important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic. You are expected to do all the questions based on this to remain competitive in IIT JEE examination. It is very important to master these concepts at early stage as this forms the basis of your preparation for IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations.

 Basic Concepts Locus has been defined as the path of a point satisfying some geometrical condition; i.e. constraint equations. The path represents a curve, which includes all the points satisfying the given condition. 

Similarly a circle can be defined as: The locus of a point which moves in such a way that its distance from a fixed point is always constant and positive. The fixed point is called the centre of the circle and the given distance the radius of the circle. In real life, when you

rotate a stone tied with one end of a string then the path followed by stone is exactly a circle whose centre is your finger an radius is length of the string. 

The equation of a circle with its centre at C(xc, yc) and radius r is: 

             (x – xc)2 + (y – yc)2 = r2 

Proof: 

          Let P(x, y) be any point on the circle. Then by the definition of the locus the constant distance is (see figure given below) 

                                             

              |PC| = r ⇒ √((x-xc) )2+(y-yc) )2 ) = r 

              ⇒ (x – xc)2 + (y – yc)2 = r2 

             which is the required equation of the circle 

Note: 

(1) If xc = yc = 0 (i.e. the centre of the circle is at origin) then equation of the circle reduce to x2 + y2 = r2 

(2) If r = 0 then the circle represents a point or a point circle. 

Equation of the circle in Various Forms 

(i) The simplest equations of the circle is x2 + y2 = r2 whose centre is (0, 0) and radius ‘r’. 

(ii) The equation (x – a)2+ (y – b)2 = r2 represents a circle with centre (a, b) and radius r. 

(iii) The equation x2 + y2 + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius √(g2+f2-c). 

(iv) Equation of the circle with points P(x1, y1) and Q(x2, y2) as extremities of a diameter is (x – x1) (x – x2) + (y – y1)(y – y2) = 0. 

Equation of a circle under Different Conditions 

       

Parametric Equation of a circle 

        Let us consider a circle of radius ‘r’ and centre at C(xc, yc) we have: 

                 (y-yc)/r = sin θ (see figure given below) 

                                    

                     ⇒ y = yc + r sin θ 

              Similarly x = xc + r cos θ 

              This gives the parametric from of the equation of a circle. 

General equation of a circle in polar co-ordinate system 

       Let O be the origin, or pole, OX the initial line, C the centre and ‘a’ the radius of the circle. 

       Let the polar co-ordinates of C be R and α, so that OC = R and ∠XOC = α . 

      Let a radius vector through O at an angle θ with the initial line cut the circle at P and Q. Let OP be r. 

                                    

     Then we have 

     CP2 = OC2 + OP2 – 2OC . OP cos COP 

     i.e. a2 = R2 + r2 – 2 Rr cos (θ – α) 

     i.e. r2 – 2 Rr cos (θ – α) + R2 – a2 = 0 …… (1) 

     This is the required polar equation. 

     Particular cases of the general equation in polar coordinates. 

Note: 

1.  Let the initial line be taken to go through the centre C. Then α = 0, and the equation becomes 

            r2 – 2Rr cos θ + R2 – a2 = 0. 

2.  Let the pole O be taken on the circle, so that 

           R = OC = α 

       The general equation the becomes 

         r2 – 2ar cos (θ – α) = 0, 

         i.e. r = 2a cos (θ – α). 

3.  Let the pole be on the circle and also let the initial line pass through the centre of the circle. In this case 

          α = 0, and R = a 

Now, the general equation reduces to the simple form r=2a cos θ 

                                

          This is at once evident from the figure given above. 

          For, if OCA were a diameter, we have 

               OP = OA cos θ, 

                r = 2a cos θ. 

    Let us consider a circle such that points P(x1, y1) and Q(x2, y2) are on it and PQ is one of the diameters of the circle. 

          If R(x, y) is any point on the circle then 

                                      

Recall: 

        ∠PRQ = π/2 (Angle subtended by diameter at any point on the circle is a right angle). 

       ⇒ QR ⊥ PR 

       ⇒ (Slope of QR) x (Slope of PR) = –1 

       ⇒ (y-y2)/(x-x2 )×(y-y1)/(x-x1 ) = – 1 

       ⇒ (x – x1) (x – x2) + (y – y1) (y – y2) = 0 

        Which gives the required equation. 

Note: 

        This equation can also be obtained considering 

               PR2 + QR2 = PQ2 

        The general from of the equation of a circle is: 

               x2 + y2 + 2gx + 2fy + c = 0 …… (1) 

               ⇒ (x + g)2 + (y + f)2 = g2 + f2 – c 

                Comparing this equation with the standard equation (x – xc)2 + (y – yc)2= r2 

       We have: 

                Centre of the circle is (–g, –f), Radius = √(g2+f2-c). 

               Equation (1) is also written as S = 0. 

Note: 

        1. If g2 + f2 – c > 0, circle is real 

        2. If g2 + f2 – c = 0, circle is a point circle. 

        3. If g2 + f2 – c < 0, the circle is imaginary. 

       4. Any second-degree equation ax2 + 2hxy + by2 + 2gx + 2fy+c=0 represents a circle only when h = 0 and a = b i.e. if there is no term containing xy and co-efficient of x2 and y2 are same, provided abc + 2fgh – af2 – bg2 – ch2 ≠ 0 

Illustration: 

         Find the centre and the radius 3x2 + 3y2 – 8x – 10y + 3 = 0. 

Solution: 

        We write the given equation as x2 + y2 – 8/3 × – 10/3 y + 1 = 0. 

        ⇒ g = -4/3, f = -5/3 , c = 1 

        Hence the centre is (4/3,5/3) and the radius is 

        √(16/9+25/9-1)=√(32/9)=(4√2)/3. 

Illustration: 

       Find the equation of the circle with centre (1, 2) which passes through the point (4, 6). 

Solution: 

      The radius of the circle is √((4-1)2+(6-2)2 )=√25 = 5. 

     Hence the equation of the circle is (x – 1)2 + (y – 2)2 = 25 

     ⇒ x2 + y2 – 2x – 4y = 20. 

Illustration: 

      A circle has radius 3 units and its centre lies on the line y = x – 1. Find theequation of the circle if it passes through (7, 3). 

Solution: 

        Let the centre of the circle be (α, β). It lies on the line y = x – 1 

       ⇒ β = α – 1. Hence the centre is (α, α – 1). 

       ⇒ The equation of the circle is (x – α)2 + (y – α + 1)2 = 9. It passes through (7, 3) 

       ⇒ (7 – α)2 + (4 – α)2 = 9 ⇒ 2α2 – 22α + 56 = 0 

       ⇒ α2 – 11α + 28 = 0 ⇒ (α – 7) = 0 ⇒ α = 4, 7. 

       Hence the required equations are 

       x2 + y2 – 8x – 6y + 6 = 0 and x2 + y2 – 14x – 12y + 76 = 0. 

Illustration: 

       Find the equation of the circle whose diameter is the line joining the points (–4, 3) and (12, –1). Find also the intercept made by it on the y-axis. 

Solution: 

     The equation of the required circle is 

     (x + 4) (x – 12) + (y – 3) (y + 1) = 0. 

     On the y-axis, x = 0 ⇒ – 48 + y2 – 2y – 3 = 0. 

     ⇒ y2 – 2y – 51 = 0 ⇒ y = 1 ± √52. 

     Hence the intercept on the y-axis = 22√52 = 4√13.

Illustration: 

        Find the equation of the circle passing through (1, 1), (2, –1) and (3, 2). 

Solution: 

       Let the equation be x2 + y2 + 2gx + 2fy + c = 0. 

       Substituting the coordinates of three points, we get 

       2g + 2f + c = –2, 

       4g – 2f + c = –5, 

       6g + 4f + c = –13. 

       Solving the above three equations, we obtain: 

       f = –1/2; g = –5/2, c = 4. 

      Hence the equation of the circle is 

       x2 + y2 – 5x – y + 4 = 0. 

Illustration: 

      Write general equation of a circle centered at a point on x-axis. 

Solution: 

     Circle is: x2 + y2 + 2gx + c = 0, g2 – c ≥ 0 

     Its centre is (–g, 0) and radius √(g2-c) 

     Or 

     (x + g)2 + (y – 0)2 = r2 

                                             Its centre is (–g, 0) and radius r. (figure given above) 

Illustration: 

         Write general equation of a circle passing through the origin. 

Solution: 

        Point (0, 0) must satisfy x2 + y2 + 2gx + 2fy + c = 0 

       ⇒ C = 0 

                                                 ∴ circle is : x2 + y2 + 2gx + 2fy = 0 (figure given above) 

Illustration: 

        Write the equation of a circle centered at x-axis at (x1, 0) and touching y-axis at the origin. (figure given below) 

Solution: 

                                    

            (x – x1)2 + (y – 0)2 = (x1) 

Illustration: 

       Write the equation of a circle passing through O (0, 0) A (a, 0) and B (0, b)? Obviously AB is the diameter of the circle. (Figure given below) 

                                        Solution: 

        (x – a) (x – 0) + (y – 0) (y – b) = 0 

Illustration: 

       Find the equation of circle shown in figure given below in polar form. 

                                  Solution: 

        OP = OA cos θ 

        r = 2a cos θ, – θ/2 ≤ θ ≤ θ/2, a is raius of circle 

Illustration: 

      Find the co-ordinates of the centre of the circle represented by 

      r = A cos θ + B sin θ. 

Solution: 

      r = A cos θ + B sin θ 

      = [A/√(A2+B2 ) cos θ +B/√(A2+B2 ) sin θ ] √(A2+B2 ) 

      = cos (θ – α) (√(A2+B2 )) 

      centre is ≡ (1/2 √(A2+B2 ),tan-1 (B/A) ) 

Note: 

      1. The equation of the circle through three non-collinear points 

               2. The circle x2 + y2 + 2gx + 2fy + c = 0 makes an intercept on x-axis if x2 + 2gx + c = 0 has real roots i.e. if g2 > c. And, the magnitude of the intercept is 2√(g2-c). 

The Position of a Point with respect to a Circle 

The point P(x1, y1) lies outside, on, or inside a circle S ≡ x2 + y2 + 2gx + 2fy + c = 0, according as S1 ≡ x1

2 + y12 + 2gx1 + 2fy1 + c > = or < 0.

Chord of a circle 

                                        

      The equation of the chord of the circle x2 + y2 + 2gx + 2fy +c=0 with M(x1, y1) as the midpoint of the chord is 

      xx1 + yy1 + g(x + x1) + f(y + y1) = x12 + y1

2 + 2gx1 + 2fy1 

      i.e. T = S1 

Illustration: 

        Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1. 

Solution: 

      Let r be the radius of the circle. Then 

      r = distance of the centre i.e. point (3, 4) from the line 5x + 12y = 1 

      = |(15+48-1)/√(25-44)|=62/13. 

      Hence the equation of the required circle is (x – 3)2 + (y – 4)2 = (62/13)2. 

      ⇒ x2 + y2 – 6x – 8y + 381/169 = 0. 

Illustration: 

       Find the co-ordinates of the point from which tangents are drawn to the circle x2 + y2– 6x – 4y + 3 = 0 such that the mid-point of its chord of contact is (1, 1). 

Solution: 

      Let the required point be (P(x1, y1). The equation of the chord of contact of P with respect to the given circle is 

     xx1 + yy1 – 3(x + x1) –2(y + y1) + 3 = 0. … (1) 

    The equation of the chord with mid-point (1, 1) is 

    x + y – 3(x + 1) – 2(y + 1) + 3 = 1 + 1 – 6 – 4 + 3 

    ⇒ 2x + y = 3. 

    Equating the ratios of the coefficients of x, y and the constant terms and solving for x1, y1 we get x1 = –1, y1 = 0. 

Chord of contact 

Let AP and AQ be tangents to circle from point P(x1, y1). Then equation of PQ is known as equation of chord of contact. 

                                

         For circle x2 + y2 = a2, it is 

              x1x + y1y = a2 

        For general circle, it is 

             x1x + y1y + g(x1 + x) + f(y1 +y) + c = 0 

Note: 

1. It is also written as T = 0 

2. The equation of chord AB [A ≡ (R cos α, R sin α); B ≡ (R cos β, R sin β)] of the circle x2 + y2 = R2 is given by 

             x cos ((α + β )/2) + y sin ((α - β )/2) = a cos ((α - β )/2) 

3. If a line y = mx + c intersects the circle x2 + y2 = a2 in two distinct points A and B then length of intercept AB = 2√((a2 (1+m2 )-c2)/(1+m2 )) 

Caution: 

The equation of a chord of contact and the equation of the tangent on a point of the circle and both given by T = 0. The difference is that while in the case of a tangent the point (x1, y1) lies on the circle. In the case of a chord of contact (x1, y1) lies outside the circle. 

Illustration: 

       Write the equations of tangents to the circle x2 + y2 = 9 and having slope 2. 

Solution: 

        Tangents with slope ‘m’ are given by y = mx + a √(1+m2 ) i.e.                          y = 2x ± 3 √(1+4). 

Note: 

       These are two parallel tangents to the circle at the end of the diameter. 

Illustration: 

     Write the equation of tangents to the circle x2 + y2 = 25 at the point (3, 4)? 

Solution: 

      Point (3, 4) lies on the circle. 

     Required equation of the tangent is : 3x + 4y= 25 using 

     x1x + y1y = a2, where (x1, y1) ≡ (3, 4) 

Illustration: 

     Write the equation of normal to x2 + y2 = 25 at (3, 4)? 

Solution: 

Recall: 

           A normal to the circle passes through the centre of the circle. 

     Normal is: y – 0 = 4/3 (x – 0) (using two point form of straight line) i.e. 3y – 4x = 0

Common Tangents (a) Direct common tangents: 

(i) The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii. 

(ii) The transverse common tangents also meet on the line of centres and divide it internally in the ratio of the radii. 

                

Notes: 

• When one circle lies completely inside the other without touching, there is nocommon tangent. 

• When two circles touch each other internally 1 common tangent can be drawn to the circles. 

• When two circles intersect in two real and distinct points, 2 common tangentscan be drawn to the circles. 

• When two circles touch each other externally, 3 common tangents can be drawn to the circles. 

• When two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn. 

       P is the point of intersection of two direct common tangents to the circles with centres C1 and C2 and radii r1, r2 respectively. C1A1, C2A2 are perpendiculars from C1 and C1 to one of the tangents (figure given below) 

                                  

∴ ΔPC1A1 and ΔPC2A2 are similar 

(C1 P)/(C2 P)=(C1 A1)/(C2 A2 )=r1/r2 i.e. P is a point dividing C1C2 externally in the ratio r1 : r2 For finding direct common tangents of two circles, find the point P dividing the line joining the centre externally in the ratio of the radii. Equation ofdirect common tangents is SS1 = T2 where S is the equation of one circle. 

Caution: 

Length C1C2 > |r1 – r2| 

(b) Transverse Common tangents 

P is the point of intersection of two transverse tangents to two non-intersecting circles with centres C1 and C2 and radii r1, and r2 respectively. Then P lies on the line joining the centres. C1A1 and C2A2 are perpendiculars from C1 and C2 to one of these tangents. (Figure given below) 

                                      

Since triangles C1A1P and C2A2P are similar. 

So (C1 P)/(C2 P)=(C1 A1)/(C2 A2 )=r1/r2 

i.e. P divides the line joining C1 and C2 internally in the ratio r1:r2 

Equation of transverse Common tangents is SS1 = T2 where S is the equation of one of the circle.

Common Tangents 

(a) Direct common tangents: 

(i) The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii. 

(ii) The transverse common tangents also meet on the line of centres and divide it internally in the ratio of the radii. 

                

Notes: 

• When one circle lies completely inside the other without touching, there is nocommon tangent. 

• When two circles touch each other internally 1 common tangent can be drawn to the circles. 

• When two circles intersect in two real and distinct points, 2 common tangentscan be drawn to the circles. 

• When two circles touch each other externally, 3 common tangents can be drawn to the circles. 

• When two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn. 

       P is the point of intersection of two direct common tangents to the circles with centres C1 and C2 and radii r1, r2 respectively. C1A1, C2A2 are perpendiculars from C1 and C1 to one of the tangents (figure given below) 

                                  

∴ ΔPC1A1 and ΔPC2A2 are similar 

(C1 P)/(C2 P)=(C1 A1)/(C2 A2 )=r1/r2 i.e. P is a point dividing C1C2 externally in the ratio r1 : r2 For finding direct common tangents of two circles, find the point P dividing the line joining the centre externally in the ratio of the radii. Equation ofdirect common

tangents is SS1 = T2 where S is the equation of one circle. 

Caution: 

Length C1C2 > |r1 – r2| 

(b) Transverse Common tangents 

P is the point of intersection of two transverse tangents to two non-intersecting circles with centres C1 and C2 and radii r1, and r2 respectively. Then P lies on the line joining the centres. C1A1 and C2A2 are perpendiculars from C1 and C2 to one of these tangents. (Figure given below) 

                                      

Since triangles C1A1P and C2A2P are similar. 

So (C1 P)/(C2 P)=(C1 A1)/(C2 A2 )=r1/r2 

i.e. P divides the line joining C1 and C2 internally in the ratio r1:r2 

Equation of transverse Common tangents is SS1 = T2 where S is the equation of one of the circle.

Enquiry: What do we understand by co-axial circles and limiting points? 

         A system of circles every pair of which has the same radical axis is called a coaxial system. 

                                        

       The centres of circles of a coaxial system, which are of zero radiuses, are called the limiting points o the coaxial system. 

         Let the equation of a system of coaxial circles be 

           x2 + y2 + 2gx + c = 0 

    Where g is a parameter and c is a constant. 

    It’s radius √(g2-c)and centre is (–g, 0) 

         If g2 – c = 0 or g = + √c, then radius become zero and for these two values of g we have two circles of zero radius whose centres are (± √c, 0). 

                                 

         These circles of zero radius are just points and according to definition given above are the limiting points of the co-axial system. 

1. If the system of circles is intersecting one, then c is negative and these limiting points are two imaginary points. 

2. If the system of circles are non intersecting then c is positive and these limiting points are both real. 

3. If c = 0, points of intersection are coincident to (0, 0) i.e. 

    Circle touch each other at (0, 0). 

    Limiting points coincide at (0, 0). 

Enquiry: When do two circles intersect orthogonally? 

        The angle of intersection between two curves intersecting at a point is the angle between their tangents drawn at that point. The curves are said to be intersecting orthogonally, if the angle between their tangents are common point is a right angle. 

        Consider two circles 

       S1 ≡ x2 + y2 + 2g1x + 2f1y + C1 = 0 

       S2 ≡ x2 + y2 + 2g2x + 2f2y + C2 = 0 

     They intersect at point P such that tangent PT1 and PT2 are at right angle (see figure given below) 

                                    

Since radius of a circle is perpendicular to the tangent. So C1P and C2P are also perpendicular. 

      ⇒ (C1C2)2 = (C1P)2 + (C2P)2 ⇒ (g1 – g2)2 + (f1 – f2)2 = r12 + r2

2 

      g12 + g2

2 – 2g1g2 + f12 + f2

2 – 2f1f2=g12 + f1

2 – C1 + g22 + f2

2 – c2 

     ⇒ 2g1g2 + 2f1f2 = C1 + C2 

 which is the required condition for the orthogonal intersection of two circles.

At what angle do the circles shown in figure intersect? 

                                           

From the triangle C1C2P it is clear that angle θ can be written as: 

           cos θ = (r12+r2

2-d2)/(2r1 r2 ) 

Enquiry: What do we understand by pole and polar? 

       Let P(x1, y1) be a fixed point and chords be drawn through this point to a fixed circles S (see figure given below). The locus of the point of intersection of tangents drawn at the end points of chords is a line which is called the polar of point P(x1, y1) (Point P is called the pole) w.r.t. the circle. So there is a fixed polar for a fixed point and a fixed pole for a fixed line. 

                                      

                The equation of polar of a fixed point P(x1, y1) with respect to the circle x2 + y2 + 2gx + 2fy + c = 0 is 

              xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 

              i.e. of the form T = 0

Family of Circles 

Enquiry: If the numbers of conditions for a circle to be drawn are less than three then what shall we get? 

       The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0. Since this equation involves three unknowns i.e. g, f and c so we need at least three conditions to get a unique circle. 

        For example, if we are given two circles and we want to determine the third circle touching both of them. We shall need one more condition. Without the condition we get the equation of family of circles which satisfies the two given conditions. Imposition of a third condition will result in the equation representing a particular circle. 

Let us now see some of the ways of the providing only two condition and equations of the family of circles resulting under these conditions. 

 1. Family of circles having a fixed centre 

           This equation is given by (x – h)2 + (y – k)2 = r2 (Figure given below) 

                                            

         Since (h, k) is fixed, so only parameter varying is r. This is one parameter family of circles, and is the equation of the family of concentric circles. Fixation of the radius will give a particular circle. 

 2. Equation of Family of circles passing through intersection of two circles S1= 0 and S2 = 0. 

       The general equation of the family of circles passing through the intersection of S1 and S2 in given by S1 + kS2 = 0. Here again we have the one-parameter (k) equation of family of circles. The particular value of the parameter gives a unique circles. 

                                          

Caution: 

  If k = –1, we get equation of common chord i.e. straight line instead of circle. 

          Let S1 ≡ x2 + y2 + 2g1x + 2f1y + c1 = 0 

               S2 ≡ x2 + y2 + 2g2x + 2f2y + c2 = 0 

     Since, point lies on both the circles, 

         ⇒ x2A + y2

A + 2g1xA + 2f1yA + c1 = 0 

           x2A + y2

A + 2g2xA + 2f2yA + c2 = 0           ⇒ x2

A + y2A + 2g1xA + 2f1yA + c1 + λ (x2

A + y2A + 2g1xA + 2f1yA + c1)=0 

        ⇒ Point A(xA, yA) lies on S1 + λ S2 = 0 ∀ λ ∈ R 

        Similarly point B(xB, yB) lies on S1 + λ S2 = 0 ∀ λ ∈ R 

∴ S1 + λ S2 = 0 is the family of circles through the intersection of S1 = 0 and S2 = 0

3. Family of Circles passing through intersection of line L and circle S: 

       This equation is given by S + λL = 0 

       The particular value of the parameter λ gives a unique circle. 

4. Family of circles touching the circle S = 0 and line L = 0 at their point of contact 

       Equation S + λL = 0, where λ is the required family. 

5. Family of circle passing through two given points A(x1, x1) and B(x2, y2) 

        

           ∴ Required family of circles is 

 

6. Family of circles touching a given line L = 0 at a point (x1, x1) on the line is (x – x1)2 + (y – x1)2 + λ L = 0, the particular value of the parameters λ gives a unique circle. 

Family of Circles 

(i) If S ≡ x2 + y2 + 2gx + 2fy + c=0 and S’ ≡ x2 + y2 + 2g’x + 2f’y + c=0 are two intersecting circles, then S + λS’ = 0, λ ≠ –1, is the equation of a family of circlespassing through the points of intersection of S = 0 and S’ = 0. 

(ii) If S = x2 + y2 + 2gx + 2fy + c = 0 is a circle which is intersected by the straight line μ = ax + by + c = 0 in two real and distinct points, then S + λμ = 0 is the equation of a family

of circles passing through the points of intersection of S = 0 and μ = 0. 

(iii) The equation of a family of circles passing through two given points (x1, x1) and (x2, y2) can be written in the form 

 

where λ is a parameter. 

(iv) The equation of the family of circles which touch the line y – x1 = m 

(x – x1) at (x1, x1) for any values of m is (x – x1)2 + (y – x1)2 + λ 

[(y – x1) – m(x – x1)] = 0. 

Notes: 

The two circles are said to intersect orthogonally if the angle of intersection of the circles i.e., the angle between their tangents at the point of intersection is 90o. 

The condition for the two circles to cut each other orthogonally is 2gg1+ 2ff1 = c + c1 where (–g, –f) and (–g1, –f1) are the centres of the respective circles, S = 0 and S1 = 0. 

Illustration: 

         Find the equation of the circle described on the common chord of the circles x2 + y2 – 4x – 5 = 0 and x2 + y2 + 8y+ 7 = 0 as diameter 

Solution: 

      Equation of the common chord is S1 – S2 = 0 

      ⇒ x + 2y + 3 = 0 

     Equation of the circle through the two circles is S1 + λS2 = 0 

     ⇒ x2 + y2 -4/(1+λ ) x+8λy/(1+λ )+(7λ -5)/(1+λ ) = 0.       Its centre (2/(1+λ ),-4/(1+λ )) lies on x + 2y + 3 = 0 

     ⇒ 2/(1+λ )-8/(1+λ ) + 3 = 0 ⇒ 2 – 8λ + 3 + 3λ = 0 ⇒ λ = 1. 

     Hence the required circle is x2 + y2 – 2x + 4y + 1 = 0.

Illustration: 

     The line Ax + By + C = 0 cuts the circle x2 + y2 + ax + by + c = 0 in P and Q. The Line A?x + B?y + C? = 0 cuts the circle x2 + y2 + a'x + b'y + c' = 0 in R and S. If P, Q, R, S are concyclic, prove that 

                                        

Solution: 

       The equation of the circle through the first line and the first circle, i.e. through P and Q is 

       x2 + y2 + ax + by + c + λ1 (Ax + By + C) = 0. ........................ (1) 

      The equation of the circle through R and S is 

      x2 + y2 + a'x + b'y + c + λ2 (A'x + B'y + C') = 0....................  (2) 

     Since P, Q, R, S are concyclic, (1) and (2) are identical. 

     ⇒ 1 = (a+λ1 A)/(a'+λ2 A' )=(b+λ1 B)/(b'+λ2 B' )=(c+λ1 C)/(c'+λ2 C' ) 

     ⇒ λ1A-λ2A' + a - a' = 0, 

       λ1B -λ2B' + b - b' = 0, 

     λ  1C-λ 2C' + c - c' = 0. 

       

Illustration: 

      Show that the circle passing through the origin and cutting the circles x2 + y2 - 2a1x - 2b1y + c = 0 and x2 + y2 - 2a2x - 2b2y + c = 0 orthogonally is 

                                      

Solution: 

      Let the equation of the circle passing through the origin be 

     x2 + y2 + 2gx + 2fy = 0. ..................... (1) 

     It cuts the given two circles orthogonally 

    ⇒ -2ga1 - 2fb1 = c1 ⇒ c1 + 2ga1 + 2fb1 = 0 ? (2) 

    and -2ga2 - 2fb2 = c2 ⇒ c2 + 2ga2 + 2fb2 = 0 ? (3) 

      

Illustration: 

     Find that member of the family of circles having centre at (2, 3) which has radius of 5 unit. 

Solution: 

Family of circles having fixed centre (2, 3) is 

(x - 2)2 + (y - 3)2 = r2 

here we want that particular circle that has radius 5 units i.e. r = 5, the required circle is, 

(x - 2)2 + (y - 3)2 = 25 

Illustration: 

       Find a circle passing through the intersection of x2 + y2 - 4 = 0 and x2 + y2 - 6x + 5 = 0 which passes through the point (2, 1)? 

Solution: 

        Family of required circles is S1 + λ S2 = 0 

       ⇒ (x2 + y2 - 4) + λ (x2 + y2 - 6x + 5) = 0 

Since the required circle passes through the point (2, 1), the previous equation is satisfied for the point (2, 1) 

     ⇒ (4 + 1 - 4) + λ (4 + 1 - 12 + 5) = 0 

     1 - 2λ = 0 ⇒ λ = 1/2 

    ∴ Equation of the required circle is 

     (x2 + y2 - 4) + 1/2 (x2 + 2y - 6x + 5) = 0 

     ⇒ x2 + y2 - 2x - 1 = 0

External and Internal Contacts of Circles 

If two circles with centres C1(x1, y1) and C2(x2, y1) and radii r1 and r2 respectively, touch each other externally, C1C2 = r1 + r2. Coordinates of the point of contact are A ≡ ((r1 r2+r2 r1)/(r1+r2 ),(r1 y2+r2 y1)/(r1+r2 )). 

                                   

The circles touch each other internally if 

C1C2 = r1 – r2. 

                                            

Coordinates of the point of contact are 

T ≡ ((r1 x2+r2 x1)/(r1+r2 ),(r1 y2+r2 y1)/(r1+r2 )). 

Illustration 11: 

       Examine whether the two circles x2 + y2 – 2x – 4y = 0 and  x2 + y2 – 8y – 4 = 0 touch each other externally or internally. 

Solution: 

        Let C1 and C2 be the centres of the circles. 

       ⇒ C1(1, 2) and C2(0, 4). Let r1 and r2 be the radii of the circles 

       ⇒ r1 = √5 and r2 = 2√5. Also C1C2 = √(1+4)=√5. 

       But r1 + r2 = 3√5 and r2 – r1 = √5 = C1C2. 

       Hence the circles touch each other internally. 

Illustration 12: 

        Find all the common tangents to the circles x2 + y2 – 2x – 6y + 9 = 0 and x2 + y2 – 6x – 2y + 1 = 0. 

Solution: 

       The centres and the radii of the circles are 

       C1 (1, 3) and r1 = √(1+9-9) = 1, C2(–3, 1) and r2 = √(9+1-1) = 3, 

       C1C2 = √20, r1 + r2 = 4 = √16 and C1C2 > r1 + r2. 

      Sense the circles are non-intersecting. Thus there will be four common tangents. 

      Transverse common tangents are tangents drawn from the point P which divides C1C2 internally in the ratio 1 : 3. 

      Direct common tangents are tangents drawn from the point Q which divides C1C2 externally in the ratio 1 : 3. 

      Coordinates of P are 

       ((1(-3)+3.1)/(1+3),(1.1+3.3)/(1+3)) i.e. (0,5/2). 

      and coordinates of Q are (3, 4). 

      Transverse tangents are tangents through the point (0,5/2). 

      Any line through (0,5/2) is 

      y – 5/2 = mx …… (1) 

      ⇒ mx – y + 5/2 = 0 

      Applying the usual condition of tangency to any of the circle, we get 

      (m.1 – 3 + 5/2 = 0)/√((m2+1) ) = 1 ⇒ (m-1/2)2 = m2 + 1 

      ⇒ – m – 3/4 = 0 or 0.m2 – m – 3/4 = 0. 

      ⇒ m = –3/4 and ∞ as coefficient of m2 is zero. 

      Therefore from (1), 

      (y-5/2)/x = m = ∞ and -3/4 ⇒ x = 0 is a tangent and 

      3x + 4y – 10 = 0 is another tangent. 

      Direct tangents are tangents drawn from the point Q(3, 4). 

      Now proceeding as for transverse tangents their equations are 

       y = 4, 4x – 3y = 0.

Power of a point with respect to a circle 

         The power of a point P(x1, y1) with respect to a given circle defined as 

         Power = PA × PB 

        Where A and b are the points on the circle where the line PAB intersects it. (See figure given below). 

                                 

          Clearly when A and B coincide to T then we have            Power = PT2 = Square of the length of tangent from point P 

         (x1, y1) to the circle.

Radical Axis 

This radical axis of two circles is the locus of a point from which the tangent segments to the two circles are of equal lengths. 

Equation of the Radical Axis 

In general S – S’ = 0 represents the equation of the Radical Axis to the two circles 

i.e. 2x(g – g’) + 2y(f – f’) + c – c’ = 0 

where S ≡ x2 + y2 + 2gx + 2fy + c = 0 

and S’ ≡ x2 + y2 + 2g’x + 2f’y + c’ = 0 

(i) If S = 0 and s’ = 0 intersect in real and distinct points then S – S’ = 0 is the equation of the common chord of the two circles. 

(ii) If S’ = 0 and S = 0 touch each other, then S – S’ = 0 is the equation of the common tangent to the two circles at the point of contact. 

           

Properties of the Radical Axes 

• The radial axis of two circles is perpendicular to the line joining the centres. 

• The radical axis of three circles taken two at a time are concurrent and the point of concurrency is known as the radical centre. 

• The radical axis of two circles bisects their direct common tangents. 

• If two circles cut a third circle orthogonally, then the radical axis of the two circles will pass through the centre of the third circle.

 Solved Examples 

Example 1: 

       Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4, x + 2y = 5. 

Solution: 

Method 1. 

      Consider the equation 

                 (x + y – 6) (2x + y – 4) + λ1 (2x + y – 4) (x + 2y – 5) + λ2 

                 (x + 2y – 5) (x + y – 6) = 0 …… (1) 

              This equation is satisfied by the points of intersection of any two of the given three lines, i.e. it is satisfied by the vertices of the triangle formed by the given lines. 

Result: 

      Now if (i) represents a circle then 

     (a) coefficient of x2 = coefficient of y2 and 

     (b) coefficient of xy = 0 

And 

     3 + 5λ1 + 3λ2 = 0 ⇒ λ1 = – 6/5 

Substituting these values in (i) and simplifying 

We get 

         x2+ y2 – 17x – 19y + 50 = 0 (Ans.) 

Which is the equation of the required circle. 

Method 2. 

       Solve the lines in pair to find the vertices of the triangle and then obtain theequation of the circle through these three points. 

Example 2: 

       Find the locus of the point of intersection of perpendicular tangents to the circle x2 + y2 = 4. 

Solution: 

Method 1. 

        y = mx + 2 √(1+m2 ) is tangent to x2 + y2 = 4 for all values of ‘m’. 

       It passes through (h, k) if 

       k = mh + 2 √(1+m2 ) 

      ⇒ (k – mh)2 = 4 (1 + m2) 

      ⇒ m2 (h2 – 4) –2h km + (k2 – 4) = 0 …… (i) 

                                    

         The roots m1 and m2 of this quadratic equation are the slopes of PT and PT’. If PT and PT’ are at right angle then m1m2 = –1. 

         From (i) 

        m1m2 = (k2-4)/(h2-4) 

       ⇒ (k2-4)/(h2-4) = –1 

       ⇒ h2 + k2 = 8 

  Locus of P(h, k) is x2 + y2 = 8

Method 2: 

            Equation of tangent at the point T(2 cos θ, 2 sin θ) 

                                  

        is 2 cos θ x + 2 sin θ y = 4 

        i.e. x cos θ + y sin θ = 2 …… (i) 

        Tangent at T’ would be perpendicular to the tangent at T 

        If ∠ TOT’ = 90o 

        i.e. ∠AOT’ = 90 + θ 

        Co-ordinates of T’ are (2 cos (90 + θ), 2 sin (90 + θ)) 

        Equation of tangent at the point T’ is 

        – 2 sin θ x + 2 cos θ y = 4 

        or, – x sin θ + y cos θ = 2 …… (ii) 

Think: 

       How to get the locus of point P(h, k)? 

Caution: 

      Do not simply square and add (i) and (ii). Though we get the required result, but that it not the right approach. 

