Conic Section

Section 1 1

Conic Section

Conic Section

PAIR OF LINES THROUGH ORIGIN Section - 1

1.1 General Equationax2 + 2hxy + by2 = 0 is a homogeneous equation of second degree that represents two straight line passingthrough the origin. The individual equations of these straight lines can be obtained by factorising the expression ax2 + 2hxy + by2 = 0 into two real linear factors. The equation represents two lines in real XY plane onlyif h2 – ab 0.

1.2. Slopes of the linesax2 + 2hxy + by2 = 0

Let m1, m2 be the slopes of the lines through origin expressed byax2 + 2hxy + by2 = 0 . . . (i)

The lines are y = m1x and y = m2x and the equation of pair is (y – m1x) (y – m2x) = 0 m1m2x

2 – xy (m1 + m2) + y2 = 0 . . . (ii)

1 2 1 2 1

2m m m m

a h b

1 22hm mb

and 1 2

am mb

We can solve for m1 and m2 to get

2 2

1 2andh h ab h h ab

m mb b

Note : Again that h2 – ab 0 is the necessary condition for the equation representing real lines.

What does the equation x2 – 5xy + 4y2 = 0 represents?

SOLUTION :

x2 – 5xy + 4y2 = 0 x2 – 4xy – xy + 4y2 = 0 (x – 4y) (x – y) = 0 The equation represent two straight lines through origin whose Equations are :

x – 4y = 0 and x – y = 0.

Illustration - 1

Conic Section

Section 12

Conic Section

1.3 Angle between the linesax2 + 2hxy + by2 = 0

If the acute angle between lines ax2 + 2hxy + by2 = 0

then

22 21 2 1 21 2

1 2 1 2

4 4( ) 4

tan1 1 |1 / |

h am m m m bm m b

m m m m a b

2

1 2tan

h aba b

Hence we also have following results.

(i) The lines ax2 + 2hxy + by2 = 0 are mutually perpendicular if and only if a + b = 0.i.e. coefficient of x2 + coefficient of y2 = 0.

(ii) The lines ax2 + 2hxy + by2 = 0 are coincident if h2 – ab = 0.

1.4 Angle bisector of linesax2 + 2hxy + by2 = 0.

Let m1, m2 be the slopes of lines ax2 + 2hxy + by2 = 0. lines are y – m1 x = 0 and y – m2 x = 0 [m1 + m2 = – 2h/b and m1m2 = a/b] bisectors of angles are :

2 22 21 22 1 1 22 2

1 2

1 1 01 1

y m x y m x m y m x m y m xm m

On simplification we get :– y2 (m1 + m2) + x2 (m1 + m2) – 2xy (1 – m1m2) = 0

2 2 1 2

1 2

2 (1 )xy m mx ym m

2 2 2 (1 / )

2 /xy a bx y

h b

The equation of Bisectors is 2 2x b xya b h

.

Conic Section

Section 1 3

Conic Section

1.5 Pair of lines perpendicular to the linesLet L1 and L2 be the lines ax2 + 2hxy + by2 = 0.Let P1 be the line Perpendicular to L1 and P2 be the line perpendicular to L2.We have to find equation of P1P2.Let L1 be y – m1x = 0 and L2 be y – m2x = 0 P1 is m1y + x = 0 and P2 is m2 y + x = 0 Pair P1P2 is (m1y + x) . (m2y + x) = 0 m1 m2 y

2 + xy (m1 + m2) + x2 = 0

2 22 0a hy xy x

b b

bx2 – 2hxy + ay2 = 0 is the equation of the pair of lines perpendicular to the pairs of lines

ax2 + 2hxy + by2 = 0.

Note : By interchanging the coefficients of x2 and y2 and reversing the sign of the xy term, we can get equationof P1P from L1L2.

Find the area formed by the triangle whose sides are y2 – 9xy + 18x2 = 0 and y = 9.

SOLUTION :y2 – 9xy + 18x2 = 0

(y – 3x) (y – 6x) = 0 The sides of the triangle are : y – 3x = 0 and y – 6x = 0 and y – 9 = 0.

By solving these simultaneousy, we get the vertices as :A (0, 0) ; B (3/2, 9) ; C (3, 9)

0 0 11 3 27Area 9 1 sq. units.2 2 4

3 9 1

Find the angle between the lines x2 + 4y2 – 7xy = 0.

SOLUTION :

21 1 1

7 1(4)22 2 33tan tan tan

1 4 5h aba b

.

Illustration - 2

Illustration - 3

Conic Section

Find the equation of pair of lines through origin which form an equilateral triangle withthe lines Ax + By + C = 0. Also find the area of this equilateral triangle.

SOLUTION :Let PQ be the side of the equilateral triangle lying on the line Ax + By + C = 0. Let m be the slope of theline through origin and making an angle of 60° with Ax + By + C = 0. m is the slopes of OP or OQ.As the triangle is equilateral, the line Ax + By + C = 0 makes an angle 60° with OP and OQ.

( / )tan 60

1

m A BAm

B

2

3 mB AB mA

. . . (i)

This quadratic will give two values of m which are slope of OP and OQ. As OP and OQ pass through origin,their equations can be taken as :

y = mx . . . (ii)

As we have to find the equation of OP and OQ, we will not find values of m but we will eliminate m between(i) and (ii) to directly get the equation of the pairs of lines : OP and OQ.

2

//

By x AB yA x

2

3 By AxBx yA

3 (B2x2 + y2A2 – 2ABxy) = (B2x2 + A2x2 + 2ABxy) (A2 – 3B2) x2 + 8ABxy + (B2 – 3A2) y2 = 0 is the pair of lines through origin making an equilateral

triangle (OPQ) with Ax + By + C = 0.

Area of 2

23 3(side)4 4 sin 60

POPQ

[where P = altitude]

22 2

2 22 22 2

3 4 1 1 | |area4 3 3 3 3 ( )

C CP PA BA B

If a pair of lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 is such that each pair bisects theangle between the other pair, prove that pq = – 1.

SOLUTION :The pair of bisectors for x2 – 2pxy – y2 = 0 is

Illustration - 4

O

Y

X

PQ

Illustration - 5

Section 14

Conic Section

2 2

1 ( 1)x y xy

p

2 2 2xyx yp

2 22 0x xy yp

As 2 22 0x xy yp

and x2 – 2qxy – y2 = 0 coincide,

we have :1 2 / 11 2 1

pq

2 2qp

pq = – 1.

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

SOLUTION :Let L1L2 be one pair and P1P2 be other, if the angle between L1P1 is equal to the angle between L2P2, thepair of bisectors of L1L2 is same as that of P1P2. Pair of bisectors of P1P2 is :

2 2

( ) ( )x y xy

a b h

2 2x y xya b h

This is same as the bisector pair of L1L2.Hence the statement is proved.

Show that the orthocentre of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and lx + my

= 1 is given by 2 22x y a bl m am hlm bl

.

SOLUTION :Let the triangle be OBC where O is origin and BC is the line lx + my = 1. OB, OC are ax2 + 2hxy + by2 = 0.

The altitude from O to BC is y – 0 = ( 0)m xl

mx – ly = 0

Let OB : y – m1x = 0 and OC : y – m2x = 0.

1

1 1

1 , mBl mm l mm

Y

X

L1

P1

OP2

L2

Illustration - 6

Illustration - 7

Y

X

C

B

O

Section 1 5

Conic Section

Slope of altitude from B to OC is – 1/m2

The equation of altitude from B is

1

1 2 1

1 1my xl mm m l mm

(l + mm1) x + m2 (l + mm1) y – (1 + m1m2) = 0 . . . (i) mx – ly + 0 = 0 . . . (ii)Solving (i) and (ii), we get orthcentre :

1 1 2 1 2 1

1(1 ) (1 ) ( ) ( )

x yl mm m m m l l mm mm l mm

2 21 2 1 2

(1 / )( )

x y a bl m l m m m lm m m

2 2 2x y a bl m bl am hlm

[using values of m1m2 and m1 + m2]

PAIR OF LINES IN GENERAL Section - 2

2.1 General formax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is the general second degree equation and represents conics (pairsof lines, parabola, hyperbola, Ellipse and circle).This equation represents a pair of lines in the XY plane if the expression on LHS can be factorized into tworeal linear factors in terms of x, y. The condition for this is :

abc + 2fgh – af 2 – bg2 – ch2 = 0 [and h2 – ab 0]

or in determinant form, 0a h gh b fg f c

[and h2 – ab 0]

Section 26

Conic Section

Prove that the equation 6x2 – xy – 12y2 – 8x + 29y – 14 = 0 represent a pair of lines. Findthe equations of each line.Illustration - 8

SOLUTION :Using the condition given above in (i), we have :

6 1/ 2 4291/ 2 12 02

294 142

a h gh b fg f c

Hence the given equation represents a pair oflines. To find the equation of each line, we haveto factories the LHS we first factories the seconddegree term.

6x2 – xy – 12y2 = 6x2 – 9xy + 8xy – 12y2

= (3x + 4y) (2x – 3y).Let the factors be 3x + 4y + c1 and 2x – 3y + c2. 6x2 – xy – 12y2 – 8x + 29y – 14

= (3x + 4y + c1) (2x – 3y + c2).Comparing the coefficients of x and y, we get :

– 8 = 3c2 + 2c1 and 29 = 4c2 – 3c1On solving for c1 and c2, we get c2 = 2 and c1 = – 7 The lines are 3x + 4y – 7 = 0 and 2x – 3y + 2

= 0.

2.2 Angle between the lines

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is = 2

1 2tan

h aba b

.

2.3 Point of intersection of lines

ax2 = 2hxy + by2 + 2gx + 2fy + c = 0 is 2 2,bg hf af ghh ab h ab

.

2.4 Equation of pair of lines joining origin to the point of intersection of lx + my + n =0 and the conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0Let the line lx + my + n = 0 cuts the curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 in pionts P and Q.The equations of pair of lines OP and OQ (where O is the origin), joining origin O to P and Q, can be foundby making the equation of curve homogeneous with the help of equation of the line lx + my + n = 0.First write the equation of curve as :

ax2 + 2hxy + by2 + (2gx + 2fy) . 1 + c (1)2 = 0

Since write the equation of line is : lx + my + n = 0 1lx my

n

Substituting the value of 1 in above equation, we get :

ax2 + 2hxy + by2 – (2gx + 2fy) 2

0lx my lx mycn n

as the equation of pair of lines OP and OQ.

Section 2 7

Conic Section

Find the equation of the lines joining the origin to the points of intersection of the line 4x –3y = 10 with the circle x2 + y2 + 3x – 6y – 20 = 0 and show that they are perpendicular.

SOLUTION :

We will make the equation of circle homogeneous by using 1 = 4 3

10x y

.

The pair of lines is 2

2 2 4 3 4 3(3 6 ) 20 010 10

x y x yx y x y

.

10x2 + 15xy – 10y2 = 0.Since the coefficient x2 + coeff. of y2 = 10 – 10 = 0. They are perpendicular.The equation can also be asked as :[Show that the chord 4x – 3y = 10 of the circle x2 + y2 + 3x – 6y – 20 = 0 subtends a right angle at origin.]

A variable chord of the circle x2 + y2 + 2gx + 2fy + c = 0 always subtends a right angle atorigin. Find the locus of the foot of the perpendicular drawn from origin to this chord.

SOLUTION :Let the variable chord be lx + my = 1 [where l, m are changing quantities]

[i.e. parameters that change with the moving chord]Let P (x1, y1) be the foot of the perpendicular from origin to the chord.If AB is the chord, the equation of OA and OB is x2 + y2 + (2gx + 2fy) (lx + my) + C (lx + my)2 = 0. x2 (1 + 2gl + cl2) + y2 (1 + 2fm + cm2) + (2gm + 2f l + 2clm) xy = 0As OA perpendicular OB, (1 + 2gl + cl2) + (1 + 2fm + cm2) = 0 . . . (i)P lies on AB. lx1 + my1 = 1 . . . (ii)

OP AB 1

11y l

x m

. . . (iii)

We have to eliminate l, m using (i), (ii), (iii).From (ii) and (iii), we get :

12 21 1

ymx y

and

12 21 1

xlx y

Now from (i), we get :2 2

1 1 1 12 2 2 2 2 2 2 2 2 21 1 1 1 1 1 1 1

2 22 0( ) ( )

gx cx fy cyx y x y x y x y

Illustration - 9

Illustration - 10

Section 28

Conic Section

2 (x12 + y1

2) + 2gx1 + 2fy1 + c = 0

The locus of P is : 2 (x2 + y2) + 2gx + 2fy + c = 0.

PARABOLA Section - 3

3.1 Section of a right Circular Cone by different planes(i) Section of a right circular cone by a plane which is passing through its vertex is a pair of straight lines,

lines always passes through the vertex of the cone.Here VA and VB are lines where V is vertex of cone.

(ii) Section of a right circular cone by a plane which parallel to its base is a circle.

V

Plane

QP OA

B

Plane

QP O

Circle

V

Section 3 9

Conic Section

(iii) Section of a right circular cone by a plane which is parallel to a generator of the cone is a parabola.

(iv) Section of a right circular cone by a plane which is not parallel to any generator and not parallel orperpendicular to the axis of the cone is an ellipse.

(v) Section of a right circular cone by a plane which is parallel to the axis of the cone is a hyperbola.

QP O

V

Plane

Parabola

Generator

QP O

V

Plane

Parabola

Hyperbola

Axis

Section 310

Conic Section

3.2 Definition of ConicConic section is defined as a locus of a point (P) in a plane which moves in such a way, that ratio of itsdistances from a fixed point (called focus of conic, say S) and from a fixed line (called directrix of conic,say L) is constant (called the eccentricity of conic, denoted by ‘e’).

i.e.PSPN = constant (eccentricity e)

In other words ‘a conic section may be defined as the locusof a point which moves such that the ratio of its distance fromfocus to its distance from directrix is a constant’.Let P (h, k), S (x1, y1) and L Ax + By + C = 0

PSPN = e PS = e PN

2 21 1 2 2

( ) ( ) Ah Bk Ch x k y eA B

2

2 2 21 1 2 2

( )( ) ( ) Ah Bk Ch x k y eA B

Hence, locus of moving point P (h, k), i.e. equation of conic is given by(A2 + B2) {(x – x1)

2 + (y – y1)2} = e2 (Ax + By + C)2

Eccentricity and Shape of Conic Section

e = 1 : Parabolae < 1 : Ellipsee > 1 : Hyperbolae = 0 : Circle (special case of ellipse)

Note : that as ‘e’ is the ratio of distances, so ‘e’ cannot be negative.