        Well, lines (i) and (ii) both pass through the point P(h, k) 

        ⇒ h cos θ + k sin θ = 2 and 

        – h sin θ + k cos θ = 2 

        .Now we want to get a relation between h and k and also eliminate  

        The best way to do this is to square and ad above two 

        Equations and we get h2 + k2 = 8 

        ⇒ Locus of P(h, k) is x2 + y2 = 8 

Method 3: 

       OP2 = OT2 + TP2 (∵ ΔOTP is a right angled triangle) 

      ⇒ OP2 = OT2 + OT’2 

      ⇒ OP2 = 2R2 

      ⇒ h2 + k2 = 2(4) = 8 

      ∴ Locus of P(h, k) is x2 + y2 = 8. 

Note: 

       x2 + y2 = 8 is director circle of the circle x2 + y2 = 4 

Example 3: 

       Find the condition that the line 3x + 44y – p = 0 is tangent to the circle x2 + y2– 4x – 6y + 9 = 0 

Solution: 

       Radius of the given circle is 2 and centre is (2, 3). So for line 3x + 4x – p = 0 to be tangent to the circle we have, 

       |(3.2+4.3-p)/√(32+42 )| = 2. 

       |18 – p| = 10 

       18 – p = 10 ⇒ p = 8 

       18 – p = –10 ⇒ p = 28 

Hence the required condition is p = 8 or p = 28. (Ans.)

Example 4 

         A circular plot of land in the form of a unit circle is to be divided into two equal parts by the arc of a circle whose centre is on the circumference of the circular plot. Show that the radius of the circular arc is 2 cos θ, where θ is given by sin 2θ – 2θ cos 2θ = π/2. 

Solution: 

        Let O be the centre of the given circular plot of radius 1 i.e. OA = OB = OC = 1 and A be any point on its circumference. Again BDC be the arc of the circle with centre A and dividing the given circle into two equal parts. Let r the radius of the new circle, then AB = AC = AD = r. 

                                          

       Let ∠AOB = θ 

      Then ∠OBA = ∠OCA = θ, and ∠AOB = (π – 2θ). 

      Now area ABDCA must be = 1/2 area of unit circle = (π (1)2)/2 = π/2 …… (i) 

              r/sin(π -2θ ) =1/(sin π ) ⇒ r = 2 cos θ. 

     Required Area APBDCQA = Area of sector ABDCA + area of sector OCQAPB – 2 area of OAB 

       ⇒ π/2 = 1/2 r2 (2θ) + 1/2 (1)2 (2π – 4θ) – 2 × 1/2 (1)2 sin (π – 2θ) 

         sin 2θ – 2θ cos 2θ = π/2 (Ans.) 

Example 5: 

       Find the equation of the circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of lines 3x + 5y = 1, (2 + c)x + 5c2 y = 1. 

Solution: 

    Solving, 3x + 5y = 1, (2 + c)x + 5c2y = 1 

 We get, 

        

   when c → 1 

Pause: 

      We will study limits in detail in module 5. 

     Now, we want to find out the equation of the circle which passes through (2, 0) and has its centre at (2/5), –1/25). 

     Equation of the circle is 

     (x-2/5)2+(y+1/25)2=(2/5-2)2+(1/25)2 

   This is the required equation of the circle.

Example 6: 

          Find the equation of circle having the lines x2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the circle, x(x – 4) + y(y – 3) = 0 

Solution: 

        The combined equation of two normal of the circle is given by 

         x2 + 2xy + 3x + 6y = 0 

        ⇒ (x + 3)(x + 2y) = 0 

        ⇒ x = –3, x = –2y 

Recall: 

      A normal to a circle always passes through the centre of the circle. Now solving these, we get the co-ordinates of the centre of the circle as (–3, 3/2); because the two normal intersect at the centre of the circle 

    The required circle just contains the circle 

           x(x – 4) + y(y – 3) = 0 

           i.e. x2 + y2 – 4x – 3y = 0 …… (i) 

    Hence the required circle will touch the circle given by (1) internally. 

    Let r be the radius of the required circle. Now the two circles given by (1) 

        = √([22+(3/2)2 ] )=5/2 and centre = (2,3/2)     Now the required circle will touch the circle (i) internally, if    We have distance between the centre of the two circles = difference between their radii. 

       ⇒ √((-3-2)2+(3/2-3/2)2 )=(r-5/2)          ⇒ r = 15/2 

  Hence the equation of the required circle is given by 

       (x + 3)2 + (y-3/2)2=(15/2)2 

       x2 + y2 + 6x – 3y – 45 = 0 (Ans.) 

Example 7: 

        A tangent is drawn to each of the circles x2 + y2 = a2, x2 + y2 = b2. Show that if these two tangents are perpendicular to each other, the locus of their point of intersection is a circle concentric with the given circles. 

Solution: 

Method 1: 

       Let P ≡ (x1, y1) be the point of intersection of the tangents PA and PB where A, B are points of contact with the circles respectively. 

      As PA is perpendicular to PB, the corresponding radii OA and OB are also perpendicular. Let ∠AOX = θ 

                                          

        ⇒ ∠BOX = 90 + θ 

  Using the parametric from of the circle we can take 

      A ≡ (a cos θ, a sin θ) 

     ∠ B ≡ (–b sin θ, b cos θ) 

 The equation of PA is 

     x (a cosθ) + y (a sin θ) = a2 

     y cos θ – x sin θ = b 

 Since P(x1, y1) lies on these tangents 

   ⇒ x1 cos θ + y1 sinθ = a and y1 cos θ – x1 sin θ = b 

. Squaring and adding above equation (we get).as θ is a variable quantity; we eliminate  

    x12 + y1

2 = a2 + b2 

⇒ locus of p is x2 + y2 = a2 + b2, which is concentric with given circles. 

Method 2: 

    OAPB is a rectangle 

        ⇒ OP2 = OA2+ AP2 

        ⇒ x12 + y1

2 = a2 + b2 

       ⇒ Locus of P(x1, y1) is x2 + y2 = a2 + b2

Example 9: 

          The circle x2 + y2 = 1 cuts x-axis at P and Q. Another circle with centre at Q and variable radius intercepts the first circle at R above x-axis and the line segment PQ at S. Find the maximum area of the triangle QSR. 

Solution: 

Method 1. 

          Equation of circle centred at Q is (x + 1)2 + y2 = μ2 

                                    

         Since point R (cos θ, sin θ) lies on this circle 

        ⇒  μ2 = (cosθ + 1)2 + (sin θ)2 = 2 + 2 cos θ 

       = 2 (2cos2 θ/2) ⇒ μ = 2 cos θ/2 

     A = Area of ΔQSR = 1/2 base × altitude 

        = 1/2 μ sin θ 

       = cos θ/2 sin θ…… (i) 

      dA/dθ = cos θ cos θ/2-1/2 sin θ/2 sin θ 

Pause: 

          We will study maxima and minima in module 5. 

          For max./min. dA/dθ = 0 

                                       

        ⇒ tan θ = 2 cot θ/2 

        ⇒ 2t/(1-t2 )=2/t (where t=tan θ/2) 

       ⇒ t = tan θ/2=1/√2 

      (d2 A)/(d2 θ ) = – sin θ cos θ/2-1/2 cos θ sin θ/2-1/2 sin θ/2 cos θ – 1/4 cos θ/2 sin θ 

       (A"(θ )(at tan θ/2=1/√2) ) = – ve          From (i) 

       Maximum area = cos θ/2 sin θ 

        = 2 sin θ/2 (cos θ/2)2 

        4/(3√3) sq. units. (Ans.) 

Method 2. 

        Equation of circle I is 

            x2 + y2 = 1 It cuts x-axis 

      where P ≡ (1, 0) and 

               Q ≡ (–1, 0) 

     Let QR = μ then equation of the circle II. 

   Centred at Q(–1, 0) and radius = μ is given by 

         (x + 1)2 + y2 = μ2 …… (1) 

  Solving it with x-axis; we get S ≡ (μ – 1, 0). 

  Also solving the two circles, we get the co-ordinates of R 

          as [(μ2/2)-1,μ/2 √((4-μ2 ) )] 

The area of ΔQRS = 1/2 QS × RL 

       = 1/2 μ μ/2 √((4-μ2 ) ) 

       = A (say) 

   Now A is max./min. means A2 is max./min. 

       Let A2 = Z. 

     Then Z = μ4/16 (4 – μ)2 

     ⇒ dz/dμ=1/4. 4μ3 – (6μ5)/16 

      (d2 z)/dμ2 = 3μ3 – 30 (μ4/16) 

  For max./min. of A i.e. max./min. of A2 or Z. we get 

       dZ/dμ = 0 ⇒ μ = √(8/3) and then 

       (d2 z)/dμ2 = 3.8/3-30/10.64/9 = –ve 

       For μ = √((8/3)) 

  Thus area is max when μ = √((8/3)) 

  Also max. area of ΔQRS 

      1/4.8/3 √((4/3))=(4/3√3) sq. units.

Example 10: 

          Distances from the origin to the centres of three circles x2 + y2 – 2λx = c2(where c is constant and λ is variable) are in G.P. Prove that the lengths of tangents drawn from any point on the circle x2 + y2 = c2 to the three circles are in G.P. 

Solution: 

     The equation of the three circles is 

         x2 + y2 – 2λx = c2 …… (1) 

   where λ = λ1, λ2, λ3. Their centres are: 

(λ       1, 0), (λ2, 0) and (λ3, 0) 

   Now distances of these points from the origin are λ1, λ2 and λ3 

       ⇒ λ1λ3 = λ22 …… (2) 

   Now, let P(h, k) be any point on the circle 

        x2 + y2 = c2, then h2 + k2 = c2 …… (3) 

   If r1, r2, r3 are the lengths of the tangents from P(h, k) on 

  The three circles, we then obtain 

      r12 = h2 + k2 – 2λ1h – c2 = c2 h – c– 22, using (3) 

      r12 = – 2λh 

  Similarly r22 = – 2λ2h 

       r32 = – 2λ3h 

     = r12 r3

2 = 4λ1 λ2 h2 = 4λ22 h2 = (r22)2 

     = r2r3 – r22 ⇒ r1, r2, r3 are in G.P.                  (Proved). 

Example 11: 

        Show that the equation x2 + y2 – 4x – ky – 5 = 0 represents (for variable k) a family of circles passing through two fixed points A and B. Find the equation of the circle belonging to this family and cutting circle x2 + y2 – 6x – 5y = 0 at right angles.

Solution: 

         x2 + y2 – 4x – ky – 5 = 0 

        (x2 + y2 – 4x – 5) + k (– y) = 0 

    This is the equation of family of circles passing though the intersection point of x2+ y2 – 4x – 5 = 0 (a circle) and a straight line putting y = 0, in x2 + y2 – 4x – 5 = 0 gives 

       x2 – 4x – 5 = 0        ⇒ x = –1, 5 

  Hence the given circle passes through two fixed points (–1, 0) and (5, 0) 

  For given family of circles 

      x2 + y2 – 4x – ky – 5 = 0 …… (1) 

      g = –2, f = – k/2, c = –5 

  One member of family (1) and circle x2 + y2 – 6x – 5y – 0 …… (2) 

  Intersect orthogonally 

  For circle (2) 

     g’ = – 3, f’ = –5/2, c’ = 0 

For two circle to be orthogonal 

    2(gg’ + ff’) = (c + c’) 

    2[(–2) (–3) + (–k/2)(5/2)] = – 5 + 0      2 [6+5K/4] = – 5 

    12 + 5K/2 = – 5      5K/2 = – 17 

    K = 34/5 

    Required equation of circle is 

    x2 + y2 – 4x + 34/5 y – 5 = 0 (Ans.)

Example 12: 

      Lines 5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 touch a circle C1 of diameter 6. If the centre of C1 lies in first quadrant, find the equation of circle C2 which is concentric with C1 and cuts intercept of length 8 on these lines. 

Solution: 

Recall: 

      If a circle touches two lines L1 and L2 then the centre of the circle lies on the angle bisectors of the lines. 

     Angle bisector of given lines are 

             (5x+12y-10)/13=±(5x-12y-40)/13 

    Taking +ve sign: y = (-5)/4 

    Taking –ve sign: x = 5 

    ⇒ Given lines L1 and L2 intersect at (5,-5/4) 

    Since the centre of C1 lies in the first quadrant, it can lie on x = 5 only. 

    Let the centre of C1 be (5, y1) 

         ⇒ |(5(5)+12y1-10)/√(52+122 )| = 3 

         ⇒ 15 + 12y1 = ± 39 

         ⇒ y1 = 2 or y1 = -54/12 (Neglect) 

         ⇒ Centre of C1 is (5, 2) 

    Since C2 is concentric with C1, its centre is also (5, 2) 

                                      

    C2 cuts intercept of length 8 on lies          5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 

        ⇒ AB = CD = 8 

Recall: 

     Perpendicular from centre bisects the chord. 

          ⇒ AN = 4 

          C2 N = 3 (given) 

    Radius of C2 = r (say) 

         r2 = (AN)2 + (C2N)2 = 16 + 9 

        ⇒ r = 5 

     Equation of C2 is (x – 5)2 + (y – 2)2 = 25 

Example 13: 

      Find the equation of the circle which passes through the point (2a, 0) and whose radical axis with respect to the circle x2 + y2 = a2 is the lines x = a/2. 

Solution: 

Recall: 

      Radical axis of two circles is S1 – S2= 0. 

      Let equation of circle is 

         x2 + y2 + 2gx + 2fy + c = 0 …… (i) 

      radical axis of circle (i) and circle x2 + y2 – a2 = 0 is given by 

           x2 + y2 + 2gx + 2fy + c = x2 + y2 – a2 

           2gx + 2fy + c + a2 = 0 …… (ii) 

      It is given that radical axis is x – a/2 = 0 …… (iii) 

      Comparing (ii) and (iii) 

      We get f = 0, (c+a2)/2g=-a/2 

           ⇒ ag + a2 + c = 0 …… (iv) 

      Circle (i) passes through (2a, 0) 

           4ag + 4a2 + c = 0 …… (v) 

      From (iv) and (v) 

           3ag + 3a2 = 0 

           ⇒ g = –a 

           ⇒ c = – (ag + a2) from (iv) 

           = –(–a2 + a2) = 0 

      Equation of circle is 

           x2 + y2 – 2ax = 0 (Ans.)

Example 14: 

        Show that x2 + y2 + 4y – 1 = 0, x2 + y2 + 6x + y + 8 = 0 and x2 + y2 – 4x – 4y – 37 = 0 touch each other. 

Solution: 

      S1 ≡ x2 + y2 + 4y – 1 = 0 

      S2 ≡ x2 + y2 + 6x + y + 8 = 0 

      S3 ≡ x2 + y2 – 4x – 4y – 37 = 0 

      C1 ≡ (0, –2), C2 ≡ (–3, –1/2), C3 ≡ (2, 2) 

      r1 = √(4+1) = √5 

      r2 = √(9+1/4-8)=√5/2 

      r3 = √(4+4+37)=√45=3√5 

      C2C3 = √((2+3)2+(2+1/2)2 )=√(25+25/4)=5/2 √5 

      C1C3 = √((2-0)2+(2+2)2 )=2√5 

                                

      r1 + r2 = 3/2 √5 = C1C2 

      r3 – r2 = 5/4 √5 = C2C3 

      r3 – r1 = 2√5 = C1C3 

Recall: 

       Two circles touch each other. 

       (i) externally if C1C2 = r1 + r2 

       (ii) internally if C1C2 = r1 – r2 

Example 15: 

       Find the four common tangents to the circles x2 + y2 – 22x + 4y + 110 = 0 and x2 + y2 – 22x – 4y – 100 = 0 

Solution: 

       S1 ≡ x2 + y2 – 22x + 4y + 100 = 0 

       S2 ≡ x2 + y2 + 22x – 4y – 100 = 0 

       C1 (11, –2), C2 (–11, 2) 

                                      

      r1 = √((11)2+(-2)2-100) = 5 

       r2 = √((-11)2+(2)2+100) = 15 

   Out of four common tangents are transverse tangents and other two direct tangents. (1) and (2) are direct common tangents while (3) and (4) are transverse common tangents. 

Recall: 

     Transverse common tangents divide line joining centres in ratio of radii internally while direct tangents divides line joining centres in ratio of radii externally. 

    Let T1, T2 divide C1C2 in ratio of r1 : r2 internally and externally respectively. 

   Co-ordinates of T1 are (15×11+5×(-11))/(15+5) and (15×(-2)+5×2)/(15+5) 

   That is T1 is the point (11/2,-1) 

   Co-ordinates of T2 are (15×11-5(-11))/(15-5) and (15×(-2)-5×2)/(15-5) 

   that is T2 is the point (22, – 4) 

   Let the equation to either of the tangents, passing through T1 be 

       y + 1 = m (x – 11/2) …… (A) 

   Then the perpendicular from the point (11, –2) on it is equal to + 5 and hence 

       (m(11-11/2)-(-2+1))/√(1+m2 ) = ± 5 

   On solving, we have m = -24/7 or 4/3 

   The required tangents through T1 are therefore 

       24x + 7y = 125, and 4x – 3y = 25 

   Similarly the equation to the tangents through T2 is 

       y + 4 = m (x – 22) …… (B) 

   where (m(11-22)-(-2+4))/√(1+m2 ) = ± 5 

   On solving, we have m = 7/24 or –3/4 

   On substitution in (B) the required equations are therefore 

   x – 24y = 250 and 3x + 4y = 50 

   The four common tangents are therefore found. (Ans.)

Example 16: 

        The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is x + y – xy + k (x2 + y2)1/2 = 0 Find k? 

Solution: 

Method 1. 

      The equation of the incircle can be put in the from (x – 2)2 + (y – 2)2 = 4 

      This implies that the inradius r = 2 …… (i) 

     Let the hypotenuse of the triangle meet OX and OY at A (a, 0) and B (0, b) respectively. 

          r = (Area of ΔAOB) / ((1/2)(Sum of sides of ΔAOB)) = ((1/2)ab) / ((1/2)(a+b+√(a2+b2 ))) 

        ∴ ⇒ 2 = ab/(a+b+√(a2+b2 )) 

   Let M (x1, y1) be the circumcentre of ΔOAB. Since Δ OAB is right angled, it’s circumcentre is the mid point of hypotenuse. 

       ∴ ⇒ 2 = (4x1 y1) / (2x1+2y1+2√(x12+y1

2 )) 

         ⇒ x1 + y1 + √(x12+y1

2 ) = x1y1 

  Hence the equation of the locus of M(x1, y1) is 

  x + y – xy + √(x2+y2 ) = x1y1     Comparing it with the given equation of the locus, we find that 

   k = 1. (Ans.) 

Method 2. 

    Equation of AB is x/a + y/b = 1 …… (1) 

   (x1, y1) ≡ (a/2,b/2), where M(x1, y1) is the circumcentre of ΔOAB i.e. midpoint of the hypotenuse. 

   (1) becomes: x / (2x1 )+y / (2y1 ) = 1 …… (2) 

   ⇒ In circle (x – 2)2 + (y – 2)2 = 4 touches line (2) 

       ⇒ (2/(2x1 )+2 / (2y1 )-1) / √((1/2x1 )2+ (1/2y1 )2 ) = –2 

       ⇒ x1 + y1 – x1 y1 + √(x12+y1

2 ) = 0 

   ∴ Locus of M(x1, y1) is 

       x + y – xy + √(x2+y2 ) = 0 

   Comparing it with the given equation of the locus, we find that k = 1. 

Note: 

   Distance from (2, 2) to the line x/2x1 +y/2y1 -1 = 0 has been taken –2, because origin and this point lies on the same side of the origin. 

Parabola      Parabola Definition

Parabola is the chief and easiest chapter in the Conic Sections of Co-ordinate Geometry in Mathematics. Let us start analyzing the thought “A point always moves such that the ratio of its distances from a fixed point and a fixed line is constant. Can we get some meaningful result? Yes we get a parabola, which follow the above rule/thought. We can represent the curve mathematically using the co-ordinates and this curve is useful in finding out the many physical/practical phenomenons. 

If we take a Right Circular Cone and cut it by a horizontal plane, we get a cross-section which is circular. If we cut this cone by planes of different orientations, different planes produce different type of curves. As all these curves are sections of a right circular cone, we call them conic sections. When we analyze these curves in detail we find that when the ratio of the distances of a point on the curve from a fixed point to its distance from a fixed line, is equal to 1 we have one type of curve, for ratio less than 1, we have second type of curve and for ratio more than 1, we have third type of curve. With this basis we can analyze these three curves. 

In this sections we’ll discuss the curve of 1st type for which the ratio is equal to 1 i.e. the distance from the fixed point is always equal to the distance from the fixed line. We call this a Parabola.

“Parabola” is one of the easiest and important chapters of Conic Sections of Co-ordinate Geometry in the Mathematics syllabus of IIT JEE, AIEEE and other engineering examinations. The chapter is important because it fetches 1-2 questions in most of the engineering examination

Index 

1. Conic Sections- Parabola Definition

2. Focal Chord

3. Chord

4. Tangent to a Parabola

5. Normal to a Parabola

6. Propositions on the Parabola

7. Solved Examples

Parabola is important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic. You are expected to do all the questions based on this to remain competitive in IIT JEE examination. It is very important to master these concepts at early stage as this forms the basis of your preparation for IIT JEE, AIEEE, DCE, EAMCET and other engineering entrance examinations. 

Conic Sections 

        Conics or conic sections are the curves corresponding to various plane sections of a right circular cone by cutting that cone in different ways. 

      Each point lying on these curves satisfies a special condition, which actually leads us towards the mathematical definition of conic sections. 

      If a point moves in plane in such a way that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line, always remains constant, then the locus of that point I called a Conic Section. 

      The fixed point is called the focus and the fixed line is called directrix of the conic. The constant ratio is called the eccentricity and is denoted by e. 

      According to the value of there are three types o conic i.e. for e = 1, e < 1 and e > 1 the corresponding conic is called parabola, ellipse and hyperbola respectively. 

A conic section or conic is the locus of a point, which moves so that its distance from a fixed point is in a constant ratio to its distance from a fixed straight line, not passing through the fixed point. 

The fixed point is called the focus. 

• The fixed straight line is called the directrix. 

• The constant ratio is called the eccentricity and is denoted by e. 

• When the eccentricity is unity i.e. e = 1, the conic is called a parabola; when e < 1, the conic is called an ellipse; and when e > 1, the conic is called a hyperbola. 

• The straight line passing through the focus and perpendicular to the directrix is called the axis of the parabola. 

• The point of intersection of a conic with its axis is called vertex. 

• The chord passing through focus and perpendicular to axis is called latus rectum. 

• Any chord of the parabola which is perpendicular to the axis is called double ordinate. 

• The straight line perpendicular to axis of the parabola passing through vertex is called tangent at the vertex. 

Axis of the conic: 

The line through focus and perpendicular to the directrix is called the axis of the conic. The intersection point o conic with axis is known as the vertex of the conic. 

Enquiry: How do we mathematically define a parabola and what are its various features? 

       The locus of the point, which moves such that its distance from a fixed point (i.e. focus) is always equal to its distance from a fixed straight line (i.e. directrix), is called parabola. 

Equation of Parabola: 

       Let S be the focus, V be the vertex, ZM be the directrix and x-axis be the axis of parabola. We require therefore the locus of a point P, which moves so that its distance from S, is always equal to PM i.e. its perpendicular distance from ZM. After appropriate configuration let S = (a, 0)

                                    

                  We have ten SP2 = PM2 

                  ⇒ (x – a)2 + y2 = (a + x)2 

                  ⇒ y2 = 4ax This is the standard equation of a parabola.

There are four common forms of parabola according to their axis, with their vertex at origin (0, 0). 

            

Let us understand some other features of a parabola. 

(a) Focal Distance: 

      The distance of a point on the parabola from its focus is called the focal distance of the point Focal distance of P = SP = x + a. 

(b) Focal Chord: 

      A chord of the parabola, which passes through its focus, is called Focal chord. 

(c) Latus Rectum: 

     The chord through focus and perpendicular to the axis of the parabola is called the latus rectum. 

     The co-ordinates of the end point of the latus rectum are (a, 2a) and (a, –2a) and length of latus rectum = |4a|. 

For horizontal parabola 

      Let us consider origin (0, 0) as the vertex A of the parabola and two equidistant points S(a, 0) as focus and Z(–a, 0) a point on the directrix now let P(x, y) be the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ, that is A, will lie on the locus of P. 

i.e. AS = AZ. 

                              

The x-axis along AS, and the y-axis along the perpendicular to AS, as A, as in the figure. Now by definition PM = PS ⇒ MP2 = PS2 

So, that, (a + x)2 = (x – a)2 + y2. 

Hence, the equation of horizontal parabola is y2 = 4ax. 

Similarly for the vertical parabola 

     Let us consider origin (0, 0) as the vertex A of the parabola and two equidistant points S(0, b) as focus and Z(0, –b) a point on the directrix now let P(x, y) be the moving point. Draw SZ perpendicular from S on the directrix. Then SZ is the axis of the parabola. Now the middle point of SZ, that is A, will lie on the locus of P i.e. AS = AZ. 

  The y-axis along AS, and the x-axis along the perpendicular to AS at A, as in the figure.

 

Now by definition PM = PS 

⇒ MP2 = PS2 

So that, (b + y)2 = (y – b)2 + x2. 

Hence, the equation of vertical parabola is x2 = 4by. 

Finding the end points of latus Rectum 

For finding the end points of latus rectum LL’ of the parabola y2 = 4ax, we put x = a as latus rectum passes through focus (a, 0) therefore we have 

                                   

                                    

      y2 = 4a2 

      ⇒ y = + 2a 

Hence the end points are (a, 2a) and (a, – 2a). 

Also LSL’ = |2a – (–2a)| = 4a = length of double ordinate through the focus S. 

Note: 

  Two parabolas are said to be equal when their latus recta are equal.

  The important points & lines related to standard Parabola 

 

Note: 

     1. The points and lines of two parabolas can be interchanged by transformations. 

       2. If a > 0 & a < 0 the parabola will be forward opening and backward opening respectively. 

       3. If b > 0 & b < 0 the parabola will be forward opening and downward opening respectively. 

The important points & lines related to shifted Parabola 

The forms of the horizontal and vertical parabola having vertex at (h, k) can be obtained by shifting the origin at (h, k) as below 

Illustration: 

      Find the vertex, axis, directrix, tangent at the vertex and the length of thelatus rectum of the parabola 2y2 + 3y – 4x – 3 = 0. 

Solution: 

     The given equation can be re-written as (y-3/4)2= 2 (x+33/32) 

     Which is of the form Y2 = 4aX. 

     Hence the vertex is (-33/32,-3/4). 

     The axis is y + 3/4 = 0 y = –3/4. 

     The directrix is X + a = 0. 

    ⇒ x + 33/32+1/2 = 0 ⇒ x = -49/32. 

    The tangent at the vertex is x + 33/32 = 0 ⇒ x = – 33/32. 

    Length of the latus rectum = 4a = 2. 

Illustration: 

        The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the

equation of the parabola and the points where it meets the coordinate axes. 

Solution: 

       Focus of the parabola is the mid-point of the latus rectum. 

      ⇒ S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is 

      y – 4 = 0/(5-3) (x – 7) ⇒ y = 4. 

     Length of the latus rectum = (5 – 3) = 2. 

    Hence the vertex of the parabola is at a distance 2/4 = .5 from the focus. We have two parabolas, one concave rightwards and the other concave leftwards. The vertex of the first parabola is (6.5, 4) and its equation is 

    (y – 4)2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0). 

    The equation of the second parabola is (y – 4)2 = –2 (x – 7.5). 

15).     It meets the x-axis at (–0.5, 0) and the y-axis at (0, 4 +

Parametric Form of a Parabola 

    Suppose that the equation of a tangent to the parabola y2 = 4ax. … (i) 

    is y = mx + c. … (ii)       The abscissa of the points of intersection of (i) and (ii) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points hence discriminant should be zero 

   ⇒ (mx – 2a)2 = m2c2 … (iii) 

   ⇒ c = a/m. 

   Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m. 

  Equation (mx + c)2 = 4ax now becomes (mx – a/m)2 = 0. 

  ⇒ x = a/m2 and y2 = 4ax 

  ⇒ y = 2a/m. 

  Thus the point of contact of the tangent y = mx + a/m is (a/m2 ,2a/m). 

 Taking 1/m = t where t is a parameter, i.e., it varies from point to point. Theparabola y2 = 4ax as a parametric form is given by the co-ordinate (at2, 2at) and we refer to it as point ‘t’.

Illustration: 

      Prove that the area of the triangle inscribed in the parabola y2 = 4ax is a2 |(t1– t2) (t2 – t3) (t3 – t1)| where t1, t2 and t3 are the vertices. 

Solution: 

    The three points on the parabola are (at12, 2at1), (at2

2, 2at2) and (at32, 2at3).

       

THE GENERAL EQUATION OF A PARABOLA

We shall now obtain the equation of a parabola when the focus is any point and the dircectrix is any line. 

Let (h, k) be the focus S and lx + my + n = 0 the equation of the directrix ZM of a parabola. Let (x, y) be the coordinates of any point P on the parabola. Then the relation, PS = distance of P from ZM, gives 

        (x – h+) + (y – k)2 = (lx + my + n)2/(l2 + m2) 

        ⇒ (mx – ly)2 + 2gx + 2fy + d = 0. 

                                

This is the general equation of a parabola. It is clear that second-degree terms in the equation of a parabola form a perfect square. 

The converse is also true, i.e. if in an equation of the second degree, the second-degree terms from a perfect square then the equation represents a parabola, unless it represents two parallel straight lines. 

Note: 

     The general equation of second degree i.e. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a parabola if Δ ≠ 0 and h2 = ab. (Δ = abc + 2fgh – af2 – bg2 – ch2). 

Special case: 

Let the vertex be (α, β) and the axis to be parallel to the x-axis. Then the equation of parabola is given by (y – β)2 = 4a (x – α) which is equivalent to x = Ay2 + By + C.

If three points are given we can find A, B and C. 

Similarly, when the axis is parallel to the y-axis, the equation of the parabola is y = A’x2 + B’x + C’. 

Illustration: 

Find the equation of the parabola whose focus is (3, –4) and directrix x – t + 5 = 0. 

Solution: 

    Let P(x, y) be any point on the parabola. Then 

    √((x-3)2 (y+4) )=|x-y+5|/√(1+1) 

   ⇒ (x – 3)2 + (y + 4)2 = (x-y+5)2/2 

   ⇒ x2 + y2 + 2xy – 22x + 26y + 25 = 0. 

   ⇒ (x + y)2 = 22x – 26y – 25. 

Illustration: 

Find the equation of the parabola having focus (–6, 6) and verte (–2, 2). 

Solution: 

       Let S(–6, –6) be the focus and A(–2, 2) the vertex of the parabola. On SA take a point K (x1, y1) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directrix of

the parabola. 

      Since A bisects SK, ((-6+x1)/2,(-6+y1)/2) = (–2, 2) 

      ⇒ – 6 + x1 = – 4, and – 6 + y1 = 4 

     Or (x1, y1) = (2, 10). 

                                

     Hence the equation of the directrix KM is y – 10 = m (x + 2). … (1) 

    Also gradient of 

     SK = (10-(-6))/(2-(-6) )=16/8 = 2; ⇒ m = (-1)/2 

    So that equation (1) becomes y – 10 = –1/2 (x – 2) 

    or x + 2y – 22 = 0 is the directrix. 

    Next, let PM be a perpendicular on the directrix KM from any point 

    P(x, y) on the parabola. 

    Form SP = PM, the equation of the parabola is 

    √({(x+6)+(y+6)2 } )=(x+2y-22)/((12+22 ) ) 

    or 5 = (x2 + y2 + 12x + 12y + 72) = (x + 2y – 22)2 

    or 4x2 + y2 – 4xy + 104x + 148 y – 124 = 0. 

    or (2x – y)2 + 104x + 148y – 124 = 0. 

 Illustration: 

If the point (2, 3) is the locus and x = 2y + 6 is the directrix of a parabola, find 

(i) the equation of the axis, 

(ii) the co-ordinates of the vertex, 

(iii) length of the latus rectum, 

(iv) equation of the latus rectum. 

Solution: 

(i) We know that the axis of a parabola is the line through the focus 

     And perpendicular to the directrix. 

     The equation of any line passing through the focus (2, 3) is 

     y – 3 = m (x – 2) ⇒ mx – y = 3 – 2m 

     If the line be perpendicular to the directrix x – 2y = 6, we have, 

     m (1/2) = – 1 ⇒ m = – 2. 

    Hence the equation of the axis is y – 3 = – 2 (x – 2) ⇒ 2x + y = 7. 

(ii) The co-ordinates of the point of intersection (say) A of the directrix x – 2y = 6 and the axis 2x + y = 7 are obtained by solving the two equations; thus they are (4, –1). Since the vertex is the middle point of A (4, –1) and the focus S(2, 3); the co-ordinates of the vertex are ((4+2)/2,(3-1)/2), i.e. (3, 1). 

(iii) Since OS = √((3-2)2+(1-3)2 )=√5, 

The length of the latus rectum = 4OS = 4√5. 

(iv) Since the latus rectum is the line through the focus parallel to the directrix, its equation is x – 2y + c = 0, where c is given by 2 – 2.3 + c = 0, i.e. c = 4.

Focal Chord 

Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax. 

Let y2 = 4ax be the equation of a parabola and (at2, 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at12, 2at1).

                                    

Then, PS and SQ, where S is the focus (a, 0), have the same slopes 

⇒ (2at-0)/(at2-a)=(2at1-0)/(at12-a) 

⇒ tt12 – t = t1 t2  (tt– t1 1 + 1) (t1 – t) = 0. 

Hence t1 = –1/t, i.e. the point Q is (a/t2, –2a/t). 

The extremities of a focal chord of the parabola y2 = 4ax may be taken as the points t and –1/t. 

Illustration: 

          Prove that the circle with any focal chord of the parabola y2 = 4ax as its diameter touches its directrix. 

Solution: 

         Let AB be a focal chord. If A is (at2, 2at), then B is (a/t2 ,-2a/t). 

Equation of the circle with AB as diameter is 

(x – at2) (x-a/t2 ) + (y – 2at) (y+2a/t) = 0. 

For x = –a, this gives (a2 (1+t2 )2)/t2 + y2 – 2ay (t-1/t) – 4a2 = 0. 

⇒ a2 (t-1/t)2 + y2 – 2ay(t – 1/t) = 0 

⇒ [y – a(t – 1/t)]2 = 0, which has equal roots. 

Hence x + a = 0 is a tangent to the circle with diameter AB. 

Illustration: 

         Find the locus of the centre of the circle described on any focal chord of aparabola as diameter. 

Solution: 

        Let the equation of the parabola be y2 = 4ax. 

Let t1, t2 be the extremities of the focal chord. Then t1 . t2 = – 1. 

The equation of the circle on t1, t2 as diameter is 

(x – at22) (x – at2

2) + (y – 2at1) (y – 2at2) = 0 

or x2 + y2 – ax (t12 + t2

2) – 2ay (t1 + t2) + a2 t12 t12 + 4a2 t1t2 = 0 

⇒ x2 + y2 – ax (t12 + t2

2) – 2ay (t1 + t2) – 3a2 = 0. (∵ t1t2 = –1) 

If (α,β) be the centre of the circle, then α = a/2 (t12+t2

2 ) If (α, β) be the centre of the circle, then α = a/2 (t1

2+t22 ) 

β = a (t1 + t2) ⇒ (t1 + t2)2 =β2/a2 ⇒ t12 + t2

2 + 2t1t2 =β2/a2 ⇒ 2α/a-2= β2/a2 

⇒ 2aα – 2a2 = β2 ⇒ β2 = 2a (α – a). 

Hence locus of (α, β) is y2 = 2a(x – a). 

Focal Distance of a Point 

The focal distance of a point P on the parabola

                                 

y2 = 4ax is the distance between the point P and the focus S, i.e. PS. Thus thefocal distance of P = PS = PM = ZN = ZA + AN = a + x. 

or 

PS = a + at2 = a(1 + t2). 

Position of a point relative to a Parabola 

Consider the parabola y2 = 4ax. 

If (x1, y1) is a given point and y12 – 4ax1 = 0, then the point lies on the parabola. But when

y12 – 4ax1 ≠ 0, we draw the ordinate PM meeting the curve in L. Then P will lie outside

the parabola if PM > LM, i.e., PM2 – LM2 > 0.

                                   

Now, PM2 = y12 and LM2 = 4ax1 by virtue of the coordinates of L satisfying the equation of

the parabola. Hence, the condition for P to lie outside the parabolabecomes y12 – 4ax1 >

0. 

Similarly, the condition for P to lie inside the parabola is y12 – 4ax1 < 0.

Chord 

Intersection of a Straight Line with a Parabola 

The combined equation of straight line y = mx + c and parabola 

y2 = 4ax gives us the co-ordinates of point(s) of their intersection. The combined equation m2x2 + 2x (mc – 2a) + c2 = 0 will give those roots. The straight line therefore meets the parabola at two points. 

Points of Intersection of a straight line with the parabola y2 = 4ax 

Points of intersection of y2 = 4ax and y = mx + c are given by (mx+c)2=4ax 

i.e. m2x2 + 2x(mc – 2a) + c2 = 0. …… (i) 

Since (i) is a quadratic equation, the straight line meets the parabola in two points, real, coincident, or imaginary. The roots of (i) are real or imaginary according as {2(mx – 2a)}2 – 4m2c2 is positive or negative, i.e. according as – amc + a2 is positive or negative, i.e. according as mc is less than or greater than a, (taking a as positive). 

Note: 

When m is very small, one of the roots of equation (i) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the vertex. It means that a line parallel to the axis of theparabola meets the parabola only in one point. 

Length of the chord 

As in the preceding article, the abscissae of the points common to the straight line y = mx + c and the parabola y2 = 4ax are given by the equation m2x2 + (2mx – 4a) x + c2 = 0. 

            

Hence, the required length of chord

              

Illustration: 

          Find the Length of the chord intercepted by the parabola y2 = 4ax from the line y = mx + c. Also find its mid-point. Solution: 

         Simply by applying the formula o length of the joining (x1, y1) and (x2, y2) we get, 

Length of the chord = √((x1-x2 )2+(y1-y2 )2 ) 

= √((x1-x2 )2+m2 (x1-x2 )2 ) 

= |x1 – x2| √(1+m2 ) = 4 √(a(a-mc) ) √(1+m2 ) 

[ ∵ x1+x2=(-2(m-2a) )/m2 and x1 x2=c2/m2 ] 

The midpoint of the chord is ((2a-mc)/m2 ,2a/m)

Tangent to a Parabola 

        Let P(x1, y1) and Q(x2, y2) be two neighbouring points on the parabola y2 – 4ax. Then the equation of the line joining P and Q is 

       y – y1 = (y2-y1) / (x2-x1 ) (x – x1) …… (1) 

      Since, points P and Q lies on the parabola, we have 

     y12 = 4ax1 …… (2) 

    y22 = 4ax2 …… (3) 

     (equation 3 and 2) give 

     y22 – y1

2 = 4a(x2 – x1) 

     ⇒ (y2-y1)/(x2-x1 )=4a/(y1+y2 ) 

     Equation of chord PQ (i.e. equation (1) becomes): 

     y – y1 = 4a/(y1+y2 ) (x – x1) …… (4) 

Our aim is to in the equation of tangent at point P. For that, let point Q approach point P i.e. x2 → x1 and y2 → y1. 

    y – y1 = 4a/(2y1 ) (x – x1) 

   ⇒ y1y = 2a (x1 + x) (using equation (2)) 

 This is the required equation of the tangent to the parabola y2 – 4ax at P(x1, y1).

Note: 

       The angle between the tangents drawn to the two parabolas at the point of their intersection is defined as the angle of intersection of two parabolas. 

Tangent at the point (x1, y1) 

     Let the equation of the parabola be y2 = 4ax. 

     Hence, value of dy/dx at P(x1, y1) is 2a/y1 and the equation of the tangent at P is 

    y – y1 = 2a/y1 (x – x1) i.e. yy1 = 2a(x – x1) + y12.

⇒ yy1 = 2a(x + x1). 

   Alternatively, we write the equation of the chord joining the points P(x1, y1) and Q(x2, y2) on the parabola y2 = 4ax. Equation of the chord is (x-x1)/(x2-x1 )=(y-y1)/(y2-y1 ) 

    or (x-x1)/(x2-x1 ) = (y-y1 )(y2+y1 )/(y22-y1

2 )= (y-y1 )(y2+y1 )/((x2-x1 ) )      or 4a(x – x) = (y – y1) (y2 + y1). 

    When the two points P and Q tend to coincide, y2 → y1 and the line PQ becomestangent to the parabola. Its equation is 4a (x – x1) = (y – y1) (2y1) = 2yy1 – 2y1

2= 2yy1 – 8ax1 or yy1 = 2a(x + x1). 

Tangent in terms of m 

    Suppose that the equation of a tangent to the parabola y2 = 4ax … (i) 

    is y = mx + c. … (ii) 

     The abscissae of the points of intersection of (i) and (ii) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points 

   ⇒ (mx – 2a)2 = m2c2 … (iii) 

   ⇒ c = a/m. … (iv) 

Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m. 

   Equation (mx + c)2 = 4ax now becomes (mx – a/m)2 = 0. 

   ⇒ x = a/m2 and y2 = 4ax ⇒ y = 2a/m. 

   Thus the point of contact of the tangent y = mx + a/m is (a/m2 ,2a/m).

Illustration: 

         Find the condition that the line y = mx + c may touch the Parabola y2 = 4ax and also find its point of contact. 

Solution: 

        Equation of Parabola is yy2 = 4ax …… (1) 

        Slope of tangent at any point is 

        dy/dx=2a/y = m (say) …… (2) 

        ⇒ y = 2a/m 

        from (1), x = a/my2 

        ⇒ point of contact is (a/my2 ,2a/m) 

        Equation of tangent is 

        y – 2a/m = m (x-a/my2 ) 

        or y = mx + a/m 

Therefore, the condition that y = mx + c touches the Parabola 

       yy2 = 4ax is c = a/m. 

Illustration: 

     Find the equation of normal to the Parabola yy2 = 4ax, having slope m. 

Solution: 

       Slope of tangent at any point is 

        dy/dx=2a/y 

        Slope of normal at that point is 

        -y/2a = m (say) 

       ⇒ Point of contact of a normal having slope ‘m’ with the Parabola 

       yy2 = 4ax is (amy2, – 2am) 

       So, equation of normal at this point is 

        y + 2am = m (x – amy2) 

       or y = mx – 2am – amy3. 

Illustration: 

       If the line 2x + 3y = 1 touches the Parabola yy2 = 4ax, find the length of its latus rectum. 

Solution: 

        Equation of any tangent to yy2 = 4ax is 

        y = mx + a/m ⇒ my2x – my + a = 0. 

        Comparing it with the given tangent 2x + 3y – 1 = 0, we find 

        my2/2=(-m)/3=a/(-1), ⇒ m = (-2)/3, a = m/3 = -2/9. 

        Hence the length of the latus rectum. 

        = 4a = 8/9, ignoring the negative sign for length. 

Alternative Solution: 

        Writing the given equation as 

        y = -2/3 x + 1/3=-2/x x-(2/9)/(-2/3), we find that a = 2/9. 

        Hence the length of the latus rectum = 4a = 8/9. 

         tangent at the point t 

         Equation of the tangent at ‘t’ is ty = x + aty2. T

         the point of intersection of the tangents at ‘t1’ and ‘t2’ is (at1t2, a(t1 + t2)). 

Illustration: 

         One the Parabola yy2 = 4ax, three points E, F, G are taken so that their ordinates are in G.P. Prove that the tangents at E and G intersect on the ordinate of F. 

Solution:                Let the points E, F, G be (at1y2, 2at1), (at2y2, 2at2), (at3y2, 2at3) respectively. Since the ordinates of these points are in G.P., t2

2 = t1t3. tangents at E and G are t1y = x + at12 and t3y = x + at3

2. Eliminating y from these equation, we get x = at1t3 = at22. Hence the point lies

on the ordinates of F.

Illustration: 

       Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.

                       

The intersection of the tangents, at these points, are the points 

(at1, t2, a(t1 + t2)}, {at2t3, a(t2 + t3)}, {at3 t1, a(t3 + 1)}. 

The area of the Δ formed by these points=1/2 a2(t1 – t2) (t2 – t3) (t3 – t1). 

Equation of the Tangents from an External Point 

         Let y = mx + a/m be any tangent to y2 = 4ax passing through the point (x1, y1). 

       Then, we have 

       y1 = mx + a/m or m2x1 – m1 + a = 0 

       If m1 and m2 are to roots of (i) then 

       m1 + m2 = y1/x1 and m1m2 = a/x1 

      Also the two tangents are y = m1x + a/m1 , and y = m2x + a/m2 

     ⇒ Their combined equation is 

      (y – m1x – a/m1 ) (y – m2x – a/m2 ) = 0 

On solving this we get 

    (y2 – 4ax) (y12 – 4ax1) = [yy1 – 2a (x + x1)]2 

    ⇒ SS1 = t2. 

     Where S = y2 – 4ax, S1 = y12 – 4x1 

Let y2 = 4ax be the equation of a parabola and (x1, y1) an external point P. Then, equation of the tangents is given by 

SS1 = t2, where S = y2 – 4ax, S1 = y12 – 4ax1, T = yy1 – 2a(x + x1). 

If the tangents from the external point (x1, y1) touch the parabola at P and Q, then PQ is the chord of contact of the tangents. 

Illustration: 

      Prove that through any given point (x1, y1) there pass, in general, twotangents to the parabola y2 = 4ax. 

Solution: 

        The equation to any tangent is y = mx + a/m. …… (1) 

       If this passes through the fixed point (x1, y1), we have 

       y1 = mx1 + a/m, i.e. m2x1 – m y1 + a = 0. …… (2) 

This is a quadratic equation and gives two values of m (real or imaginary). Corresponding to each value of m we have, two different tangents. The roots of (2) are real and different if y1

2 – 4ax1 > 0, i.e. If the point (x1, y1) lies outside the curve. The roots are equal, if y12 –

4ax1 = 0 i.e. if the point (x1, y1) lies on the curve. In this case the two tangent merge into one. The two roots are imaginary if y1

2 – 4ax1 < 0, i.e. if the point (x1, y1) lies within the curve.

Chord of Contact 

      The chord joining the points of contact of the tangents on the parabola from an external point is called the chord of contact. 

     Let the tangent drawn from the point P(x1, y1) touch Parabola at Q(x2, y2) and R(x3, y3) then QR is the chord of contact of the point P(x1, y1) with respect to y2= 4ax. 

            The equation of tangents at Q and R are 

           yy2 = 2a(x + x2) …… (1) 

           yy3 = 2a(x + x3) …… (2) 

                     

     Since (ii) and (iii) pass through (x1, y1) so we have 

         y1y2 = 2a(x1 + x2) …… (3) 

        y1y3 = 2a(x1 + x3) …… (4) 

From (ii) and (iv) we find that the points Q(x2, y2) and R(x3, y3) lie on yy1 = 2a (x2+ x1), which being of first degree in x and y represents a straight line. Hence the equation of the chord of contact of P(x1, y1) is 

       yy1 = 2a (x + x1) and is of the form T = 0. 

Equation of the chord of contact of the tangents drawn from a point (x1, y1) to theparabola y2 = 4ax is T = 0, i.e. yy1 – 2a(x + x1) = 0. 

Note: 

      The equation of the chord of the parabola y2 = 4ax with mid point 

        (x1, y1) is T = S1. 

Illustration: 

           Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5, –7). 

Solution: 

       Here (x1, y1) = (5, –7), and y2 = 12x = 4ax ⇒ a = 3. 

      The equation of the chord is S1 = T 

      or y12 – 4ax1 = yy1 – 2a(x + x1) or (–7)2 – 12.5 = y(–7) – 6 (x + 5). 

      Or 6x + 7y + 19 = 0. 

Normal to a Parabola

Normal at the point (x1, y1) 

The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1). Since the slope of the tangent = 2a/y1, slope of the normal is –y1/2a. Also it passes through (x1, y1). 

Hence, its equation is y – y1 = -y1/2a (x – x1). … (i) 

Normal in terms of m 

In equation (i), put -y1/2a = m so that y1 = –2a and x1 = (y12)/4a = am2, then the equation

becomes y = mx – 2am – am3 

where m is a parameter. Equation (ii) is the normal at the point (am2, –2am) of theparabola. 

Note: 

If this normal passes through a point (h, k), then k = mh – 2am – am3. 

For a given parabola and a given point (h, k), this cubic in m has three roots say m1, m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes are m1, m2, m3. For this cubic, we have m1+ m2+ m3 = 0, m1 m2 + m2 m3 + m3 m1 = (2a – h)/a, m1 m2 m3 = –k/a. 

If we have an extra condition about the normals drawn from a point (h, k) to a given parabola y2 = 4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we can get the locus of (h, k). 

Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the normals from a given point is zero. These points are called Co-Normal Points.

Illustration: 

          Find the locus of the point of intersection of two normals to a parabolawhich are at right angles to one another. 

Solution: 

         The equation of the normal to the parabola y2 = 4ax is 

         y = mx – 2am – am3. 

        It passes through the point (h, k) if 

        k = mh – 2am – am3 => am3 + m(2a – h) + k = 0. … (1) 

 Let the roots of the above equation be m1, m2and m3. Let the perpendicularnormals correspond to the values of m1 and m2 so that m1 m2 = –1. 

        From equation (1), m1 m2 m3 = -k/a. Since m1 m2 = –1, m3 = k/a. 

        Since m3 is a root of (1), we have a (k/a)3+k/a (2a – h) + k = 0. ⇒ k2 + a(2a – h) + a2 = 0 

        ⇒ k2 = a(h – 3a). 

        Hence the locus of (h, k) is y2 = a(x – 3a).

Normal at the point t 

The normal, being perpendicular to the tangent at (at, 2at) is given by y = –tx + 2at + at3. 

Note: 

            If normal at the point at1 meets the parabola again at the point at2, then at2 = – at1 – 2/at1 . 

            Point of intersection of the normals to the parabola y2 = 4ax (at12, 2at1) and (at2t2,

2at2) is (2a+a(t12+t2

2+t1t2), a t1t2(t1+t2)). 

Illustration: 

          Prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus. 

Solution: 

         If the normal to the parabola y2 = 4ax at P(at1t2, 2at1) meets it again at the point t2, then we have t2 = – t1 – 2/t1 .

If the abscissa and the ordinates of P be equal, then at12 = 2at1 

⇒ t1 = 2 (rejecting t1  t= 0) 2 = – 2 – 1 = – 3 

The co-ordinates of P and Q are therefore (4a, 4a) and (9a, – 6a) respectively. 

The focus is the point S (a, 0). 

Slope of PS = ¾ and slope of QS = – ¾. 

⇒ ∠PSQ = right angle. Hence the result.

Illustration: 

               Find the locus of the middle points of the normal chords of the parabolay2 = 4ax. 

Solution: 

               Equation of the normal chord at any point (at2, 2at) of the parabola is 

               y + tx = 2at + at3. … (1) 

               Equation of the chord with mid point (x1, y1) is T = S1 

               or yy1 – 2a(x + x1) = y12 – 4ax1 or yy1 – 2ax = y1

2 – 2ax1. … (2) 

                   Since equations (1) and (2) are identical, 1/y1 =t/(-2a)=(2at+at3)/t = 2a + ((-2a)/y1 )2 

               or -(y12)/2a + x1 = 2a + 4a3/(y1

2 ) or x1 – 2a = (y12)/2a+4a3/(y1

2 ) 

               Hence the locus of the middle point (x1, y1) is 

               x – 2a = y2/2a+4a3/y2 . 

Illustration: 

              P and Q are the points t1 and t2 on the parabola y2 = 4ax. If thenormals to the parabola at P and Q meet at R, (a point on the parabola), show that t1t2 = 2. 

Solution: 

               Let the normals at P and Q meet at R(at2, 2at). 

               Then t = – t1 – 2/t1 and t = – t2 – 2/t2 . 

               Therefore t1 + 2/t1 = t2 + 2/t2 ⇒ (t1 – t2) = 2(t1-t2 )/(t1 t2 ) ⇒ t1t2 = 2. 

Illustration: 

              Find the equations of the normals to the parabola y2 = 4ax at the extremities of its latus rectum. If the normals meet the parabola, again at P and Q, prove that PQ = 12a. 

Solution: 

              The ends of the latus rectum are (a, 2a) and (a, –2a). The equations of the normals

to the parabola at these points are (put t = 1 and –1) 

               y + x = 3a and y – x = 3a. 

         These      lines meet the parabola again at P(9a, 6a) and Q(9a, 6a) respectively. 

              ⇒ PQ = 6a + 6a = 12a. 

Propositions on the Parabola 

                   

(i)  The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix. 

             The tangent at P (at2, 2at) is ty = x + at2. 

            It meets the x-axis at T(–at2, 0). 

            Hence ST = a (1 + t2). 

            Also, SP = √(a2 (1+t2 )2+4a2 t2 ) = a(1 + t2) = ST, so that 

            ∠MPT = ∠PTS = ∠SPT ⇒ TP bisects ∠SPM. 

(ii) The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus. 

            Let P(at2, 2a), be a point on the parabola y2 = 4ax. 

            The tangent at P is ty = x + at2. 

      Point of intersection of the tangent with the directrix x + a = 0 is (–a, at – a/t).

            Now, slope of SP is (2at-0)/(at2-a)=2t/(t2-1) 

             and slope of SK is (at-a/t-0)/(-a-a)=-(t2-1)/2t 

            ⇒ (Slope of the SP).(Slope of SK) = –1. 

             Hence SP is perpendicular to SK i.e. ∠KSP = 90°. 

(iii) Tangents at the extremities of any focal chord intersect at right angles on the directrix. 

            Let P(at2, 2at) and P(at12, 2at1) be the end points of a focal part on the parabola. Then t.t1 = –1. Equations of the tangents at the point P and the point P’ are ty = x + at2 and t1y = x + at12 respectively. 

        Let these tangents intersects at a point (h, k). Then h = att1 and k = a(t + t1). 

             Since the tangents are perpendicular, tt1 = – 1 ⇒ h – a. 

            Hence the locus of the point (h, k) is x = –a which is the equation of the directrix. 

(iv) Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the vertex. 

             Equation of the perpendicular to the tangent ty = x + at2 … (1) 

            From the focus (a, 0) is tx + y = at. … (2) 

           and (2) intersect at x = 0 which is the equation of the tangent at the vertex.

Pole and polar of a conic 

        The locus of the point of intersection of tangents drawn at the extremities of the chord of the conic drawn through a point is called the polar of that point with respect to the conic. This point itself is called the pole. 

       Equation of the polar of a point (x1, y1) with respect to the parabola y2 = 4ax.

       Let us draw the chord QR from the point P(x1, y1) and if the tangents drawn from point Q and R meet at the point T(h, k), required locus of (h, k) is polar. Since QR is the chord of contact of tangents from (h, k), it’s equation is 

                    ky = 2a(x + h) 

      This straight line passes through the point (x1, y1), we have 

                 ky1 = 2a(x1 + h) …… (1) 

   Since the relation (1) is true, it follows that point (h, k) always lies on the line. 

                yy1 = 2a(x + x1) …… (2) 

  Hence (2) is the equation to the polar of pole (x1, y1) 

Co-normal Points: 

       The three points on the parabola, the normals at which pass through a common point, are called the co-normal points. 

Diameter: 

      The locus of the middle point of a system of parallel chords of a parabola is called its diameter. 

  

       Let the parabola be 

             y2 = 4ax. …… (i) 

             let y = mx + c …… (ii) 

be a system of parallel chords to (i) for different chords, only c varies, m remains constant. 

          y2 = 44a (y – c)/m 

         my2 – 4ay + 4ac = 0 

let y2 and y3 be the roots of (iii), then y2 and y3 are the ordinates of the points where (ii) cuts (i) 

          from (iii), y2 + y3 = 4a/m 

Also, if (x1, y1) be the midpoint of the chord them 

         y1 = (y2-y3)/2=2a/m 

     ∴ Locus of (x1, y1) is y = 2a/m, which is the equation of the diameter. 

Note: 

     y = 2a/m is a straight line parallel to the axis of the parabola.

Solved Examples 

Example 1: 

        Find the equation of the parabola whose focus is (3, –4) and directrix is the line parallel to 6x – 7y + 9 = 0 and directrix passes through point (3/2,2). 

Solution: 

       Let (x, y) be any point on the parabola. Then by definition, the distance between (x, y) and the focus (3, – 4) must be equal to the length of perpendicular from (x, y) on directrix. So first we will find the equation of the directrix. 

     The line parallel to 6x – 7y + 9 = 0 is 

     6x – 7y + 6 = 0 …… (1) 

     Since directrix passes through (3/2,2), this point wil satisfy equation 91) hence 

     6 (3/2) – 7 (2) + k = 0 

    ⇒ K = – 9 + 14 = 5. 

    Equation of directrix is 6x – 7y + 5 = 0 

    Now by definition of parabola, 

   √({(x-3)2+(y+4)2 } )= (6x-7y+5)/√(62+72 ) 

   ⇒ 85 {(x – 3)2 + (y + 4)2} = (6x – 7y + 5)2 

   ⇒ 49x2 + 36y2 + 84xy – 570x + 750y + 2100 = 0 

Example 2: 

       Find the equation of the parabola whose directrix makes an isosceles right angled triangle of area 4 square units with the axis in the 3rd quadrant and focus is on the line y = x, 2 units away from the origin. 

Solution: 

      First we find the equation of directrix. Let the directrix form the isosceles triangle OAB with OA = OB = a.

                                  

    Then according to the given condition, 

     (Δ OAB) = 4 

     ⇒ 1/2 a2 = 4 

     ⇒ a = ± 2√2                       {∵the triangle in 3rd quadrant∴=-2√2} 

Therefore the co-ordinate, of A and B are (–2√2, 0) and (0, –2√2) respectively. 

    So, equation of directrix 

    (y – 0) = ((0+2√2)/(-2√2-0)) (x + 2√2) ⇒ x + y + 2√2 = 0 

    Now the focus S is on line y = x and 2 units away from the origin i.e. 

    OS = 2 ⇒ point (√2, √2) by definition of parabola, we have 

   √((x-√2)2+(y-√2)2 )=|(x+y+2√2)/√(12+12 )| 

   ⇒ x2 + y2 – 8√2x – 8√2y – 2xy = 0

Example 3: 

          The x and y co-ordinate of any point P are expressed as x = (v cos α) t, y = (v sin α) t – 1/2 gt2, where t is parameter and v, α, g are constants, show that the locus of the point P(x, y) is a parabola. Find the coordinate of the vertex of thisparabola. 

Solution: 

       We are given x = (v cos α) t …… (1) 

       y = (v sin α) t – 1/2 gt2 …… (2) 

       We will eliminate t from above two relations, 

       From (1) : = x/(v cos α ) 

       Put this value in (2) 

  From equation (3) it is clear that this is a equation of parabola whose vertex is the point with the co-ordinates 

((v2 sin α cos α )/g,(v2 sin2 α )/2g) 

Example 5: 

        Find the equation of the common tangents to the parabola y2 = 32x and x2 = 108y. 

Solution: 

       The equation of the tangent to the parabola y2 = 4ax, is 

       y = mx + a/m …… (1) 

      The equation of tangent to the parabola 

      y2 = 32x …… (2) is 

      y = mx + 8/m …… (3) 

      If this line given by (3) is also a tangent to the parabola 

      x2 = 108y, then (3) meets x2 = 108y …… (4) 

      in two coincident points 

     ⇒ Substituting the value of y from (3) in (4) we get 

      x2 = 108 [mx+(8/m) ] 

     ⇒ mx2 – 108m2x – 864 = 0 

The roots of this quadratic are equal provided b2 = 4ac. 

    i.e. (– 108 m2)2 = 4m (–864) 

   ⇒ m = (-2)/3 (m ≠ 0, from geometry of curves) 

Substituting their value of m in (3), the required equation is 

     y = (-2)/3 x + 8/(–2/3) 

     y = (-2)/3 x – 12 

     ⇒ 2x + 3y + 36 = 0 

Example 6: 

        Find the equation of the tangent to the parabola y2 = 6x, which is parallel to the line cutting intercepts 3 and 4 on x and y axis respectively. 

Solution: 

     The line cutting intercepts 3 and 4 on x and y-axis respectively is, 

     x/3+y/4 = 1 

    ⇒ 4x + 3y = 12 …… (1) 

    Slope of line (1) is –4/3       The equation of the parabola is y2 = 4 (3/2) x …… (2) 

    The equation of any tangent to (2) is 

    y = mx + (3/2m) …… (3) 

    for all value of m … 

   If this tangent is parallel to line (1), then 

    m = (-4)/3 

   Hence from (3) required equation of the tangent is 

    y = (-4)/3 x + (3/2×(-3)/4) 

   ⇒ y = (-4)/3 x-9/8 

   ⇒ 32x + 24y + 27 = 0

Example 7: 

         Find the angle of intersection of the parabola y2 = 8x and x2 = 27y. 

Solution: 

      The given parabolas are 

      y2 = 8x …… (1) 

     and x2 = 27y …… (2) 

   Solving (1) and (2) we get 

      (x2/27)2 = 8x 

     ⇒ x4 = 5832x 

     ⇒ x4 – 5832x = 0        ⇒ x(x3 – 5832) = 0 

     ⇒ x = 0, x = 18 

Substituting these values of x in (2) we get y = 0, 12 

The point of intersection are (0, 0), (18, 12) 

At the point (0, 0) 

     y2 = 8x ⇒ dy/dx=4/y ⇒ dy/dx|(0,0) = ∞          x2 = 27y ⇒ dy/dx=2x/27 ⇒ dy/dx|(0,0) = 0 

    ⇒ the two curves intersect at the point (0, 0) at right angle. 

At the point (18, 12) 

     y2 = 8x ⇒ dy/dx|(18,12) =4/12=1/3 = m1 (say) 

    x2 = 27y ⇒ dy/dx=2x/27 ⇒ dy/dx|(18,12)= (2×18)/27=4/3 = m2 (say) 

   Let Θ be the angle at which the two curves intersect at the point (18, 12) 

   Then tan Θacute = |(m2-m1)/(1+m1 m2 )|=3/13 

    Θacute = tan-1 (3/13) 

Example 8: 

       Prove that (x + a)2 = (y2 – 4ax), is the locus of the point of intersection of the tangents to the parabola y2 /4.= 4ax, which includes an angle  

Solution: 

     Let two tangent to the parabola y2 = 4ax …… (1) 

     be yt1 = x + at12 …… (2) 

    and yt2 = x + at22 …… (3) 

Let the point of intersection of the tangent be (x1, y1) then solving equation (1) and (2) we get, 

     x1 = at1 t2 

    y1 = a(t1 + t2) 

Also the slope of these tangents are 1/t1 and 1/t2 

   ∴ If α be the angle between these two tangents then 

     tan α = + ((m1-m2)/(1+m1 m2 ))=(±(1/t1 -1/t2 ))/(1+1/t1 ×1/t2 ). 

   ⇒ tan α = ± ((t2-t1)/(1+t2 t1 )) 

we are given α = π/4 

   ∴ tan π/4 = 1 = ± ((t2-t1)/(1+t2 t1 )) 

   ⇒ (1 + t1t2)2 = (t2 – t1) 

   ⇒ {1+(x1/a)2 } = (t1 – t2)2 – 4t1t2 = (y1/a)2-4x1/a 

   ⇒ (x1 + a)2 = y12 – 4ax1 

   ⇒ Required locus of (x1, y1) is (x + a)2 = (y2 – 4ax) 

Example 9: 

        Prove that normal at one end of latus rectum of a parabola is parallel to the tangent at the other end. 

Solution: 

      Let the parabola be y2 = 4ax …… (1) 

     The end points of latus rectum are (a, 2a) & (a, –2a) and 

     The equation of the normal to (1) at (a, 2a) is 

     (y – 2a) = – (x – a) 

    ⇒ y = – x + 3a …… (2) 

   The equation of the tangent to (1) at (a, –2a) 

    y + 2a = – (x – a) 

    ⇒ y = – x – a 

and from (2) and (3), we find that the slope of normal to one end of the latus rectum is equal to the slope of tangent at other end of tangent to the other end. Hence the required fact is proved.

Example 10: 

        Find the length of the focal chord of parabola y2 = 4ax whose one end point is P(at2, 2at) 

Solution: 

     Let Q (at12, 2at) be the other end of this focal chord. 

    The equation of the line joining (at2, 2at) and (at12, 2at1) is 

     (y – 2at) = (2at1-2at)/(at12-at2 ) (x – at2) 

     If this passes through the focus (a, 0), then 

    –2at = 2/(t1+t) (a – at2) 

     ⇒ – t (t + t1) = 1 – t2 

     ⇒ tt1 = –1 

     ⇒ t1 = – 1/t 

    Then Q is the point (a/t2 ,(-2a)/t) 

    The length of the focal chord PQ 

    = a |t+1/t| √((t-1/t)2+4) 

    a (t-1/t)2 

Example 11: 

         Prove that the locus of the point of intersection of two mutually perpendicular tangents one to each of the parabola y2 = 4a (x + a) and y2 = 4b (x + b), is a line parallel to y-axis. 

Solution: 

      The given parabolas are 

      y2 = 4a(x + a) …… (1) 

      and y2 = 4b(x + b) …… (2) 

Any tangent to (1) is y = m (x + a) + a/m …… (3) 

Similarly any tangent to (2) is y = m’ (x + b) + b/m’ …… (4) 

       mm’ = – 1 

      or m’ = -1/m 

∴ (4) becomes y = –{(x + b)/m} – bm …… (5) 

The required locus is obtained by eliminating m between (3) and (5). 

For this subtracting (5) from (3), we get 

     0 = x (m+1/m)+a(m+1/m)+b(m+1/m) 

     ⇒ x + a + b = 0 

   This is the required locus which is parallel to y axis. 

Example 12: 

      If two tangents to a parabola intercept a constant length on any fixed tangent, find the locus of their point of intersection. 

Solution: 

    Let yt = x + at2 …… (1) 

    be a fixed tangent to the parabola y2 = 4ax …… (2) 

    Let the other two tangent to (2) be 

    yt1 = x + at12 …… (3) 

   And yt2 = x + at22 …… (4) 

      The point of intersection of (1) with (3) and (4) are P {att1, a(t1 + t2)} and Q{at t2, a (t + t2)} 

     Given PQ is constant or PQ2 = constant 

     ⇒ a2t2 (t1 – t2)2 + a2(t1 – t2)2 = constant 

     ⇒ a2(t2 + 1) (t1 – t2)2 = constant 

     ⇒ (t1 – t2)2 = constant since t is constant as (1) is a fixed tangent. 

     ⇒ (t1 – t2)2 = c (say) …… (5) 

    Let (x1, y1) be the point of intersection of (3) and (4), 

    then x1 = at1t2 and y1 = a(t1 + t2) …… (6) 

    We know that 

    (t2 – t1)2 = (t1 + t2)2 

    From equation (5), (6) and 97) we get 

     c = (y1/a)2- 4x1/a 

    or y12 = 4x1 a – a2c = 4a (x1+1/4 ac) 

    ⇒ the locus of (x1, y1) is 

   y2 = 4a (x+1/4 ac) 

  ⇒ The required locus is a parabola whose latus rectum is 4a i.e. equal to latus rectum of y2 = 4ax.

Example 13: 

        Find the locus of the poles of normals to parabola y2 = 4ax. 

Solution: 

        Any normal to the parabola y2 = 4ax is …… (1) 

        y = mx – 2am – am3 …… (2) 

       Let (x1, y1) be the pole of (2) with respect to (1), then (2) is the polar of (x1, y1) w.r.t (1) 

      i.e. 

      yy1 = 2a (x + x1) 

      comparing (2) & (3), we get 

      2a/m=y1/1=(2ax1)/(-2am-am3 ) 

      Hence we get 

      x1 = –2a – am2 …… (4) 

      and y1 = 2a/m …… (5) 

      Eliminating m between (4) & (5) we get 

      y12 (x1 + 2a) + 4a3 = 0 

      ∴ The required locus of (x1, y1) is 

      (x + 2a) y2 + 4a3 = 0 

Example 14: 

         Find the locus of the point P if the perpendicular from that point P upon its polar with respect to parabola, is of constant length. 

Solution: 

      Polar of P (x1, y1) with respect to the parabola y2 = 4ax is 

      y1y = 2a (x1 + x) 

      or y1y = 2ax – 2ax1 = 0 …… (1) 

      We are given that the distance of P (x1, y1) from line (1) is constant, say 

      ⇒ |y12-2ax1-2ax1 |/√(y1

2+(2a)2 ) = λ (constant) 

      ⇒ Locus of (x1, y1) is 

      (y2 – 4ax)2 = λ2 (y2 + 4a2) 

Example 15: 

             Find the locus of the point, from which the three normals to the parabola y2 = 4ax cut the axis at points whose distance from the vertex are in A.P. 

Solution: 

          Any normal to the parabola y2 = 4ax is y = mx – 2am – am3 … (1) 

          If (1) passes through (x1, y1) then 

          y1 = mx1 – 2am – am3 

          ⇒ am3 + m (2a – x1) + y1 = 0 … (2) 

           m1 + m2 + m3 = 0 … (3) 

          m1m2 + m2m3 + m3m1 = ((2a-x1 ))/a … (4) 

          m1 m2 m3 = (-y1)/a … (5) 

        Also, the normal with slope m1, i.e. y = m1x – 2am1 – am13 cuts axis of theparabola at

the point A(2a + am12, 0) similarly, normals with slope m2 and m3 cut the axis at B(2a +

am22, 0) and c(2a + am3

2, 0). 