N P (Moving point)

S (Fixed point of )Focus

L (Fixed line or )Directrix

Section 3 11

Conic Section

Find the locus of a point, which moves such that its distance from the point (0, –1) is twiceits distance from the line 3x + 4y + 1 = 0.Illustration - 11

SOLUTION :Let P (x1, y1) be the point, whose locus isrequired.Its distance from (0, –1) = 2 × Its distance fromthe line 3x + 4y + 1 = 0,

2 21 1( 0) ( 1)x y

1 12 2

|3 4 1|2(3 4 )

x y

2 21 1 1 15 ( 1) 2 | 3 4 1|x y x y

Squaring and simplifying, we have2 21 1 125 ( 2 1)x y y

= 4 (9x12 + 16 y1

2 + 1 +24x1y1 + 6x1 + 8y1)

2 21 1 1 1 111 39 96 24 18x y x y x y

21 0 Hence the locus of (x1, y1) is

11x2 + 39y2 + 96xy + 24x1 – 18y – 21 = 0.

What conic does the equation 25 [x2 + y2 – 2x + 1] = (4x – 3y + 1)2 represent?Illustration - 12

SOLUTION :The given equation is

25 [x2 + y2 – 2x + 1] = (4x – 3y + 1)2 . . . (i)Write the right hand side of this equation, so thatit appears in perpendicular distance form to get :

R.H.S. = 2

22 2

4 3 1(4 3 1) 25(4 3 )

x yx y then equation (i) can be re-written as

22 2

2 24 3 125[( 1) ( 0) ] 25

4 3

x yx y

2 22 2

| 4 3 1|( 1) ( 0)4 3

x yx y

Here e = 1.Thus the given equation represents a parabola. Itmay noted that (1, 0) is the focus and 4x – 3y + 1= 0 is the directrix of the parabola.

3.3 Conditions for Second Degree Equation to Represent Various Conic SectionThe equation of conic represented by the general equation of second degree in two variables is given by

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0The discriminant of the above equation is represtned by

2 2 22a h g

abc fgh af bg ch h b fg f c

Section 312

Conic Section

Case I : When = 0In this case above equation represents the Degenerate conic. The nature of conic is given below :

Condition Shape = 0 and h2 = ab A pair of coincident straight lines= 0 and h2 > ab A pair of real straight lines= 0 and h2 < ab A pair of Imaginary straight line

Case II : When 0In this case above equation represents the Non-degenerate conic. The nature of conic is given below :

Condition Shape 0, h = 0, a = b a circle 0, h2 = ab a parabola 0, h2 < ab an Ellipse 0, h2 > ab a Hyperbola 0, h2 > ab and a + b = 0 a rectangular hyperbola

What conic does 13x2 – 18xy + 37y2 + 2x + 14y – 2 = 0 represent?Illustration - 13

SOLUTION :Compare the given equation with

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 a = 13, h = – 9, b = 37, g = 1, f = 7,

c = – 2then = abc + 2f gh – af 2 – bg2 – ch2

= (13) (37) (–2) + 2 (7) (1) (–9) – 13 (7)2– 37 (1)2 + 2 (–9)2

= – 962 – 126 – 637 – 37 + 162= – 1600 0

and alsoh2 = (–9)2 = 81 and ab = 13 × 37 = 481

Here ab – h2 = 400 > 0So we have h2 < ab and 0. Hence the givenequation represents an ellipse.

Section 3 13

Conic Section

If the equation of conic 2x2 + xy + 3y2 – 3x + 5y + = 0 represent a single point, then findthe value of .Illustration - 14

SOLUTION :For single point

h2 < ab and = 0Comparing the given equation with

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

then 1 3 52, , 3, , ,2 2 2

a h b g f c

Here 2 1 236 04 4

h ab

Given conic represents a single point if = 0.and = abc + 2f gh – af 2 – bg2 – ch2

5 3 1 25(2) (3) ( ) 2 22 2 2 4

9 13 04 4

15 25 276 04 2 4 4

23 23 0

4 = 4.

3.4 Definition of ParabolaA parabola is the locus of a point which moves in a plane 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).

3.5 Standard Parabola (y2 = 4ax)The general form of standard parabola is :

y2 = 4ax, [where ‘a’ is a constant]

3.6 Important terms related with standard Parabola y2 = 4ax(i) The straight line passing through the focus and perpendicular to the directrix is called the Axis of the

Parabola. The parabola is symmetrical about its axis (i.e. y = 0 line). It means if (x, y) lieson parabola, then (x, –y) also lies on other sides of the axis.

Y

PL

M

dire

ctrix

x

+ a

= 0

– a O S (a, 0) axis

L

y = 4ax2

x = – a

N

P

Q

Q

X

Section 314

Conic Section

(ii) The point which bisects every chord of the conic passing through it, is called the Centre of theParabola.

(iii) The points of intersection of the conic section and the axis is (are) called Vertex. Veretex is point A(0, 0) i.e. the origin is the vertex of the parabola

(iv) The fixed point is known as Focus (denoted by S).(v) The fixed line is known as Directrix.(vi) A chord passing through the focus is known as Focal Chord.(vii) The straight line through focus and perpendicular to directrix is known as Axis.(viii) The focal chord which is perpendicular to the axis is known as Latus Recturm. As double ordinate

LL passes through the foucs. Since focus S (a, 0) the equation of the latus rectum of the parabola isx = athen solving x = a and y2 = 4axthen we get y = ± 2aHence the co-ordinates of the extremities of the latus rectum are L (a, 2a) and L (a, – 2a) respec-tively.Since LS = LS = 2a

Length of latus rectum LL = 2 (LS) = 2 (LS) = 4a.

(ix) If Q be the point of the parabola, draw QN perpendicular to the axis of parabola and produced tomeet the curve again at Q, then QQ is called a double ordinate.If abscissa of Q is h then ordinate of Q, y2 = 4ah

or 2y ah (for first quadrant)

and ordinate of Q is 2y ah (for fourth quadrant)

Hence co-ordinates of Q and Q are ( , 2h ah ) and ( , 2 )h ah respectively..

(x) The Parametric Equations : From the equation of the parabola y2 = 4ax, we can write

22y x ta y where ‘t’ is a parameter. Then

y = 2at and x = at2

The equations x = at2 and y = 2at are called parametric equations :

The point (at2, 2at) is also referred to as the point ‘t’.

Note : Co-ordinates of any point on the parabola y2 = 4ax, may be taken as (at2, 2at).

Section 3 15

Conic Section

3.7 Other forms of Parabola

y2 = 4 ax y2 = – 4ax x2 = 4 ay x2 = – 4 ay Parabola Parabola Parabola Parabola

towards right towards left opening upwards opening downwardsVertex (0, 0) (0, 0) (0, 0) (0, 0)Focus (a, 0) (– a, 0) (0, a) (0, – a)Equation of axis y = 0 y = 0 x = 0 x = 0Equation of directrix x + a = 0 x – a = 0 y + a = 0 y – a = 0Length of latus rectum 4a 4a 4a 4aExtremities of Latus rectum (a, ± 2a) (–a, ± 2a) (± 2a, a) (± 2a, –a)Equation of latus rectum x – a = 0 x + a = 0 y – a = 0 y + aEquation of tangent at vertex x = 0 x = 0 y = 0 y = 0Parametric co-ordinates (at2, 2at) (–at2, 2at) (2at, at2 ) (2at, –at2 )

Figure

Find the coordinates of focus, vertex, equation of directrix, tangent at vertex, axis andlength of latus of rectuum for the following parabolas.

(i) y2 = 6x (ii) x2 = 8y (iii) y2 = – 16x (iv) x2 = – 12y

SOLUTION :

(i) 2 34 · ·2

y x 32

a

Focus : (a, 0) 3 , 02

Vertex : (0, 0)

Directrix : x = – a x = 3

2

Tangent at vertex x = 0

Axis : y = 0 Length of latus recturm = 4a = 6(ii) x2 = 8y = 4.2 . y a = 2

Focus : (0, a) (0, 2) Vertex : (0, 0)Directrix : y = – a y = – 2 Tangent at vertex y = 0Axis : x = 0 Length of latus recturm = 8

M

X X

Y

Y

A S

L

L P

Z

Y

Y

X XM

ZAS

PL

L

Y

Y

Z M

A

L L

PXX

S M

Y

Y

XX

Z

A

P

LS

L

Illustration - 15

Section 316

Conic Section

(iii) y2 = – 16x = – 4.4x a = 4Focus : (– a, 0) (– 4, 0) Vertex : (0, 0)Directrix : x = a x = 4 Tangent at vertex y = 0Axis : y = 0 Length of latus recturm = 16

(iv) x2 = – 12y = – 4.3y a = 3Focus : (0, – a) (0, – 3) Vertex : (0, 0)Directrix : y = a y = 3 Tangent at vertex y = 0Axis : x = 0 Length of latus recturm = 4a = 12

3.8 General Form of ParabolaFinding equation of parabola when focus and line of directrix are givenAssume that the focus as S (h, k), line of directrix as ax + by + c = 0 and point P as (x1, y1) (, )whose locus is parabola.As we know that for parabola,

PS = PN PS2 = PN2

2

2 2 1 11 2 2 2

( ) ( ) ax by cx h y ka b

. . . (i)

Simplifying the equation (i), and then replacing ‘x1’ by ‘x’ and ‘y1’ by ‘y’, we will get the required equationto parabola. The simplified form of general equation of parabola would look line.

(bx – ay)2 + 2gx + 2fy + c = 0. Where g f and c are real constants.

Note : The second degree terms in the general equation of parabola forms a perfect square.

Find the equation of the parabola whose focus is at (–1, –2) and the directrix is the straightline x – 2y + 3 = 0.

SOLUTION :Let P (x, y) be any point on the parabola whose focus isS (–1, –2) and the directrix x – 2y + 3 = 0.Draw PM perpendicular from P (x, y) on the directrixx – 2y + 3 = 0. Then by definition

SP = PM (SP)2 = (PM)2

Illustration - 16

x –

2y +

3 =

0

M P (x, y)

S (–1, –2)

Section 3 17

Conic Section

22 2

2 2| 2 3 |( 1) ( 2)(1) ( 2)

x yx y

5 (x2 + y2 + 2x + 4y + 5) = (x2 + 4y2 – 4xy+ 6x – 12y + 9) 4x2 + y2 + 4xy + 4x + 32y + 16 = 0.

Find the points as the parabola y2 = 8x whose focal distance is 8.Illustration - 17

SOLUTION :Comparing y2 = 8x with y2 = 4ax 4a = 8 a = 2We know focal distance of (at2, 2at)= a (1 + t2) = 8 1 + t2 = 4

3t

Points on the parabola are (at2, 2at) where 3t . The required points are ( 6, 4 3 ) and ( 6, 4 3 ).

Y

A S X

P (x , y )11

Focaldistance = 8

x +

2 =

0

Z

Dire

ctrix

M

Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax sothat one angular point is at the vertex.Illustration - 18

SOLUTION :Let ABC be the inscribed equilateral triangle,with one angular point at the vertex V of theparabola.

y2 = 4ax . . . (i)Let the length of the side of equilateral triangle= l AB = BC = CA = l The co-ordinates of B is

(l cos 30°, l sin 30°)

i.e., 3 ,

2 2l l

. As B lies on (i), then we get :

2 342 2l la

8 3l a

Y

B

l

l

30°30°AX X

C

Y

Section 318

Conic Section

Find the vertex, focus, latus rectum, axis and the directrix of the parabola x2 + 8x + 12y +4 = 0.Illustration - 19

SOLUTION :The equation of parabols is

x2 + 8x + 12y + 4 = 0 . . . (i) (x + 4)2 – 16 + 12y + 4 = 0 (x + 4)2 = 12 – 12y (x + 4)2 = – 12 (y – 1)Let x + 4 = X, y – 1 = Y . . . (ii) X2 = – 12Y . . . (iii)Comparing it with X2 = – 4aY, we get : a = 3Also Vertex of (iii) is (0, 0)i.e. X = 0, Y = 0Using (ii)x + 4 = 0, y – 1 = 0 x = – 4, y = 1

Vertex of (i) is (–4, 1).Focus of (iii) is (0, – 3)i.e., X = 0, Y = – 3Using (ii), x + 4 = 0, y – 1 = – 3

x = – 4, y = – 2 focus of (i) is (–4, –2) and

latus rectum = 4a = 12.Equation of axis of (iii) is X = 0 Equation of axis of (i) is x + 4 = 0Equation of directrix of (iii) is

Y = 3 y – 1 = 3 y – 4 = 0 Equation of directrix of (i) is y – 4 = 0.

A parabola whose vertex is 11 3,8 2

and focus is 15 3,8 2

. Find its length of latus

recturm.

Illustration - 20

SOLUTION :V is mid-point of N and F.

15

11 88 2

78

and3

3 22 2

32

Directrix is line passing through N and perpendicu-lar to VF.

Equation of directrix is 7

8x

Equation of parabola becomes2 215 3 7

8 2 8x y x

(8x +15)2 + 16 (2y – 3)2 = (8x + 7)2

64x2 + 240x + 225 + 64y2 – 192y + 144 = 64x2 + 112x + 49

y2 – 3y + 2x + 5 = 0Axis is perpendicular to directrix and passing through

focus and hence equation of axis is given by 32

y .

Length of latus rectum = 2 × Distance of focus from

directrix 15 72 28 8

.