        OA, OB and OC are in A.P. (given) where O is origin or vertex of the parabola. 

       ⇒ 2OB = OA + OC … (6) 

       ⇒ 2m22 = m1

2 + m32 

       ⇒ (m1 + m2 + m3)2 – 2 (m1 m2 + m2 m3 + m3 m1) 

       = 0 – 2 ((2a-x1)/a) 

       ⇒ m22 = 2/3 ((x1-2a)/a) … (7) 

       From equation (3), we have 

       m22 = (m1 + m3)2 

           = m1

2 + m32 + 2 m1 m3 

      = 2m22 + 2 (-y1/am2 ) (using 6 and 5) 

     ⇒ m32 = -2y1/a … (8) 

       Cubing equation (7) and squaring equation (8) we get 

      27 ay12 = 2(x1 – 2a)3 

      ∴ Locus of point (x1, y1) is 

      27 ay2 = 2(x – 2a)3

Example 16: 

       Show that the locus of the mid point of all focal chords, of a parabola is also a parabola. 

Solution: 

Method 1: 

      Let the parabola be given by 

      y2 = 4ax 

     Then its focus is (a, 0). Let (x1, y1) be the mid point of a chord of the given parabola. Then it equation is 

      S1 = T 

     or y12 – 4ax1 = y.y1 – 2a (x + x1) 

     It passes through the focus (a, 0) of (1), then 

      y12 – 4ax1 = 0.y1 – 2a (a + x1) 

     ⇒ y12 = 2a(x1 – a) 

     ∴ The required locus of (x1, y1) is 

      ⇒ y2 = 2a (x – a), which is a parabola. 

Method 2: 

      Equation of chord AB is 

      y (t1 + t2) = 2(x + at1t2) … (1) 

      It passes through the focus (a, 0) … (2) 

      ⇒ 0 = 2(a + at1 + t22) ⇒ t1t2 = –1 

      ⇒ 2h = a(t12 + t2

2) and k = a (t1 + t2) … (3) 

      ⇒ 2h = a((t1 + t2)2 – 2t1t2) 

      = a ((k/a)2+2) using (2) and (3) 

      ⇒ k2 = 2a (h – a) 

      ∴ Locus of M (h, k) is 

      y2 = 2a (x – a) 

Example 17: 

          A tangent to the parabola y2 + 12x = 0 cuts the parabola y2 = 4ax at P and Q. Find the locus of middle points of PQ. 

Solution: 

      Any tangent to the parabola y2 = –4bx is y = mx – b/m 

      y = mx – 3/m 

     Let (x1, y1) be the mid point of PQ, where P and Q are point of intersection o line (1) and y2 = 4ax 

     Equation of chord PQ is 

      S1 = T 

      y12 – 4ax1 = y1y – 2a(x + x1) 

      y1y – 2ax – y12 + 2 ax1 = 0 …… (2) 

     Equation (1) can be written as 

      my – m2 x + 3 = 0 …… (3) 

     equation (2) and (3) represent the same line 

     ⇒ m/y1 =m2/2a=3/(2ax1-y12 ) …… (4) 

     ⇒ m = 2a/y1 (from 4) 

     Again, from (4), we get 

      m/y1 =3/(2ax1-y12 ) 

     ⇒ 2a/y1 =3/(2ax1-y12 ) 

     ⇒ Locus of (x1, y1) is 

      4a2x = y2 (3 + 2a) 

Example 18: 

          The normal at any point P of the y2 = 4ax meets the axis in G and to thetangent at the vertex at H. If A be the vertex and the rectangle AGQH be completed, prove that the locus of Q is x3 = 2ax2 + ay2. 

Solution: 

      Any normal to the parabola y2 = 4ax … (1) 

      is y = mx – 2am – am3 … (2) 

     This normal (2) meets the axis y = 0 of (1) in G and the tangent at the vertex i.e. x = 0 in H 

     ∴ The coordinates of G and H are (2a + am2, 0) and (0, –2am – am3) respectively. Also the vertex A is (0, 0). Let Q be (x1, y1) 

     Given that AGQH is a rectangle. AQ and GH are its diagonals and therefore there mid points are same. Now the mid point of AQ is (1/2 x1,1/2 y1 ) and that of GH is 

      [1/2 (2a+am2+0) 1/2 (0-2am-am3 ) ] 

     i.e. [1/2 (2a+am2 )-1/2 (2am+am3 ) ] 

     ⇒ The mid points coincide so we have 

      1/2 x1 = 1/2 (2a + am2), 1/2 y1 = -1/2 (2am + am3) 

     or x1 = 2a + am2, y1 = – (2am + am3) 

     The required locus of Q is obtained by eliminating m between these. 

     Now y12 = (m2a2) (2 + m2)2 

     = a (m2a) [2+am2/a]2 

     = a (x1 – 2a) (x1/a)2 

     ⇒ ay12 = (x1 – 2a) x1

2 

     So the locus of Q is 

     ⇒ ay2 + 2ax2 = x3

EllipseEllipse is one of the easiest topics in the Conic Sections of Co-ordinate Geometry in Mathematics.

"Ellipse" is defined as the locus of a point which moves such that the ratio of its distance (Eccentricity) from a fixed point (Locus) and a fixed line (Directrix) is less than one i.e. a point moves such that its distance from a fixed point is always less than the distance from a fixed line, we get a different types of curve for one value of eccentricity, which are similar for all values of eccentricity less than one. Thus curve looks like a circle but it is not exactly a circle. Rather it is more like the edges of an egg. And if we plot the movement of the Earth and other planets around the Sun, it is the same curve satisfying the above condition of eccentricity less than one. This beautiful curve has been named as "Ellipse".

In this chapter we will discuss in detail the nature/properties of this beautiful and important curve. As you will see, the curve is symmetrical about two axes. We will study the standard form of ellipse where the X and y-axes will be taken as these axes. The main emphasis in this chapter should be on learning the properties ofellipse. The judgement of using parametric co-ordinates, which can reduce the complexity of the problem, should also be learnt.

Topics Covered under Ellipse are:-

1. Basic Concepts

2. Tangent and Normal

3. Propositions of an Ellipse

4. Solved Examples of Ellipse

Basic ConceptsIf we slice an egg obliquely there will appear a typical curve by its edge. We find the similar but larger curve if we trace the curve of the movement of earth around the sun. Our mathematicians and scientist, named this curve as the ellipse. Ellipse one of the conic sections is obtained by cutting one nappe of cone with a plane that does not pass through the vertex.

Definition

An ellipse is locus of a point, which moves in a plane such that the ratio of its distances from a fixed point and a fixed line is constant and always less than one.

     In other words "Ellipse" is a conic for which the eccentricity e < 1. Let S be the focus of ellipse, P any point on the ellipse and PM perpendicular distance of the directrix from P, then

     SP/PM = e < 1

     Let S be the focus and ZM be the directrix of the ellipse. Let be itseccentricity.

We draw SZ perpendicular to the directrix and divide SZ internally and externally in the ratio e : 1 and let A and A' be the internal and external point of division.

 

 

      Then we have SA = e AZ                                ...... (1)

       And SA' = e A'Z                                            ...... (2)

       .·. A and A' lie on the ellipse.

 

Let AA' = 2a and take O the midpoint of AA' as origin. Let P(x, y) be any point on theellipse referred to OA and OB as co-ordinate axis.

         Then from figure it is evident that

                AS = AO - OS = a - OS

                AZ = OZ - OA = OZ - a

                A'S = A'O + OS = a + OS

                A'Z = OZ + OA' = OZ + a

Substituting these values in (1) and (2), we have

                a - OS = e (OZ - a)                                    ...... (3)

                a + OS = e (OZ + a)                                  ...... (4)

        Adding (3) and (4), we get

                2a = 2 e OZ

        Or     OZ = a/e                                                 ...... (5)

        Subtracting (3) from (4), we get

             2 OS = 2ae => OS = ae                                ...... (6)

             .·. The directrix MZ is x = OZ = a/e and the co-ordinate of the focus S are (OS, 0) i.e. (ae, 0). Now as P(x, y) lies on the ellipse.

        So we get

             SP = e PM or SP2 = e2 PM2

             (x - ae)2 + y2 = e2 [OZ - x co-ordinate of P]2

        => (x - ae)2 + y2 = e2 [a/e - x]2 = (a - ex)2      ...... (7)

        => x2 + a2e2 - 2axe + y2 = a2 + e2x2 - 2aex

        or  x2/a2 + y2/a2(1-e2) = 1    [Dividing each term by a2 (1 - e2)]

        or  x2/a2 + y2/b2 = 1 where b2 = a2 (1 - e2)

    This is the standard equation of an ellipse, O is called the centre of the ellipse, AA' and BB' are called the major and minor axes of ellipse (where b < a).

    There exists a second focus and second directrix for the curve. On the negative side of the origin take a point S', which is such that SO = S'O = ae and another point Z' such that ZO = OZ' = a/e.

 

              Draw Z'K' perpendicular to ZZ' and PM' perpendicular to Z'K'

       The equation (7) may also be written in the form

               (x + ae)2 + y2 = (a + ex)2

       =>    S'P2 = e2 (PM')2

       Hence, any point P on the curve is such that its distance from S' is e times to its distance from Z'K' so we should have obtained the same curve, if we had started with S' as focus, a Z'K' as directrix and the same eccentricity.

 

Pause:

      We have considered a > b, now if we consider b > a, what will be the shape of the ellipse x2/a2 +y2/b2? In this case the major axis AA' of the ellipse is along the y-axis and is of length 2b. See figure.

 

       The minor axis of BB' = 2a. The foci S and S' are (0, be) and (0, -be) respectively. The directrix are MZ and M'Z' given by y = + b/e, respectively. Also here a2 = b2 (1 - e2).

Note:

Let P(x1, y1) be any point. This point lies outside, on or inside the ellipse (8) according as x2

1/a2 + y21/b2 = 1 > 0 or = 0 or < 0.

Central Curve

A curve is said to be a central curve if there is a point, called the centre, such that every chord passing through it is bisected at it.

Latus rectum:

        The length of a chord through the focus and at right angle to the major axis of the ellipse is known as the latus rectum of the ellipse.

 

        There being two foci of an ellipse, there are two rectum, which are of equal length.

                yL = b2/a

        .·. The length of latus rectum LSL' = 2b2/a

 

Notes:

The major axis AA' is of length 2a and the minor axis BB' is of length 2b.

The foci are (-ae, 0) and (ae, 0).

The equations of the directrices are x = a/e and x = -a/e.

The length of the semi latus rectum = b2/a.

Circle is a particular case of an ellipse with e = 0.

Focal Distance of a Point

Since S'P = ePN', SP = ePN,

S'P + SP = e(PN + PN')

 

 

=      e (NN') = e(2a/e) = 2a

=>    the sum of the focal distances of any point on the ellipse is equal to its major axis.

 

Another definition of an ellipse

                Let ellipse be x2/a2 +y2/b2 = 1                                               ...... (i)

                Its foci S and S' are (ae, 0) and (-ae, 0). The equation of its directrices MZ and M'Z' are x = a/e and x = -a/e respectively. Let P(x1, y1) be any point on (i)

                Now SP = e PM = e NZ = e (OZ - ON) = e[(a/e)-x1] = a - ex1

                and S'P = ePM' = e (Z'N) = e (OZ' + ON) = e[(a/e) + x1] = a + ex1

 

     

             .·. SP + S'P = 2a = AA'

So by this property an ellipse can also be defined as "the locus of a point which moves such that the sum of its distances from two fixed point is always constant.

Other Forms

If in the equation x2/a2 +y2/b2 = 1, a2 < b2, then the major and minor axis of theellipse lie

along the y and the x-axis and are of lengths 2b and 2a respectively. The foci become

(0, + be), and the directrices become y = + b/e where e = √(1-a2/b2 ). The length of the

semi-lactus rectum becomes a2/b2.

If the centre of the ellipse be taken at (h, k) and axes parallel to the x and the y-axes, then

the equation of the ellipse is (x-h)2/a2 +(y-k)2/b2 = 1.

Let eh equation of the directrix of an ellipse be ax + by + c = 0 and the focus be (h, k).

 

Let the eccentricity of the ellipse be e(e < 1).

If P(x, y) is any point on the ellipse, then

PS2 = e2 PM2

          

 

=>  (x - h)2 + (y - k)2 = e2(ax+by+c)2/(a2 + b2 ), , which is of the form

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ... (*) where

Δ = abc + 2fgh + af2 - bg2 - ch2 ≠ 0, h2 < ab.

These are the necessary and sufficient conditions for a general quadratic equation given by (*) to represent an ellipse.

 

Position of a Point Relative to an Ellipse

The point P(x1, y1) is outside or inside the ellipse x2/a2 + y2/b2  =  1, according as the quantity ((x1

2)/a2 +(y12)/b2 -1)  is positive or negative.

 

Parametric Equation of an Ellipse

Clearly, x = a cosθ, y = b sinθ satisfy the equation x2/a2 +y2/b2   = 1 for all real values of θ.

Hence, the parametric equations of the ellipse  x2/a2 +y2/b2 = 1 are x = a cosθ, y = b sinq where θ is the parameter.

Also (a cos θ, b sin θ) is a point on the ellipse  x2/a2 +y2/b2 = 1 for all values of θ(0 < θ < 2Π).

The point (a cosθ, b sinθ) is also called the point θ. The angle θ is called the eccentric angle of the point (a cosθ, b sinθ) on the ellipse.

Draw a circle with AA' (the major axis) as the diameter. This circle is called the auxiliary circle of the ellipse. The equation of the circle is x2 + y2 = a2. Any point Q on the circle is (a

cosθ, a sinθ). Draw QM as perpendicular to AA' cutting the ellipseat P. The x-co-ordinate of P = CM = a cosθ.

      =>  y-co-ordinate of P is b sinθ

      =>  P ≡ (a cosθ, b sinθ).

 

Illustration:

Find the centre, the length of the axes and the eccentricity of the ellipse 2x2 + 3y2 - 4x - 12y + 13 = 0.

Solution:

        The given equation can be written as 2(x - 1)2 + 3(y - 2)2 = 1

        =>(x-1)2/(1/2)+(y-2)2/(1/3)   = 1 => The centre of the ellipse is (1, 2).

        The major axis = 2. 1√2 = √2.

        The minor axis = 2.1/√3=2/3 => e2 = 1 – 1/2 = 1/3 => e = 1/√3.

 

Illustration:

Find the equation of the ellipse whose foci are (2, 3), (-2, 3) and whose semi minor axis is of length √5.

Solution:

       Here S is (2, 3), S' is (-2, 3) and b = √5.

        => SS' = 4 = 2ae → ae = 2.

        But b2 = a2 (1 - e2) => 5 = a2 - 4 => a = 3.

        Centre C of the ellipse is (0, 3).

        Hence the equation of the ellipse is (x-0)2/9+(y-3)2/5   = 1

        => 5x2 + 9y2 - 54y + 36 = 0.

 

Illustration:

Find the equation of the ellipse having centre at (1, 2), one focus at (6, 2) and passing through the point (4, 6).

Solution:

With centre at (1, 2) the equation of the ellipse is (x-1)2/a2 +(y-2)2/b2 = 1. It passes through the point (4, 6).

=>     9/a2 +16/b2 = 1.                ...... (1)

Distance between the focus and the centre = (6 - 1) = 5 = ae

=>     b2 = a2 - a2e2 = a2 - 25.    ...... (2)

Solving for a2 and b2 from the equations (1) and (2), we get a2 = 45 and b2 = 20.

Hence the equation of the ellipse is (x-1)2/45+(y-2)2/20 = 1.

  

 Illustration:

Find the equation of the ellipse (in standard form) having latus rectum 5 andeccentricity 2/3.

Solution:

        Let the ellipse be x2/a2 +y2/b2  = 1 with a > b.

        Latus rectum = 5 = 2b2/a => 2b2 = 5a.                                ...... (1)

        Also b2 = a2 (1 - e2) = a2(1-4/9) = 5a2/9

        => 5a/2 = 5a2/9  =>  a  = 9/2 and hence b2 =  5/2a = 45/4.

The equation of the ellipse, in the standard form, is thus x2/(81/4)+y2/(45/4) = 1.

 

Illustration:

        Find the equation of the ellipse, which cuts the intercept of length 3 and 2 on positive x and y-axis. Centre of the ellipse is origin and major and minor axes are along the positive x-axis and along positive y-axis.

Solution:

               x2/a2 +y2/b2 = 1                                       ...... (1)

              According to the given condition the ellipse (1) passes through (3, 0) and (0, 2), so we have.

             9/a2 = 1 => a2 = 9

        And 4/b2 = 1 => b2 = 4

        Therefore, the equation of the ellipse is x2/9 + y2/4 = 1

 

Illustration:

        Obtain the equation of an ellipse whose focus is the point (-1, 1) whose directrix is the line passing through (2, 5) having the unit gradient and whose eccentricity is ½.

Solution:

        Let P(x, y) be any point on ellipse.

        Its focus is S (-1, 1).

        Let the directrix be y = x + c                            ...... (1)

                (·.· gradient m = 1)

        Line (1) passes through (2, 5) so,

                5 = 2 + c => c = 3

        The directrix is y = x + 3

                => x - y + 3 = 0                                    ...... (2)

         Now let PM be the perpendicular from P, drawn to its directrix

(2). By definition of ellipse SP = e PM

                or SP2 = e2 PM2

        => (x + 1)2 + (y - 1)2 = (1/2)2 [(x-y+3)/√((12+12 ))]2

        => 8[(x + 1)2 + (y - 1)2] = (x - y + 3)2

        => 7x2 + 7y2 + 2xy + 10x - 10y + 7 = 0,

        This is the required equation of ellipse.

Tangent and Normal        The ellipse is x2/a2 +y2/b2 = 1                         ...... (1)

Let P(x1, y1) and Q(x2, y2) be two points on the ellipse. The equation of the line PQ is,

            y - y1 = ((y2-y1)/(x2-x1 )) (x - x1)                 ...... (2)

        since points P and Q lies on (1), we get

       (y2-y1)/(x2-x1 )=(-b2 (x2-x1 ))/(a2 (y2-y1 ) )

        So (2) becomes

        y - y1 =(-b2 (x2-x1 ))/(a2 (y2-y1)) (x - x1)

 

As point Q approaches towards point P along the ellipse, the line PQ tends to thetangent at P. So, by substituting x1 and y1 for x2 and y2 in the above equation, we have the equation of the tangent at P as

y - y1 = (-b2 (2x1 ))/(a2 (2y1)) (x - x1)

=>  (xx1)/a2 + yy1/b2 =(x12)/a2 +(y`

1)/b2 = 1     [as P lies on (1)]

Hence the equation of the tangent to x2/a2+y2/b2= 1 at P(x1,y1) is xx1/a2 + yy1/b2= 1.            ...... (3)

Equation of the tangent in terms of 'm'

Let the line y = mx + c                                                ....... (4)

Touch the ellipse x2/a2 +y2/b2 = 1                                ...... (5)

Eliminating y between (4) and (5), we get;

        x2 (b2 + a2m2) + 2a2mcx + a2(c2 - b2) = 0             ...... (6)

If (4) touches (5) then the roots f (6) must be coincident i.e. D = 0

        i.e. (2a2mc)2 = 4(b2 + a2m2) a2(c2 - b2)

solving this we get c = +√(a2 m2+b2)

So the equation of tangent is y = mx + √(a2 m2+b2) for all real m    ......(7)

From the equation (6) and (7) we get the point of contact as  ((+_a2 m)/√(a2

m2+b2 ),(±b2)/√(a2 m2+b ))

Equation of the Tangent to an Ellipse

*  Let the equation of the ellipse be x2/a2 +y2/b2 = 1  . Slope of tangent at

(x1, y1) = dy/dx(x1,y1)=-b2/a2 x2/y1

Hence the equation of the tangent at (x1, y1) is y - y1 = (-b2 x1)/(a2 y1)(x - x1)

=>    (xx1)/a2 +(yy1)/b2 = 1.

*  Equation of tangent at the point q i.e. (a cosθ, b sinθ) is obtained by putting x1= a cosθ, y1 = b sinθ

=>  (x cosθ )/a+(y sinθ )/b = 1.

 

Equation of the Tangent in Terms of its slope; Condition of Tangency

To find the condition that the line y=mx+ c may touch the ellipse x2/a2 +y2/b2 = 1.

Tangent to the ellipse at (a cosθ, b sinθ) is (x cosθ)/a+(y sinθ)/b = 1.

If y - mx = c is also a tangent to the given ellipse at 'q', then comparing coefficients, we get

(cosθ)/am = (sinθ)/b=1/c or 1/c =(sinθ)/b=(cosθ)/(-am)=±√((sin2θ+cos2θ)/(b2+(-am)2))=±1/√(a2 m2+b2 )

=>   c = ±√(a2 m2+b2 ).

Therefore, the equation of a tangent to the ellipse x2/a2 +y2/b2 = 1 is

        y = mx ±√(a2 m2+b2 ) for all values of m.

 

Illustration:

       Find the locus of the point of intersection of the tangents to the ellipse x2/a2+y2/b2 = 1 (a > b) which meet at right angles.

Solution:

The line y = mx ±√(a2 m2+b2) is a tangent to the given ellipse for all m. Let is passes through (h, k).

=>   k - mh = √(a2 m2+b2 ) => k2 + m2h2 - 2hkm = a2m2 + b2

=>   m2 (h2 - a2) - 2hkm + k2 - b2 = 0.

If the tangents are at right angles, then m1m2 = -1.

=> (k2-b2)/(h2-a2 ) = - 1 => h2 + k2 = a2 + b2.

Hence the locus of the point (h, k) is x2 + y2 = a2 + b2 which is a circle. This circle is called the Director Circle of the ellipse.

Note:

The locus of the point of the intersection of two perpendicular tangents to anellipse is a circle known as the director circle.

 

Illustration:

Prove that the locus of the mid-points of the intercepts of the tangents to theellipse x2/a2 + y2/b2 = 1 = 1, intercepted between the axes, is a2/x2 +b2/y2 = 4.

Solution:

 The tangent to the ellipse at any point (a cosθ, b sinθ)(x cosθ)/a+(y sinθ )/b = 1.

 Let it meet the axes in P and Q, so that P is (a secθ, 0)

 and Q is (0, b cosecθ). If (h, k) is the mid-point of PQ, then h = (a secθ)/2.

 => cosθ = a/2h     and k = (b cosecθ)/2 => sinθ = b/2k.

 Squaring and adding, we get a2/4h2 +b2/(4k2 ) = 1

 Hence the locus of (h, k) is a2/x2 +b2/y2 = 4.

 

Illustration:

Prove that the product of the lengths of the perpendiculars drawn from the foci to any tangent to the ellipse x2/16+y2/9 = 1 is equal to 9.

Solution:

For the given ellipse a = 4, b = 3 and hence 9 = 16 (1 - e2)

=> e = √7/4. The foci are thus located at (√7,0) and (-√7,0).

Equation of a tangent to the given ellipse is

       y = mx + √(16m2+9)            (as a = 4, b = 3).

Lengths p1 and p2 of the perpendiculars drawn from the foci are

        p1 = (√(16m2+9)+√7 m)/(1+m2 ) and p2 = (√(16m2+9)-√7 m)/(1+m2)

=>    p1p2 = (16m2+9-7m2)/(1+m2) = 9(1+m2)/(1+m2) = 9.

Note:

The product of lengths of the perpendiculars drawn from the foci to any tangent to the ellipse x2/a2 +y2/b2 = 1 is b2.

 

Equation of the Normal to an Ellipse

The normal to a curve is a line perpendicular to the tangent to curve through eh point of contact.

.·. The slope of normal at point (x1, y1) = a2y1/b2x1 and so its equation is (x-x1)/((x1/a2))=(y-y1)/((y1/b2 ) ).

 

*  To find the equation of the normal to the ellipse x2/a2 +y2/b2 = 1 at (x1, y1):

Equation of the tangent at (x1, y1) is (xx1)/a2 +(yy1)/b2 = 1

=>     Slope of the normal is a2/b2 +y1/x1; => equation of the normal is

        y - y1 = a2/b2 +y1/x1 (x - x1) =>(x-x1)/(x1/a2)=(y-y1)/(y1/b2 ).

*  Equation of the normal at (a cosθ, b sinθ) is (x-a cosθ)/((a cosθ)/a2) = (y-b sinθ )/((b sinθ)/b2).

=> ax secθ - by cosecθ = a2 - b2.

ax1 secθ - by1 cosecθ = a2 - b2.

or ax1((1+t2)/(1-t2)) - by1 ((1+t2)/2t) = a2 - b2, where t = tan θ/2.

On simplification, this equation gives

by1t4 + 2(ax1 + a2 -b2)t3 + 2(ax1 - a2 + b2)t - by1 = 0.

This is a 4th degree equation in t which gives, in general, four values fo t. Hence from a fixed point four normals can be drawn to the given ellipse.

 

Illustration:

If the normals to the ellipse x2/a2 + y2/b2 = 1 at the points (x1, y1), (x2, y2) and (x3, y3) are

concurrent, prove that   = 0.

Solution:

        The equation of the normal to the given ellipse at (x1, y1) is

                a2xy1 - b2yx1 - (a2 - b2)x1y1 = 0.                 ...... (1)

        Similarly the normals at (x2, y2) and (x3, y3) are

                a2xy2 - b2yx2 - (a2 - b2)x2y2 = 0.                 ...... (2)

                a2xy3 - b2yx3 - (a2 - b2)x3y3 = 0.                 ...... (3)

Eliminating a2x, b2y and (a2 - b2) from (1), (2) and (3), we find that the three lines are

concurrent if    =  0.

Tangent and Normal

Illustration:

If the normals at the end of a latus rectum of the ellipse x2/a2 +y2/b2 = 1 passes through the extremity of a minor axis, prove that e4+e2-1=0.

Solution:

        Equation the normal to te given ellipse at

        (ae,b2/a) is (x-ae)/(ae/a2) = (y-b2/a)/(b2/ab2).

        Since it passes through (0, - b).

        - a2 = - ab - b3

        => a2 = ab + a2 (1 - e2)

        or b = ae2 => b2 = a2e4

        or a2 (1 - e2) = a2e2 Þ a4 + e2 - 1 = 0.

 

Enquiry:    How many tangents can be drawn from a point on an ellipse?

We know that y = mx + √(a2 m2+b2) is a tangent to the ellipse x2/a2 +y2/b2 = 1 for real values of m. If this tangent passes through a point (x1, y1) we have y1 = mx1 + √((a2 m2+b2)).

or m2(x12 - a2) - 2x1y1m + (y1

2 - b2) = 0, which being a quadratic equation in m gives two values of m. Thus from a point two tangents corresponding two values of m) can be drawn to an ellipse.

  

Pair of Tangents and Chord of Contact

From a fixed point (x1, y1) in general two tangents can be drawn to an ellipse. The equation of the pair of tangents drawn to the ellipse x2/a2 +y2/b2 = 1 is given by (x2/a2 +y2/b2-1)((x1

2)/a2 +(y12)/b2-1) = (xx1/a2 + yy1/b2-1)2.

 

 

 

Tangent and Normal

In symbols we write SS1 = T2, where

S ≡ x2/a2 +y2/b2 = 1,  S1≡ (x12)/a2 +(y1

2)/b2 – 1 and  T ≡ xx1/a2 + yy1/b2 – 1

*  If from the point P(x1, y1) tangents PQ and PR be drawn to the ellipse x2/a2+y2/b2 = 1, then the line joining the points of contact Q and R is called the chord of contact. Equation of the chord of contact is xx1/a2 + yy1/b2 - 1 = 0 or T = 0.

*  Equation of the Chord Joining the Points (α cosα, β sinα), (α cosβ, β sinβ) is

      x/a cos((α+β)/2)+ y/b sin((α+β)/2) = cos((α-β)/2)

*  Equation of a chord which is bisected at the point (x1, y1) is xx1/a2 + yy1/b2 -1 = (x12)/a2 +

(y12)/b2 - 1 or = S1

*  Length of the chord ... (from package).

 

To find the length of the chord intercepted by the ellipse x2/a2 +y2/b2 = 1 on the straight line y = mx + c.

Points of intersection of the ellipse and the line are given by x2/a2 +(mx+c)2/b2 = 1

i.e. (a2 m2 + b2)x2 + 2a2 cmx + a2(c2 - b2) = 0                ...... (1)

Therefore the straight line meets the ellipse in two points (real, coincident or imaginary).

If (x1, y1) and (x2, y2) be the points of intersection, the length of the chord is

        √((x1-x2 )2+(y1-y2)2 )=√(1+m2 )= |x1 - x2|             ...... (2)

(since y1 - y2 = m (x1 - x2))

where x1 and x2 are the roots of the equation (1), and

        x1 + x2 = -(2a2cm)/(a2 m2 + b2), x1 x2 (a2 (c2 - b2))/(a2 m2 + b2) so that

(x1 - x2)2 = (x1 + x2)2 - 4x1 x2 = (4a2 c2 m2)/(a2 m2 + b2)2 -(4a2(c2 - b2))/((a2m2

+ b2))=(4a2 b2 (a2 m2 + b2 - c2))/(a2 m2 + b2)2.

Hence the length of the chord is √(((1+m2)4a2 b2 (a2 m2 + b2 - c2))/(a2 m2 + b2)2).

i.e. 2ab/(a2 m2 + b2)2) √((1+m2))(a2) m2)+b2)-c2))).

=> e cos(α+β)/2 = cos(α-β)/2

=> e [cos α/2 cos β/2 + sin α/2 sin β/2] = cos α/2 cos β/2 + sin α/2 sin β/2

=> e [1 - tan α/2 tan β/2] = 1 + tan α/2 tan β/2 => tan α/2 . tanβ/2 = (e-1)/(e+1).

If the chord passes through (-ae, 0) then tanα/2 . tanβ/2 =  (e+1)/(e-1).

 

Tangent and Normal

Illustration:

The tangent and normal at a point P on an ellipse meet the minor axis at A and B. Prove that AB subtends a right angle at each of foci.

Solution:

The equations of the tangents and normal at a point P(x1, y1) on the ellipse x2/a2+ y2/b2 = 1 are

xx1/a2 + yy1/b2 = 1                                    ...... (1)

and (x-x1)/((x1/a2))+(y-y2)/((y1/b2)) = 1       ...... (2)

Also the minor axis is y-axis i.e. x = 0

Solving (1) and x = 0, we have A (0, b2/y1)

Solving (2) and x = 0, we have B (0, y1 - a2y1/b2)

Let S(ae, 0) be one of the foci of the ellipse. Then the slope of SA = ((b2/y1) - 0)/(0-ae)= b2/aey1 = m1                                              (say)

And the slope of SB = ([y1 (a2/b2 ) y1 ]-0)/(0-ae)

   = y1/ae ((a2-b2 ))/b2= (y1 a2 e2)/(aeb2) [·.· b2 = a2(1 - e2)]

   = aey1/b2 = m2                              (say)

                Evidently m1m2 = -1 => SA | SB

i.e. AB subtends a right angle at S(ae, 0). Similarly we can show that AB subtends a right angle at the other focus S'(-ae, 0)

 

Illustration:

Prove that the locus of the middle points of the portion of tangent to the ellipsex2/a2 +y2/b2 = (a+b) between the axis is the curve a/x2 + b/y2 = 4/((a+b)).

Solution:

Any tangent to the ellipse  x2/a2 +y2/b2 = 1 , is

y = mx + √(a2 m2 + b2)

Therefore the tangent to given ellipse is

y = x + √(a(a+b) m2+b(a+b))                  ...... (1)

(1)  Meets x-axis at {-√(a(a+b) m2+ b(a+b) )/m,0}

(2)  Meets y-axis at {0,√(a(a+b) m2+b(a+b))}

Let (x1, y1) be the mid points of the portion of the tangent intercepted between the axes, then

x1 = 1/2 [{-√(a(a+b) m2 + b(a+b))/m} + 0]

y1 = 1/2 [0+{√(a(a+b) m2+b(a+b))} + 0]

=> 4x12 = (a(a+b) m2+ b(a+b))/m2

and 4y12 = a(a + )m2 + b(a + b)

Eliminating m, we get

a(a+b)/(4x12) - b(a+b)/(4y1

2) = am2/(am2+b) + b/(am2+b) = 1

 .·. The locus of (x1, y1) is a/x2 + b/y2 = 4/(a+b)  = 1

 

Illustration:

If PS1Q and PS2R be two focal chord of the ellipse whose two foci are S1 and S2and the eccentric angle of P is 'θ' then show that the equation of chord QR is x/a cos θ + y/b.(1+e2)/(1-e2) sin θ + 1 = 0.

Solution:

Let Q be (a cos α, b sin α) and R be (a cos β, b sin β) then the equation of the chord QR is

x/α cos(α+β)/2 + y/b sin(α+β)/2 = cos(α-β)/2

which on simplifying becomes

x/a (1-tan α/2 tan β/2)+y/b (tan α/2+tan β/2) = 1 + tan α/2 tan β/2 ..... (1)

Also, PQ and PR are focal chord thus

tan θ/2 tan α/2 = (e-1)/(e+1) and tan θ/2 tan β/2 = (e+1)/(e-1)

(From previous illustration)

On substituting the values of tan α/2  and β/2  in (1), we get

 

Tangent and Normal

Illustration:

Show the locus of middle points chord of the ellipse  x2/a2 +y2/b2 = 1 which subtend right angle at the centre is x2/a4 + y2/b4 =(1/a2 +1/b2 ) (x2/a2 + y2/b2)2.

Solution:

Let (x1, y1) be the middle point of chord PQ, then its equation is

         T = S1 or 1/a2 + yy1/b2 =(x1/b2)/a2 +(y1/b2)/b2            ...... (1)

Since the origin 'O' is the centre so the equation of pair of lines OP and OQ can be obtained by homogenizing the equation of the ellipse x2/a2 +y2/b2 = 1, with the help of (1), thus

  x2/a2 +y2/b2 = (1)2 = 

or ((x12)/a2 +(y1

2)/b2)2 (x2/a2 +y2/b2 ) = (x2 x12)/a4 +(y2 y1

2)/b4 + (2xyx1 y1)/(a2b2).

It represents a pair of perpendicular lines if

1/a2 ((x12)/a2 +(y1

2)/b2 )2 - (x12)/a4 + 1/b2 ((x1

2)/a2 +(y12)/b2)2-(y1

2)/b4 = 0.

or, (x12)/a4 +(y1

2)/b4 = (1/a2 +1/b2)(x2/a4 +y2/b4)2.

So the locus of (x1, y1) is

x2/a4 + y2/b4 = (1/a2 + 1/b2 ) (x2/a2 +y2/b2 )2.

Propositions on Ellipse:

Auxiliary circle of an ellipse

Auxiliary circle of an ellipse which is a circle described on the major axis of anellipse as its diameter.

Let the ellipse be

        x2/a2 + y2/b2 =1                                              ...... (1)

Then the equation of its auxiliary circle is x2 + y2 = a2    ...... (2)

Take a point P(x1, y1) on (1).

Through P, draw a line perpendicular to major axis intersecting major axis in N and auxiliary circle in P'.

 

Let P' be (x1, y2). Then we have (x12)/a2 + (1

2)/b2 = 1 and 12 + y12 = a2

From these two relations, we get:

       (y22)/(y1

2)=(a2-12)/(b2 {1-(x1

2/a2)}) = a2/b2

        =>   y2/1 = a/b

Now, let OP' make an angle f with the major axis of the ellipse (1), P' being the corresponding point of P on the auxiliary circle of the ellipse. Then Φ is called the eccentric angle of the point P. From the figure it is evident that if (x1, y1) are the co-ordinates of P, then  x1 = ON = OP' cosΦ = a cosΦ. Also P(x1, y1) is a point on (1).

So solving(x12)/a2 +(1

2)/b2 =1 for y1, we get y1 = b sin Φ

 

Therefore (a cos Φ, b sin Φ) are the parametric co-ordinates of any point P on theellipse, where Φ is the eccentric angle of P.

Propositions on Ellipse:

Note:

1.    Tangent to the ellipse at the point:

The equation of the line PQ reduces to that of tangent when x/a cos Φ + y/b sin Φ = 1  ...... (a)

2.     Normal to the ellipse at the point 'f':

Equation of normal can be derived by using the formula of equation of straight line passing through point (a cosΦ, b sinΦ) and perpendicular to tangent (a) i.e. (y - b sin Φ) = (a sin Φ/b cos Φ) (x - a cos Φ).

=> a x sec Φ - b y cosec Φ = a2 - b2

 

Diameter of an ellipse

The locus of the middle points of a system of parallel chords of an ellipse is called the diameter of the ellipse.

Let y = mx + c                                            ...... (1)

be the equation of a system of parallel chords of the ellipse x2/a2 + y2/b2 = 1        ...... (2)

 

In (1) m is constant and c varies from chord to chord. Let K (x1, y1) be the midpoint of a chord PQ of this system.

Eliminating y between (1) and (2) we get

(a2m2 + b2)x2 + 2a2mcx + a2(c2 - b2) = 0        ...... (3)

=> x2 + x3 = (-2a2 mc)/(a2 m2+b2 )

But x1 = (x2+x3)/2 =(-a2 mc)/(a2 m2+b2)

Or c = (-x(a2 m2+b2))/(a2 m2)

Also K (x1, y1) is a point on (1) so

        y1 = mx1 + c => y1= -b2x1/a2m

.·. The locus of K (x1, y1) is y = -b2x/a2m which is a diameter of the ellipse x2/a2 + y2/b2 = 1.

 

Note:

1.  Conjugate Diameters: Two diameters of an ellipse which bisects chords parallel to each other are called conjugated diameters. Therefore the diameters y = mx and y = m1x of the ellipse x2/a2 +y2/b2 = 1 are conjugate if mm1 = -b2/a2.

2.  In an ellipse, the major axes bisects all chords parallel to the minor axes and vice-versa, therefore major axes and minor axes of an ellipse are conjugate diameters but they do not satisfy the condition mm1 = -b2/a2 and are the only perpendicular conjugate diameters.

3.   Equi-conjugate diameter:

 If the length of two conjugate diameters of ellipse be equal then they are called equi conjugate diameters.

       The equation of equi conjugated diameters are x2/a2 ± y2/b2 = 1.

 

4.  The eccentric angles of the ends of a pair of conjugate diameters of an ellipsediffer by a right angle i.e., if one end of a diameter (PQ) is P(a sin Φ, b cos Φ).

5.  The sum of the squares of any two conjugate semi-diameters of an ellipse is constant and equal to the sum of the squares of the semi-axis of the ellipse i.e. OP2 + OP2 = a2 + b2.

6.  The produce of the focal distances of a point on an ellipse is equal to the square of the semi-diameters, which is conjugate to the diameter through the point.

7.  The tangents at the ends of a pair of conjugate diameters of an ellipse form a parallelogram and the area of the parallelogram is constant and is equal to the product of the axis i.e. equal to 4ab.

 

Director circle of an ellipse

The director circle is the locus of the point of intersection of pair of perpendicular tangents to an ellipse.

Two perpendicular tangents of ellipse x2/a2 + y2/b2 = 1 are

       y - mx = √(a2 m2+b2)                               ...... (1)

and   my + x = √(a2+b2 m2)                             ...... (2)

To obtain the locus of the point of intersection (1) and (2) we have to eliminate m squaring and adding (1) and (2), we get

(y - mx)2 + (my + x)2 = (a2m2 + b2) + (a2 + b2m)

=> x2 + y2 = a2 + b2, which is the equation of the director circle.

 

Enquiry:    How to find equation of a chord of an ellipse whose mid point is (x1, y1)?

Let the ellipse be x2/a2 + y2/b2 = 1                   ...... (1)

Any line through (x1, y1) is y - y1 = m(x - x1)        ...... (2)

Eliminating y between (1) and (2) we get

        x2 (b2 + a2m2) + 2ma2 (y1 - mx1) x + a2 [(y1 - mx1)2 - b2] = 0

If x2 and x3 are the roots of this equation then

        x2 + x3 = -2ma2 (y1 - mx1)/(b2 + a2m2)

But (x1, y1) the mid point of the chord.

        .·. x1 = (x2+x3)/2=(-ma2 (y1-mx1 ))/(b2+a2 m2)

        => m = (-b2 x1)/(a2 y1)

        .·. From (2) and (3) the equation of the chord where mid point is (x1, y1) is

        y - y1 = ((-b2 x1)/(a2 y1)) (x - x1)

=>    xx1/a2 + yy1/b2 =(x12)/a2 +(y1

2)/b2

Or     T = S1 where T = xx1/a2 + yy1/b2 -1

and   S1 = (x12)/a2 + (y1

2)/b2 -1

Pole and polar of an ellipse

 The locus of the points of intersection of tangents drawn at the point extremities of the chords passing through a fixed point is called the polar of that fixed point and the fixed point is called the pole.

 

        Equation of the polar of P(x1, y1) with respect to the ellipse

         x2/a2 + y2/b2 = 1 is

         xx1/a2 + yy1/b2 =1

 

Illustration:

         How to find out pole of the line lx + my + n = 0 w.r.t. the ellipse x2/a2+y2/b2 = 1.

Solution:

         Let (x1, y1) be the pole of line lx + my + n = 0.     ...... (1)

         w.r.t. the ellipse x2/a2 + y2/b2 = 1                     ...... (2)

         Now the polar of (x1, y1) w.r.t. (2) is

        xx1/a2 + yy1/b2 = 1                            ...... (3)

        Since (1) and (3) represents the same polar, so comparing them we have

        (x1/a2)/l+(y1/b2)/b2 = (-1)/n

        or     x1 = (-a2 l)/n     y1 = (-b2 m)/n

     .·. The required pole is ((-a2 l)/n,(-b2 m)/n)

Solved Examples of Ellipse:Example 1:

Find the points on the ellipse x2 + 3y2 = 6 where the tangent are equally inclined to the axes. Prove also that the length of the perpendicular from the centre on either of these tangents is 2.

Solution:

The given ellipse is x2 + 3y2 = 6

Or x2/6+y2/2 = 1                                                      ...... (1)

If the coordinates of the required point on the ellipse (1) be (√6 cos Φ,√2 sin Φ) then the tangent at the point is x/√6 cos Φ + y/√2 sin Φ = 1      ...... (2)

Slope of (2) = (-cos Φ)/√6×√2/(sin Φ )=(-√2)/√6 cot Φ

As the tangent are equally inclined to the axes so we have

  -1/√3 cot Φ = + tan 45o = + 1

  .·. tan Φ = + 1/√3

  The coordinates of the required points are

  (±√6×√3/2, ±√2×1/2) and (±√6×√3/2, +-√2×1/2)

  = (±(3√2)/2,±1/√2) and (±(3√2)/2,+-1/√2)

  Again the length of perpendicular from (0, 0) and (2),

  = (√6.√2)/√(2 cos2Φ + 6 sin2Φ)=(2√3)/√((2,3/4)+(6.1/4) )=(2√3)/√3 = 2

Example 2:

If P be a point on the ellipse x2/a2 + y2/b2 = 2/c whose ordinate is √2/c, prove that the angle between the tangent at P and SP is tan-1 (b2/ac), where S is the focus.

Solution:

The given ellipse x2/a2 + y2/b2 = 2/c            ...... (1)

If (x',√(2/c)) be the coordinates of the given point P on the ellipse (1).

Then the tangent at P will be:(xx')/a2 +(yy')/b2 = 1

   (xx')/(a2(2/c) )+(yy' (√(2/c)))/(b2 (2/c))= 1

The slope of tangent at P is (-b2 (2/c) x')/(a (2/c)(√2(2/c)))= m1 (say)

If S be the focus, then slope of PS = = y'/(x'+ae)=√(2/c)/(x'+a√(2/c)e)= m2 (say)

If angle between the focal distance SP and tangent at P is θ, then

tan θ = (m2-m1)/(1+m2 m1).

(a2 (2/c)+b2 x'2+ aeb2 x' √(2/c))/((a2 x'+a3 e√(2/c)-b2 x')√(2/c))

point (x',√(2/c)) lies on ellipse (1), we have

      b2 x'2 + a2(2/c)=(2a2 b2)/c

and ·.· a2 - b2 = a2 e2 so we have

(-(2a2 b2)/c + aeb2 x' √(2/c))/((a2 e2 x'+a2 e√(2/c)) √(2/c)) = (√(2/c) ab2 (√(2/c)

a+ex'))/(a2 e(√(2/c a+ex')) √(2/c)) = b2/ae.

.·. θ = tan-1 b2/ae. Hence proved.

Example 3:

If P, Q, R are three points on the ellipsex2/a2 +y2/b2 = 1 whose eccentric angles are θ, Φ and Ψ then find the area of ΔPQR.

Solution:

The coordinates of the gives points P, Q, R on the ellipse x2/a2 +y2/b2 = 1, will be (a cos θ, b sin θ), (a cos Φ, b sin Φ) and (a cos Ψ, b sin Ψ). Area of triangle PQR formed by these points

   = 1/2 [x1y2 - x2y1 + x2y3 - x3y2 + x3y1 - x1y3]

   = 1/2 [ab cos θ sin Φ - ab sin θ cos Φ + ab cos Φ sin Ψ - ab sin Φ cos Ψ + ab in θ cos Ψ - ab cos θ sin Ψ]

   = 1/2 ab [2 sin(Φ-θ)/2 cos (Φ-θ)/2 + 2 sin (Ψ-Φ)/2 cos (Ψ-Φ)/2 + 2 sin (θ-Φ)/2 cos (θ-Φ)/2]

   = ab sin(θ-Φ)/2 (cos(θ+Φ+2Ψ)/2 cos(Φ-θ)/2)

   = 2ab sin (θ-Φ)/2 cos (Φ-Ψ)/2 cos (Ψ-θ)/2.

 

Example 4:

Find the locus of the extremities of the latus recta of all ellipses having a given major axis 6a.

Solution:

Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae                                   ...... (1)

And y = SL = b2/3a =(9a2(1-e2))/3a = 3a (1 - e2)      ...... (2)

Eliminating the variable 'e' from (1) and (2) we get eh locus of L.

Hence putting the value of e from (1) and (2), we get

        y = 3a(1-x2/9a2)

=> x2 = 9a2 - 3ay

=> x2 = 3a(3a - y), which is clearly a parabola. Similarly we can show that the locus of L' is x2 = 3ay(y + 3a) which is again a parabola.

Example 5:

Find the equation of the normals at the end of the latus rectum of the ellipsex2/a2 +y2/b2 = c2 and find the condition when each normal through one end of the minor axis.

Solution:

    The ellipse x2/a2 + y2/b2 = c2

    => x2/(a2 c2+y2/(b2 c2) = 1                                     ...... (1)

    Then the end point of the latus rectum is (ace,(b2 c)/a)

    The normal at this point will be

     (x-ace)/(ace/(a2 c2)) = (y-b2 c/a)/((b2 c/a)/(b2 c2))

    => ((x-ace))/e ac = (y-(b2 c)/a) ac

    => (x-ace)/e = y-(b2 c)/a

    If this normal passes through (0, - bc), then, we have

     (-ace)/e = -bc-b2/a c

    => a = b + a(1 - e2)

    => b - ae2 = 0

    => b/1-e2 => b2/a2 = e4

    => e4 + e2 = 1. This is required condition.

 

Example 6:

The circle x2 + y2 = 4 is concentric with the ellipse x2/7 + y2/3 = 1; prove that the common tangent is inclined to the major axis at an angle 30o and find its length.

Solution:

  The ellipse  x2/7+y2/3=1          ...... (1)

  The equation to the circle is

         x2 + y2 = 4                               ...... (2)

  As the line y = mx + √(a2 m2+b2)

  i.e. y = mx + √(7m2+3)              ...... (3)

is always the tangent on the ellipse. If this is also a tangent on the circle (2) then length of perpendicular from the centre (0, 0) on the line (1) must be equal to radius of circle i.e. 2.

Hence,  √(7m2+3)/√(1+m2) = 2

       => 7m2 + 3 = 22 (1 + m2)

       => 7m2 - 4m2 = 4 - 3

       => m2 = 1/3

       => m = + 1/√3

Hence the common tangent to the two curves is inclined at an angle of tan-1 (+1/√3) i.e. 30o to the axis.

Note:

We can also prove the above result by using the fact that the line = mx + √(7m2+3) will be tangent to x2 + y2 = 4 if discriminent of x2 + (m + √(7m2+3))2 = 4 is zero.

Let P and Q be the points of contact of the common tangent with ellipse and circle respectively and O be the common centre of the two, then PQ = √(OP2-OQ2)[ ⁄CPQ = 90o]

The coordinates of P are [(-a2m)/√(a2 m2 + b2), b2/√(a2 m2 + b2)]

     i.e. [(-7/√3)/√(16/3),3/√(16/3)]

i.e.[(-7)/4,(3√3)/4]

     and coordinate of O are (0, 0)

     So, OP = √(((-7)/4)2+ ((3√3)/4)2) = √(19/4)

     As OQ = r = 2

     .·. PQ = √(OP2-OQ2 ) = √(19/4-4) = √3/2

 

Example 7:

If q be the angle between CP and normal at point P, on the ellipse  a2 x2 + b2 y2 = 1, then find out tan θ and prove that its greatest value is (b2-a2)/2ab. C is centre of ellipse and P is any point on ellipse.

Solution:

The equation of the ellipse be

x2/(1/a2)+y2/(1/b2) = 1             ..... (1)

If θ be the angle between the normal at P = (1/a cos Φ,1/b sin Φ and PC where C is the centre of the ellipse given by (1) equation to the normal PG is

    x/a sec Φ - y/b cosec Φ = 1/a2 -1/b2

    => bx sec Φ - ay cosec Φ = (b2 - a2)/ab

Its slope = (b sec Φ)/(a cosec Φ) = b/a tan Φ = m1 (say)

The slope of PC = (1/b sin Φ)/(1/a cos Φ) = a/b tan Φ = m2 (say)

tan θ = (m1-m2)/(1+m1 m2) = (a/b tan Φ -a/b tan Φ)/(1+(b/a) tan Φ (a/b) tan Φ) = ((b2-a2 )tan Φ)/ab(1-tan2 θ)

       (b2-a2)/2ab.(2tanΦ)/(1-tan2 Φ)

tan θ = (b2 - a2)/2ab sin 2Φ

The value of tan θ will be maximum when sin 2Φ is maximum, sin 2Φ is maximum i.e. sin 2Φ is 1. Therefore the greatest value of tan θ is (b2-a2)/2ab.

 

Example 9:

Find the locus of the point of intersection of the two straight lines (x tan α)/a - y/b + tan x = 0 and x/a+(y tan α)/b where a is fixed angle. Also find the eccentric angle of the point of intersection.

Solution:

Equation of the lines are given as

   (x tan α)/a-y/b + tan α = 0                ...... (1)

   x/a+(y tan α)/b - 1 = 0                      ...... (2)

To find the locus of the point of intersection, we have to eliminate the variable 'tan a' from (1) and (2), so by (2),

   y/b=1/(tan α) (1-x/a) and by (1)

   y/b = tan α (1+x/a)

   Multiplying we get

   y/b)2-(1-x/a)(1+x/a)

   => x2/a2 +y2/b2 = 1

This is the equation of an ellipse

Again solving (1) and (2), we get

    x = a(1-tan2 α)/((1+tan2 α))

    x = a(1-tan2 α)/(sec2 α)

Let the abscissa of the point of intersection be a cos f, then

    x = a cos φ = a(1-tan2α)/(sec2 α)

    => cos φ = (1-tan2 α)/(sec2 α)

    => (1-cos φ )/(1+cos φ)=(sec2α -(1-tan2α))/(sec2α +(1-tan2α ) )

(By components & dividendo)

    = (2 tan2 α)/2 = tan2α

    => (2 sin2φ/2)/(2 cos2 φ) = tan2 α

    => tan2 φ/2 = tan2 α

    => tan φ/2 = tan α

Hence φ = 2α

Example 10:

If TP and TQ are perpendiculars upon the axes from any point T on the ellipsex2/a2+ y2/b2 = 4. Prove that PQ is always normal to fixed concentric ellipse.

Solution:

     The ellipse is given by x2/(4a2 )+ y2/(4b2 ) = 1         ...... (1)

If the co-ordinates of T on the ellipse be (2a cos f, 2b sin f) an TP and TQ perpendiculars on x-axis and y-axis respectively, the co-ordinates of P and Q will be (2a cos f, 0) and (0, 2b sin f) respectively.

Now equation to PQ is

     y - 0 = (2b sin φ -0)/(0-2a cos φ)(x - 2a cos φ)

     => x/(2a cos φ)+y/(2b sin φ) = 1

     => x/a sec φ + y/b cosec φ = 2                              ...... (2)

Now equation to the normal at point (A cos φ, B sin φ) with respect to any other concentric ellipse

     x2/A2 +y2/B2 = 1 is

     Ax sec φ - By cosec φ = A2 - B2                            ...... (3)

As (2) and (3) are similar, on comparing then we have,

     A/(1⁄2a)+B/(1⁄2b) = A2 - B2                                    ...... (4)

Solving the two equations given by (4), we get

     B = (2a2b)/(a2-b2) and A = (-2 ab2)/(a2-b2 )

So the line (2) i.e. x/2a sec φ + y/2 cosec φ = 1 is a normal to the fixed ellipsex2/A2 +y2/B2.

Where A = (-ab2)/(a2-b2)and B =(a2 b)/(a2-b2).

 

Example 11:

If the straight line y = 2x + 2 meet the ellipse 2x2 + 3y2 - 6, prove that equation to the circle, described on the line joining the points of intersection as diameters, is 7x2+ 7y2 + 12x - 4y - 5 = 0.

Solution:

        The line is given as

                y = 2x + 2                                                 ...... (1)

        and the ellipse is given as

                x2/a2 +y2/b2 = 1                                        ...... (2)

        Solving (1) and (2), we gets

                x2/3+(2x+2)2/2 = 1

                => x2/3 + 2(x2 + 2x + 1) = 1

                => 7x2 + 12x + 6 = 3

                => 7x2 + 12x + 6 = 3

        Let x1 and x2 be the two roots of this equation.

                x1 + x2 = -12/7                                            ...... (3)

        and   x1x2 = 3/7                                                   ...... (4)

Let y1 and y2 be the corresponding ordinates for the abscissa x1 and x2; so the co-ordinates of the points of intersection will be (x1, y1) and (x2, y2). As these lie on line

        y = 2(x + 1)

We have y1 = 2(x1 + 1) and y2 = 2(x2 + 1)

Where y1 + y2 = 2(x1 + x2) + 4                                     ...... (5)

        y1y2 = 4(x1 + 1) (x2 + 1)                                     ...... (6)

 

The equation to the circle drawn with the line joining (x1, y1) and (x2, y2) as diameter is

     (x - x1) (x - x2) + (y - y1) (y - y2) = 0

     => x2 + y2 - x(x1 + x2) - y(y1 + y2) + x1x2 + y1y2 = 0

     => x2 + y2 - x(x1 + x2) y[2(x1 + x2) + 4] + x1x2 +......+ 4[x1x2 + (x1 + x2) + 1] = 0

     (By (5) and (6))

     Putting the values from (3) and (4)

             x2 + y2 - x (-12/7) - y[2(-12/7)+4] + 3/7 + 4[3/7+(-12/7)+1] = 0

             = 7x2 + 7y2 + 12x - 4y - 5 = 0. Hence proved.

 

Example 12:

If the product of the perpendiculars from the foci upon the polar of P be constant and equal to c2. Find the locus of P.

Solution:

Suppose the equation to the ellipse x2/a2 + y2/b2 = 1. The co-ordinate of foci are (ae, 0) and (-ae, 0).

Let the co-ordinates of P be (h, k). Then polar of P is

        xh/a2 +yk/b2 = 1

Or b2xh + a2yk - a2b2 = 0                                             ...... (1)

If P1 and P2 be the lengths of the perpendiculars on the line (1) from (ae, 0) respectively are

        P1 = (b2 hae-a2 b2)/√(b2 h2 + a4 k2)

And   P2 = (-a2 b2-b2 hea)/√(b2 h2+a4 x2)

        .·. P1P2 = (-b2 h2 a2 e2+a4 b4)/(b4 h2+a4) = c2 (By hypothesis)

        => a4b4 - b4h4a2e2 = c2b4h2 - c2a4k2

        => b4h2 (c2 + a2e2) + c2a4k2 = a4b4

Generalizing the locus of the point P(h, x) is

        b4 x2 (c2 + a2 e2) + c2 a4 y2 = a4 b4

Example 13:

    Chords of ellipse x2/a2 +y2/b2 = 1 always touch another concentric ellipse  x2/α2+ y2/β2 = 1, show that the locus of their poles is (α2 x2)/a2 + (β2 y2)/b2 = 1.

Solution:

Let (x1, y1) be the pole of a chord of the ellipse

         x2/a2 +y2/b2 = 1                                                ...... (1)

Then the equation of this cord is the same as the polar of (x1, y1) with respect to (1)

i.e.   xx1/a2 + yy1/b2 = 1                                                        ...... (2)

If (2) touches the ellipse x2/α2 + y2/β2 = 1, then

      (b2/y1)2 = α2{-{b2x1/a2y1}}2 + β2

      => b4/y21 = (α2b4x2

1/a4y21) + β2

      => (α2 x12)/a4 +(β2 y1

2)/b4 = 1

.·. The locus of (x1, y1) is (α2 x2)/a2 +(β2 y2)/b4 = 1

Hence proved.

 

Example 14:

If the straight line y = x tan θ + √((a2 tan2θ +b2)/2), θ being the angle of inclination, intersects the ellipse x2/a2 +y2/b2 = 1. Then prove that the straight lines joining the centre to their point of intersection are conjugate diameters.

Solution:

      The equation of the ellipse be

            x2/a2 +y2/b2 = 1                                            ...... (1)

      and equation to the line is given as

              y = x tan θ + √((a2 tan2θ +b2)/2), θ being angle of inclination.

      We can write this equation as,

              y = mx +  √((a2 m2 +b2)/2)

      => ((y-mx)/√2)/√(a2 m2 + b2 ) = 1                          ...... (2)

To get the equation to the lines joining the point of intersection to the origin, making (1) homogeneous with the help of (2), we have

    x2/a2 + y2/b2 = [((y-mx)/√2)/√(a2 m2 + b2 )]2

              = 2((y2 + m2 x2 - 2mxy))/(a2 m2 + b2)

      => (b2 x2 + a2 y2) (a2m2 + b2) = 2a2 b2(y2 + m2x2 - 2mxy)

            y2 a2(a2m2 - b2) + 4m2 - b2xy - b2x2 (a2m2 - b2) = 0

      => y2 + (4 mb2)/((a2 m2+b2)) xy -b2/a2 x2 = 0         ...... (3)

This equation represents two straight lines y = m1x and y = m2x then the combined equation will be y2 - (m1 + m2)xy + m1m2x2 = 0.

Comparing (3) and (4); we get

                m1m2 = -b2/a2

which is the condition of diameter to be conjugate. Hence the lines are the conjugate diameters.

Example 15:

The eccentric angles of two points P and Q on the ellipse Φ1, Φ2. Find the area of the parallelogram formed by the tangents at the ends of diameters through P and Q.

Solution:

        The ellipse is

                x2/a2 +y2/b2 = 1

        Equation to the tangent at the points P and Q are

                x/a cos Φ1 + y/b sin Φ1 = 1                                           ...... (1)

        and

                x/a cos Φ2 + y/b sin Φ2 = 1                                           ...... (2)

Solving (1) and (2), we will have the coordinates of the point of intersection. Multiplying (1) by sin Φ2 and (2) by sin Φ1 and subtracting, we get

        x/a (sin Φ2 cos Φ1 - cos Φ2 sin Φ1) = sin Φ2 - sin Φ1

=> x/a sin (Φ2 - Φ1) = 2 sin (Φ2 - Φ1)/2cos (Φ2 + Φ1)/2

.·. x = a (cos((Φ1 - Φ2)/2))/(cos(Φ1 - Φ2)/2); y = b (sin(Φ1 + Φ2)/2)/(cos (Φ1 - Φ2)/2)

Above are co-ordinates of the point of intersection L of tangents at p and Q, i.e. at Φ1 and Φ2. Putting Φ1 = p + Φ1 in above we get the co-ordinates of the point of intersection M of tangent at Q and P' as

        

Area of the parallelogram LMNO = 4ΔCLM

       = 4 . 1/2 (x1 y1 - x2 y2)

       = 2 ab/(sin (Φ1 - Φ2)/2 cos (Φ1 - Φ2)/2) [ - cos2 (Φ1 + Φ2)/2-sin (Φ1 + Φ2)/2]

       = (-4ab)/(sin(Φ1 - Φ2))

       = - 4ab cosec (Φ1 - Φ2) Area can't be (-) ve

So the area = 4ab cosec (Φ1 - Φ2)

HyperbolaWe have studied earlier about parabolas and ellipses as two conic sections. In this chapter we will study another conic section called hyperbola which can be obtained by cutting a

right circular come at both the nappe by a plane. Thus it has two branches, one on each nappe.

Hyperbola is the locus of a point which moves in a plane such that it distance from a fixed point is e(>1) times its distance from a fixed straight line. It is symmetrical about two axes and one branch is the reflection of other about one of the axes. Since, we have studied ellipse in the previous chapter, it will be easier to understand this chapter. We will give you a proportion by which you can easily get the formulae for hyperbola if you know the formulae for ellipse.

Most of the properties of hyperbola are similar to those of the ellipse. We will introduce the concept of asymptotes. You will also learn about rectangular hyperbolas and conjugate hyperbola.

The rectangular hyperbola can be very simply represented in a parametric form. It is advisable that this fact should always be kept in the fore while solving problems on and related to rectangular hyperbola.

Wishing you “All the Best” for the preparation Hyperbola with askIITians.com.

Topics Covered: 

Basic Concepts of Hyperbola

Relation between Focal Distances

Parametric Coordinates

Important Properties of Hyperbola

Examples Based on Hyperbola

Ellipse Vs Hyperbola

Propositions of a Hyperbola

Examples on Propositions of a Hyperbola

Rectangular hyperbola

Intersection of a Circle and a Rectangular Hyperbola

Conjugate hyperbola

Examples on finding locus of point

Solved Examples on Hyperbola Part-I

Solved Examples of Hyperbola (Part-II)

Solved Examples of Hyperbola Part-III

Basic ConceptsAs we have studied earlier that by slicing a cone with a plane in different orientations, we obtain conic sections. The hyperbola is obtained by cutting a right circular cone at both the nappes by a plane. The hyperbola is to be thought of a single curve consisting of two branches, one on each nappe. Definitions A hyperbola is the locus of a point which moves in a plane such that its distance from a fixed point (called the focus) is e ( >1) times its distance from a fixed straight line (called the directrix). The hyperbola is a conic section for which the eccentricity (e) is greater than unity. Standard Equation Let S be the focus and ZM the directrix of a hyperbola. 

 Since e > 1, we can divide SZ internally and externally in the ratio e : 1; let the points of division be A and A’ as in the figure. Let AA’ = 2a and be bisected at C. Then, SA = e.AZ, SA’ = e.ZA’ ⇒ SA + SA’ = e(AZ + ZA’) = 2ae i.e., 2SC = 2ae or SC = ae. Similarly by subtraction, SA’ – SA          = e(ZA’ – ZA) = 2e.ZC ⇒ 2a = 2eSC ⇒ SC = a/e. Now, take C as the origin, CS as the x-axis, and the perpendicular line CY as the y-axis. Then, S is the point (ae, 0) and ZM the line x = a/e. Let P(x, y) be any point on the hyperbola. Then the condition PS2 = e2.(distance of P from ZM)2 gives (x – ae)2 + y2 = e2 (x – a/e)2 or x2(1 – e)2+ y2 = a2(1 – e2) i.e. x2/a2 – y2/a2(e2–1) = 1.                                      …… (i)

 Since e > 1, e2 – 1 is positive. Let a2 (e2 – 1) = b2. Then the equation (i) becomes x2/a2 – y2/a2 = 1. The eccentricity e of the hyperbola x2/a2 – y2/a2 = 1 is given by the relation e2 = (1 + b2/a2). Since the curve is symmetrical about the y-axis, it is clear that there exists another focus S’ at (–ae, 0) and a corresponding directrix Z’M’ with the equation x = –a/e, such that the same hyperbola is described if a point moves so that its distance from S’ is e times its distance from Z’M’. • The points A and A’ where the straight line joining the two foci cuts the hyperbola are called the vertices of the hyperbola. • The straight line joining the vertices is called the transverse axis of the hyperbola, its length AA’ is 2a. • The middle point C of AA’ possesses the property that it bisects every chord of the hyperbola passing through it. It can be proved by taking P(x1, y1) as any point on the hyperbola. If (x1, y1) lies on the hyperbola then so does P(–x1, –y1) because the hyperbola is symmetrical about the x and the y axes. Therefore PP’ is a chord whose middle point (0, 0), i.e. the origin C. On account of this property the middle point of the straight line joining the vertices of the hyperbola is called the centre of the hyperbola. • The straight line through the centre of a hyperbola which is perpendicular to the transverse axis does not meet the hyperbola in real points. If B and B’ be the points on this line such that BC=CB’=b, the line BB’ is called the conjugate axis. • A latus rectum is the chord through a focus at right angle to the transverse axis. • The length of the semi-latus rectum can be obtained by putting x = ae in the equation of the hyperbola. Thus y = b √a2e2/a2–1 = b√e2–1 = b.b/a = b2/a.

Foci and Directrices:• Since the curve is symmetrical about x-axis, therefore there exists another focus at point (–ae, 0) of the hyperbola. (Similar to ellipse).

• Corresponding to these foci, there are two directrices whose equations are x = a/e and x = –a/e.

Relation between Focal Distances

 The difference of the focal distances of a point on the hyperbola is constant. PM and PM’ are perpendiculars to the directrices MZ and M’Z’ and PS’ – PS = e(PM’ – PM) = eMM’ = e(2a/e) = 2a = constant. 

Another Definition of Hyperbola A hyperbola can be defined in another way; Locus of a moving point such that the difference of its distances from two fixed points is constant, would be a hyperbola. Transverse and Conjugate axes: The points A(a, 0) & A’(–a, 0) are called the vertices of the hyperbola and the line AA’ joining the vertices is called the transverse axis and the line perpendicular to it, through the centre (0, 0) of the hyperbola is called conjugate axis.   Centre: Any chord of the hyperbola through C will be bisected at C (by symmetry), therefore C is called the centre of hyperbola. Thus hyperbola is a central curve.

Latus rectum: The chord of a hyperbola through one of the foci and at right angle to the transverse axis is called the latus rectum: ∴ If 2l be the length of the latus rectum, then the co-ordinates of one of its extremities is (ae, l). ∴ The point (ae, l) lies on the hyperbola x2/a2 – y2/b2, so we have         e2 (l2/b2) = 1 ⇒ l2 = b2(e2–1) = b2a2(e2–1)/a2 = b4/a2 (? b2 = a2(e2 – 1) ⇒ l = b2/a2

 ∴ The length of the latus rectum = 2 b2/a2

  Relative Position of a Point with respect to the Hyperbola 

The quantity x12/a2 – y1

2/b2 = 1 is positive, zero or negative, according as the point (x1, y1) lies within, upon or without the curve.

 

 Parametric Coordinates We can express the coordinate of a point of the hyperbola x2/a2 – y2/b2 = 1 in terms of a single parameter, say θ. 

In the adjacent figure OM = a sec θ and PM = b tan θ. Thus any point on the curve, in parametric form is x = a secq, y = b tanθ. 

In other words, (a sec θ, b tan θ) is a point on the hyperbola for all values of θ. The point (a secθ, b tanθ) is briefly called the point ‘θ’.

Important Properties of Hyperbola

 Since the fundamental equation of the hyperbola only differs from that of the ellipse in having –b2 instead of b2, it will be found that many propositions for the hyperbola are derived from those for the ellipse by changing sign of b2. Some results for the hyperbola x2/a2–y2/b2 = 1 are

 • The tangent at any point (x1, y1) on the curve is xx1/a2 – yy1/b2 = 1. • The tangent at point ‘θ’ is x secθ/a – y tanθ/b = 1. • The straight line y = mx + c is a tangent to the curve, if c2 = a2 m2 – b2. In other words, y = mx + √a2m2–b2 touches the curve for all those values of m when m > b/a or m < –b/a. • Equation of the normal at any point (x1, y1) to the curve is x–x1/x1/a2 = y–y1/y1(–b)2. • The equation of the chord through the

points θ1 and θ2 is  • The equation of the normal at θ is ax cosθ + by cotθ = a2 + b2. • Through a given point, four normals (real or imaginary) can be drawn to a hyperbola. • The tangent drawn at any point bisects the angle between the lines, joining the point to the foci, whereas the normal bisects the supplementary angle between the lines. • Equation of the director circle is x2 + y2 = a2 – b2. That means if a2 > b2, there would exist several points such that tangents drawn from them would be mutually perpendicular. If a2 < b2, no such point exist. For a2 = b2, centre is the only point from which two perpendicular tangents (asymptotes) to the hyperbola can be drawn. • From any point (x1, y1) in general two tangents can be drawn to hyperbola. The equation of the pair of tangents is         (x1/a2 – y2/b2 – 1) (x1

2/a2 – y12/b2 – 1) (xx1/a2 – yy1/b2 – 1)2 or SS1 = T2

   The equation of the chord of contact is xx1/a2 – yy1/b2 – 1 = 0 or T = 0. • The equation of the chord bisected at the point (x1, y1) is         xx1/a2 – yy1/b2 – 1 = x1

2/a2 – y12/b2 – 1 or T = S1.