N V( , ) F

y = 32_

–158

, 32_–11

8, 3

2_

Directrix

Y

X

Section 3 19

Conic Section

If a double ordinate y2 = 4ax is of length 8a, then prove that triangle formed by doubleordinate and lines joining vertex with end points of double ordinate is a right-angled triangle.Illustration - 21

SOLUTION :(, 4a) lies on y2 = 4ax

16a2 = 4a = 4a

Slope of 14 0 14 0

aOA ma

Slope of 24 0 14 0

aOA ma

m1 m3 = – 1 'OA OA

Hence, OAA is right-angled triangle.

O

A ( , 4a)

4a

( , 0)4a

A ( , – 4a)

Y

X

Find the equation of the parabola with its vertex at (3, 2) and its focus at (5, 2).Illustration - 22

SOLUTION :Let Vertex A (3, 2) and focus is S (5, 2).

Slope of 2 2 05 3

AS

(which is parallel to x-axis)

The equation is of the form(y – k)2 = 4a (x – h)

(y – 2)2 = 4a (x – 3)as (h, k) is the vertex (3, 2)

a = distance between the focus and the vertex

= 2 2(5 3) (2 2) = 2

Hence the required equation is(y – 2)2 = 8 (x – 3)

y2 – 8x – 4y + 28 = 0

(3, 2)A S (5, 2)

Y

O X

Section 320

Conic Section

Show that the focal chord of parabola y2 = 4ax makes an angle with the x-axisis of length 4a cosec2 .Illustration - 23

SOLUTION :Let P (at1

2, 2at1) and Q (at22, 2at2) be the end

points of a focal chord PQ which makes an angle with the axis of the parabola. Then

PQ = a (t2 – t1)2

= a ((t2 + t1)2 – 4 t1t2)

= a ((t2 + t1)2 + 4) . . . (i)

[as t1t2 = – 1]

tan = slope of PQ

= 2 12 22 1

2 2at atat at

2 1

2tant t

t2 + t1 = 2 cot . . . (ii)Substituting the value of t2 + t1 from (ii) in (i) then

PQ = a (4 cot2 + 4) = 4a cosec2 .

A LINE AND A PARABOLA Section - 4

4.1 Intersection of a line and parabolaLet the parabola be y2 = 4ax . . . (i)and the given line be y = mx + c . . . (ii)Eliminating y from (i) and (ii), then

(mx + c)2 = 4ax m2x2 + 2x (mc – 2a) + c2 = 0 . . . (iii)This equation, being quadratic in x, gives two values of x. Shows that every straight line will cut the parabolain two points may be real, coincident or imaginary according asDiscriminant of (iii) >, = , < 0i.e., 4 (mc – 2a)2 – 4m2c2 <, =, < 0 4a2 – 4amc >, =, < 0 a >, =, < mc . . . (iv)

4.2 Condition of TangencyIf the line (ii) touches the parabola (i), then equation (iii) has equal roots Discriminant of (iii) = 0 4 (mc – 2a)2 – 4m2c2 = 0 – 4amc + 4a2 = 0

, 0ac mm

. . . (v)

So, the line y = mx + c touches the parabola y2 = 4ax if acm

(which is condition of tangency).

Section 4 21

Conic Section

Substituting the value of c from (v) in (ii) then

, 0ay mx mm

. . . (vi)

Hence the line ay mxm

will always be a tangent to the parabola y2 = 4ax.

4.3 The point of Contact

Substituting acm

is equation (iii), then

22 2 2 . 2 0a am x x m a

m m

2

2 2 2 0am x axm

2

0amxm

0amxm

2ax

m

Substituting this value of x is ay mxm

2a ay m

mm

2aym

Hence, the point of contact is 22,a amm

.

Note : If m = 0 then equation (iii) gives – 4ax + c2 = 0 = 2

4cxa

which gives only one value of x and so every line

parallel to X-axis cuts the parabola only in one real point.

4.4 Position of point (x1, y1) with respect to a parabolay2 = 4axLet P (x1, y1)

Now P will lie outside, on or inside the parabola

y2 = 4ax according as

(y12 – 4ax1) >, = , < 0

Y

(on)P

P (Inside)X

(Outside) P

Section 422

Conic Section

4.5 Parametric Relation between the co-ordinates of the ends of a focal chordLet y2 = 4ax be parabola, if PQ be a focal chord.If P (at1

2, 2at1) and Q (at12, 2at2)

Since PQ passes through the focus S (a, 0) Q, S, P are collinear Slope of PS = Slope of QS

1 22 21 2

0 0 2at atat a a at

1 22 21 2

0 21 1

t tt t

2 21 2 2 1( 1) ( 1)t t t t t1 t2 (t2 – t1) + (t2 – t1) = 0

t2 – t1 0 t1 t2 + 1 = 0

t1 t2 = – 1 or 21

1tt

which is required relation.

Note : If one extremity of a focal chord is (at12, 2at1) then the other extremity (at2

2, 2at2) becomes 2 11

2,a att

by virtue of relation (1).

Prove that the straight line lx + my + n = 0 touches the parabola y2 = 4ax if ln = am2.Illustration - 24

SOLUTION :The given line is lx + my + n = 0

l ny xm m

Comparing this line withy = Mx + c . . . (i)

andl nM cm m

The line (i) will touch the parabolay2 = 4ax, if

acM

cM = a

n l am m

ln = am2.

Alternative Method :The given line lx + my + n = 0 . . . (i)and the parabola y2 = 4ax . . . (ii)Substituting the value of x from (i) i.e.,

n myxl

in (ii) then

(We should not substituting the value of y from (i) in(ii) since y is quadratic substituting the value of xsince x is linear)

2 4 n myy al

ly2 + 4amy + 4an = 0 . . . (iii)Since equation (i) touches the parabola (ii) then rootsof equation (iii) must be coincident and condition forthe same is “B2 = 4AC”.i.e. (4am)2 = 4. l. 4an am2 = ln ln = am2

Section 4 23

Conic Section

Show that the line x cos + y sin = p touches the parabola y2 = 4ax if p cos + a sin2 = 0 and that the point of contact is (a tan2 , – 2a tan ).Illustration - 25

SOLUTION :The given line is

x cos + y sin = p y = – x cot + p cosec Comparing this line with y = mx + c m = – cot and c = p cosec since the given line touches the parabola

acm

cm = a

(p cosec ) (– cot ) = a a sin2 + p cos = 0

and point of contant is 22,a amm

i.e.

22,

cotcota a

(a tan2 , – 2a tan ).

Prove that the line 1x yl m touches the parabola y2 = 4a (x + b) if m2 (l + b) + al2 = 0.Illustration - 26

SOLUTION :The given parabola is

y2 = 4a (x + b) . . . (i)Vertex of this parabola is (– b, 0).Now shifting (0, 0) at (–b, 0) then

x = X + (– b) and y = Y + 0 x + b = X and y = Y . . . (ii)from (i), Y 2 = 4aX . . . (iii)

and the line 1x yl m reduces to 1X b Y

l m

1 X bY ml

1m bY X ml l

. . . (iv)

The line (iv) will touch the parabola (iii), if

21 b a m bm l a

ml l ll

m2 (l + b) + al2 = 0

Alternative Method :The given line and parabola are

1x yl m . . . (i)

and y2 = 4a (x + b) . . . (ii)respectively.Substituting the value of x from (i),

i.e., 1 yx lm

in (ii)

then 2 4 1 yy a l bm

2 4 4 ( ) 0aly y a l bm

. . . (iii)

Since the line (i), touches the parabola (ii) then theroots of equation (iii) are equal

24 4.1{ 4 ( )} 0al a l b

m

2

2 ( ) 0al l bm

al2 + m2 (l + b) = 0

m2 (l + b) + al2 = 0.

Section 424

Conic Section

Prove that the semi-latus rectum of the parabola y2 = 4ax is the harmonic mean betweenthe segments of any focal chord of the parabola.Illustration - 27

SOLUTION :Let parabola be y2 = 4axIf PQ be the focal chord then if P (at2, 2at)

then 22,a aQtt

Length of latus rectum LL = 4a

Semi latus rectum = 1 (4 )2

a = 2a

If sections of focal chord are k1 and k2

then k1 = SP = PM = a + at2 = a (1 + t2)

and2

2 2 2(1 )a a tk SQ QN a

t t

Harmonic Mean of k1 and 1 22

1 2

2k kkk k

=

2 1

21 1k k

= 2

2 2

21

(1 ) (1 )t

a t a t

= 2 21 a

a

= Semi latus rectum.

M

N

x +

a =

0

Z Ak2

Q at 2 –2a

t 2,

S (a, 0) X

k1

P (at , 2at)2

Y

L

L

TANGENT AND NORMAL TO THE PARABOLA Section - 5

5.1 Equation of Tangent(i) Point Form :

Equation of tangent to all standard Parabolas at the point (x1, y1) are

Note : The equation of tangent at (x1, y1) can also be obtained by replacing x2 by xx1, y2 by yy1, x by 1

2x x

,

y by 12

y y.

Section 5 25

Conic Section

(ii) Parametric Form :Equation of tangent to all standard parabolas at ‘t’.

(iii) Slope Form :Equation of tangent to all standard parabolas in slope (m) form :

Note :(i) The condition for y = mx + c to be a tangent to the parabola

y2 = 4ax is : c = am

(ii) The point of intersection of tangents drawn at two different pointsof contact (at1

2, 2at) and (at22, 2at2) is :

P [at1t2, a (t1 + t2)].(iii) The angle between tangents drawn at two different points of

contacts (at12, 2at) and (at2

2, 2at2) is :

1 2 1

1 2tan

1t t

t t

.

(iv) The length of focal chord having parameters t1 and t2 for its end points is a (t2 – t1)2.

(v) If l1 and l2 are the length of segments of a focal chord of a parabola, then its latus recturm is 1 2

1 2

4l ll l .

slope = 1/t 1

slope = 1/t2

t1

t2

P

Section 526

Conic Section

5.2 Equation of Normal(i) Point Form :

Equation of normal to all standard parabolas at the point (x1, y1).

(ii) Parametric FormThe equation of normal of all Standard Parabolas at ‘t’.

(iii) Slope FormThe equation of normal, point of contact, and condition of normality in terms of slope (m) of allstandard parabolas.

Section 5 27

Conic Section

5.3 Co-normal Points :In general, three normals can be drawn from a point to a parabolaand their feet, points where they meet the parabola, are called co-normal points.Let P (h, k) be any given point and y2 = 4ax be a parabola.The equation of any normal to y2 = 4ax is y = mx – 2am – am3

If it passes through (h, k) thenk = mh – 2am – am3

am3 + m (2a – h) + k = 0 . . . (i)This is a cubic equation in m. So, it has three roots, say m1, m2 and m3

m1 + m2 + m3 = 0, m1m2 + m2m3 + m3m1 = (2 )a h

a

. . . (ii)

Hence for any given point P (h, k) (i) has three real or imaginary roots. Cooresponding to each of thesethree roots, we have one normal passing through P (h, k). Hence in total, we have three normals PA, PB andPC drawn through P to the parabola.Points A, B, C in which the three normals from P (h, k) meet the parabola are called co-normal points.

Note :

(i) If the normal drawn at ponit 2P (at , 2at) intersects parabola again inQ (at1

2, 2at1), then

12t tt

.

(Note the PQ is normal to the curve at P and not at Q).

(ii) The point of intersection of normals drawn at P (t1) and Q (t2) is :

A [2a + a (t12 + t2

2 + t1t2), – at1t2 (t1 + t2)].

BA

P (h, k)C

Y

Y

X X

P(t)

Q(t )1

P

Q

A

Section 528

Conic Section

(a) Find the equation of the tangents drawn to y2 + 12x = 0 from the point (3, 8). (b) Find the equation of tangents to the parabola y2 = 4x + 5 which is parallel to the

line y = x + 7.

Illustration - 28

SOLUTION :(a) y2 + 12x = 0 y2 = – 12x. a = –3.

Let tangent be = y = mx – 3m .

Since tangent passes through (3, 8), 8 = 3m – 3m

3m2 – 8m – 3 = 0 (m – 3) (3m + 1) = 0

m = 3, m = –13

Tangent is are y = 3x – 1 andy = – 3x

+ 9.

(b) Any line || to given is y = 2x +.If it is tangent to the parabola then it will meet itin two coincident points.Eliminating x, we get : y2 – 5 = 2 (y – ) y2 – 2y + 2 – 5 = 0. Roots are equal. b2 – 4ac = 0 4 – 4 (2 – 5) = 0 = 3 y = 2x + 3

Prove that the two parabolas y2 = 4ax and x2 = 4by intersect at an angle1/3 1/3

12/3 2/3

32 ( )

a btana b

.

Illustration - 29

SOLUTION :Angle between two curves is equal to the anglebetween the tangents at their common point.

Solving the two curves, i.e. y = 2

4xb

and y2 = 4ax.

We get (0, 0) and (4a1/3 b1/3 4a2/3 b1/3).At (0, 0), two tangents are x axis and y-axis Angle is 90°.At (4a1/3 b2/3, 4a2/3 b1/3), for y2 = 4ax,

Slope m1 = 1

2ay =

1/3

1/32ab

.

for x2 = 4by, slope = m2 =1

2xb =

1/3

1/32ab

.

Angle between tangents

= tan–1 1 2

1 21m m

m m

= tan–1 1/3 1/3

2/3 2/33

2 ( )a b

a b.

Section 5 29

Conic Section

Show that the locus of a point, such that two of the three normals drawn from it to theparabola y2 = 4ax are perpendicular is y2 = a (x – 3a).Illustration - 30

SOLUTION :Let P (x1, y1) be the point from where normalsAP, BP, CP are drawn to y2 = 4ax.Let y = mx – 2am – am3 be one of these normals.P lies on it y1 = mx1 – 2am – am3.Slopes m1, m2, m3 of AP, BP, CP are roots of thecubic

y1 = mx1 – 2am – am2. am3 + (2a – x1) m + y1 = 0 m1 + m2 + m3 = 0

11 2 2 3 3 1

2a xm m m m m ma

11 2 3

ym m ma

As two of the three normals are perpendicular, wetake m1m2 = – 1. (i.e. we assume AP perpendicu-lar BP).To get the locus, we have to eliminate m1, m2, m3.