 • Equation of the chord the points (a sec θ, b tan θ) and (a sec ?, b tan ?) is          x/a cos θ–?/2 – y/b sin θ+?/2 = cos θ+?/2.

Examples Based on Hyperbola

 Illustration:

Find the equation of the hyperbola whose directrix is 2x + y = 1, focus is (1, 1) and eccentricity is √3.

Solution:

Let S(1, 1) be focus and P(x, y) be any point on the hyperbola. From P draw PM perpendicular to the directrix then PM =2x+y–1/√22+12 = 2x+y–1/√5

Also from the definition of the hyperbola, we have

        SP/PM = e ⇒ SP = ePM

⇒ √(x–1)2+(y–1)2 = √3 (2x+y–1/√5)

⇒ (x – 1)2 + (y – 1)2 = 3 (2x+y–1)2/5

⇒ 5[(x2+1–2x)+(y2 + 1 – 2y)]=3(4x2 + y2 + 1 + 4xy – 4x – 2y)

⇒ 7x2 – 2y2 + 12xy – 2x – 4y – 7 = 0

 

 

 Illustration:

Find the directrix, foci and eccentricity of the hyperbola

       ax2 – y2 = 1

Solution:

The given hyperbola is

        ax2 – y2 = 1

or     x2/(1/a) – y2/1 = 1                                                         …… (1)

which is of the form(x2/a2) – (y2/b2) = 1

Here a2 = 1/a, b2 = 1

If e be the eccentricity of the hyperbola, then

         b2 = a2(e2 – 1)

⇒ 1 =  (e2 – 1)

⇒ a = (e2 – 1)

or e2 = a + 1 or e =√a+1

Also the foci are given by (+ ae, 0)

∴ The required foci are

    (+ 1/√a √(a+1), 0)

or (+ √a+1/a, 0)

And the directricies are given by x = + (a/e)

⇒ x = + [1/√a/√a+1]                    (? a = 1/√a, e = 1/√a+1)

   x = + 1/√a(a+1)

 Illustration:

Find the locus of a point, the difference of whose distances from two fixed points is constant.

Solution:

Let two fixed points be S (ae, 0) and S’ (–ae, 0). Let P(x, y) be a moving point such that

SP – S’P = Constant = 2a (say).

Then √[(x–ae)2 + (y–0)2] – √[x+ae]2+(y–0)2 = 2a

⇒ √[(x–ae)2+y2] = + 2a + √[(x+ae)2+y2

⇒ (x – ae)2 + y2 = 4a2 + (x + a2)2 + y2 + 4a √[(x–ae)2+y2]

⇒ (x – ae)2 – (x + ae)2 – 4a2 = + 4a √[(x–ae)2+y2]

⇒ –4aex – 4a2 = + 4a √[(x–ae)2+y2]

⇒ (ex + a)2 = (x + ae)2 + y2

⇒ (e2 – 1)x2 – y2 = a2(e2 – 1)

⇒ x2/a2–y2/b2 = 1 taking b2 = a2(e2 – 1)

This is a hyperbola.

Illustration:

If A, B, C are three points on the hyperbola xy = c2 and AC is perpendicular to BC, prove that AB is parallel to the normal to the curve at C.

Solution:

Let the three points A, B, C respectively be (ct1, c/t1), (ct2, c/t2) and (ct3, c/t3). Since AC is perpendicular to BC,

        (c/t3–c/t1/ct3–ct1) = – 1 ⇒ t1t2 = –1                                 …… (1)

Normal to the curve at C (ct3, c/t3) is

        y = xt32 + 2/t3 (1–t3

4) and its slope is t32 = –1/t1t2              …… (2)

Slope of AB = (c/t2–c/t1/ct2–ct1) = – 1/t1t2 = t32

⇒ AB is parallel to the normal at C.

 Illustration:

Find the equation of the hyperbola the distance between whose foci is 16, whose eccentricity is √2 and whose axis is along the x-axis centre being the origin.

 Solution:

We have b2 = a2(e2 – 1) = a2 ⇒ b = a.

Also 2ae = 16 ⇒ ae = 8 ⇒ a = 4√2.

Hence the equation of the required hyperbola is

    x2/32 – y2/32 = 1 ⇒ x2 – y2 = 32.

 Illustration:

The perpendiculars drawn from the centre of a hyperbola x2/a2 – y2/b2 = 1 upon the tangent and normal at any point of the hyperbola meet them in Q and R. Find the locus of Q and R.

 

 

 

 

 Solution:

Tangent at any point P(a sec θ, b tan θ) is  sec θ – y/b  tan θ = 1. … (1)

Equation of the line through centre (origin) perpendicular to (1) is y = –a sin θ/b x

⇒ sin θ = –by/ax

Eliminating θ from (1), we get x/a cos θ – y/b cos θ (–by/ax) = 1.

⇒ x2 + y2 = ax cos θ ⇒ (x2 + y2)2 = a2x2(1 – b2y2/a2x2)

Or (x2 + y2)2= a2x2 – b2y2, which is the locus of Q.

Normal at the point P (a sec θ, b tan θ) is ax cos θ + by 

cot θ = a2 + b2                                                         … (2)

Equation of the line perpendicular to (2) drawn from the centre is 

      y = bx/a sin θ                                                     … (3)

Form (2) and (3),

sin θ = bx/ay and ax √1–b2y2/a2x2 + by √1–b2x2/a2y2 . ay/bx = a2 + b2

⇒ (x2 + y2)2 (a2y2 – b2x2) = (a2 + b2)2 x2y2, which is the locus of R.

Ellipse Vs Hyperbola

Enquiry:    How do we get formulae for hyperbola if we know the formulae for ellipse?

 Most of the results obtained in the case of the ellipse x2/a2 – y2/b2 = 1 hold good for the hyperbola x2/a2 – y2/b2 = 1, when only the sign of b2 is changed. The proofs of these results can be derived exactly in the same manner as they were derived for ellipse. So let us see some final results.

1. Tangent at (x1, y1) to the hyperbola x2/a2 – y2/b2 = 1 is xx1/a2 – yy1/b2= 1 i.e. T = 0. 2. Equation of tangent in terms of ‘m’ is y = mx + √(a2m2–b2). 3. Equation of the normal at (x1, y1) to the hyperbola is x–x1/(x1/a2)–y–y1/(y1/b2). 4. Equation of pair of tangents drawn from point (x1, y1) to the hyperbolax2/a2 – y2/b2 = 1 is given by SS1 = T2

     Where S = x2/a2 – y2/b2 = 1     S1 = x1

2/a2 – y12/b2 = 1

     T = xx1/a2 – yy1/b2 = 1 5. The Chord of Contact of tangents from (x1, y1) to the hyperbola x2/a2 – y2/b2 = 1 is given by T = 0 i.e. xx1/a2 – yy1/b2 = 1. 6. The Polar of Pole (x1, y1) to the hyperbola x2/a2 – y2/b2 = 1 is given by T = 0 i.e. xx1/a2 – yy1/b2 = 1. 

7. The equation of Chord of hyperbola x2/a2 – y2/b2 = 1 whose middle point is (x1, y1) is given by T = S1 i.e. x1

2/a2 – y12/b2 = xx1/a2 – yy1/b2 = 1.

   Pause: Try to get above results yourself using traditional methods similar to the ellipse.  Enquiry: Can we represent a hyperbola mathematically in form of one parameter and is there any geometrical significance of that parameter like eccentric angle in case of ellipse? Yes. Before that let us understand the concept of the auxiliary circle of a hyperbola. The circle described on the transverse axis of hyperbola as its diameter is called its auxiliary circle. 

 We know that the line AA’ joining the vertices A(a, 0) and A’(–a, 0) of the hyperbola x2/a2 – y2/b2 = 1 is called the transverse axis. ∴ The equation of the auxiliary circle, described on AA’ as diameter, is        (x – a) [x – (–a)] + (y – 0)] + (y – 0)(y – 0) = 0 or     x2 + y2 = a2

 Now let us draw the foot N of any ordinate NP of the hyperbola draw a tangent NU to this circle, and join CU. The         CU = CN cos NCU i.e.    x = CN = a sec NCU The angle NCU is therefore the angle ?. Also NU = CU tan ? = a tan ? 

So that NP : NU = b : a So the ordinate of the hyperbola is therefore in a constant ratio to the length of the tangent drawn from its foot to the auxiliary circle. When it is desirable to express the co-ordiantes of any point of the curve in terms of one parameter than we use x = a sec ?, and y = b tan ?  Note:

This angle ? is not so important an angle for the hyperbolas the eccentric angle is for the ellipse.

Propositions of a Hyperbola:

Diameter of a hyperbola The locus of the middle point of a system of parallel chords of a hyperbola is called its diameter. The equation of the diameter is y = b2x/(a2m), where m is the slope of the system of parallel chords.  Note:  Conjugate Diameters: Two diameters of a hyperbola which bisect chords parallel to each other are called conjugate diameters. ∴ The diameters y = mx and y = m1x of the hyperbola x2/a2 – y2/b2 = 1 are conjugate if mm1 = b2/a2. Director circle of a hyperbola The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola. Equation of the director circle of the hyperbola x2/a2 – y2/b2 = 1 is x2 + y2 = a2 – b2 i.e. a circle whose centre is origin and radius is √(a2–b2).

  

 

 

Note:If b2 < a2, this circle is real. If b2 = a2, the radius of the circle is zero, and it reduces to a point circle at the origin. In this case the centre is only point from where tangents at right angle can be drawn to the hyperbola. If b2 > a2, the radius of the circle is imaginary, so that there is no such circle, and so no tangents at right angles can be drawn to the circles. Asymptote A line, which is tangent to the hyperbola at infinity, but which is not itself at infinity, is called the asymptote of the curve. To find the equation of the asymptotes of the hyperbola x2/a2 – y2/b2 = 1. Let y = mx + c be an asymptotic to the given hyperbola. Then eliminating y, the abscissas of the points of intersection of y = mx + c or x2(b2 – a2m2) – 2a2mcx – a2(b2 + c2) = 0                 …… (1) If the line y = mx + c is an asymptote of the hyperbola then it touches the hyperbola at infinity i.e. both the roots of the equation (1) are infinite and for this we must have b2 – a2m2 = 0 and 2a2mc = 0. Hence we get m = + (b/a) and c = 0. ∴ The asymptotes are y = + (b/a) x Or x2/a2 + y2/b2 = 0  ∴ Their combined equation is (x/a+y/b)(x/a–y/b) = 0 or xx2/a2 – y2/b2 = 0which shows that the equation of the asymptote differs from that of the hyperbola in the constant term only. Also the angle between the asymptotes is 2 tan–1 (b/a). The lines x2/a2+y2/b2=0 are also asymptotes to the conjugate hyperbolax2/a2–y2/b2=1. 

 

 

 

Remarks: • The equation of the hyperbola and that of its pair of asymptotes differ by a constant. For example, if S = 0 is the equation of the hyperbola, then the combined equation of the asymptotes is given by S + K = 0. The constant K is obtained from the condition that the equation S + K = 0 represents a pair of lines. Finally the equation of the corresponding conjugate hyperbola is S + 2K = 0. 

 

• Any line drawn parallel to the asymptote of the hyperbola would meet the curve only at one point.

Illustration:

Find the hyperbola whose asymptotes are 2x – y = 3 and 3x + y – 7 = 0 and which passes through the point (1, 1). 

 

 Solution: The equation of the hyperbola differs from the equation of the asymptotes by a constant. ⇒ The equation of the hyperbola with asymptotes 3x + y – 7 = 0 and 2x – y = 3 is (3x + y – 7) (2x – y – 3) + k = 0. It passes through (1, 1) ⇒ k = –6. Hence the equation of the hyperbola is (2x – y – 3)(3x + y – 7) = 6.  

 

Illustration: Find the angle between the asymptotes of the hyperbola x2/a2–y2/b2 = 1, then length of whose latus rectum is 4/3 and which passes through the point (4, 2).  

 

Solution: We have 2b2/a = length of the latus rectum = 4/3 ⇒ 3b2 = 2a Also, the hyperbola passes through the point (4, 2). Hence 16/a2 – 4/b2 = 1 ⇒ 16/a2 – 6/a = 1 Or a2 + 6a – 16 = 0 ⇒ (a – 2)(a + 8) = 0 ⇒ a = 2 ⇒ b2 = 4/3. The asymptotes of the given hyperbola are y = + b/a x or y + 1/√3 x. If θ1 and θ2 are the angles which the asymptotes make with the positive x-axis, then         tan θ1 =  ⇒ θ1 = π/6 and tan θ2 = –1/√3 ⇒ θ2 = –π/6. Hence the angle between the asymptotes = π/3.  

 

Illustration: Prove that the chords of the hyperbola x2/a2–y2/b2 = 1, which touch its conjugate hyperbola are bisected at the point of contact.  

 

Solution: Let P(x1, y1) be the mid-point of the chord of the given hyperbola, so that the equation of the chords is xx1/a2–yy1/b2 = x1

2/a2–y12/b2.                …… (1)

 

If touches the conjugate hyperbola x2/a2–y2/b2 = 1, then x2/a2–1/b2 [xx1/a2–x1

2/a2+y12/b2]2.b4/y1

2 + 1 = 0 will have equal roots. Simplifying, we find that x2/a2 [y1

2/b2–x12/a2]+2xx1/a2[x1

2/a2–y12/b2]–[y1

2/b2–x12/a2]2+y1

2/b2= 0 has equal roots so that 4x1

2/a2 [x12/b2–y1

2/a2]+4[y12/b2–x1

2/a2][[y12/b2–x1

2/a2]2–y12/b2]= 0

 or, x1

2/a2 [x12/a2–y1

2/b2]–[x12/a2–y1

2/b2]2 + y12/b2 = 0 or (x1

2/a2–y12/b2)(x1

2/a2–x1

2/a2+y12/b2)+y1

2/b2= 0 or x1

2/a2 – y12/b2 + 1= 0 ⇒ (x1, y1) lies on the conjugate hyperbola.

 Hence the chord (1) touches the conjugate hyperbola at its midpoint (x1, y1).  

 Alternative solution:

 Any tangent to the conjugate hyperbola x2/a2–y2/b2 = –1 is         x = my + √b2m2–a2.                                           …… (2) If this is same as the chord (1), then m = a2y1/b2x1 and hence         a4/x1

2 [x12/a2 – y1

2/b2]2 = b2m2 – a2 = b2a4y12/b4x1

2 – a2

 Or [x1

2/a2 – y12/b2]2 = y1

2/b2 – x12/a2 or x1

2/a2 – y12/b2 = –1

 ⇒ (x1, y1) lies on the conjugate hyperbola. 

⇒ the chord (1) touches conjugate hyperbola and is bisected at the point of contact.

Rectangular hyperbola

 If the asymptotes of a hyperbola are at right angles to each other, it is called a rectangular hyperbola.

A hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The angle between asymptotes of the hyperbola x2/a2 – y2/b2 = 1, is 2 tan–1 (b/a). This is a right angle if tan–1 b/a = π/4, i.e., if b/a = 1 ⇒ b = a. • The equation of the rectangular hyperbola referred to its transverse and conjugate axes as axes of coordinates is therefore:              x2 – y2 = a2.                                               …… (1) • Equation referred to asymptotes as coordinate axes: To obtain this equation we rotate the axes of reference through –45o. Thus we have to write              x/√2 + y/√2 for x and –x/√2 + y√2 for y.     The equation (i) becomes      (1/2) (x + y)2 – (1/2)(x – y)2 = a2 i.e. xy = ½ a2 or xy = c2 where c2 = a2/2.  Equation:         We know that the asymptote of hyperbola                  x2/a2 – y2/b2 = 1                                          …… (1)                 are given by y = + (b/a) x                                …… (2)         If θ be the angle between the asymptotes, then                 θ = tan–1 (m1–m2/1+m1m2) = tan–1 ((b/a)–(–b/a)/1+(b/a–b/a))                 tan–1 (2(b/a)/1–(b2/a2)) = 2 tan–1 (b/a)         But if the hyperbola is rectangular, then θ =π/2                 i.e., π/2 = 2 tan–1 (b/a)                 or tan (π/2) = b/a ⇒ b = a         ∴ From (1) the equation of the rectangular hyperbola is x2 – y2 = a2

  It should be noted that: 

(i)     In a hyperbola b2 = a2 (e2 – 1). In the case of rectangular hyperbola (i.e., when b = a) result become a2 = a2(e2 – 1) or e2 = 2 or e =√2         i.e. the eccentricity of a rectangular hyperbola =√2 (ii)    In case of rectangular hyperbola a = b i.e., the length of transverse axis = length of conjugate axis. 

        Hence it is also called an equilateral hyperbola.

Intersection of a Circle and a Rectangular Hyperbola

A rectangular hyperbola and a circle meet in four points. The mean of these four points is the middle point of the centres of the hyperbola and that of the circle. Let the rectangular hyperbola be xy = c2 and the equation of the circle be x2 + y2 + 2gcp + 2fy + k = 0. Any point on the hyperbola is (cp, c/p). If it lies on the circle, then c2p2 + c2/p2 + 2gcp + 2fc/p + k = 0. ⇒ c2p4 + 2gcp3 + kp2 + 2fcp + c2 = 0. This is fourth degree equation in p, which has four roots. Hence the circle and the hyperbola intersect in four points. If p1, p2, p3, p4 are the roots of this equation, then         p1 + p2 + p3 + p4 = –2gc/c2 = –2g/c ⇒ cp1 + cp2 + cp3 + cp4 = 2g ⇒ x1+x2+x3+x4/4 = –g/2 Also 1/p1 + 1/p2 + 1/p3 + 1/p4 = p1p2p3/p1p2p3p4 = –2fc/c2/c2/c2 = –2f/c ⇒ c/p1 + c/p2 + c/p3 + c/p4 = –2f ⇒ y1+y2+y3+y4/4 = –f/2. Hence the mean of the four points is (–g/2, –f/2) which is the mid-point of the centre of the hyperbola and that of the circle.  Illustration: 

A circle and a rectangular hyperbola meet in four points A, B, C and D. If the line AB passes through the centre of the circle, prove that the centre of the hyperbola lies at the mid-point of CD.  Solution: The line AB passes through the centre of the circle. Hence AB is the diameter of the circle and the mid-point of AB is the centre of the circle. Let the co-ordinates of A, B, C, D be respectively (x1, y1) (x2, y2), (x3, y3) and (x4, y4). Let the centres of the hyperbola and the circle be (h, k) and (g, f). Then x1+x2x3+x4/4 = h+g/2. But g = x1+x2/2 ⇒ 2g+x3/x4/4 ⇒ h+g/2 ⇒ x3+x4/2 = h Similarly y3+y4/2 = k. Hence (h, k) is the mid-point of CD. Enquiry:    As the asymptotes of a rectangular hyperbola are mutually perpendicular, can we find the equation of a hyperbola whose asymptotes are the co-ordinate axes? Let transverse and conjugate axes as axes of co-ordinates (X’OX and Y’OY in the above figure), the equation of rectangular hyperbola is X2 – Y2 = a2                                                …… (1) Also we know that the asymptotes of a rectangular hyperbola are at right angles to each other. 

 Let Ox and Oy be the asymptotes, each making as angle of π/4 with the co-ordinate axes. Rotate the axes through as angle of π/4 with the co-ordinate

axes. To find the equation of the rectangular hyperbola referred to asymptotes as axes. We have to substitute for X and Y in (1) the xcos(–π/4) – ysin(–π/4) and ycos(–π/4) + xsin(–π/4) respectively. i.e. x+y/√2 and y–x/√2 respectively. So from (1) equation of the rectangular hyperbola referred to asymptotes as axes is (x+y/√2)2 = (y–x/√2)2 = a2 ⇒ xy = a2/2 ⇒ xy = c2 where 2c2 = a2

 The shape of the above rectangular hyperbola referred to asymptotes as co-ordinate axes is as shown is the adjacent figure.  Note: 1. Parametric co-ordinates of any point on the rectangular hyperbola xy = c2 is (ct, c/t) where t is the parameter. 

2. Equations of tangent and normal at any point (ct, c/t) on the rectangular hyperbola xy = c2 are x + yt2 = 2ct and xt3 – yt – ct4 + c = 0 respectively.

Conjugate hyperbola

A hyperbola whose transverse and conjugate axes respectively are the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola. The asymptotes of these two hyperbolas are also the same. Equation of a conjugate hyperbola is x2/a2 – y2/b2 = 1         …… (1)

 of the given hyperbola x2/a2 – y2/b2 = 1 Its transverse and conjugate axes are along y and x axes respectively.  

 

 

Note: 1.Any point on conjugate hyperbola (2) is (a tan q, b sec q) 2.The equation of the conjugate hyperbola to xy = c2 is xy = –c2. 3.By comparing the equations of hyperbola x2/a2 – y2/b2 = 1.    asymptotes x2/a2 – y2/b2 = 0    and conjugate hyperbola x2/a2 – y2/b2 = 1 we find that: (a) The equation of the hyperbola and asymptotes differ by the same constant by which the equations of the asymptotes and the conjugate hyperbola differ. (b) Hyperbola + Conjugate hyperbola = 2 (Asymptotes). (check yourself). 4.The tangents drawn at the points, where a pair of conjugate diameters meets a hyperbola and its conjugates form a parallelogram, whose vertices lie on the asymptotes and whose area is constant. (prove yourself). 5.If a pair of conjugate diameters of hyperbola meet the hyperbola and its conjugate in P, P’ and D, D’ respectively, then the asymptotes bisect PD and PD’. (Prove yourself). 

 

 

Illustration: If e1 and e2 are the eccentricities of the hyperbola x2/a2 – y2/b2 = 1 and its conjugate hyperbola, prove that e1

–2 + e2–2 = 1.

 

 

 

 

Solution: The eccentricity e1 of the given hyperbola is obtained from        b2 = a2(e1

2 – 1).                                                  …… (1) The eccentricity of e2 of the conjugate hyperbola is given by        a2 = b2(e2

2 – 1).                                                  …… (2) Multiply (1) and (2), we get,        1 = (e1

2 – 1)(e22 – 1) Þ 0 = e1

2 e22 – e1

2 – e22

 

 

⇒ e1–2 + e2

–2 = 1.

Example:

 C is a centre of the hyperbola x2/a2–y2/b2 = 1 and the tangent at any point P meets asymptotes in the point Q and R. Find the equation to locus of the centre of the circle circumscribing the triangle CQR.   Solution:  This equation to the hyperbola is given as              x2/a2–y2/b2 = 1                                      …… (1)   

  Let P any point on it as (a sec ?, b tan ?), then the equation of tangent at P is             x/a – y/b sin ? = cos ?                              …… (2)     The equation to the asymptotes to (1) are             x/a = y/b                                                …… (3)     and   x/a = – y/b                                             …… (4)     Solving (2) and (3), we get the coordinates of Q as             (acos?/1–sin?, bcos?/1–sin?)     Solving (2) and (4), we get the coordinates of R as             (acos?/1+sin?, –bcos?/1+sin?) 

    Let O be the centre of the circle passing through C, Q and R having its coordinates as (h, k). Then clearly OC = OQ     ⇒ h2 + k2 = (h–acos?/1–sin?)2 + (k–bcos?/1–sin?)2     ⇒ h2 + k2 = (h–acos?/1–sin?)2 + (k–bcos?/1–sin?)2     ⇒ h2 + k2 = h2 + k2 + (a2 + b2) cos2?/(1–sin?)2 – (2ah + 2bk) cos?/1–sin?     ⇒ 2(ah + bk) = (a2 + b2) cos?/1–sin?                              …… (5)     Similarly OC = OR     Hence h2 + k2 = (h–acos?/1+sin?)2 + (k+bcos?/1+sin?)2

     Which on simplification as in the last case, given    

         2(ah –bk) = (a2 + b2) cos?/1–sin?                                   …… (6)     to get the locus of the point O we have to eliminate f from (5) and (6), so multiplying the two we get             4(a2h2 – b2k2) = (a2 + b2) cos2?/1–sin2? = (a2 + b2)2

    for (h, k), we get the required locus as           4(a2x2 – b2y2) = (a2 + b2)2

  Example:  A straight line is drawn parallel to the conjugate axis of a hyperbola meets it and the conjugate hyperbola in the points P and Q. Find the locus of point of intersection of tangents at P and Q.   Solution:     Let the equation to the hyperbola be                     x2/a2–y2/b2 = 1                                      …… (1)    

 and its conjugate hyperbola be                     y2/b2–x2/a2 = 1                                      …… (2)     Let p be any point (a sec ?, b tan ?) on P. The equation of the line parallel to the conjugate axis (1) i.e. y-axis passes through P will be x = a sec ?                                                                 …… (3)     The line (3) will cut the conjugate hyperbola (2) at Q where x = a sec ? and hence y = b √(1+sec2?), therefore the coordinates of Q will be {asec?, b√(1+sec2?)}     Now the equation to the tangent to (1) at P is             x/a–y/b sin ? = cos ?     or     x/a – cos ? = y/b sin ?                                      …… (4)     and the equation to the tangent to (2) at Q is     

        y/b √(1+sec2?) sec ? = 1     ⇒ x/a + cos ? = y/b √(1+cos2?)                                  …… (5)     on squaring and adding (4) and (5), we have              2x2/a2 + 2 cos2 ? = y2/b2 [(1 + cos2 ?) + sin2 ?] = 2 y2/b2

    ⇒ cos2 ? = y2/b2 – x2/a2

     Putting the value of cos ? in (5) we get              x/a + √(y2/b2–x2/a2) = y/b √(1+y2/b2–x2/a2)     Squaring we have,              x2/a2 + y2/b2 – x2/a2 + 2 x/a √(y2/b2 – x2/a2) = y2/b2 (1+y2/b2–x2/a2)    

 ⇒ 2x/a √(y2/b2–x2/a2) = y2/b2(y2/b2–x2/a2)     ⇒ y2/b2 √(y2/b2–x2/a2) = 2x/a     ⇒ y4/b4 (y2/b2–x2/a2) = 4x2/a2

     There is the required locus.   Example:  From a point A, perpendiculars AB and AC are drawn to two straight lines OB and OC. If the area OBAC is constant, find the locus of A.   Solution:  Let the bisectors of the angles BOC be taken as axis. So the equations of OB and OC are respectively.             x cos α + y sin α = 0     and   x cos α – y sin α = 0            where α = 1/2 ∠BOC   

  Take any point A as (h, k); then     AB = Perpendicular from A on OB             = hcosα+ksinα/√cos2α+sin2α = h cos α + k sin α      …… (1)     and similarly     AC = Perpendicular from A on C             = h cos α – k sin α                                              …… (2)     The equation to AB will be             (h – x) sin α + (y – k) cos α = 0    ⇒ y cosα – x sinα + h sin α – k cos α = 0                       …… (3)  

  Similarly the equation AC will be             (h – x) sin α – (y – k) cos α = 0    

    Now OB = Perpendicular distance of (3) from (0, 0).             = 0–0+hcosα+ksinα/√cos2α+sin2α             = h sin α – k cos α     Similarly OC = perpendicular distance of (0, 0) from (4)             = h sin α + k cos α.  

   Now the area of quad. OBAC = ΔOAB + ΔOAC             = 1/2 OB × AB + 1/2 OC × AC             = 1/2 [h sinα – kcosα][h cosα + k sinα] + 1/2 [h cosα – k sinα] [h sinα + k cosα]             = (h2 – k2) sinα.cosα = constant = S (say)     ⇒ h2 – k2 = {s/sina, cosa} which is again constant = a2 (say)    

Therefore the locus of the point (h, k) will be x2 – y2 = a2, which is hyperbola.

Solved Examples on Hyperbola Example:         Show that the line 4x – 3y = 9 touches the hyperbola 4x2 – 9y2 = 27.  Solution: We know that if the line y = mx + c touches the hyperbola x2/a2–y2/b2 = 1, then c2 = a2m2 – b2

 Here the hyperbola is         x2/(27/4)–y2/(27/9) = 1 i.e. here a2 = 27/4                               b2 = 27/9 = 3 And comparing 4x – 3y = 9 with y = mx + c, we get         ‘m’ = 4/3,              ‘c’ = –3 ∴ a2m2 – b2 = (27/4)(4/3)3 – 3 = 12 – 3 = 9 = (–3)2

  or a2m2 – b2 = c2

  Hence the given line touches the given hyperbola. Example: Prove that the mid points of chords of the hyperbola x2/a2–y2/b2 = 1 parallel to the diameter y = mx lie on the diameter a2my = b2x Solution:         The hyperbola is x2/a2–y2/b2 = 1                                            …… (1) The equation of any chord parallel to the diameter          y = mx is y = mx + c                                          …… (2) Eliminating y between (1) and (2), we get          x2/a2–(mx+c)2/b2 = 1 ⇒ x2(b2 – a2m2) – 2a2m cx – a2(b2 + c2) = 0 ⇒ b2x3 = a2my3

 ∴ The locus of (x3, y3) is          b2x = a2my                 Hence proved.  Example: 

Prove that the angle subtended b any chord of a rectangular hyperbola at the centre is the supplement of the angle between the tangents at the end of the chord. Solution: Let the equation of the hyperbola be x2 – y2 = a2 and P and Q be any two points on it such that their coordinates are respectively (a sec ?1, a tan ?1) and (a sec ?2, a tan ?2) and C be the centre of the hyperbola. Equation of the line PC is y – 0 = atan?1–0/asec?1–0 (x – 0)         ⇒ y = x sin ?1                                                  …… (1) Similarly equation to QC will be y = x sin ?2                    …… (2) If a be the angle between PC and QC, then         tan α = sin?1–sin?2/1+sin?1sin?2                          …… (3) Again the equation to the tangent at P is         x a sec ?1– y a tan ?1 = a2

         y = x/sin?1 – acos?1/sin?1                                  …… (4) Similarly the equation to the tangent at Q2 is         y = x/sin?2 – a cos?2/sin?2

 If b be the angle between the tangents at f1 and f2, then         tan ß = 1/sin?2 1/sin?2/1+1/sin?1 1/sin?2 = sin?2–sin?1/1+sin?1sin?2

                 = –(sin?1–sin?2)/1+(sin?1sin?2) ⇒ tanß = –tanα ⇒ tanß = tan (π – α)                    (By (3))

⇒ ß = π – α                                 Hence proved.

Example:

The normal to the hyperbola 16x2 – 9y2 = 144 meets the axes in M and N. MP and NP and are drawn at right angles to the axes. Find the locus of P. Solution:         The equation to the hyperbola is                 x2/a2 – y2/b2 = 1                                     …… (1) Let L be any point on it having the coordinates (3 sec ?, 4 tan ?) then the equation to the normal at this point will be given by                 3x sin ? + 4y = (9 + 16) tan ?                    …… (2) Let this normal cut the axis of x at M whose coordinates are (x, 0) and the axis of y at N whose coordinates are (0, y) solving (2) with y = 0, we get         x = 9+16/3cos? = 25/3cos?                                …… (3) 

 Similarly solving (2) with x = 0, we get         y = (9+16)tan?/4 = 25tan?/4                               ……(4) If PM and PN be the lines parallel to the axes, the coordinates of P = (x, y) will be clearly given by (3) and (4). The required locus of P will be obtained by eliminating ? from (3) and (4). Using the fact sec2 ? – tan2 ? = 1, we get         9x2 – 16y2 = 625 Example:

 Prove that a circle can be drawn through the foci of a hyperbola and the points at which any tangent meets the tangents at the vertices of the hyperbola. Solution: Recall: If four points (x1, y1), (x2, y2), (x3, y3) and (x4, y4) are concyclic then         

              

 These points are (ae, 0), (–ae, 0), (a, b tan ?/2) and (–a, –b cot f/2)         

              

 apply : R1 → R1 – R2 and expand along R1, we get 

        = (–2ae)  = (–2aeb) [(a2 e2 – a2 – b2) cot ?/2 + 9a2e2 – b2 – a2) tan ?/2] = 0 ∴ points S, S’, R and Q are conclycic. Example: 

On a level plain the crack of the rifle and the thud of the ball striking the target are heared at the same instant, prove that the locus of the hearer is a hyperbola. Solution: Suppose A to be the target and B to be the firing point and let the hearer be at a point P. If V1 and V2 be the velocity of the sound and bullet respectively, then the time taken by the bullet in reaching from B to the target A = BA/v2 and the time taken by the sound in reaching from A to P = AP/v1. Again the time taken by the sound in reaching from B to P = BP/v1. As by hypothesis the sound reaches the hearer at P, simultaneously, the time taken by the sound to reach upto the position P from then target together with the time taken by the bullet to reach the target must be same as the time taken by the sound in reaching from the rifle upto P. Hence BA/v2 + AP/v1 = BP/v1

 ⇒ BA/v2 = BP/v1 – AP/v1 = 1/v1 (BP – AP) ⇒ BP – AP = v1/v2 AB As v1, v2 and AB are constants, hence         BP –AP = Constant Therefore the locus of the point P is the hyperbola having foci at A and B and transverse axis equal to v1/v2 AB

Example:

Prove that the locus of the pole of a chord of the hyperbola which subtends a right angles at the vertex, is, x = a2–b2/a2+b2.   Solution: The coordinates of the vertex are (a, 0). Transferring the origin to this point, the equation of the hyperbola x2/a2–y2/b2 becomes;         (x+a)2/a2– y2/b2 = 1 

⇒ x2/a2–y2/b2 = –2x/a ⇒ b2x2 – a2y2 = –2ab2x                                         …… (1) The equation to the polar of (h, k) w.r.t hyperbola is given by         b2hx – a2hky = a2b2                                             …… (2) After transformation the equation (2) becomes           b2(x + a)h – a2yk = a2b2

  or     b2hx – a2yk = a2b2 – ab2h                                     …… (3) The equation of the lines joining the points of intersection of the hyperbola and the chord to the origin is obtained by making (1) homogeneous with the help of (3). Hence on simplification, this equation becomes         (a2b2 – ab2h) (b2x2 – a2y2) = –2ab2x(b2xh – a2yk) If they are at right angles, the sum of the coefficients of x2 and y2 must be zero; hence         b2 – a2 + 2b2h/a–h = 0 generalizing for (h, k), we get the required locus as         x = a a2–b2/a2+b2

   Example: Find the locus of intersection of tangent to a hyperbola, which meet at a constant angle ß.   Solution: Let the equation to the hyperbola be          x2/a2–y2/b2 = 1                                             …… (1) 

Equation to any tangent to (1) is         y = mx + √a2m2–b2

 If the tangent passes through a point (h, k) when we must have         k = mh + √(a2m2–b2) or     m2(h2 – a2) – 2mhk + (k2 + b2) = 0                        …… (2) Let m1 and m2 be the two roots of this equation. ⇒ m1 = tanθ1 and m2 = tanθ2, we have            tanθ1 + tanθ2 = 2hk/h2–a2

            tanθ1 × tanθ2 = k2+b2/h2–a2

 and as (tanθ1 – tanθ2)2 = 4h2k2–4(k2+b2)(h2–a2)/(h2–a2)2.            = 4(a2k2–b2h2+a2b2)/(h2–a2)2

 If the two tangents met at an angle ß, clearly ß = (θ1 – θ2). Hence cotß = cot(θ1 – θ2)          = 1/tan(θ1–θ2) = 1+tanθ1tanθ2/tanθ1–tanθ2

 ⇒ cot2ß = (1+tanθ1tanθ2)2/(tanθ1–tanθ2)2 = (h2+k2+b2–a2)2/4(a2k2–b2h2+a2b2)2

 Simplifying, the required locus is         (x2 + y2 + b2 – a2)2 = 4cot2b(a2y2 – b2x2 + a2b2)2

   Example: Find the equation to the hyperbola whose asymptotes are the straight lines x + 3y – 1= 0 and 2x – y + 7 = 0, and which passes through the point (1, 2).   Solution: Equation to the asymptotes are given as

          x + 3y – 1 = 0    and                                           …… (1)          2x – y + 7 = 0                                                    …… (2) (1) and (2) may be given by          (x + 3y – 1)(2x – y + 7) = 0                                 …… (3) As the equation to the hyperbola will differ from (3) only by a constant, it may be given by         (x + 3y – 1)(2x – y + 7) =  λ                                  …… (4) (where λ is a constant) (1, 2) lies on the curve given by (4), we have         (1 + 6 – 1)(2 – 2 + 7) = λ ⇒ λ = 42 Hence the equation to the hyperbola will be         (x + 3y – 1)(2x – y + 7) = 42 ⇒ 2x2 – xy + 6xy + 7x – 3y2 + 21y – 2x + y – 7 = 42   ⇒ 2x2 – 3y2 + 5xy + 5x + 22y – 49 = 0

3D GeometryThe moving power of mathematical invention is not reasoning but imagination. – A.DEMORGAN

We constantly describe both the shapes and positions of three dimensional objects. For example, one simple way to describe a 3D object is to approximate its shape as a mesh of triangles. Each triangle is de?ned by three vertices, and the positions of each of these vertices will have to be described by three coordinates [x,y,z]t.