11 2 2 3 3 1

2a xm m m m m ma

13 3

21 ( ) a xm ma

2

1 121 y a xa a

[using m1m2m3 = – y1/a and m1m2 = – 1]

a2 + y12 = – 2a2 + ax1

y12 = a (x1 – 3a)

y2 = a (x – 3a) is the required locus.

A

B

C

P (x , y )11

Suppose that the normals drawn at three different points on the parabola y2 = 4x passthrough the point (h, k). Show that h > 2.Illustration - 31

SOLUTION :Let the normal(s) be y = mx – 2am – am3. Theypass through (h, k). k = mh – 2am – am3.The three roots m1, m2, m3 of this cubic are theslope of the three normals. Taking a = 1, we get :

m3 + (2 – h) m + k = 0 m1 + m2 + m3 = 0

m1m2 + m2m3 + m3m1 = 2 – h m1m2m3 = – k.

[as m1, m2, m3 are real, m12 + m2

2 + m32 > 0]

(and not all are zero) (m1 + m2 + m3)

2

– 2 (m1m2 + m2m3 + m3m1) > 0 0 – 2 (2 – h) > 0 h > 2.

Section 530

Conic Section

If the normals to the parabola y2 = 4ax at three points P, Q and R meet at A and S be thefocus, prove that SP.SQ.SR = a (SA)2.

SOLUTION :Since the slopes of normals are not involved but the coordinates of P, Q, R are important,we take the normal as :

tx + y = 2at + at3.Let A (h, k) t1, t2, t3 are roots of : t (h) + (k) = 2at + at3

i.e. at3 + (2a – h) t – k = 0.

t1 + t2 + t3 = 0. 1 2 2 3 3 12a ht t t t t t

a

t1t2t3 = k/aRemember that distace of point P(t) from focus and from directrix is SP = a (1 + t2). SP = a (1 + t1

2), SQ = a (1 + t2), SR = a (1 + t3)2

SP.SQ.SR = a3 {t12 + t2

2 + t32 + (t1

2 t22 + t2

2 t32 + t3

2 t12) + (t1

2 + t22 + t3

2) + 1}

We can see that : 2 2 2 21 2 3 1 2 3 1 2

(2 )( ) 2 0 2 a ht t t t t t t ta

and also 22 21 2 1 2 1 2 2 32 ( ) ( )t t t t t t t t 22using : 2 a a ab

2

2 21 2 1 2 3 2

2 2a ht t t t t ta

= 2 2

1 2 32 2(2 ) (2 )2 (0)a h a ht t t

a a

2 2

32 2

(2 ) 2 4. . 1k a h h aSP SQ SR aaa a

= a {(h – a)2 + k2} = a SA2

Show that the tangent and the normal at a point P on the parabola y2 = 4ax are thebisectors of the angle between the focal radius SPand the perpendicular from P on the directrix.

SOLUTION :Let P (at2, 2at), S (a, 0).Equation of SP is :

22 00 ( )aty x aat a

Illustration - 32

P(t )1

AQ(t )2

R(t )3

Illustration - 33

Section 5 31

Conic Section

2tx + (1 – t2) y + (–2at) = 0 . . . (i)

Equation of PM is :y – 2at = 0 . . . (ii)

Angle bisectors of (i) and (ii) are :2

2 2 2

2 2 (1 ) 20 1 4 (1 )

y at tx t y at

t t

2

22 (1 ) 22

1tx t y aty at

t

ty = x + at2 and tx + y = 2at + at3

tangent and normal at P are bisectors of SPand PM.

M

Y

SO– a

P

X

Alternate Method :Let the tangent at P meet X axis in Q.As MP is parallel to X-axis, MPQ = PQSNow we can find SP and SQ.

2 2 2( ) (0 2 ) (1 )SP a at at a t

Equation of PQ is ty = x + at2

Q (–at2, 0)

2 2( ) 0 (1 )SQ a at a t

SP = SQ SPQ = SQP = MPQ.Hence PQ bisects SPM.It obviously follows that normal bisects exteriorangle.

M

Y

P

Q – a O S

x = – a

X

In the parabola y2 = 4ax, the tangent at the point P, whose abscissa is equal to the latusrectum meets the axis in T and the normal at P cuts the parabola again in Q. Prove that PT : PQ = 4 : 5.Illustration - 34

SOLUTION :Latus rectum = xp = 4a.Let P (at2, 2at) at2 = 4a t = ± 2.We can do the problem by taking only one of thevalues.Let t = 2 P (4a, 4a) tangent at P is 2y = x + 4aT lies on X-axis, T (– 4a, 0)

2 2(8 ) (4 ) 4 5PT a a a

Let us now find PQ.If normal at P(t) cuts parabola again at Q(t1), then

t1 = – t – 2/t t1 = – 2 – 2/2 = – 3 Q (9a, – 6a)

2 225 100 5 5PQ a a a

PT : PQ = 4 : 5.

Section 532

Conic Section

A variable chord PQ of y2 = 4ax subtends a right angle at vertex. Prove that the locus ofthe point of intersection of normals at P, Q is y2 = 16a (x – 6a).Illustration - 35

SOLUTION :Let the coordinates of P and Q be (at1

2, 2at1)and (at2

2, 2at2) respectively.As OP and OQ are perpendicular, we can have :

1 22 21 2

2 0 2 0 10 0

at atat at

t1t2 = – 4 . . . (i)Let the point of intersection of normals drawn at Pand Q be (x1, y1).

Using the result of point of intersection of normalwe get :

x1 = 2a + a (t12 + t2

2 + t1t2) . . . (ii)and y1 = – a t1t2 (t1 + t2) . . . (iii)Eliminating t1 and t2 from (i), (ii) and (iii), we get :

y12 = 16a (x1 – 6a)

The required locus is y2 = 16a (x – 6a).

The normal at a point P to the parabola y2 = 4ax meets the X axis in G. Show that P andG are equidistant from focus.

SOLUTION :Let the coordinates of the point P be (at2, 2at) The equation of normal at P is : tx + y = 2at + at3

The point of intersection of the normal with X-axis is G (2a + at2, 0).

SP = a (1 + t2) and 2 2 2 2( ) (1 ).SG a at O a t SP = SGHence P and G are equidistant from focus.

Tangents to the parabola y2 = 4ax drawn at points whose abscissas are in the ratio 2 : 1.Prove that the locus of their point of intersection is y2 = [1/2 + –1/2]2 ax.

Illustration - 36

Illustration - 37

SOLUTION :Let the coordinates of the two points on which thetangents are drawn be (at1

2, 2at1) and (at22, 2at2).

As the abscissas are in the ratio 2 : 1, we get :2

2122

atat

t1 = t2. . . . (i)

Let the point of intersection of two tangents beM (x1, y1).

Using the result of intersection of Tangents, we get :

M (x1, y1) [at1 t2, a (t1 + t2)] x1 = at1 t2 . . . (ii)and y1 = a (t1 + t2) . . . (iii)Eliminate t1 and t2 from equations (i), (ii) and (iii) toget :

y12 = [1/2 + –1/2]2 ax1

The required locus of M is :y2 = [1/2 + –1/2]2 ax.

Section 5 33

Conic Section

Find the equation of common tangent to the circle x2 + y2 = 8 and parabola y2 = 16x.Illustration - 38

SOLUTION :Let ty = x + at2 (where a = 4) be a tangent to pa-rabola which also touches circle. ty = x + 4at2 and x2 + y2 = 8

have only one common solution. (ty – 4t2)2 + y2 = 8

has equal roots as a quadratic in y.

(1 + t2) y2 – 8t3y + 16t4 – 8 = 0 has equalroots.

64t6 = 64t6 + 64t4 – 32 – 32t2

t2 + 1 – 2t4 = 0 t2 = 1, – 1/2 t = ± 1 the common tangents are

y = x + 4 and y = – x – 4.

Through the vertex O of the parabola y2 = 4ax, a perpendicular is drawn to any tangentmeeting it at P and the parabola at Q. Show that OP . OQ = constant.Illustration - 39

SOLUTION :Let ty = x + at2 be the equation of the tangent. OP = perpendicular distance of tangent from

origin

2

21

atOPt

Equation of OP is y – 0 = – t (x – 0) y = – tx

Solving y = – tx and y2 = 4ax, we get

24 4,a aQ

tt

2 2

24 2

16 16a aOQt t

OP.OQ = 4a2.

Prove that the circle drawn on any focal chord as diameter touches the directrix.

SOLUTION :Let P (t1) and Q (t2) be the ends of a focal chord.Using the result of chord of contact of P (t1) and Q (t2) passes through focus if t1t2 = – 1, we get :Equation of circle with PQ as diameter is :

(x – at12) (x – at2

2) + (y – 2at1) (y – 2at2) = 0 [using diametric form of equation of circle]For the directrix to touch the above circle, equation of circle and directrix must have a unique solution i.e.Solving x = – a and circle simultaneously, we get

a2 (1 + t12) (1 + t2

2) + y2 – 2ay (t1 + t2) + 4a2 t1t2 = 0This quadratic in y has discriminant = D = B2 – 4AC D = 4a2 (t + t2)

2 – 4a2 [(1 + t12) (1 + t1

2) + 4t1t2] = 0 [using t1t2 = – 1] circle touches x = – a. circle touches the directrix.

Illustration - 40

Section 534

Conic Section

Find the equation of the common tangents to the parabola y2 = 4ax and x2 = 4by.Illustration - 41

SOLUTION :The equation of any tangent in terms of slope (m)to the parabola y2 = 4ax is

ay mxm

. . . (i)

If this line is also tangent to the parabola x2 = 4aythen (i) meets x2 = 4by in two coincident points.Substituting the value of y from (i) in x2 = 4bywe get

2 4 ax b mxm

2 44 0abx bmxm

The roots of this quadratic are equal provided“B2 = 4AC”

i.e., 2 4( 4 ) 4.1. abbmm

16b2m3 + 16ab = 0, m 0 m3 = – a/b m = – a1/3 / b1/3

Substituting the value of m is (i) the requiredequatuion is

1/3 1/3

1/3 1/3a aby xb a

1/3

2/3 1/31/3

ay x a bb

a1/3 x + b1/3 y + a2/3 b2/3 = 0

The tangents to the parabola y2 = 4ax make angle 1 and 2 with x-axis. Find the locus oftheir point of intersection if cot 1 + cot 2 = c.Illustration - 42

SOLUTION :Let the equation of any tangent to the parabolay2 = 4ax is

y = mx + (a/m).Let (x1, y1) be the point of intersection of thetangents to y2 = 4ax then (i) passes through x1, y1). y1 = mx1 + (a/m) m2x1 – my1 + a = 0Let m1 and m2 be the roots of this quadratic equa-tion then

m1 + m2 = y1/x1 and m1m2 = a/x1

tan 1 + tan 2 = y1/x1 andtan 1 tan 2 = a/x1 . . . (ii)

Now cot 1 + cot 2 = c (given)

1 2

1 1tan tan

c

1 2

1 2

tan tantan tan

c

[From (ii)]

1 1

1

//

y x ca x

y1 = ac

The required locus is y = ac (which is a line parallelto x-axis)

Section 5 35

Conic Section

Find the centre and radius of the smaller of the two circles that touch the parabola 75y2

= 64 (5x – 3) at (6/5, 8/5) and the x-axis.Illustration - 43

SOLUTION :Equation of tangent at

6 8,5 5

y – 85 =

1815

64 5 6150 5y

xy

3y – 4x = 0.

Equation of family of circles touches 3y – 4x at

6 8,5 5

is 2 28 6

5 5y x

+ k [3y – 4x]

= 0

x2 + y2 – 12 45

k

x + 1635

k

y + 64 3625 25

= 0

As required circle touches x-axis,

g2 = c 26 2

3k

= 4 6

5 + 2k = 2

k = 25 and k = –

85

required circle x2 + y2 – 4x – 2y + 4 = 0 andx2 + y2 + 4x – 8y + 4 = 0.

required circle having smaller radiusx2 + y2 – 4x – 2y + 4 = 0.

Find the locus of the point of intersection of the tangents to the parabola y2 = 4ax whichinclude an 45°.Illustration - 44

SOLUTION :P (at1

2, 2 at1) and Q (at22, 2at2). T (x1, y1)

x1 = a t1t2 . . . (i) and y1 = a (t1 + t2).

As PTQ = 45°, tan 45° = 1 2

1 21m m

m m

= 1 2

1 2

1 1

11

t t

t t

= 2 1

1 21t t

t t

(t2 – t1)2 = (1 + t1 t2)

2

(t2 + t1)2 – 4 t1t2 = (1 + t1t2) . . . (iii)

Replace values of t1 + t2 and t1 t2 from (i) and (ii) in(iii) to get

21 12 4y x

aa =

211 xa

Required locus y2 – 4ax = (x + a)2

PT

Q

Y

X

Section 536

Conic Section

IMPORTANT RESULTS IN PARABOLA Section - 6

6.1 Equation of Chord of ContactLet us consider a point R whose coordinates are (x1, y1) andwhich lies outside the parabola. From point R, two tangents aredrawn to the parabola which meet parabola in points P and Q.The line joining P and Q is known as chord of contact of theparabola and its equation is :

yy1 = 2a (x + x1)

Note : The above result is same as equation of tangent drawn to parabola at (x1, y1).

6.2 Equation of chord of a parabola whose mid-point is givenLet us consider a chord PQ of the parabola whose mid-pointis R (x1, y1). The equation of this chord PQ in terms of coor-dinates of R is :

S1 = T[where S1 = y1

2 – 4ax1 and T = yy1 – 2a (x + x1)].