In this chapter, we shall study the direction cosines and direction ratios of a line joining two points and also discuss about the equations of lines and planes in space

under different conditions, angle between two lines, a line and a plane, shortest distance between two skew lines and distance of a point from a plane.

Lower level geometry of planes and graphical two dimensional coordinates moves into a three dimensional space in three dimensional geometry.

Topics Covered: 

Rectangular Co ordinate system in space

Direction cosines of a line

Direction Ratios

Parallel Lines

Projection of a Line

Theory of 3D Plane

Theory of 3D Straight Line

Shortest Distance between Two Non Intersecting Lines

Theory of Sphere

Solved Problems of 3D Geometry Part I

Solved Problems of 3D Geometry Part II

Solved Problems of 3D Geometry Part III

Solved Problems of 3D Geometry Part IV

Solved Problems of 3D Geometry Part V

Solved Problems of 3D Geometry Part VI

 RECTANGULAR COORDINATE SYSTEM IN SPACE

  Let ‘O’ be any point in space

and   be three lines perpendicular to each other. These lines are known as coordinate axes and O is called origin. The planes XY, YZ, ZX are known as the coordinate planes.

Coordinates of a Point in Space:

 

Consider a point P in space. The position of the point P is given by triad (x, y, z) where x, y, z are perpendicular distance from YZ-plane, ZX-plane and XY-plane respectively.

 

 

 If We assume i, j, k unit vectors along OX, OY, OZ respectively, then position vector of point P is xi + yj + zk or simply (x, y, z).

                                                                                         

•         x-axis = {( x, y, z) | y = z = 0}

 

•         y-axis = {(x, y, z) | x = z = 0}

 

•         z-axis = {(x, y, z) | x = y = 0}

 

•         xy plane = {(x, y, z) | z = 0}

 

•         yz plane = {(x, y, z) | x = 0}

 

•         zx plane = {(x, y, z) | y = 0}

 

•         OP = √x2 + y2 + z2

  

Shifting the Origin:

 

 

 

Shifting the origin to another point without changing the directions of the axes is called the translation of axes.                 

 

Let the origin O be shifted to another point O' (x', y', z') without changing the direction of axes. Let the new coordinate frame be O'X'Y'Z'. Let P (x, y, z) be a point with respect to the coordinate frame OXYZ. 

 

Then, coordinate of point P w.r.t. new coordinate frame O'X'Y'Z' is (x1, y1, z1), where 

 

x1 = x – x', y1 = y – y', z1 = z – z'

 

                                                                                                         

  

Example -1:      If the origin is shifted (1, 2, –3) without changing the directions of the axes then find the new coordinates of the point (0, 4, 5) with respect to new frame.

 

Solution:       x' = x – x1,      where (x1, y1, z1)  is the shifted origin

 

                        y' = y – y1

 

                        z' = z – z1

 

                        x' = 0 – 1 = –1

 

                        y' = 4 – 2 = 2

 

                       z' = 5 + 3 = 8

 

∴ The coordinates of the point w.r.t. to new coordinate frame is (-1, 2, 8).

 

Note:

 

•         Distance between the points P(x1, y1, z1) and Q (x2, y2, z2) is

 

          √(x1 – x2)2 + (y1 – y2)2 + (z1 – z2)2

 

•         The point dividing the line joining P(x1, y1, z1) and Q(x2, y2, z2) in m : n ratio is 

 

       (mx2 – nx1 / m + n, my2 – ny1 / m + n, mz2 – nz1 / m + n) where m + n ≠ 0 ., 

 

•         The coordinates of centroid of a triangle having vertices A (x1, y1, z1), B (x2, y2, z2) and C (x3, y3, z3) is G (x1 + x2 + x3 / 3, y1 + y2 + y3 / 3, z1 + z2 + z3 / 3).

 

Example -2:      Find the coordinates of the point which divides the line joining points (2, 3, 4) and (3, –4, 7) in ratio 3 : 5.

 

Solution:       Let the coordinates of the required point be (x, y, z), then

 

                        x = 2(3) + 3(5) / 3 + 5 = 21/8

 

                        y = 3(3) – 4(5) / 3 + 5 = – 11/8 

 

                        z = 4(3) + 7(5) / 3 + 5 = 47/8

 

                        Hence the required point is (21/8, –11/8, 47/8).

 

Example -3:      Prove that the three points A (3, –2, 4), B (1, 1, 1) and C (–1, 4, –2) are collinear.

 

Solution:       The general coordinates of a point R which divides the line joining A (3, –2, 4) and B (1, 1, 1) in the ratio μ : 1 are (μ + 3 / μ + 1, μ – 2 / μ + 1, μ + 4 / μ + 1)       ……(1)

 

                        If C (–1, 4, –2) lies on the line AB, then for some value of m the coordinates of μ the point R will be the same as those of C.

 

                        Let x-coordinate of point R = x-coordinate of point C. 

 

                        Then, μ + 3 / μ + 1 = –1 => μ = –2

 

                        Putting μ = –2 in (1) the coordinates of R are (–1, 4, –2) which are also the coordinates of C.

 

                        Hence the points A, B, C are collinear.

DIRECTION COSINES OF A LINEIf α, β, γ be the angles which a given directed line makes with the positive directions of the co-ordinate axes, then cosα, cosβ, cosγ are called the direction cosines of the given line and are generally denoted by l, m, n respectively.

Thus, l = cosα, m = cosβ and n = cosγ

By the definition it follows that the direction cosine of the axis of x are respectively cos0°, cos90°, cos 90°i.e. (1, 0, 0).

Similarly direction cosines of the axes of y and z are respectively (0, 1, 0) and (0, 0, 1).

                                                                                                               Relation between the Direction Cosines:                               

 

Let OP be any line through the origin O which has direction cosines l, m, n.

Let P º (x, y, z) and OP = r

Then OP2 = x2 + y2 + z2 = r2                                   …. (1)

From P draw PA, PB, PC perpendicular on the coordinate axes, so that

OA = x, OB = y, OC = z.

Also, ∠POA = α, ∠POB = β and ∠POC = γ.

From triangle AOP, l = cosα = x/r  => x = lr

Similarly y = mr and z = nr

Hence from (1)

Dig: Relation between direction cosines

 

r2(l2 + m2 + n2) = x2 + y2 + z2 = r2 => l2 + m2 + n2 = 1

Note:

If the coordinates of any point P be (x, y, z) and l, m, n be the direction cosines of the line OP, O being the origin, then (lr, mr, nr) will give us the co-ordinates of a point on the line OP which is at a distance r from (0, 0, 0).

Direction Ratios:If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios. They are also called direction numbers or direction components.

Hence by definition, we have

1/a = m/b = n/c = k (say)

=> l = ak, m = bk, n = ck => k2(a2 + b2 + c2) = l2 + m2 + n2 = 1

=> k = ± 1 / √a2 + b2 + c2 = ± 1/√Σa2

 l = ± a/√Σa2.   Similarly m = ± b/√Σa2 and n = ± n/√Σa2

where the same sign either positive or negative is to be chosen throughout.

Example: If 2, – 3, 6 be the direction ratios, then the actual direction cosines are 2/7, –3/7, 6/7.

 

Note:

•     Direction cosines of a line are unique but direction ratios of a line in no way unique but can be infinite.

Parallel Lines: Since parallel lines have the same direction, it follows that the direction cosines of two or more parallel straight lines are the same. So in case of lines, which do not pass through the origin, we can draw a parallel line passing through the origin and direction cosines of that line can be found.

CXmhaU -4:      Find the direction cosines of two lines which are connected by the relations l–5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0.

hb :     The given relations are

 

                        l–5m + 3n = 0 => l = 5m – 3n                                  ……(1)

 

                        and 7l2 + 5m2 – 3n2 = 0                                          ……(2)

 

                        Putting the value of l from (1) in (2), we get 

 

                        7(5m – 3n)2 + 5m2 – 3n2 = 0

 

                        or, 180m2 – 210mn + 60n2 = 0 or, (2m – n)(3m – 2n) = 0

 

                        ∴ m/n = 1/2  or 2/3

 

                        when m/n = 1/2 i.e. n = 2m

 

                        ∴ l = 5m – 3n = –m or 1/m = –1

 

                        thus  and  = –1 giving

 

                        or, 1/–1 = m/1 = n/2 = √(l2 + m2 + n2) / √{(–1)2 + l2 + 22} = 1/v6

 

                        So, direction cosines of one line are –1/v6, 1/v6, 2/v6

 

                        Again when m/n = 2/3 or n = 3m/2

 

                        ∴ l = 5m – 3.3m/2 = m/2 or 1/m = 1/2

 

                        Thus, m/n = 2/3 and i/m = 1/2 giving i/1 = m/2 = n/3 = 1/v12 + 22 + 32 = 1/v14

 

                        ∴ The direction cosines of the other line are 1/v14, 2/v14, 3/v14.

Direction Cosine of a Line joining two given Points:  

 

The direction ratios of line PQ joining P (x1, y1, z1) and Q(x2, y2, z2) are x2 – x1 = a(say), y2 – y1 = b (say) and z2 – z1 = c (say).

 

Then direction cosines are

 l = (x2 – x1) / √∑(x2 – x1)2, m = (y2 – y1) / √∑(x2 – x1)2, n = (z2 – z1) / √∑(x2 – x1)2

 

                                                                                                                        Alt txt: direction cosine of a line

 

CXmhaU -5:      Find the direction ratios and direction cosines of the line joining the points A(6, –7, –1) and B(2, –3, 1).

 

hb :      Direction ratios of AB are (4, – 4, – 2) = (2, – 2, – 1)

 

            a2 + b2 + c2 = 9

  

            Direction cosines are (± 2/3, ± 2/3, ± 1/3).  

 

Angle between two Lines:

  

Let θ be the angle between two straight lines AB and AC whose direction cosines are given whose direction cosines are l1, m1, n1 and l2, m2, n2 respectively, is given by cosq = l1l2 + m1m2 + n1n2

 

If direction ratios of two lines are a1, b1, c1 and a2, b2, c2 are given, then angle between two lines is given by

 

cos θ =                                                 alt txt: angle between two lines

 

Particular Results:

We have, sin2θ = 1 – cos2θ

            = (i12 + m12 + n1

2)(i22 + m22 + n2

2) – (l1l2 + m1m2 + n1n2)2

 

            = (l1m2 – l2m1)2 + (m1n2 – m2n1)2 + (n1l2 – n2l1)2

 

            => sinθ = ± √Σ(l1m2 – l2m1)2 .

 

Condition of perpendicularity:

 

If the given lines are perpendicular, then θ = 900 i.e. cos θ = 0

 

=> l1l2 + m1m2 + n1n2 = 0 or a1a2 + b1b2 + c1c2 = 0 .

 

Condition of parallelism:

 

If the given lines are parallel, then q = 00 i.e. sin θ = 0

 

Þ (l1m2 – l2m1)2 + (m1n2 – m2n1)2 + (n1l2 – n2l1)2 

 

which is true, only when

 

l1m2 – l2m1 = 0, m1n2 – m2n1 = 0 and n1l2 – n2l1 = 0 => I1/l2 = m1/m2 = n1/n2

 

Similarly, a1/a2 = b1/b2 = c1/c2.

 

CXmhaU -6:      Show that two lines having direction ratios –1, 3, 2 and 2, 2, –2 are perpendicular.

 

hb :     a1a2 + b1b2 + c1c2 = (–1)(2) + (3)(2) + (2)(–2) = –2 + 6 – 4 = 0

 

            ∴ lines are perpendicular

Projection of a Line:

Projection of the line joining two point P (x1, y1, z1) and Q (x2, y2, z2) on another line whose direction cosines are l, m, n is

AB = l(x2 – x1) + m(y2 – y1) + n(z2 – z1)

  

Perpendicular Distance of a Point from a Line:

 

Let AB is straight line passing through point A (a, b, c) and having direction cosines l, m, n.

AN = projection of line AP on straight line AB

      = l(x – a) + m(y – b) + n(z – c)

and AP = √(x–a)2 + (y–b)2 + (z–c)2

∴ perpendicular distance of point P

       PN = √AP2 – AN2

 

 

CXmhaU -7:      Find out perpendicular distance of point P (0, –1, 3) from straight line passing through A (1, –3, 2) and having direction ratios 1, 2, 2.

 

hb :                 Direction cosines of the line is 1/√12 + 22 + 22, 2/√12 + 22 + 22, 2/√12 + 22 + 22

 

                        i.e. .1/3, 2/3, 2/3

 

                        ∴ PN = l(x – a) + m(y – b) + n(z – c)        1/3 = (0 – 1) + 2/3 (–1 + 3) + 2/3(3 – 2) = 5/3

 

                        AP = √(0–1)2 + (–1 + 3)2 + (3–2)2 = √6

 

                        ∴ Perpendicular distance PN = √AP2 – PN2 = √6–25/9 = √29/3.

  

Area of a Triangle

 

 

 

 

 

So, area of ΔABC is given by the relation Δ2 = Δx2 + Δy

2 + Δz2

 

THE PLANE 

Definition:   Consider the locus of a point P(x, y, z). If x, y, z are allowed to vary without any restriction for their different combinations, we have a set of points like P. The surface on which these points lie, is

called the locus of P. It may be a plane or any curved surface. If Q be any other point on it’s locus and all points of the straight line PQ lie on it, it is a plane. In other words if the straight line PQ, however small and in whatever direction it may be, lies completely on the locus, it is a plane, otherwise any curved surface.

Equation of Plane in Different Forms:

General equation of a plane is ax + by + cz + d = 0

Equation of the plane in Normal form is lx + my + nz = p where p is the length of the normal from the

origin to the plane and (l, m, n) be the direction cosines of the normal.

The equation to the plane passing through P(x1, y1, z1) and having direction ratios

(a, b, c) for its normal is a(x – x1) + b(y – y1) + c (z – z1) = 0

The equation of the plane passing through three non-collinear points (x1, y1, z1),

(x2, y2, z2) and (x3, y3 , z3) is   = 0

The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is x/a + y/b + z/c =

1 (a b c ≠ 0)

Equation of YZ plane is x = 0,           equation of plane parallel to YZ plane is x = d.

           Equation of ZX plane is y = 0,           equation of plane parallel to ZX plane is y = d.

           Equation of XY plane is z = 0,           equation of plane parallel to XY plane is z = d.

Four points namely A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) will be coplanar if one point

lies on the plane passing through other three points.

 

CXmhaU -8: Find the equation to the plane passing through the point (2, -1, 3) which is the foot of the perpendicular drawn from the origin to the plane.

         

hb :     The direction ratios of the normal to the plane are 2, -1, 3.

 

                        The equation of required plane is 2(x –2) –1 (y + 1) + 3 (z –3) = 0

 

                        => 2x – y + 3z –14 = 0

 

Angle between the Planes

 

Angle between the planes is defined as angle between normals of the planes drawn from any point to the planes.

 

Angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0

is 

 

Note:

 

•         If a1a2 +b1b2 +c1c2 = 0, then the planes are perpendicular to each other.

 

•         If a1/a2 = b1/b2 = c1/c2 then the planes are parallel to each other.

 

CXmhaU -9: Find angle between the planes 2x – y + z = 11 and x + y + 2z = 3.

 

hb :    

 

 

 

CXmhaU -10:           Find the equation of the plane passing through (2, 3, –4), (1, –1, 3) and parallel to x-axis.

 

hb :     The equation of the plane passing through (2, 3, –4) is

 

                        a(x – 2) + b(y – 3) + c(z + 4) = 0                              ……(1)

 

                        since (1, –1, 3) lie on it, we have

 

                        a + 4b – 7c = 0                                                          ……(2)

 

                        since required plane is parallel to x-axis i.e. perpendicular to YZ plane i.e.

 

                        1.a + 0.b + 0.c = 0 Þ a = 0 Þ 4b – 7c = 0 => b/7 = c/4

 

                        ∴ Equation of required plane is 7y + 4z = 5.

 

Perpendicular Distance:

The length of the perpendicular from the point P(x1, y1, z1) to the plane ax + by + cz + d = 0 is |ax1 + by1 + cz1 + d / √a2 + b2 + c2|.

Family of Planes:

Equation of plane passing through the line of intersection of two planes u = 0 and v = 0 is u + λv = 0.

Intersection of a Line and Plane:

If equation of a plane is ax + by + cz + d = 0, then direction cosines of normal to this plane are a, b, c. So angle between normal to the plane and a straight line having direction cosines l, m ,n is given by cos θ = al + bm + cn / √a2 + b2 + c2.

Then angle between the plane and the straight line is π/2 – θ.

Plane and straight line will be parallel if al + bm + cn = 0

Plane and straight line will be perpendicular if a/l = b/m = c/n.

 

CXmhaU -11:           Find the equation of plane passing through the intersection of planes 2x – 4y + 3z + 5= 0, x + y + z = 6 and parallel to straight line having direction cosines (1, –1, –1).

hb :     Equation of required plane be

                        (2x – 4y + 3z + 5) + λ(x + y – z – 6) = 0

 

                        i.e. (2 + λ)x + (–4 + λ)y + z(3 – λ) + (5 – 6λ)     = 0

 

                        This plane is parallel to a straight line. So, al + bm + cn = 0

 

                        1(2 + λ) + (–1)(–4 + λ) + (–1)(3 – λ) = 0 i.e. λ = –3

 

                        ∴ Equation of required plane is –x – 7y + 6z + 23 = 0

 

                        i.e. x + 7y – 6z – 23 = 0.

Bisector Planes of Angle between two Planes: 

The equation of the planes bisecting the angles between two given planes a1x +b1y +c1z +d1 = 0 and a2x + b2y + c2z +d2 = 0 is

 

 

THE STRAIGHT LINE

Straight line in three dimensional geometry is defined as intersection of two planes. So general equation of straight line is stated as the equations of both plane together i.e. general equation of straight line is a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0              ……(1)

So, equation (1) represents straight line which is obtained by intersection of two planes.

Equation of Straight Line in Different Forms:

Symmetrical Form:  

       •   Equation of straight line passing through point P (x1, y1, z1) and whose direction cosines are l, m, n is x–x1 = y – y1/m = z – z1 / n

Equation of straight line passing through two points P (x1, y1, z1) and Q (x2, y2, z2) isx–x1 / x2 – x1 = y–y1 /

y2 – y1 = z – y1 / z2 – z1

 

Note:      The general coordinates of a point on a line is given by (x1 + lr, y1 + mr, z1 + nr) where r is distance between point (x1, y1, z1) and point whose coordinates is to be written.

CXmhaU 10.   Find the equations of the straight lines through the point (a, b, c) which are

            (a) parallel to z-axis

            (b) perpendicular to z-axis

10. (i) Equation of straight lines parallel to z-axis have α = 900, β = 900, γ = 00

 

            => l = 0, m = 0, n = 1

 

            Therefore equation of straight line is parallel to z-axis and passing through (a, b, c) is x – a / 0 = y – b / 0 = z – c / 1

 

(ii)        equation of straight lines perpendicular to z-axis

 

            let they make α, β angle with x and y axes respectively.

 

            Then equation of straight lines perpendicular to z axis and passing through (a, b, c) is x–a / cos α = y – b / sin α = z – c / 0

 

            => x – a / l = y – b / m = z – c / 0.

  

CXmhaU -12:           Find the coordinates of the point where the line joining the points (2, –3, 1) and (3, –4, –5) cuts the plane 2x + y + z = 7.

  

hb :     The direction ratios of the line are 3 – 2, –4 – (–3), –5 – 1 i.e. 1, –1, –6

 

                        Hence equation of the line joining the given points is

 

                        x–2 / 1 = y + 3 / –1 = z – 1 / – 6 = r (say)                                               

 

                        Coordinates of any point on this line are (r + 2, –r – 3, –6r + 1)

 

                        If this point lies on the given plane 2x + y + z = 7, then

 

                        2(r + 2) + (–r – 3) + (–6r + 1) = 7 => r = –1

 

                        Coordinates of the point are (–1 + 2, –(–1) – 3, –6(–1) + 1) i.e. (1, –2, 7).

 

Note:

If equation of straight line is given in general form, it can be changed into symmetrical form. The method

is described in following Example.

 

CXmhaU -13:           Find in symmetrical form the equations of the line

 

                        3x + 2y – z – 4 = 0= 4x + y – 2z + 3.

  

hb :     The equation of the line in general form are

 

                        3x + 2y – z – 4 = 0, 4x + y – 2z + 3 = 0                  ……(1)

 

                        Let l, m, n be the direction cosines of the line. Since the line is common to both the planes, it is perpendicular to the normals to both the planes.

 

                        Hence 3l + 2m – n = 0, 4l + m – 2n = 0

 

                        Solving these we get,

 

                        1/–4+1 = m–4+6 = n/3–8 i.e. 1/–3 = m/2 = n/–5 = 1/√(–3)2 + 22 + (–5)2 = 1/√38

 

                        So, direction cosines of the line are –3/√38, 2/√38, –5/√38

 

                        Now to find the coordinates of a point on a line. Let us find out the point where it meets the plane z = 0. Putting z = 0 in the equation given by (1), we have

 

                        3x + 2y – 4 = 0, 4x + y + 3 = 0

 

                        solving these, we get x = –2, y = 5

 

                        So, one point of the line is (–2, 5, 0)

 

                        ∴ equation of the line in symmetrical form is

  

                         i.e. .

 

 Shortest Distance between two non Intersecting Line: Two lines are called non intersecting lines if they do not lie in the same plane. The straight line which is perpendicular to each of non-intersecting lines is called the line of shortest distance. And length of shortest distance line intercepted between two lines is called length of shortest distance.  

Method: Let the equation of two non-intersecting lines be

 

            x–x1 / l1 = y–y1/m1 = z–z1/n1 = r1 (say)                                         ……(1)

 

            And x–x2 / l2 = y–y2/m2 = z–z2/n2 = r2 (say)                                  ……(2)

 

Any point on line (1) is P (x1 + l1r1, y1 + m1r1, z1 + n1r1) and on line (2) is

 

Q (x2 + l2r2, y2 + m2r2, z2 + n2r2).

 

Let PQ be the line of shortest distance. Its direction ratios will be

 

[(l1r1 + x1– x2– l2r2), (m1r1 + y1– y2– m2r2), (n1r1 + z1– z2– n2r2)]

 

This line is perpendicular to both given line. By using condition of perpendicularity we obtain 2 equations in r1 and r2.

 

So by solving these, values of r1 and r2 can be found. And subsequently point P and Q can be found. The distance PQ is shortest distance.

 

The shortest distance can be found by PQ =  .

 

Note:

If any straight line is given in general form then it can be transformed into symmetrical form and we can

further proceed.

 

CXmhaU -14:           Find the shortest distance between the lines , x–3/3 = y –8/–1 = z–3/1, x + 3/–3 = y+7/2 = z–6/4 . Also find the equation of line of shortest distance.

 

hb :     Given lines are x–3/3 = y–8/–1 = z–3/1 = r1 (say)                 ……(1)

 

                         = x+3 / –3 = y + 7 / 2 = z – 6 / 4 = r2 (say)                ……(2)

 

                        Any point on line (1) is P (3r1 + 3, 8 – r1, r1 + 3) and on line (2) is

 

                        Q (–3 – 3r2, 2r2 – 7, 4r2 + 6).

 

                        If PQ is line of shortest distance, then direction ratios of PQ

 

                        = (3r1 + 3) – (–3 – 3r2), (8 – r1) – (2r2 – 7), (r1+ 3) – (4r2 + 6)

 

                        i.e. 3r1 + 3r2 + 6, –r1 – 3r2 + 15, r1 – 4r2 – 3

 

                        As PQ is perpendicular to liens (1) and (2)

 

                        ∴ 3(3r1 + 3r2 + 6) – 1(–r1 – 2r2 + 15) + 1(r1 – 4r2 + 3) = 0

 

                        => 11r1 + 7r2 = 0                                                                    ……(3)

 

                        and –3(3r1 + 3r2 + 6) + 2(–r1 – 2r2 + 15) + 4(r1 – 4r2 + 3) = 0

 

                        i.e. 7r1 + 11r2 = 0                                                                   ……(4)

 

                        On solving equations (3) and (4), we get r1 = r2= 0.

 

                        So, point P (3, 8,3) and Q (–3, –7, 6)

 

                        ∴ Length of shortest distance PQ = √{(–3–3)2 + (–7–8)2 + (6–3)2} = 3√30

 

                        Direction ratios of shortest distance line is 2, 5, –1

 

                        ∴ Equation of shortest distance line x–3/2 = y–8/5 = z–3/–1.

 

 THE SPHERE A sphere is a locus of a point which moves in space such that its distance from a fixed point is constant. Fixed point is called centre of sphere and constant distance is called radius of sphere.

Equation of Sphere in Different Forms:

If centre of sphere is (a, b, c) and radius is r, then equation of sphere is

      (x – a)2 + (y – b)2 + (z – c)2 = r2.

If centre of sphere is origin and radius is r, then x2 + y2 + z2 = r2.

General form: The general equation of a sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0

Centre of sphere = (–u, –v, –w), radius = √μ2 + v2 + w2 – d.

Diameter form: Equation of a sphere whose extremities of diameter are A (x1, y1, z1) and B (x2, y2, z2) is

(x – x1) (x – x2) + (y – y1) (y – y2) + (z – z1) (z – z2) = 0.

CXmhaU 13. Find the equation of the sphere which passes through the points (1, –3, 4), (1, –5, 2) and (1, –3, 0) and whose centre is on the plane x + y + z = 0.  

hb :     Let equation of the sphere be

            x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 

            its centre is (- u, - v, - w) which is on x + y + z = 0

            => u + v + w = 0                                                                     … (1)

            it passes through (1, - 3, 4) => 2u - 6v + 8w + d = - 26         … (2)

            (1, - 5, 2) => 2 u - 10 v + 4 w + d = - 30                                … (3)

            and it passes through (1, - 3, 0) => 2 u - 6 v + d = - 10        … (4)

            solving these four equations we get,

            u = - 1, v = 3, w = - 2 and d = 10

            Therefore required equation of the sphere is

            x2 + y2 + z2 - 2 x + 6 y - 4 z + 10 = 0. 

 

CXmhaU -15:           Find the equation of the sphere whose centre is (2, –3, 4) and which passes through the point (1, 2, –1).

hb :     Radius of sphere = √{(2–1)2 + (–3–2)2 + (4+1)2} = √51

                        ∴ Equation of the sphere is (x – 2)2 + (y + 3)2 + (z – 4)2 = (√51)2

                        i.e. x2 + y2 + z2 – 4x + 6y – 8z – 22 = 0.

 

SOLVED PROBLEMS

OBJECTIVE

1.                     The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l2 + m2 + n2 = 0 is

 

                      (A) π/3                                                   (B) 2π/3

 

                     (C) π/4                                                   (D) None of these

 

hb :                 Eliminating n between the two relations, we have

 

                        l2 + m2 – (l + m)2 = 0 or 2lm = 0 => either l = 0 or m = 0

 

                        if l = 0, then m + n = 0 i,e. m = – n

 

                        => l/0 = m/1 = n/–1, giving the direction ratios of one line.

 

                        If m = 0, then l + n = 0 i.e. l = – n

 

                        => l/0 = m/1 = n/–1, giving direction ratios of the other lines.

 

                        The angles between these lines is

 

                        

 

2.                     The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:

 

                        (A) x – 2y + 11 = 0                             (B) x + 2y + 11 = 0

 

                        (C) x + 2y – 11 = 0                             (D) x – 2y – 11 = 0

 

hb :                 Equation of the required plane is (x + y + z – 6) + λ(2x + 3y + z + 5) = 0

 

                        i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0

 

                        This plane is perpendicular to xy plane whose equation is z = 0

 

                        i.e. 0 . x + 0 . y + z = 0

 

                        ∴ By condition of perpendicularity

 

                        0.(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0 i.e. λ = –1

 

                        ∴ Equation of required plane is

 

                        (1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) = 0 or x + 2y + 11 = 0.

 

3.                     The coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0 are :

 

                         (A) (–3/61, 4/61, 6/61)                             (B) (3/61, –4/61, 6/61)

 

                        (C) (–3/61, –4/61, 6/61)                          (D) (3/61, 4/61, 6/61)

 

hb :                 The equation of the plane is 3x + 4y – 6z + 1 = 0                      ……(1)

 

                        The direction ratios of the normal to the plane (1) are 3, 4, –6. So equation of the line through (0, 0, 0) and perpendicular to the plane (1) are

 

                        x/3 = y/4 = z/–6 = r (say)                                                                        ……(2)

 

                        The coordinates of any point P on (2) are (3r, 4r, –6r). If this point lie on the plane (1), then

 

                        3(3r) + 4(4r) – 6(–6r) + 1 = 0 i.e. r = –1/61

 

                        Putting the value of r coordinates of the foot of the perpendicular P are (–3/61, –4/61, 6/61).

 

4.                     The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line x/2 = y/3 = z/–6 is:

 

                        (A) 7 unit                                            (B) 4 unit

 

                       (C) 1 unit                                            (D) 2 unit

 

hb :                 Here we are not to find perpendicular distance of the point from the plane but distance measured along with the given line. The method is as follow:

 

                        The equation of the line through the point (1, –2, 3) and parallel to given line is

 

                        x–1/2 = y+2/ 3 = z–3/–6 = r (say)

 

                        The coordinate of any point on it is (2r + 1, 3r – 2, –6r + 3).

 

                        If this point lies in the given plane then

 

                        2r + 1 – (3r – 2) + (–6r + 3) = 5 Þ –7r = –1 or r = 1/7

 

                        ∴ point of intersection is (9/7, –11/7, 15/7)

 

                        ∴ The required distance

 

                        = the distance between the points (1, –2, 3) and  (9/7, –11/7, 15/7)

 

                        = √(1–9/7)2 + (–2+11/7)2 + (3–15/7)2 = 1/7 = √49 = 1 unit.

 

5.                     The image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0 is :

 

            (A) (–3, 5, 2)                                       (B) (3, 2, 5)

 

            (C) (–5, 3, –2)                                     (D) (–2, 5, 3)

 

hb :                 As it is clear from the figure that PQ will be perpendicular to the plane and foot of this perpendicular is mid point of PQ i.e. N.

                        So, direction ratios of line PQ is 2, –1, 1

                        => Equation of line PQ = x–1/2 = y–3/–1 = z–4/1 r (say)

                        Any point on line PQ is (2r + 1, –r + 3, r + 4)

                        If this point lies on the plane, then

                        2(2r + 1) – (–r + 3) + (r + 4) + 3 = 0 => r = –1

                        ∴ coordinate of foot of perpendicular N = (–1, 4, 3)

                        As N is middle point of PQ

 

                        ∴ –1 = 1+x1/2, 4 = 3+y1/2, 3 = 4+z1/2 

 

                        => x1 = –3, y1 = 5, z1 = 2

 

                        ∴ Image of point P (1, 3, 4) is the point Q (–3, 5, 2).

 

6.                     The equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) and has its radius as small as possible is :

 

            (A) 3(x2 + y2 + z2) + 2(x + y + z) – 1= 0

 

(B) 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0

 

            (C) (x2 + y2 + z2) – 2(x + y + z) – 1= 0

 

            (D) (x2 + y2 + z2) + 2(x + y + z) – 1= 0

 

hb :                 Let equation of sphere be given by

 

                        x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0                   ……(1)

 

                        As sphere passes through points (1, 0, 0), (0, 1, 0) and (0, 0, 1). So we have

 

                        1 + 2u + d = 0, 1 + 2v + d = 0, 1 + 2w + d = 0

 

                        On solving u = v = w = –1/2 (d + 1)

 

                        If r is the radius of the sphere, then

 

                        r = √u2 + v2 + w2 – d

 

                        r2 = 3/4 (d + 1)2 – d = m (say)

 

                        for r to be minimum

 

                        dμ/dd = 0 => 3/4.2(d + 1) – 1 = 0 or d = –1/3

 

                        Also, d2μ/dd2 = 3/2 positive at d = –1/3

 

                        Hence m is minimum at d = –1/3

 

                        So, substituting value of d we have u = v = w = –1/3

 

                        ∴ equation of the sphere

 

                        x2 + y2 + z2 – (x + y + z) – 1/3 = 0 => 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0.

 

7.                     A point moves so that the ratio of its distances from two fixed points is constant. Its locus is a:

 

            (A) plane                                            (B) st. lines

 

            (C) circle                                             (D) sphere

  

hb :                 Let the coordinates of moving point P be (x, y, z). Let A (a, 0, 0) and B (–a, 0, 0) be two fixed points. According to given condition

 

                        AP/BP = constant = k (say) => AP2 = k2BP2

 

                        or, (x – a)2 + (y – 0)2 + (z – 0)2 = k2{(x + a)2 + (y – 0)2 + (z – 0)2}

 

                        => (1 – k2)(x2 + y2 + z2) – 2ax(1 + k2) + a2(1 – k2) = 0

 

                     ∴ required locus is x2 + y2 + z2 – 2a(1+k2)/(1–k2) + a2 = 0. Which is a sphere.

 

8.         The ratio in which yz–plane divides the line joining (2, 4, 5) and (3, 5, 7)

 

                         (A) -2 : 3                                              (B) 2 : 3

 

                        (C) 3 : 2                                                (D) -3 : 2

 

hb :     Let the ratio be λ : 1

 

             x-coordinate is 3λ+2 / λ+1 = 0 => λ = –2/3 .

 

                        Hence (A) is the correct answer.

 

9.                     A line makes angles α, β, γ, δ with the four diagonals of a cube then cos2α + cos2β + cos2γ + cos2δ =

 

                         (A) 1                                                     (B) 4/3

 

                        (C) 3/4                                                  (D) 4/5

 hb                   The direction ratios of the diagonal  (1, 1, 1)

                        Direction cosine are (1/√3, 1/√3, 1/√3)

                        Similarly direction cosine of AS are (–1/√3, –1/√3, –1/√3)

                        BP are (1/√3, 1/√3, –1/√3) 

                        CQ are (1/√3, –1/√3, 1/√3)

 

                        Let l, m, n be direction cosines of the line

 

                        cosα = l+m+n/√3, cosβ = l–m–n/√3, cosγ = l+m–n/√3, cosδ = l–m+n/√3

 

                        cos2a + cos2β + cos2γ + cos2δ = 4(l2 + m2 + n2) / 3 = 4/3 ( since l2 + m2+ n2 = 1)

 

                        Hence (B) is the correct answer.

10.        The points (0, -1, -1), (-4, 4, 4), (4, 5, 1) and (3, 9, 4) are

 

                        (A) collinear                                         (B) coplanar

 

                        (C) forming a square                            (D) none of these

 

hb :     Equation of the plane passing through the points (0, -1, -1), (-4, 4, 4) and (4, 5, 1)

is   = 0                      …. (1)

 

           The point (3, 9, 4) satisfies the equation (1).

 

            Hence (B) is the correct answer.

 

11.        A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. The locus of the point common to plane through A, B, C parallel to coordinate planes is

 

                        (A) ayz + bzx + cxy = xyz                     (B) axy + byz + czx = xyz

 

                        (C) axy + byz + czx = abc                     (D) bcx + acy + abz = abc

 

hb :     Let the equation to the plane be x/α + y/β + z/γ = 1

 

                        => x/α + y/β + z/γ = 1 (the plane passes through a, b, c)

 

                        Now the points of intersection of the plane with the coordinate axes are A(α, 0, 0), B(0, β, 0) & C(0, 0, γ)

 

                        => Equation to planes parallel to the coordinate planes and passing through A, B & C are x = α, y = β and z = γ.

 

                        ∴ The locus of the common point is

 

                        a/x + b/y + c/z = 1 (by eliminating α, β, γ from above equation)

 

                        Hence (A) is the correct answer.

 

12.                    Consider the following statements:

 

                        Assertion (A): the plane y + z + 1 = 0 is parallel to x-axis.

 

                        Reason (R): normal to the plane is parallel to x-axis.

 

                        Of these statements:

 

                        (A) both A and R are true and R is the correct explanation of A

 

                        (B) both A and R are true and R is not a correct explanation of A

 

                        (C) A is true but R is false

 

                        (D) A is false but R is true

 

hb :                 Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0

 

                        but normal to the plane will be perpendicular to x-axis.

 

                        Hence (C) is the correct answer.

 

13.                    The equation of the plane containing the line x = α/l, y = β/m and z =γ/n is a(x – α) + b(y – β) + c(z – γ) = 0, where al + bm + cn is equal to

 

                        (A) 1                                                     (B) –1

 

                        (C) 2                                                     (D) 0

 

hb :                 Since straight line lies in the plane so it will be perpendicular to the normal at the given plane. Since direction cosines of straight line are l, m, n and direction ratios of normal to the plane are a, b, c. So, al + bm + cn = 0.

 

                        Hence (D) is the correct answer.

 

14.                    The shortest distance between the two straight lines x–4/3 / 2 = y+6/5 / 3 = z–3/2 / 4 and 5y+6/8 = 2z–3/9 = 3x–4/5 is

 

                        (A) √29                                                  (B) 3

 

                        (C) 0                                                     (D) 6√10

 

hb :                 Since these two lines are intersecting so shortest distance between the lines will be 0.

 

                        Hence (C) is the correct answer.

 

15.                    A straight line passes through the point (2, –1, –1). It is parallel to the plane 4x + y + z + 2 = 0 and is perpendicular to the line x/1 = y/–2 = z–5/1. The equations of the straight line are

 

                        (A) x–2/4 = y+1/1 = z+1/1                      (B) x+2/4 = y–1/1 = z–1/1

 

                        (C) x–2/–1 = y+1/1 = z+1/3                   (D) x+2/–1 = y–1/1 = z–1/3

 

hb :                 Let direction cosines of straight line be l, m, n

 

                        ∴ 4l + m + n = 0

 

                        l – 2m + n = 0

 

                        => l/3 = m/–3 = n/–9 => l/–1 = m/+1 = n/3

 

                        ∴ Equation of straight line is x–2/–1 = y+1/1 = z+1/3.

 

                        Hence (C) is the correct choice.

 

16.                    If centre of a sphere is (1, 4, –3) and radius is 3 units, then the equation of the sphere is

 

                        (A) x2 + y2 + z2 – 2x – 8y + 6z + 17 = 0                (B) 2(x2 + y2 + z2) – 2x – 8y + 6z + 17 = 0

 

                        (C) x2 + y2 + z2 – 4x + 16y + 12z + 17 = 0           (D) x2 + y2 + z2 + 2x + 8y – 6z – 17 = 0

 

hb :                 Equation of sphere will be (x – 1)2 + (y – 4)2 + (z + 3)2 = 9

 

                        Hence (A) is the correct answer.

  

17.                    If equation of a sphere is 2(x2 + y2 + z2) – 4x – 8y + 12z – 7 = 0 and one extremity of its diameter is (2, –1, 1), then the other extremity of diameter of the sphere will be

 

                        (A) (2, 9, –13)                                        (B) (0, 9, 7)

 

                        (C) (0, 5, 7)                                           (D) (2, 5, –13)

 

hb :                 The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then

 

                        x1+2/2 = 1, y1–1/2 = 2, z1+1/2 = –3

 

                        ∴ Required point is (0, 5, 7).

 

                        Hence (C) is the correct answer.

 

18.       The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (1, – 2, – 2), (0, 2, 1) is

 

            (A) (2/3, 1/3, 2/3)                                                (B) (2/3, –1/3, –2/3)

 

            (C) (2/3, 1/3, –2/3)                                             (D) (–2/3, –1/3, –2/3)

 

hb :     Let direction ratios of the required line be

 

            Therefore a - 2 b - 2 c = 0

 

            And 2 b + c = 0

 

            => c = - 2 b

 

            a - 2 b + 4b = 0 => a = - 2 b

 

            Therefore direction ratios of the required line are <- 2b, b, - 2b> = <2, - 1, 2>

 

            direction cosines of the required line

 

             = 

 

19.       The points (4, 7, 8), (2, 3, 4), (–1, –2, 1) and (1, 2, 5) are :

 

            (A) the vertices of a parallelogram            (B) collinear

 

            (C) the vertices of a trapezium                  (D) concyclic

 

hb :     Let A Ξ (4, 7, 8), B Ξ (2, 3, 4), C Ξ (- 1, - 2, 1), D Ξ (1, 2, 5)

 

            Direction cosines of AB Ξ (2/6, 4/6, 4/6) = (1/3, 2/3, 2/3)

 

            Direction cosines of CD Ξ (–2/6, –4/6, –4/6)

 

            = (1/3, 2/3, 2/3)

 

            So, AB parallel to CD

 

            Direction cosines of AD Ξ (3/√43, 5/√43, 3/√43)

 

            Direction cosines of BC Ξ (–3/√43, –5/√43, –3/√43)  

 

            = (3/√43, 5/√43, 3/√43)

 

            so, AD is parallel to BC.

 

            Therefore ABCD is a parallelogram.

 

20.       The equation of the plane parallel to the plane 4x – 3y + 2z + 1 = 0 and passing through the point (5, 1, – 6) is :

 

            (A) 4x - 3y + 2z - 5 = 0                                (B) 3x - 4y + 2z - 5 = 0

 

            (C) 4x - 3y + 2z + 5 = 0                               (D) 3x - 4y + 2z + 5 = 0

  

hb :     Equation of the plane parallel to the plane 4x - 3y + 2z + 1 = 0 is of the form of              

 

            4x – 3y + 2z + k = 0 again it passes through (5, 1, - 6)

 

            so, 20 - 3 - 12 + k = 0 Þ k = - 5

  

            Therefore required equation is 4x - 3y + 2z - 5 = 0. 

 

21.       A plane is passed through the middle point of the segment A (–2, 5, 1), B (6, 1, 5) and is perpendicular to this line. Its equation is :

 

            (A) 2x - y + z = 4                                           (B) 2x - y + z = 4

 

            (C) x - 3y + z = 5                                         (D) x - 4y + 2z = 5

 

hb :     Mid-point of A (- 2, 5, 1) and B (6, 1, 5) is (2, 3, 3)

 

            direction ratios of the line joining A and B is <2, - 1, 1>

 

            Therefore equation of the line perpendicular to AB and passing through (2, 3, 3) is

 

            2(x – 2) – 1(y – 3) + 1(z – 3) = 0 => 2x - y + z = 4

  

22.       A plane meets the co-ordinates axes in A, B, C such that the centroid of triangle ABC is (a, b, c). The equation of the plane is :

 

            (A) x/a + y/b + z/c = 3                                           (B) x/a + y/b + z/c = 1

 

            (C) x/a + y/b + z/c = 2                                           (D) None of these

  

hb :     The plane meets the co-ordinate axes at A, B, C such that centroid of the triangle ABC is (a, b, c)

 

            so, the plane cuts X-axis at (3 a, 0, 0)

 

            So, X-intercept = 3 a 

 

            The plane cuts Y-axis at (0, 3 b, 0)

 

            => Y-intercept = 3 b

 

            the plane cuts Z-axis at (0, 0, 3 c)

 

            => Z-intercept = 3 c

 

            Therefore required equation is x/3a + y/3b + z/3c = 1

 

            => x/a + y/b + z/c = 3.

  

23.       The radius of the sphere (x + 1)(x + 3) + (y – 2)(y – 4) + (z + 1)(z + 3) = 0 is:

 

            (A) √2                                                            (B) 2

 

            (C) √3                                                            (D) 3

  

hb :     (x + 1) (x + 3) + (y - 2) (y - 4) + (z + 1)(z + 3) = 0 is the given equation of sphere.

 

            So, end points of the diameter are

 

            (- 1, 2, - 1) and (- 3, 4, - 3)

 

            radius = √(–2+1)2 + (3–2)2 + (–2+1)2 = √3

  

24.       The sum of the direction cosines of a straight line is

 

            (A) zero                                                          (B) one

 

            (C) constant                                                  (D) none of these

 

hb :     cos α = l, cos β = m, cos γ = n

 

            sum of direction cosines cos α + cos β + cos γ

 

            = l + m + n

 

            which is constant.

   

25.       The angle between straight lines whose direction cosines are (1/2, –1/2, 1/√2) and(1/√3, 1/√3, –1/√3) is

 

            (A) cos–1 (2/√3)                                 (B) cos–1 (1/√6)

 

            (C) cos–1 (–1/√6)                               (D) none of these

 

hb :     

 

26.       Which one of the following is best condition for the plane ax + by + cz + d = 0 to intersect the x and y axes at equal angle

 

            (A) |a| = |b|                                                     (B) a = –b

 

            (C) a = b                                                         (D) a2 + b2 = 1

 

hb :     The plane a x + b y + c z + d = 0 intersects x and y axes at equal angles

 

            Therefore |cos α| = |cos β|

 

            => |l| = |m|

 

            => |a| = |b|.

 

27.       The equation of a straight line parallel to the x-axis is given by

 

            (A) x–a/1 = y–b/1 = z–c/1                                (B) x–a/0 = y–b/1 = z–c/1

 

            (C) x–a/0 = y–b/0 = z–c/1                                (D) x–a/1 = y–b/0 = z–c/0

 

hb :     Equation of straight line parallel to x-axis is

 

           x–a/1 = y–b/0 = z–c/0

 

            because l = cos α= 1, m = cos β = cos π/2 = 0, n = cos γ = 0. 

 

28.       If P (2, 3, –6) and Q (3, –4, 5) are two points, the direction cosines of line PQ are

 

            (A) –1/√171, –7/√171, –11/√171                         (B) 1/√171, –7/√171, 11/√171

 

            (C) 1/√171, 7/√171, –11/√171                             (D) –7/√171, –1/√171, 11/√171

 

hb :     P ≡ (2, 3, - 6), Q ≡ (3, - 4, 5)

 

            direction ratios = <1, - 7, 11>

 

            direction cosines = (1/√1+49+121, 7/√1+49+121, 11/√1+49+121)

 

            = (1/√171, –7/√171, 11/√171)

 

29.       The ratio in which yz–plane divide the line joining the points A(3, 1, –5) and B(1, 4, –6) is

 

            (A) –3 : 1                                                        (B) 3 : 1

 

            (C) –1 : 3                                                       (D) 1 : 3

 

hb :     A ≡ (3, 1, - 5), B ≡ (1, 4, - 6)

 

            Therefore 3+λ/λ+1 = 0 => λ = – 3

 

            Therefore required ratio is - 3 : 1

 

30.       A straight line is inclined to the axes of x and z at angels 450 and 600 respectively, then the inclination of the line to the y-axis is

 

            (A) 300                                                            (B) 450

 

            (C) 600                                                            (D) 900

 

hb :     A straight line is inclined to the axes of x and z at an angle 450 and 600

 

            l2 + m2 + n2 = 1

 

            => m2 = 1/4

 

            => m = 1/2

 

            => angle made by 600

 

31.       The angle between two diagonals of a cube is

 

            (A) cos θ = √3/2                                              (B) cos θ = 1/√2

 

            (C) cos θ = √1/3                                              (D) none of these

 

hb :     cos θ = –a2 + a2 + a2 / √a2 + a2 + a2 √a2 + a2 + a2 of side is a

 

            = 1/3.

 

32.       Given that A (3, 2, –4), B (5, 4, –6) and C (9, 8, –10) are collinear. The ratio in which B divides AC

 

            (A) 1 : 2                                                          (B) 2 : 1

 

            (C) –1 : 2                                                        (D) –2 : 1

 

hb :     9λ + 3/λ+1 = 5 => 9α - 5α = 2

 

            => λ = 1/2.

 

33.       If P1P2 is perpendicular to P2P3, then the value of k is, where P1(k, 1, –1), P2(2k, 0, 2) and P3(2 + 2k, k , 1)

 

            (A) 3                                                               (B) –3

 

            (C) 2                                                               (D) –2

 

hb :     Direction ratios of P1 P2 = - 1, 3>

 

            direction ratios of P2 P3 = <2, k, - 1>

 

            Therefore 2 k - k - 3 = 0

 

            => k = 3.

 

34.       A is the point (3, 7, 5) and B is the point (–3, 2, 6). The projection of AB on the line which joins the points (7, 9, 4) and (4, 5, –8) is

 

            (A) 26                                                             (B) 2

 

            (C) 13                                                             (D) 4

 

 hb :    Distances of the line joining (7, 9, 4) and (4, 5, - 8) is < 3/13, 4/13, 12/13 >

 

            Therefore required projection is 26/13 = 2 (B)

 

Exercise # 1 

 

35.       The shortest distance of the point from  the x-axis is equal to(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.1)                                    

 

            (A) √x12 + y1

2                                   (B) √x12 + z1

2

 

            (C) √y12 + z1

2                                   (D) None of these

 

Ans.    (C)

 

hb . Foot of perpendicular drawn from P to x-axis will have its coordinates as (x, 0, 0).

 

 Required distance = √y12 + z1

2 

 

36.       The point of intersection of the xy plane and the line passing through the points and  is:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.4)   

 

            (A)       (–13/5, 23/5, 0)                                   (B)       (13/5, 23/5, 0)

 

            (C)       (13/5, –23/5, 0)                                   (D)       (–13/5, –23/5, 0)

 

Ans.    (B)

 

hb . Direction ratios of AB are 2, -3, 5.

 

Thus equation of AB is, x–3/2 = y–4/–3 = z–1/5

  

For the point of intersection of this line with xy–plane, we have

 

Z = 0

=> x–3/2 = y–4/–3 = –1/5

=> x=3 –2/5 = 13/5, y = 4 + 3/5 = 23/5

   

Hence, the required point is (15/5, 23/5, 0)

 

37.       The projections of the line segment AB on the coordinate axes are –9, 12, -8 respectively. The direction cosines of the line segment AB are:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.5)       

 

            (A) –9/17, 12/17, –8/17                    (B) –9/289, 12/289, –8/289

 

            (C) –9/√17, 12/√17, –8/√17              (D) None of these

 

Ans.    (A)

 

            Length of segment AB = √81 + 144 + 64 = 17

 

            Thus direction cosines of AB are

 

            .–9/17, 12/17, –8/17

 

38.       The direction cosines of two mutually perpendicular lines are l1,m1,n1 andl2,m2,n2. The direction cosines of the line perpendicular to both the given lines will be:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.6)                                    

 

            (A)       l1+ l1, m1+m2, n1+n2                 (B)       l2 – l2, m1– m2, n1+n2

 

            (C)       l1 l2, m1m2, n1n2                       (D)      m1n2 – m2n1, n1 l2 – n2, l1m2 – l2m1

 

Ans.    (D)

 

            Let the direction cosine of the required line be l, m and n.

 

            We must have,

 

            ll1 + mm1 + nn1 = 0, ll2 + mm2 + nn2 = 0

  

            

 

            We have l1l2 + m1m2 + n1n2 = 0

 

            Thus

 

            Σ(m1n2–n1m2)2 = (Σl2)2 (Σl2)2 – Σl1l2 = 1

 

            => l = m1n2 – n1m2, m = n1 l2 – n2 l1,

 

            n = l1m2 – l2m1

 

39.       A directed line segment angles α,β,γ with the coordinate axes. The value ofΣcos2α is always equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.7)

 

            (A)       -1                                                         (B)       1

 

            (C)       -2                                                         (D)       2

 

Ans.    (A)

 

            Σcos2α = Σ(2cos2α–1)

 

            = 2Σl2–3

 

            = –1

 

 

 

40.       The locus represented by xy + yz = 0 is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.9)

 

            (A)       A pair of perpendicular lines         (B)       A pair of parallel lines

 

            (C)       A pair of parallel planes                  (D)       A pair of perpendicular planes

 

Ans.    (D)

 

            xy + yz = 0

 

            => x(y+z) = 0

 

            => x=0, y+z = 0

 

            Thus it represents a pair of planes

 

            X = 0, y + z = 0

 

            that are clearly mutually perpendicular.

 

41.       The plane 2x – 3y + 6z – 11 = 0 makes an angle (a) with x-axis. The value of ‘a’ is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.10)         

 

            (A)       √3/2                                                     (B)       √2/3

 

            (C)       2/7                                                      (D)       3/7

 

Ans.    (C)

 

            If ‘θ’ be the angle between the plane and x–axis, then

 

            sin θ 2/√4+9+36 = 2/7

 

            => θ = sin–1 (2/7)            => a = (2/7)

  

42.       The planes x + y = 0, y + z = 0 and x + z = 0:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.12)        

 

            (A)       meet in a unique point                   (B)       meet in a unique line

 

            (C)       are mutually perpendicular            (D)       none of these

 

Ans.    (B)

 

            Clearly, given planes have a common line of intersection namely the z-axis.

  

43.       The equation of a plane passing through (1, 2, -3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.13)   

 

            (A)       3x + y + 5/3 z = 0                (B)       4x + y + 2z = 0

 

            (C)       3x – y + z/3                        (D)       x + y + z = 0

 

Ans.    (D)

 

            Let the required plane be

 

            Ax + by + cz = 0.

 

            We have 3a – 5b + 2c = 0, a + 2b – 3c = 0

 

            => a/15–4 = b/2+9 = c/6+5

 

            => a : b : c = 11 : 11 : 11

 

            Thus plan is        x + y + z = 0

 

44.       The direction ration of a normal to the plane passing through (1, 0, 0), (0, 1, 0) and making an angle π/4 with the plane x + y = 3 are:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.15)

 

            (A)       (1, √2, 1)                                  (B)       (1, 1,√2)

 

            (C)       (1,1,2)                                      (D)       (√2, 1, 1)

 

Ans.    (B)

 

            Let the plane be x/a + y/b + z/c = 1

 

            => 1/a = 1, 1/b = 1

 

            .=> a = b = 1

 

            Also, 

 

            => c = ± 1/√2

 

            Thus direction rations are

 

            (1,1,√2)   or     (1,1,–√2)

 

45.       The equation of a plane passing through the line of intersection of the planes x + y + z = 5, 2x – y + 3z = 1 and parallel to the line y = z = 0 is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.16)         

 

            (A)       3x – z = 9                                           (B)       3y – z = 9

 

            (C)       x – 3z = 9                                           (D)       y – 3z = 9

 

Ans.    (B)

 

            Plane will be in the form

 

            (x + y + z – 5) + a(2x – y + 3z) –1) = 0 i.e., x(1 + 2a) + y(1 – a) + z(1 + 3a) = 5 + a

 

            It is parallel to the line y = z = 0.

 

            Since, (1 + 2a) = 0

 

            ∴ a = – 1/2

 

            Thus required plane is

 

            3/2y – 1/2 z = 9/2

 

            i.e.,    3y – z = 9

 

46.       The angle between lines whose direction cosines are given by l + m + n = 0, l2 + m2 – n2 = 0, is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.17)         

 

            (A)       π/2                                                        (B)       π/3

 

            (C)       π/6                                                        (D)       None of these

 

Ans.    (D)

 

            .l + m + n = 0, l2 + m2 – n2 = 0

 

            We also have

 

            l2 + m2 + n2 = 1

 

            => 2n2 = 1

 

            => n = 1/√2, – 1/√2

 

            Also,    l2 + m2 = n2 = (–(l+m))2

 

            => lm = 0 and l + m = ± 1/√2

 

            Hence, direction cosines are lines are

 

            (1/√2, 0, –1/√2), (–1/√2, 0, 1/√2)

 

            (0, 1/√2, –1/√2), (0, –1/√2, 1/√2)

 

            Angle between these lines in both cases is zero.

 

47.       Centroid of the tetrahedron OABC, where , ,  and O is the origin is (1, 2, 3). The value of  is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.3Q.19)                                  

 

            (A)       75                                                        (B)       80

 

            (C)       121                                                     (D)       None of these

 

Ans.    (A)

 

            We have

 

                        4 = a + 1 + 2 + 0,

 

            => a = 1,

 

                        8 = 2 + b + 1 + 0

 

            => b = 5,

 

                        12 = 3 + 2 + c + 0.

 

            => c = 7,

 

            ∴ a2 + b2 + c2 = 1 + 25 + 49 = 75

 

48.       The equation of the plane passing through the points (2, -1, 0), (3, -4, 5) and parallel to the line 2x = 3y = 4z is:(Ref.P.K.Sharma_Three Dimen._P.C6.3Q.20)    

 

            (A)       125x – 90y – 79z = 340                   (B)       32x – 21y – 36z = 85

 

            (C)       73x + 61y – 22z = 85                       (D)       29x – 27y – 22z = 85

 

Ans.    (D)

 

            Give line is 2x = 3y = 4z

 

            i.e.,    x/6 = y/4 = z/3

 

            Let the plane be

 

            Ax + by + cz = 1.

 

            We have

 

                        6a + 4b + 3c = 0

 

                        2a – b = 1

 

                        3a – 4b + 5c = 1.

 

            => a = 29/85, b = 27/85, c = –22/85

 

            Thus equation of plane is

 

            29x – 27y – 22z = 85.

49.       A plane passes through the point . If the distance of this plane from the origin is maximum, then it’s equation is:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.1)       

 

            (A)       x + 2y – 3z + 4 = 0                            (B)       x + 2y + 3z = 0

 

            (C)       2y - x + 3z = 0                                    (D)       x - 2y + 3z = 0

 

Ans.    (B)

 

            Clearly in this case OA will be a normal to the plane.

 

            Direction cosine of segment OA are

 

            1/√14, 2/√14, 3/√14

 

            and  .OA = √1 + 4 = 0 = √14

 

            Thus the equation of plane is

 

            x + 2y + 3z = 14.

 

50.       The shortest distance of the plane 12 + 4y + 3z = 327, from the sphere , is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.3)          

 

            (A)       39 units                                              (B)       26 sq. units

 

            (C)       13 units                                              (D)       None of these

 

Ans.    (C)

 

            Center and radius of given sphere are (–2, 1, 3) and 13 unit respectively.

 

            Now, distance of center of the sphere from the given plane

 

            = |–24 + 4 + 9 – 327| / √122 + 42 + 32 = 26 units

 

            ∴Shortest distance = (26 – 13) = 13 units.

  

51.       The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.5)           

 

            (A)       (a + a’)(b + b’)(c + c’)                        (B)       aa’ + cc’ + 1 = 0

 

            (C)       aa’ + bb’ + cc’ + 1 = 0                       (D)       (a + a’) (b + b’) (c + c’) + 1 = 0

 

Ans.    (B)

 

            Give lines are

 

            x–b/a = y = z – d/c and x–b' / a' = y = z–d'/c'

 

            These lines will be mutually perpendicular, provided

 

                        a.a' + 1.1' + c.c' = 0

 

            .=> a.a' + c.c' + 1= 0

  

52.       The straight lines x–2/1 = y–3/1 = z–4/–k and x–1/k = y–4/2 = z–5/1, will intersect provided:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.6)           

 

            (A)       k = {3, -3}                                            (B)       k = {0, -1}

 

            (C)       k = {-1, 1}                                            (D)       k = {0, -3}

 

Ans.    (D)

 

            Any point on the first line can be takes as

 

            P1 = (r1 + 2, r1 + 3, – kr1+4)

 

            These lines will intersect if for some r1 and r2 we have

 

            r1 + 2 = kr2 + 1,

 

            r1 + 3 = 2r2 + 1,           

 

            –kr1 + 4 = r2 + 5,

 

            ∴ r1 + kr2 + 1 = 0, r1 = 2r2 + 1,           

 

            => r2 = 2/k–2, r1 = k+2 / k–2

 

            putting these values in the last condition, we get

 

                        k2 + 3k = 0

 

            .=> k = {–3,0}

  

53.       The plane ax + by + cz = d, meets the coordinate axes at the points, A, B and C respectively. Area of triangle ABC is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.7)   

 

            (A)       d2√a2 + b2 + c2 / |abc|               (B)       d2√a2 + b2 + c2 / 2|abc|

 

            (C)       d2√a2 + b2 + c2 / 4 |abc|            (D)       None of these

 

Ans.    (B)

 

            A = (d/a, 0, 0), B = (0, d/a, 0), c = (0,0, d/a)

 

            Area of triangle OAB = Δ1 = 1/2 d2/|ab|

 

            Area of triangle OBC = Δ2 = 1/2 d2/|bc|

 

            Area of triangle OAC = Δ3 = 1/2 d2/|ac|

 

            If area of triangle ABC be Δ, then

 

 

54.       Equation of the plane passing through (-1, 1, 4) and containing the line x–1/3 = y–2/1 = z/5,  is:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.9)           

 

            (A)       9x – 22y + 2z + 23 = 0                     (B)       x + 22y + z = 25

 

            (C)       9x + 22y - 3z = 1                               (D)       22y – 9x + z = 35

 

Ans.    (D)

 

            Equation of any plane containing the line x–1/3 = y–2/1 = z/5 will be

 

            a(x–1) + b (y–z) + cz = 0

 

            where,    3a + b + 4c = 0                                                                              …. (i)

 

            It is given that plane passes through (–1, 1, 4).

 

            ∴         –2a – b + 4c = 0                                                                                …. (ii)

 

            From (i) and (ii), we get

 

            a/–9 = b/22 = c/1.

 

            Thus the equation of required plane is,

 

                        –9(x – 1) + 22(y – 2) + z = 0

 

            i.e.,      22y – 9x + z = 35

 

55.       Equation of the plane containing the lines x/1 = y–2/3 = z+4/–1 and x–4/2 = y/3 = z/1 is,:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.10)        

 

            (A)       x + y – 4z = 6                                     (B)       x - y + 4z = 6

 

            (C)       x + y + 4z = 6                                     (D)       None of these

 

Ans.    (D)

 

            Equation of any plane containing the line x/1 = y–2/3 = z+4/–1 is ax + b (y–2) + c (z+4) = 0  

 

            where a + 3b – c = 0

 

            This plane will also contain the second line if  

 

                        2a – 3b + c = 0

 

            and     4a b(0 – 2) + c(0 + 4) = 0

 

            Solving these equation, we get

 

                        a = 0, b = 0, c = 0.

 

            That means the given lines are non–coplanar.

 

56.       Equation of the plane such that foot of altitude drawn from (-1, 1, 1) to the plane has the coordinate (3, -2, -1), is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.11)       

 

            (A)       x + y + z = 0                                       (B)       4x - 3y – 2z = 20

 

            (C)       3x + y – z = 8                                     (D)       4x + 3y – z = 7

 

Ans.    (B)

 

            Clearly, the direction ratio of the plane are

 

                        4, –3, –2

 

            Thus equation of plane will be

 

                        4x – 3y – 2z = d

 

            It will necessarily pass through (3, –2, –1)

 

            i.e,,      d = 12 + 6 + 2 = 20

 

            Thus the equation of plane is

 

                        4x – 3y – 2z = 20.

 

57.       The distance of the point (-1, 2, 6) from the line x–2/6 = y–3/3 = z+ 4/–4, is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.13)         

 

            (A)       7 units                                                (B)       9 units

 

            (C)       10 units                                              (D)       12 units

 

Ans.    (A)

 

            Any point on the line is

 

                        P = (6r1 + 2, 3r1 + 3, –4r1 –4).

 

            Direction ration of the line segment PQ, where Q = (–1, 2, 6), are

 

                        6r1 + 3, 3r1 + 1, – 4r, – 10.

 

            If ‘P’ be the foot of altitude drawn from Q to the given line, then

 

                        6(6r1 + 3) + 3(3r1 + 1) + 4(4r1 + 10) = 0.

 

            =>        r1 = –1.

 

            Thus, P = (–4, 0, 0)

 

            ∴ Required distance = √9 + 4 + 36

 

                                                    = 7 units.

 

58.       The point of intersection of the lines x+1 / 3 = y + 3/5 = z + 5 / 7 and x–2/1 = y–4/3 = z–6/5 is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.14)         

 

            (A)       (1/3, –1/3, –2/3)                  (B)       (1/2, –1/2, –3/2)

 

            (C)       (1/2, 1/2, 1/2)                     (D)       (1/3, 1/3, 1/3)

 

Ans.    (B)

 

            Any point on the first line is

 

                        (3r1 – 1, 5r1 – 3, 7r1 – 5)

 

            and any point on the second line is

 

                        (r2 + 2, 3r2 + 4, 5r2 + 6).

 

            At the point of intersection, we must have

 

                        3r1 – 1 = r2 + 2,

 

                        5r1 – 3 = 3r2 + 4.

 

                        7r1 – 5 = 5r1 + 6.

 

            Thus,   r1 = 1/2, r2 = –3/2

 

            Hence required point is  i.e. (1/2, –1/2, –3/2)

59.       The shortest distance between the line x + y + 2z – 3 = 2x + 3y + 4z – 4 = 0 and the z-axis is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.15)         

 

            (A)       1 unit                                                  (B)       2 units

 

            (C)       3 units                                                (D)       4 units

 

Ans.    (B)

 

            We have,

 

            x + y + 2z – 3 = 0, x + 2z – 2 + 3/2 y = 0

 

            Solving these equations, we get

 

                        Y = –2.

 

            Thus required shortest distance is 2 units.

 

60.       The length of projection, of the line segment joining the points (1, -1, 0) and (-1, 0, 1), to the plane 2x + y + 6z = 1, is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.17)  

 

            (A)       √255/61                                                (B)       √237/61

 

            (C)       √137/61                                               (D)       √155/61

 

Ans.    (B)

 

            Let A = (1, –1, 0), B = (–1, 0, 1).

 

            Direction rations of segment AB are

 

                        2, –1, –1.

 

            If ‘θ’ be the acute angle between segment AB and normal to plane,

 

             .

 

            Length of projection

 

                        = (AB) sin θ

 

                        =  √6.√1–9/246 = √237/61 units

 

61.       Reflection of the line x–1/–1 = y–2/3 = z–3/1 in the plane x + y + z = 7 is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.18)                                  

 

            (A)       x–1/3 = y–2/1 = z–4/1                         (B)       x–1/–3 = y–2/–1 = z–4/1

 

            (C)       x–1/–3 = y–2/1 = z–4/–1                    (D)       x–1/3 = y–2/1 = z–4/1

 

Ans.    (C)

 

            Given line passes through (1, 2, 4) and this point also lies on the given plane.

 

            Thus required line will be in the form of x–1/l = y–2/m = z–4/n.

 

            Any point on the given line is

 

                        (–r1 + 1, 3r1 + 2, r1 + 4).

 

            If r1 = 1, this point becomes P = (0, 5, 5).

 

            Let Q = (a, b, c) be the reflection of ‘P’ in the given plane, then

 

                        a/2.1 + b+5/2.1 + 5+c/2.1 = 7

 

            i.e,       a + b + c = 4,

 

            and     a/1 = b–5/1 = c–5/1 = λ(say)

 

            => a = λ, b = 5 + λ, c = 5 + λ

 

            => 10 + 3λ = 4

 

            => λ = –2

 

            Thus, Q = (–2, 3, 3)

 

            Hence direction rations of reflected line are

 

                        –3, 1, –1

 

            Thus it’s equation is

 

                        x–1/–3 = y–2/1 = z–4/–1