6.3 Equation of the pair of tangentsLet us consider a point R whose coordinates are (x1, y1) andwhich lies outside the parabola. From point R, two tangentsare drawn to the parabola which meet parabola in points Pand Q. The equation of pair of tangets RP and RQ is :

SS1 = T2

[where S = y2 – 4ax, S1 = y12 – 4ax1 and T = yy1 – 2a (x + x1)].

6.4 Pole and PolarLet P be any point inside or outside a parabola. Suppose anystraight line drawn through P intersects the parabola at Q and R.Then the locus of the point of intersection of the tangents to theparabola at Q and R is called the polar of the given point P withrespect to the parabola and the point given called the pole of thepolar.

Q(x , y )11

Chord ofContactP (h, k)

R (x , y )22

Q(x , y )22

P(x , y )11

R(x , y )33

T (h, k)Q

R

P (x , y )11

(h, k)T

TPolar

R

Q

Q

R

PoleP (x , y )11

Section 6 37

Conic Section

6.5 Equation of the PolarThe equation of the polar of a point (x1, y1) with respect to the parabola y2 = 4ax is :

yy1 = 2a (x + x1)

6.6 Properties of Pole and Polar(i) If the polar of P (x1, y1) passes through Q (x2, y2), then the polar of Q (x2, y2) goes through P (x1, y1)

and such points are said to be Conjugate Points.(ii) If the pole of a line ax + by + c = 0 lies on the another line a1x + b1y + c1 = 0 then the pole of the

second line will lie on the first and such lines are said to be Conjugate Lines.(iii) Polar of the focus is the directrix

Since the focus is (a, 0) Equation of polar of y2 = 4ax is y . 0 = 2a (x + a) x + a = 0 which is the directrix of the parabola y2 = 4ax.

(iv) Any tangent is the polar of its point of contactIf the point P (x1, y1) be on the parabola, its polar andtangent at P are identical. Hence the tangent is the polarof its own point of contact.

6.7 DiameterThe locus of the middle points of a system of parallel chords of a parabola is called a diameter and in caseof a parabola this diameter is shown to be a straight line which is parallel to the axis of the parabola.

6.8 Equation of the DiameterLet (x1, y1) be the middle point of the chord y = mx + c ofthe parabola y2 = 4axThen T = S1 yy1 – 2a (x + x1) = y1

2 – 4ax1

Slope = 1

2a my

2akm

Hence locus of the mid point is y = 2am .

It can be easily derived that the locus is a line parallel to the axis of the parabola.

R

P

R

(x , y )11

QQ

P (x , y )11

y = mx + c

M (h, k)A

Q (x , y )22

X

Section 638

Conic Section

Prove that the area of the triangle formed by the tangents drawn from (x1, y1) to y2 = 4axand their chord of contact is (y1

2 – 4ax1)3/2/2a.

SOLUTION :Equation of QR (chord contact is)

yy1 = 2a (x + x1) yy1 – 2a (x + x1) = 0 [as PM = Length of perpendicular from P (x1, y1) on QR]

PM = 1 1 1 12 21

| 2 ( ) |

( 4 )

y y a x x

y a

= 21 1

2 21

( 4 )

( 4 )

y ax

y a

[Since P (x1, y1) lies outside the parabola y1

2 – 4ax1 > 0]

Now Area of PQR = 1 · ·2

QR PM

= 2

2 2 2 1 11 1 1 2 2

1

( 4 )1 1· · ( 4 ) ( 4 )2 | | ( 4 )

y axy ax y aa y a

= (y12 – 4ax1)

3/2 / 2a , if a > 0.

Find the locus of the mid points of the chords of the parabola y2 = 4ax which subtend aright angle at the vertex of the parabola.

Illustration - 45

P (x , y )11

R

Q

Illustration - 46

SOLUTION :Let P (h, k) be the mid point of a chord QR of theparabola y2 = 4ax, then equation of chord QR is

T = S1 yk – 2a (x + h) = k2 – 4ah yk – 2ax = k2 – 2ah . . . (i)

Q

90°(0, 0) A P (h, k)

R

If A is the vertex of the parabola. For combinedequation of AQ and AR, Making homogenous of y2

= 4ax with the help of (i) y2 = 4 ax . 1

2

2242

yk axy axk ah

y2 (k2 – 2ah) – 4akxy + 8a2x2 = 0Since QAR = 90°. Co-efficient of x2 + Co-efficient of y2 = 0 k2 = 2ah + 8a2 = 0Hence the locus is P (h, k) is y2 – 2ax + 8a2 = 0.

Section 6 39

Conic Section

Show that the locus of the middle points of normal chords of the parabola y2 = 4ax isy4 – 2a (x – 2a) y2 + 8a4 = 0.

Illustration - 47

SOLUTION :Equation of the normal chord at any point (at2, at)of the parabola y2 = 4ax isy + tx = 2at + at3 . . . (i)But if M (x1, y1) be its middle point, its equationmust be also

T = S1

yy1 – 2a (x + x1) = y12 – 4ax1

yy1 – 2ax = y12 – 2ax1 . . . (ii)

As Equations (i) and (ii) are identical. Comparingthem

3

21 1 1

1 22 2t at at

y a y ax

From first two relations 1

2aty

. . . (iii)

From last two relations3

21 1

22 2t at at

a y ax

2

21 12 22

y ax a ata

22

1 1

1

2 222

y ax aa aa y

[from equation (iii)]

2 2 31 1 1

21

2 2 42

y ax ay aa y

R (at , 2at)2

M (x , y )11

Q

Show that the locus of the poles of normal chords of y2 = 4ax is (x + 2a) y2 + 4a3 = 0.Illustration - 48

SOLUTION :Given parabola y2 = 4ax . . . (i)Equatoin of normal of (i) is

y = mx – 2am – am3 . . . (ii)Let (x1, y1) be the pole of (ii) with respect to (i),then (ii) is the polar of (x1, y1) with respect to (i)i.e.,

yy1 = 2a (x + x1) yy1 = 2ax + 2ax1 . . . (iii)

Comparing (ii) and (iii) we get

1 13

221 2y axa

m am am

Then we get y1 = 2a/m . . . (iv)and x1 = 2a – am2 . . . (v)

Eliminating m between (iv) and (v) we get

x1 = – 2a – a 2

1

2ay

(x1 + 2a) y12 + 4a3 = 0

Required locus of (x1, y1) is (x + 2a) y2 + 4a3

= 0.

Section 640

Conic Section

The polar of a point with respect to y2 = 4bx touch x2 = 4ay; show that the locus of thispoint is rectangular hyperbola.Illustration - 49

SOLUTION :Let the point be (h, k) then the polar of (h, k) withrespect to y2 = 4bx is ky = 2b (x + h)

2 2b bhy xk k

. . . (i)

Comparing (i) with y = mx + c

2 2andb bhm ck k

Since equation (i) touch the parabola x2 = 4ay

c + am2 = 0 22 2 0bh ba

k k

hk + 2ab = 0 Locus of (h, k) is xy = – 2abwhich is rectangular hyperbola.

In the parabola y2 = 4ax, prove that the locus of the poles of all chords of constant length2c is (y2 – 4ax) (y2 + 4a2) = 4a2c2.Illustration - 50

SOLUTION :Let PQ be the chord of the parabola y2 = 4ax.If (h, k) be the pole of line PQ, then (h, k) be thepoint of intersection of tangents at P and Q to theparabola y2 = 4ax we have

h = at1t2 and k = a (t1 + t2)

1 2ht ta

and 1 2kt ta

. . . (i)

X X

Y

Y

Q(at , 2at )2 2

2

R{at t , a (t + t )}1 2 1 2

P (at , 2at )1 12

Given | PQ | = 2c (PQ)2 = 4c2

(at12 – at2

2)2 + (2at1 – 2at2)2 = 4c2

a2 (t1 – t2)2 {(t1 + t2)

2 + 4} = 4c2

a2 {(t1 + t2)2 – 4t1t2} [t1 + t2)

2 + 4] = 4c2

2 2

2 22 2

4 4 4k h ka caa a

(k2 – 4ah) (k2 + 4a2) = 4a2c2

Hence locus of (h, k) is(y2 – 4ax) (y2 + 4a2) = 4a2c2.

Prove that the locus of the poles of the chords of the parabola which are at a constantdistance b from the vertex is y2 + 4a2 (1 – x2/b2) = 0.Illustration - 51

SOLUTION :Let pole be (x1, y1). Equation of chord Equationof polar of (x1, y1) is yy1 = 2a (x + x1). yy1 – 2ax – 2ax1 = 0.Its distance from (0, 0) is b.

1

2 21

| 2 |

4

ax

a y = b

4a2 x12 = b2 (4a2 + y1

2). Locus of (x1, y1) is 4a2x2 = b2 (4a2 + y2)

Section 6 41

Conic Section

Prove that the locus of the mid-point of the portion of the normal to the parabola y2 = 4axintercepted between the curve and the axis is another parabola. Find its vertex and length of latus rectum.Illustration - 52

SOLUTION :Normal to the parabola y2 4ax at P (at2, 2at)is y + tx = 2at + at3

Intercept with x-axis a (2a + at2, 0)Let (x1, y1) be the mid-point of PG. 2x1 = at2 + 2a + at2 and 2y1 = 2at.

Replace t = 1ya in (1) to get y1

2 = a (x1 – a).

required lows is y2 = a (x – a).Vertex (a, 0). (also rectum of y2 = a (x – a)

= Latus rectum of y2 = ax = 44a

= a.

Tangents are drawn to the parabola y2 =4ax from a point P such that its chord of contactsubtends an angle of 60° at the vertex. Prove that the locus of the point P is 4 (y2 4ax) = 3 (x + 4a)2.Illustration - 53

SOLUTION :Let P (x1, y1), A (at1

2, 2at1)and B (at2

2, 2at2) x1 = at1t2 . . . (i)and y1 = a (t1 + t2) . . . (ii)

P

A

B

O60°

Y

X

S (OA) = 1

2t

, S (OB) = 2

2t

tan 60° = 1 2

1 2

2 2

2.21

t t

t t

3 (t1t2 + 4)2 = 4 (t1 – t2)2

3 (t1t2 + 4)2 = 4 [(t1 + t2)2 – 4t1t2] . . . (iii)

Replace values of t1 + t2 and t1t2 from (i) and (ii)in (iii) to get

4 (y12 – 4ax1) = 3 (x1 + 4a)2

Required locus is 4 (y2 – 4ax) = 3 (x + 4a)2

Through the vertex O of a parabola y2 = 4ax chords OP and OQ are drawn at right anglesto one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find thelocus of the mid-point of PQ.

Illustration - 54

SOLUTION :Let P (at1

2, 2at) and Q (at22, 2at2).

As POQ = 90°, S (OP) × S (OQ) = –1 t1t2 = – 4 . . . (i)

Equation of chord PQ y = 1 2

2t t

(x + at1t2)

P

O

Q

Y

X

Section 642

Conic Section

Intercept of chord PQ with x-axis (– a t1t2, 0) (4a, 0) (4, 0) as a = 1.

Chord intersect x-axis at a fixed point.Let (x1, y1) be the mid-point of PQ. 2x1 = a (t1

2 + t22)

2x1 = a [(t1 + t2)2 – 2 + t2] . . . (ii)

Also 2y1 = 2a (t1 + t2) y1 = a (t1 + t2) . . . (iii)Replace values of t1 + t2 and t1t2 from equations (i)and (iii) in (ii) to get : 2ax1 = y1

2 + 8a2

required locus y2 = 2 (x – 4) [as a = 1]

If from the vertex of the parabola y2 = 4ax a pair of chords be drawn to perpendicular toeach other and with these chords as adjacent sides a rectangle is completed, then prove that the locus of thevertex of the father end of the rectangle is the parabola y2 = 4a (x – 8a).

Illustration - 55

SOLUTION :Let P (at1

2, 2at1) and Q (at22, 2at2)

As OP OQ, t1t2 = –4 . . . (i)

P

OR (x , y )1 1

Q

Y

X

Mid-point of OR mid point of PQ

1 1,2 2x y

2 21 2

1 2, ( )2

at at a t t

x1 = a (t1

2 + t22) = a [(t1 + t2)

2 – 2t1t2]andy1 = 2a (t1 + t2) . . . (iii)Replace values of t1 t2 and t1 + t2 from (i) and (iii)in (ii) to get y1

2 = 4a (x1 – 8a) Required lows y2 = 4a (x – 8a)

ELLIPSE Section - 7

7.1 DefinitionEllipse is the locus of a point which moves such that its distance from a fixed point called focus is in aconstant ratio to its distance from a fixed line called directrix. This ratio is called eccentricity and is denotedby e. For an ellipse, eccentricity is less than unity (e < 1).

7.2 Equation of Ellipse in standard formTo find equation of Ellipse is standard form, take axis of Ellipse along x-axis as shown in the figure.

SS A

k

A

a–aX

Y

C

Section 7 43

Conic Section

Let y-axis intersect axis of Ellipse at C such that C is known as centre of the Ellipse.Let A (a, 0) and A (–a, 0) be two points on the ellipse. These are points where ellipse meets its axis. Aand A are known as vertices of the ellipse. The directrix is a line that is to the axis. So directrix is parallelto y-axis. Let equation of directrix be x = k. Let S be the focus of the Ellipse. As S lies on axis, its y-coordinate is zero.Let its x-coordinate be b. i.e. S (b, 0).It is necessary to find S and equation of directrix before we can find equation of ellipse in standard form.

Find S and Directrix :As A lies on ellipse, ratio of distance of A from S and distance of A from directrix is e.

i.e.SA eKA

a b ek a

a – b = ke – ae . . . (i)

As A lies on ellipse, ratio of distance of A from S and distance of A from directrix is e.

i.e.SA eKA

SA = e KA b + a = ea + ek . . . (ii)

Add (i) and (ii) to getake

Subtract (ii) from (i) to get, b = ae.

Equation of directrix is axe

and focus of standard ellipse is S (ae, 0).

Equation of Standard Ellipse :Ellipse is locus of a point that moves in such a way that ratio of its distance from focus and directrix is e.Let moving point be P (x, y). Focus of the standard ellipse is S (ae, 0) and equation of directrix is

axe

.

Using definition of ellipse, we get

PS ePT

PS = e PT

2 2( ) ( 0) ax ae y e x

e

Take square of both sides to get :2

2 2 2( ) ax ae y e xe

x2 + a2 e2 – 2aex + y2 = e2 x2 + a2 – 2aex. (1 – e2) x2 + y2 = a2 (1 – e2)

S A X

Y

C

P

ae_

A

k

Section 744

Conic Section

2 2

2 2 2 1(1 )

x ya a e

2 2

2 2 1x ya b

[where b2 = a2 (1 – e2)]

Therefore, standard equation of ellipse is2 2

2 2 1x ya b

[where b2 = a2 (1 – e2)].

Note : If e = 0 then b2 = a2 (1 – 0) b2 = a2

then equation of ellipse 2 2

2 2 1x ya b

changes in circle i.e., x2 + y2 = a2.

7.3 Second focus and second directrixIn the previous section, we found the focus and the directrix of the standard ellipse.

We get focus as S = (ae, 0) and equation of directrix as axe

.

From figure, we can observe that ellipse possesses mirror symmetry about y-axis.So there exists another point on left of y-axis whose properties are similar to that of focus. This is calledas second focus of the ellipse. Its coordinates are (– ae, 0).

Similarly, ellipse has a second directrix in left half whose equation is half whose equation is axe

. This

second directrix is mirror image of the first in y-axis hence its equation is axe

i.e. .axe

.

Equation of ellipse using second focus and second directrixLet P be a general point on ellipse. The second focus is S (– ae, 0) and equation of second directrix is

axe

.

S A X

Y

CA S

(ae, 0)

x = ae_

Section 7 45

Conic Section

Using definition of hyperbola, we getPS ePT

PS = e PT

2 2( ) ax ae y e x

e [using distance of a point from a line formula]

Squaring both sides, we get

(x + ae)2 + y2 = 2

2 ae xe

x2 + a2 e2 + 2aex + y2 = e2 x2 + a2 + 2aex (1 – e2) x2 + y2 = a2 (1 – e2)

2 2

2 2 2 1(1 )

x ya a e

2 2

2 2 1x ya b

[where b2 = a2 (1 – e2).]

Same equation of ellipse in standard form.This shows that a standard ellipse has two focii and two directrix as we get same equation using both ofthem.

Hence, ellipse has 2 focii i.e. S (ae, 0) and S = (– ae, 0) and their corresponding directrix are

anda ax xe e

.

7.4 Important terms related to Ellipse(i) Axes :

In figure shown, the distances AA = 2a and BB = 2b are called the major and minor axes of theellipse. Since e < 1 and b2 = a2 (1 – e2), therefore a > b AA > BB.

S A

k

X

Y

C

P

ae_

A S–

T

S A X

Y

CA S(ae, 0)(–ae, 0)

x = ae_x = – ae

_

Section 746

Conic Section

AA = 2a is the major axes and BB = 2b is the minor axes of the standard ellipse 2 2

2 2 1x ya b

.

(ii) Vertices :

The vertices of the standard ellipse 2 2

2 2 1x ya b

are points where line joining two focii S and S

meets the ellipse. In the figure, A and A are two vertices of the ellipse where coordinatesof A and A are (a, 0) and (– a, 0) respectively.

A (a, 0) and A (– a, 0) are two vertices of the ellipse 2 2

2 2 1x ya b

.

(iii) Centre :The mid point of the major axes AA is called as the centre of the ellipse. It is generally denoted asC. All chords of the ellipse passing through C and they are bisected at C.

C (0, 0) is the centre of the ellipse 2 2

2 2 1x ya b

. It is mid point of the major axes AA.

(iv) Foci :

The two foci of the standard ellipse 2 2

2 2 1x ya b

are S and S where their coordinates are (ae, 0)

and (– ae, 0) respectively.

S (ae, 0) and S (– ae, 0) are two focii of the ellipse 2 2

2 2 1x ya b

.

(v) Directrix :

The directries of the standard ellipse 2 2

2 2 1x ya b

are two lines whose equations are

anda ax xe e

.

S A X

Y

CA SZ

B P

B P L

N

L1

L1 L

Z

M

Section 7 47

Conic Section

anda ax xe e

are two directrices of the ellipse

2 2

2 2 1x ya b

.

(vi) Eccentricity :From definition of ellipse, eccentricity e of the ellipse is the ratio of the distance of any piont P on theellipse from focus and distance of P from the directrix. It is a real number that is less than 1.

We know, for standard ellipse 2 2

2 2 1x ya b

, b2 = a2 (1 – e2)

i.e. 1 – e2 = 2

2ba

2

221 be

a

2

2(2 )1(2 )

bea

2

minor axis1major axis

e

For ellipse 2 2

2 2 1 x ya b

, b2 = a2 (1 – e2) 2 2

2 2(minor axes)1 1(major axes)

bea

2

2(minor axes)1(major axes)

e is the eccentricity of the ellipse.

(vi) Double ordinate :If P be a point on the ellipse, draw PN perpendicular to the axes of ellipse and produced to meetthe curve again at P. The PP is called a double ordinate.

Coordinates of P and P (if h is abscissa of P) are 2 2 2 2, and ,b bh a h h a ha a

.

(vii) Latus rectum :The double ordinates LL and L1L1 are latus rectum of the ellipse. These lines are perpendicular tomajor axis AA and through the foci S and S respectively.

Length of latus rectum LL = L1L1 = 22b

a and end points of latus-rectum are

2 2, ; ,b bL ae L ae

a a

;2 2

1 1, ; ,b bL ae L aea a

(viii) Focal distance of a point :

Let P be a point on the ellipse whose coordinates are (x, y).The distances of point P from two focii S and S are called as focal distances.

PS = e PM and PS = e PM

Section 748

Conic Section

where PM and PM are distances of point from directrices axe

and axe

respectively..

[refer figure]

aPS e x a exe

. . . (i)

and aPS e x a exe

. . . (ii)

a – ex and a + ex are focal distances of point P from S (ae, 0) and (– ae, 0) respectively.Add (i) and (ii) to get :

PS + PS = 2a = Major axis.Thus, the sum of the focal distances of a point on the ellipse is constant and is equal to the length ofits major axis.

Another definition of ellipse :Ellipse is also defined as the locus of a point that moves in such a way that sum of the distances ofpoint P from focii S and two S is constant and is equal to the length of the major axis.

(ix) Parametric Equation of the Ellipse :The circle described on the major axis of an ellipse asdiameter is called the Auxiliary Circle of the ellipse.

Let the equation of ellipse be2 2

2 2 1x ya b

Equation of its auxiliary circle isx2 + y2 = a2 [as AA is diameter of the circle]

Let Q be a point on the auxiliary circle x2 + y2 = a2 such that QP produced is perpendicular to thex-axis.Then P and Q are the Corresponding Points on the ellipse and the auxiliary circle respectively.Let QCA = [0 < 2]i.e., the Eccentric Angle of P on an ellipse is the angle which the radius (or Radius Vector) throughthe corresponding point on the auxiliary circle makes with the major axis. Q (a cos , a sin ) Now x-co-ordinate of P is a cos

Let y-coordinate of P is y. As (a cos , y) lies on the ellipse 2 2

2 2 1x ya b

, we get

2 2 2

2 2cos 1a ya b

y2 = b2 sin2

Q

P

N S A

(a, 0)A

(–a, 0)CX X

Section 7 49

Conic Section

y = ± b sin [as P is in I quadrant] y = b sin Co-ordinates of P are (a cos , b sin ) x = a cos , y = b sin is the parametric equations of the ellipse.

This point (a cos , b sin ) is also called the point ‘’.

(x) General Equation of Ellipse :In this section, we will find general equation whose focus is the point (h, k) and directrix is lx + my +n = 0 and whose eccentricity is e.Let P (x, y) be a general point on the ellipse.From definition of ellipse, ratio of distance of P from focus S (h, k) and distance of P from directrix

lx + my + n = 0 is constant = e.

i.e.PS ePT

22 2

2 2

| |( ) ( ) lx my nx h y k el m

Squaring both sides, we get2 2

2 22 2

( )( ) ( ) e lx my nx h y kl m

which is the general equation of the ellipse.

7.5 The shape of Ellipse 2 2

2 2+ = 1x ya b

, when b > a

AA = minor-axisBB = major-axisDirectrices are = y = ± b/eFoci are S (0, be) and S (0, – be)and a2 = b2 (1 – e2)Latus Rectum = LL = 2a2/b(We take a < b only if it is particularly mentioned in any problem.Usually all problems involve a > b).

Yy = b/e

B (0, b)

S

OA A (a, 0)

y = –b/eB

S

X

Section 750

Conic Section

7.6 Difference between the ellipse

Section 7 51

Conic Section

A LINE AND AN ELLIPSE Section - 8

8.1 Intersection of a line and an Ellipse

Let the ellipse be2 2

2 2 1x ya b

. . . (i)

and the given line be y = mx + c . . . (ii)Eliminating y from equation (i) and (ii), we get :

2 2

2 2( ) 1x mx c

a b

(a2m2 + b2) x2 + 2mca2x + c2a2 – a2b2 = 0 . . . (iii)Above equation being a quadratic in x, gives two values of x. If shows that every straight line will cut theellipse in two points, may be real, coincident or imaginary according toDiscriminant of (iii) >, =, < 0i.e., 4m2 c2a4 – 4 (a2m2 + b2) (c2a2 – a2b2) >, =, < 0 – a2b2c2 + a4b2m2 + a2b4 >, = , < 0 a2m2 + b2 > , =, < c2 . . . (iv)

8.2 Condition of TangencyIf the line (ii) touches the ellipse (i), then equation (iii) has equal roots. Discriminant of (iii) = 0

c2 = a2m2 + b2 or 2 2 2c a m b . . . (v)

So, the line y = mx + c touches the ellipse 2 2

2 2 1x ya b

if c2 = a2m2 + b2

[which is condition of Tangency]Substituting the value of c from (v) in (ii), we get

2 2 2y mx a m b

Hence the line 2 2 2( )y mx a m b will always be tangents to the ellipse.

8.3 Point of ContactWhen Discriminant of (iii) is zero, roots are equal.Equal roots of a quadratic equation

2Coefficient of

2 Coefficient ofxxx

= 2 2 2 22

2 2 2 2 2 212

2 ( )m a a m bm ca

a m b a m b

. . . [using (v)]

Section 852

Conic Section

2 2

2 2 2

a m a mxca m b

Replace x in (i) to get :4 2 2

2 2 21· 1a m y

c a b

2 2 2 2 2 2 2

2 2 2 2( )1y a m c a m b

b c c c

2byc

Hence the point of contact is 2 2

,a m bc c

.

Find the equation of an ellipse whose focus is (–1, 1), ecentricity is 12 and the directrix is

x – y + 3 = 0.

Illustration - 56

SOLUTION :Let P (x, y) be any point on the ellipse whose focusis S (– 1, 1) and the directrix is x – y + 3 = 0.

P (x, y)M

x –

y +

3 =

0

S (–1, 1)

Draw PM perpendicular from P (x, y) on the directrix x – y + 3 = 0. Then by definition

SP = ePM (SP)2 = e2 (PM)2

2

2 2 1 3( 1) ( 1)4 2

x yx y

8 (x2 + y2 + 2x – 2y + 2)= x2 + y2 + 9 – 2xy + 6x – 6y

7x2 + 7y2 + 2xy + 10x – 10y + 7= 0

which is the required equation of the ellipse.

Section 8 53

Conic Section

If the angle between the straight lines joining foci and the end of minor axis of the ellipse2 2

2 2 1x ya b

is 90°, find its eccentricity..

Illustration - 57

SOLUTION :

The equation of the ellipse is 2 2

2 2 1x ya b

.

The ends of minor axis are B (0, b) and B (0, – b).If the eccentricity of the ellipse is e, then the foci areS (ae, 0) and S (– ae, 0).

B (0, b)

A AS (–ae, o) (ae, o)

90°

B (0, –b)

SC

Slope of BS is 10

0b bm

ae ae

and slope of BS is 20

0b bm

ae ae

The angle between BS and BS is 90°,

m1m2 = – 1 1b bae ae

b2 = a2e2

a2 (1 – e2) = a2e2 1 – e2 = e2

2e2 = 1 12

e .

Prove that the straight line lx + my + n = 0 touches the ellipse 2 2

2 2 1x ya b

if a2l2 + b2m2 =

n2.

Illustration - 58

SOLUTION :The given lines is lx + my + n = 0

l ny xm m

Comparing this line withy = Mx + c . . . (i)

lMm

andncm

The line (i) will touch the ellipse2 2

2 2 1x ya b

if c2 = a2M2 + b2

2 2 2

22 2

n a l bm m

a2l2 + b2m2 = n2

Section 854

Conic Section

For what value of does the line y = x + touches the ellipse 9x2 + 16y2 = 144.Illustration - 59

SOLUTION :Equation of ellipse is 9x2 + 16y2 = 144

2 2

116 9x y

Comparing this with 2 2

2 2 1x ya b

then we get a2 = 16 and b2 = 9 and comparing theline y = x + with y = mx + c

m = 1 and c = If the line y = x + touches the ellipse

9x2 + 16y2 = 144, thenc2 = a2m2 + b2

2 = 16 × 12 + 9 2 = 25 = ± 5

The following represents an ellipse, 25 (x2 – 6x + 9) + 16y2 = 400. Find the centre and thefoci of the ellipse. How should the axes be transformed so that the ellipse is represented by the equation

2 21

25 16x y

.

Illustration - 60

SOLUTION :We can write the given ellipse as :

2 2

2 2( 3)

4 5x y

= 1. 2 2

2 2X Ya b

= 1.

As a2 < b2, major axis is y-axis.

Centre is (3, 0). e2 = 1 – 2

2ab

= 1 – 1625

e = 35

Foci are X = 0, Y = ± a e = (3, ± 3)In order to transform the given equation to the form

2 2

25 16x y

= 1, the origin should be shifted to the

point (3, 0) and then the axis be turned through aright angle.

If S and S1 are the foci of an ellipse and P a point on it, if e is the eccentricity of the ellipse,prove that :

1 11 1 12 2 1

etan PSS tan PS Se

.

Illustration - 61

SOLUTION :Apply sine rule in PSS1 to get :

sinSP = 1

sinS P

=

1sin [ ( )]

S S

1sin sin

SP S P

= 2

sin ( )ae

sin ( )

sin sin

= e

2 sin cos

2 2

2 sin cos2 2

= e

Section 8 55

Conic Section

e = cos

2

cos2

.

P

S C S1

Y

X

Apply compendo-dividedo, to get11

ee

= tan 2

tan 2

. Hence proved.

Find the eccentricity, foci, latus rectum and directrices of the ellipse 2x2 + 3y2 = 6.Illustration - 62

SOLUTION :The equation of the ellipse can be written as :

2 21

3 2x y

On comparing the above equation of ellipse withthe standard equation of ellipse, we get :

3a and 2b We know that : b2 = a2 (1 – e2)

2 = 3 (1 – e2) e = 1/3Using the standard results, foci are

(ae, 0) and (– ae, 0) foci are (1, 0) and (– 1, 0)Latus rectum = 2b2/a = 4/3.Directrices are x = ± a/e x = ± 3.

EQUATION OF TANGENT NORMAL AND CHORD OF CONTACT Section - 9

9.1 Equation of tangent(i) Point form : The equation of tangent drawn to the ellipse at (x1, y1) is :

1 12 2 1xx yy

a b

(ii) Parametric form : The equation of tangent drawn to the ellipse at the point (a cos, b sin ) is :

cos sin 1x ya b

(iii) Slope form : The equation of tangent of slope m can be takes as :

2 2 2y mx a m b , where point of contact is 2 2

2 2 2 2 2 2,a m b

a m b a m b

Section 956

Conic Section

9.2 Important Results :(i) The equations of two parallel tangents of slope m that can be drawn to the ellipse are :

2 2 2y mx a m b

(ii) The condition for y = mx + c to be a tangent to ellipse is c2 = a2m2 + b2 and point of contact is

2 2,a m b

c c

.

(iii) Tangents drawn to ellipse from a point (h, k)

Let 2 2 2y mx a m b be the tangent(s).As the tangents pass through (h, k), we get :

2 2 2k mh a m b

m2 (h2 – a2) – 2 hkm + k2 – b2 = 0This quadratic in m will give slopes m1, m2 of two tangents

drawn from (h, k) to the ellipse.

Note : The equation of the tangents to the ellipse at point P (a cos 1, b sin 1) and Q (a cos 2, b sin 2) are

1 1cos sin 1x ya b

and 2 2cos sin 1x ya b

and these two tangents intersect at the point

1 2 1 2

1 2 1 2

cos sin2 2,

cos cos2 2

a b

9.3. Equation of normal(i) Point form : The equation of normal drawn to the ellipse at (x1, y1) is :

2 22 2

1 1

a x b y a bx y

(ii) Parametric form : The equation of normal drawn to the ellipse at the point (a cos, b sin ) is :

2 2cos sin

ax by a b

(iii) Slope form : The equation of normal of slope m can be takes as : 2 2

2 2 2

( )

( )

m a by mxa b m

.

(h, k)

m2

m1

Section 9 57

Conic Section

Note : The point of intersection of normals to the ellipse 2 2

2 2 1x ya b

at the points

(a cos 1, b sin 1) and (a cos 2, b sin 2) are

1 2 1 22 2 2 2

1 2 1 21 2 1 2

cos sin( ) ( )2 2cos cos , sin sin

cos cos2 2

a b a ba b

9.4 Equation of Chord of Contact(i) Point formLet us consider a point R whose coordinates are (x1, y1) andwhich lies outside the ellipse. From point R, two tangents aredrawn to the ellipse which meet ellipse in points P and Q. Theline joining P and Q isknown as chord of contact of the ellipse

and its equation is : 1 12 2 1xx yy

a b

Note : The above result is same as equation of tangent drawn to ellipse at (x1, y1).

(ii) Parametric formEquation of chord joining the points P () and Q () is :

cos sin cos2 2 2

x ya b

9.5 Equation of Chord of Contact of the ellipse whose mid-point is givenLet us consider a chord PQ of the ellipse whose mid-point is R (x1, y1).The equation of this chord PQ in terms of coordinates of R is :

S1 = T,

where2 21 1

1 2 2 1x ySa b

and 1 12 2 1xx yyT

a b .

9.6 Position of a point (h, k) with respect to an ellipse

Let ellipse be2 2

2 2 1x ya b

Let P (h, k)

P (h, k)R (x , y )22

Q (x , y )11

Chord ofContact

Q (x , y )22

R (x , y )33

(x , y )11P

Section 958

Conic Section

Now P will lie outside, on or inside the ellipse2 2

2 2 1x ya b

according as 2 2

2 2 1 , , 0h ka b

.

9.7 Director Circle

The locus of the point of intersection of the tangents to an ellipse 2 2

2 2 1x ya b

which are perpendicular to

each other is called Director circle.

Let any tangent in terms of slope of ellipse 2 2

2 2 1x ya b

is :

2 2( )y mx a m b

It is passes through (x1, y1)

2 2 21 1 ( )y mx a m b

(y1 – mx1)2 = a2m2 + b2

y12 + m2 x1

2 – 2mx1 y1 = a2m2 + b2

m2 ( x12 – a2 ) – 2y1 x1m + y1

2 – b2 = 0it is quadratic equation in m let slope of two tangents are m1 and m2

2 21

1 2 2 21

1y bm mx a

[as tangents are perpendicular]

– x12 + a2 = y1

2 – b2 x12 + y1

2 = a2 + b2

Hence the locus of (x1 y1) is x2 + y2 = a2 + b2.

If the normal at a point P () to the ellipse 2 2

114 5x y

intersect it again at Q (2). Showthat cos = – 2/3.

B

Y

P (on)

P (inside)

P (outside)

A XX A

B

Y

C

P

90°B

A A

BC

Illustration - 63

SOLUTION :The equation of normal at P () is :

2 2cos sin

ax by a b

As Q (a cos 2, b sin 2) lies on it, we can have :2 2( cos 2 ) ( sin 2 )

cos sina ba b a b

2

2 2 2 2(2 cos 1) 2 coscos

a b a b

Put a2 = 14, b2 = 5 in the above equation to get :

14 (2 cos2 – 1) – 10 cos2 = 9 cos 18cos2 – 9cos – 14 = 0 (6 cos – 7) (3 cos + 2) = 0

Section 9 59

Conic Section

cos = 7/6 (reject) or cos = – 2/3 Hence cos = – 2/3.

If the normal at end of latus rectum passes through the opposite end of minor axis, findeccentricity.Illustration - 64

SOLUTION :The equation of the normal at L (ae, b2/a) isgiven by :

2 22 2

2 /a x b y a bae b a

2 2x a by

e a

. . . (i)

According to the question, normal passes throughopposite end of latus reaction i.e. it passes through(0, –b) 0/e + b = (a2 – b2)/a

a2 – b2 – ab = 0Using b2 = a2 (1 – e2), we get

a2 e2 – ab = 0 (0, b) satisfies (i) b = ae2

a2 e4 = a2 (1 – e2) [using : b2 = a2 (1 – e2)]

e4 = 1 – e2 2 5 12

e

5 12

e

Show that the locus of the foot of the perpendicular drawn from the centre of the ellipse

2 2

2 2 1x ya b

on any tangent is (x2 + y2)2 = a2 x2 + b2 y2.

Illustration - 65

SOLUTION :

Let the tangent be 2 2 2y mx a m b .Draw CM perpendicular to tangent and letM (x1, y1).M lies on tangent,

2 2 21 1y mx a m b . . . (i)

Slope (CM) = – 1/m

1

1

1yx m

1

1

xmy

. . . (ii)

Replace the value of m from (ii) into (i) and takesquare to get :

(x12 + y1

2)2 = a2x12 + b2

y12

Hence the required locus is :(x2 + y2)2 = a2 x2 + b2 y2.

Y

M (x , y )11

A C A X

Section 960

Conic Section

The tangent at a point P on ellipse 2 2

2 2 1x ya b

cuts the directrix in F. Show that PF

subtends a right angle at the corresponding focus.

Illustration - 66

SOLUTION :Let P (x1, y1) and S (ae, 0)

The equation of tangent at P is : 1 12 2 1xx yy

a b

To find F, we put x = a/e in the equation of thetangent.

1 1

2 2 1ax yya e b

Y

C S

P (x , y )11

x = a/e

FX

2

1

1

( )ae x byaey

2

1

1

( ), ae x baFe aey

2

1

1

( ) 1slope ( ) ae x bSF aaey aee

. . . (i)

1

1

0slope ( ) ySPx ae

. . . (ii)

From (i) and (ii) :slope (SF) × slope (SP) = – 1 SF and SP are perpendicular.Hence PF subtends a right at the focus.

Note : You can also solve this question by assuming P in terms of parametric coordinates i.e. P (a cos , b sin )

Show that the normal of ellipse 2 2

2 2 1x ya b

at any point P bisects the angle between

focal radii SP and SP.

Illustration - 67

SOLUTION :Let PM be the normal and P (x1, y1) equation of normal PM is

2 22 2

1 1

xa yb a bx y

P (x , y )11

M SS

Y

X

We will try to show that : S P MSSP MS

M is the point of intersection of normal PM withX-axis. Put y = 0 is normal PM to get

2 221

12( ) , 0 , 0a b xM e x

a

MS = ae – e2 x1 and MS = e2 x1 + ae

Section 9 61

Conic Section

1 1

1 1

( )( )

e a ex a exMS SPMS e a ex a ex SP

[using result given in section 1.1]

Note : You can also solve this question by assuming P in terms of parametric coordinates i.e. P (a cos , b sin )

A tangent to an ellipse 2 2

2 2 1x ya b

touches it at a point P in the first quadrant and meets

the axes in A and B respectively. If P divides AB is 3 : 1, find the equation of tangent.

Illustration - 68

SOLUTION :Let the coordinates of the pointP (a cos, b sin) the equation of the tangent at P is :

cos sin 1x ya b

. . . (i)

The coordinates of the points A and B are :

, 0 0,cos sin

a bA and B

BP (a cos , b sin )

A

Y

X

By section formula, the coordinates of P are

3, ( cos , sin )

4cos 4 sina b a b

cos4 cos

a a

and3 sin

4 sinb b

1cos2

and3sin

2

= 60°For equation of tangent, replace the value of in(i).

The equation of tangent is : 2

2yx

a b .

If the chord through the points P and Q whose eccentric angles are and on the ellipse

2 2

2 2 1x ya b

passes through a focus S and S is the other focus, prove that

(i) SP.SP = a2 sin2 + b2 cos2 (ii) the perimeter of SPQ = 4a

(iii) 2 2 2 2 21 1( ) ( ) ( )2 2

a b cos a cos (iv)1 1 (1 ) 02 2 (1 )

etan tane

Illustration - 69

Section 962

Conic Section

SOLUTION :(i) (a + aex) (a – ex) = a2 – e2x2 = a2 – e2a2

cos2 = a2 – (a2 – b2) cos2 = a2 sin2 + b2 cos2

(ii) Perimeter (S PQ)= SP + S Q + PQ= S P + (PS + SQ) + QS= (S P + PS) + (SQ + S Q)= 2a + 2a = 4a.

P

S S

Q

Y

X

(iii) Equation of chord

xa cos sin cos

2 2 2yb

As it passes through (ae, 0)

e = cos

2

cos2

11

ee

= cos cos

2 2

cos cos2 2

11

ee = tan 2

tan 2

tan 2

tan 2

+ 11

ee

= 0

If the normal at point P of ellipse 2 2

2 2 1x ya b

with centre C meets major and minor axes

at G and g respectively, and if CF be perpendicular to normal, Prove that PF. PG = b2 and PF.Pg = a2.

Illustration - 70

SOLUTION :If PM is tangent to the ellipse at point P,then CMPF is a rectangle. CM = PF . . . (i)Let the coordinates of point P be (a cos, b sin).

Y

M

P (acos , bsin )

C GF

X

g

The equation of normal at P is :

2cos sin

ax by a b

The point of intersection of the normal at P withX-axis is

2 2( ) cos , 0a bGa

The point of intersection of the normal at P with Y-axis is

2 2( ) sin0, b agb

Using distance formula,

2 2 2 2cos sinbPG b aa

. . . (ii)

Section 9 63

Conic Section

and 2 2 2 2cos sinaPg b ab

. . . (iii)

From (i), PF = CM = distance of centre of the ellipsefrom the tangent at P

2 2 2 2 2 2

2 2

1

cos sin cos sin

ab

b aa b

. . .(iv)

Multiplying (ii) and (iv), we get :PF . PG = b2 [Hence proved]

Multiplying (iii) and (iv), we get :

PF . Pg = a2 [Hence proved]

Prove that the chord of contact of tangents drawn from the point (h, k) to the ellipse

2 2

2 2 1x ya b

will subtend a right angle at the centre, if 2 2

4 4 2 21 1h k

a b a b . Also, find the locus of (h, k).

Illustration - 71

SOLUTION :The equation of chord of contact of tangents drawn

from P (h, k) to the ellipse 2 2

2 2 1x ya b

is

2 2 1hx kya b

. . . (i)

The equation of the straight lines CA and CB isobtained by making homogeneous ellipse

2 2

2 2 1x ya b

with the help of (i),

22 2

2 2 2 2x y hx kya b a b

2 2

2 24 2 4 2 2 2

1 1 2 0h k hkx y xya a b b a b

. . . (ii)

but given ACB = 90° Co-efficient of x2 + co-efficient of y2 = 0

2 2

4 2 4 21 1 0h k

a a b b

2 2

4 4 2 21 1h k

a b a b

Hence locus of (h, k) is 2 2

4 4 2 21 1x y

a b a b .

A

B

(0, 0)90°CP(h, k)

Section 964

Conic Section

Prove that the locus of the middle points of normal chords of the ellipse 2 2

2 2 1x ya b

is

the curve22 2 6 6

2 2 22 2 2 2 ( )x y a b a b

a b x y

.

Illustration - 72

SOLUTION :Let (h, k) be the middle point of any chord of anellipse. Then its equation is

T = S1

2 2

2 2 2 21 1xh yk h ka b a b

2 2

2 2 2 2xh yk h ka b a b

. . . (i)

If (i) is a normal chord, then it must be of the form

ax sec – by cosec = a2 – b2

P (a cos , b sin )

(h, k)

Thus the equations (i) and (ii) represent the samenormal chord of the ellipse with its middle point(h, k).Hence they are identical and comparing their co-efficients, we get

2 22 2 2 2

2 2/ /sec cosec ( )

h kh a k b a ba b a b

2 2

2 23

2 2cos( )

h ka ba

h a b

. . . (iii)

and

2 2

2 23

2 2sin( )

h ha bb

k a b

Show that the locus of the middle points of chords of an ellipse. Which pass through a fixedpoint, is another ellipse.Illustration - 73

SOLUTION :Let P (x1, y1) be the middle point of any chord AB

of the ellipse 2 2

2 2 1x ya b

then equation of chord

AB isT = S1

2 2

1 1 1 12 2 2 21 1xx yy x y

a b a b

2 2

1 1 1 12 2 2 2

xx yy x ya b a b

. . . (i)

But it passes through a fixed point Q (h, k) its co-ordinates must satisfy equation (i),

P (x , y )11

B

A

Q (h, k)

Section 9 65

Conic Section

2 2

1 1 1 12 2 2 2

hx ky x ya b a b

this can be re-written as

2 2

2 21 1

2 2 2 212 24

h hx yh k

a b a b

Hence locus of P (x1, y1) is2 2

2 2

2 2 2 212 24

h kx yh k

a b a b

Its obviously an ellipse with centre at ,2 2h k

and

axes parallel to co-ordinates axes.

Find the locus of the point the chord of contact of tangents from which to the ellipse2 2

2 2 1x ya b

.

(i) Subtends a right angle at the centre of the ellipse. (ii) Touches the circle x2 + y2 = c2.

SOLUTION :(i) Let the point from which tangents are drawn be (x1, y1)

Equation of chord of contact 1 12 2

xx yya b

= 1. It subtands a right at the centre of the ellipse.

Homogenising the equation of ellipse, we get equation of pair of lines,2 2

2 2x ya b

=2

1 12 2

xx yya b

2 22 21 1 1 1

2 4 2 2 2 421 1x xy x y yx y

a a a b b b

= 0

Since the lines are , coeff. of x2 + coeff. of y2 = 0 2 21 1

2 4 2 41 1x y

a a b b = 0

(ii) Chord of contact 1 12 2

xx yya b

= 1 touches x2 + y2 = c2.

2 21 14 4

1

x ya b

= c 2 21 14 4

x ya b

= 21c

Locus is 2 2

4 4 21x y

a b c

Illustration - 74

Section 966

Conic Section

Two sides of a triangle inscribed in the ellipse 2 2

2 2 1x ya b

are parallel to two given

straight lines. Prove that its third side touches a certain ellipse.

SOLUTION :Let vertices of the ABC be points on the ellipse whose eccentric angles are 1, 2, 3 resp.

Chord AB is xa cos 1 2 1 2( ) ( )sin

2 2yb

= cos 1 2( )

2

ba

cot 1 2( )2

= constant 1 + 2 = 2 (assume)

Similarly 2 + 3 = 2 1 – 3 = 2 ( – )

Chord CA is xa cos 1 3 1 3( ) ( )sin

2 2yb

= cos 1 3( )

2

cos sinx ya b

cos( )

cosx ya b

sin = 1 tangent to ellipse :2 2

2 2 2 2x y

a b

= 1

Find the locus of the mid-points of chords of ellipse, the tangents at the extremities ofwhich intersect at right angles.

SOLUTION :Let point of Int. of tangents be P (x1, y1)As tangents are , (x1, y1) lies on x2 + y2 = a2 [director circle] x1

2 + y12 = a2 . . . (i)

Equation of chord of contact is 1 12 2

xx yya b

= 1 . . . (ii)

Equation of chord whose mid-point is (h, k) is

2 2hx kya b

= 2

2ha

+ 2

2kb

. . . (iii)

From (ii) and (iii) on comparing, we get (x1, y1)

Replace (x1, y1) in (i) to get Locus 2 2

2 2x ya b

(a2 + b2) = x2 + y2

Illustration - 75

Illustration - 76

Section 9 67

Conic Section

A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove thatthe tangents at P and Q of the ellipse x2 + 2y2 = 6 are at right angles.

SOLUTION :Chord of contact of

2 2

6 3x y

= 1 is 6 3hx ky

= 1 . . (i)

Equation of any tangent to2 2

4 1x y

= 1 is 2x

cos + y sin = 1. . . . (ii)

Compare (i) and (ii), eliminate and get locus of (h, k)i.e. x2 + y2 = 9 (i.e. a2 + b2)i.e. director circle of 2nd ellipse.

If (1/2, 2/5) be the middle-point of the chord of the ellipse 2 2

125 16x y

, then prove that its

length is 7 415 .

SOLUTION :

Equation of chord with 1 2,2 5

as midpoint is T1 = S1 i.e. 4x + 5y = 4

Solve chord and ellipse to get 12 164, , 3,5 5

Length = 7 415 .

Any tangent to an ellipse is cut by the tangents at the ends of the major axis in T and T.Prove that circle on TT as diameter passes through foci.

SOLUTION :Consider a point P on the ellipse whose coordinates are (a cos, b sin)The equation of tangent drawn at P is :

cos sin 1x ya b

. . . (i)

Illustration - 77

Illustration - 78

Illustration - 79

Section 968

Conic Section

The two tangents drawn at the ends of the major axis are x = a and x = – a.Solving tangent (i) and x = a we get :

(1 cos ), , tansin 2

bT a a b Solving tangent (i) and x = – a we get :

(1 cos )' , , cotsin 2

bT a a b

Circle on TT as diameter is :x2 – a2 + (y – b tan /2) (y – b cot /2) = 0 [using diametric form of equation of circle]

Put x = ± ae, y = 0 in LHS to get :

a2 e2 – a2 + b2 = 0 = RHS

Hence foci lie on this circle.

A normal inclined at 45° to the x-axis is drawn to the ellipse 2 2

2 2 1x ya b

. It cuts major

and minor axes in P and Q. IF C is centre of ellipse, show that are (CPQ) = 2 2 2

2 2( )2 ( )a b

a b

.

SOLUTION :Consider a point M on the ellipse whose coordinates are (a cos, b sin)The equation of normal drawn at M is :

2 2cos sin

ax by a b

As the normal makes an angle 45° with X-axis, slope of normal = tan 45°.

sintan 45cos

ab

tan ba

TP

T

AOA

Y

X

Illustration - 80

M

C

Q

P

Y

X

Section 9 69

Conic Section

2 2

sin b

a b

and

2 2cos a

a b

. . . (i)

The point of intersection of the normal with X-axis is 2 2

cos , 0a bPa

2 2

cosa bCPa

. . . (ii)

The point of intersection of the normal with Y-axis is 2 2

0, sinb aQb

2 2

sinb aCQb

. . . (iii)

1Ar ( )2

CPQ CP CQ

Using (ii) and (iii),2 2 21 ( )Ar ( ) sin cos

2a bCPQ

ab

Using (i),2 2 2

2 21 ( )Ar ( )2

a bCPQa b

.

If the portion of the line x cosa + y sina = p intercepted by the ellipse 2 2

2 2 1x ya b

sub-

tends a right angle at the centre of the ellipse, prove that the line touches a circle of radius 2 2ab

a b

concentric with the ellipse.SOLUTION :

The line x cos + y sin = p intersects ellipse in two pointsA and B.Make equation of ellipse homogeneous with the above line.

2 2

2 2x ya b

= 2cos sinx y

p

is the homogenise d eqution of degree-2.

Illustration - 81

A

O

Y

X

B

Section 970

Conic Section

Above homogenised equation represents pair of lines OA and OB.

As angle substended at centre is 2

, we get :

Coeff. x2 + coeff. of y2 = 2 2

2 2 2 21 cos 1 sin

a p b p

= 0

21p

= 2 21 1

a b p =

2 2ab

a b

A circle of radius r is concentric with the standard ellipse. Prove that the common tangent

is inclined to the major axis at an angle 2 2

12 2

r btana r

.

Illustration - 82

SOLUTION :General Tangent to ellipse :

y = mx + 2 2 2a m b .

As this is also tangent to the circle, apply conditionof tangency.

i.e.2 2 2

21

a m b

m

= r

Y

X

a2m2 + b2 = r2 (1 + m2)

m2 = tan2 = 2 2

2 2r ba r

= ± tan–1 2 2

2 2r ba r

2 2

12 2tan r b

a b

and

2 2

12 2tan r b

a r

[where is inclination of common tangent with x-axis]Negative value of means angle measured inclockwise direction.

Section 9 71

Conic Section

P and Q are two point on the ellipse 2 2

125 9x y

such that the sum of their ordinates is 3.

Prove that the locus of the point of intersection of tangents at P and Q is 9x2 + 25y2 = 150y.

SOLUTION :

Elipse is 2 2

25 9x y

= 1. Let point of intersection of tangents at P and be R (x1, y1).

Let P (5 cos , 3 sin ) and Q (5 cos , 3 sin ) be two points on ellipse.

Tangents at these points be 5x

cos + 3y

sin = 1 and 5x

cos + 3y

sin = 1

These tangents intersect at R (x1, y1) R (x1, y1) will satisfy both of them.

1 1cos sin 15 3x y

. . . (i)

and 1 1cos sin 15 3x y

. . . (ii)

Multiply (i) by sin and (ii) by sin and subtract,

15x

sin ( – ) = sin – sin

15x

2 cos 2

sin 2

= 2 sin 2

cos2

15x

cos 2

= cos 2

. . . (iii)

Similarly 13y

cos 2

= sin 2

. . . (iv)

Also, we are given that :

3 sin + 3 sin = 3 2 sin 2

cos 2

= 1 . . . (v)

On dividing (iv) by (v), 16y

= sin2 2

. . . (vi)

From (iii) and (v) on dividing, we get :

110x

= sin 2

cos 2

21

100x

= sin2 21 sin

2 2

21

100x

= 1 116 6y y

9x12 + 25y1

2 = 150 y1

Replacing x1 by x, y1 by y1, we get locus as 9x2 + 25y2 = 150y.

Illustration - 83

Section 972

Conic Section

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

2 2 1x ya b

to the

tangent drawn at a point P on the ellipse, if F1 and F2 are two foci of the ellipse then show that:

22 2

1 2 2( ) 4 1 bPF PF ad

.

SOLUTION :

Tangent at P () is xa cos +

yb sin = 1.

d = 2 2

2 2

1

cos sina b

21

d =

2 2

2 2cos sin

a b

2

2bd

= 2

2ba

cos2 + sin2

1 – 2

2bd

= cos2 – 2

2ba

cos2 = 2

22cos 1 b

a

= cos2e2.

4a2 2

21 bd

= 4a2 cos2 e2 . . . (i)

PF1 = a + ex , PF2 = a – ex PF1 – PF2 = 2ex = 2 ea cos (PF1 – PF2)

2 =4a2e2 cos2 . . . (ii)

Combining (i) and (ii), (PF1 – PF2)2 = 4a2

2

21 bd

. Hence proved.

PA and PB are tangents drawn from a point P to the ellipse 2 2

2 2 1x ya b

. The area of the

triangle formed by the chord of contact AB and axes of co-ordinates is constant. Prove that locus of P is ahyperbola.

Illustration - 84

Illustration - 85

SOLUTION :Let P (x1, y1). Chord of contact w.r.t. P is

1 12 2

xx yya b

= 1. It meets the axes in A

2

1, 0a

x

and B 2

10, b

y

.

Section 9 73

Conic Section

A

B

O

P (x , y ) 1 1 Y

X

Area of right angled triangle OAB

= 12 OA . OB =

12

2 2

1 1

a bx y = k (constant)

x1y1 = 2 2

2a b

k Required locus is xy = c2 which is a hyper

bola.

Note : You will study hyperbola later in this module.

Any ordinate NP of an ellipse meets the auxiliary circle in Q. Prove that the locus of theintersection of the normals at P and Q is circle x2 + y2 = (a + b)2.

SOLUTION :Let P (a cos , b sin ) Q (a cos , a sin ), where P lies on ellipse and Q on auxillary

Normal at P cos sin

ax by

= a2 – b2 . . . (i)

Normal at Q to the circle = y = x tan . . . (ii)Solve (i) and (ii) to get x = (a + b) cos and y = (a + b) sin .Square and add to get x2 + y2 = (a + b)2 as the required locus.

Prove that the tangents drawn to the ellipse 2 2x y a b

a b at the points where it is cut

by any tangent to the ellipse 2 2

2 2 1x ya b

intersect at right angles.

Illustration - 86

Illustration - 87

SOLUTION :

Let PQ be any tangent to ellipse 2 2

2 2x ya b

= 1.

which cuts the ellipse 2 2

( ) ( )x b

a a b b a b

= 1 at

P and Q.

Let A (h, k) be point of intersection of tangents

drawn to 2 2

( ) ( )x y

a a b b a b

= 1 at P and Q.

P

QA (h, k)

Section 974

Conic Section

As tangents at P and Q are at right angle, (h, k) willlie on director circlei.e. (h, k) satisfies x2 + y2 = a (a + b) + b (a + b) = (a + b)2 . . . (i)PQ is also chord of contact w.r.t. (h, k)Equation of PQ is

( ) ( )hx ky

a a b b a b

= 1

It is tangent to 2 2

2 2x ya b

= 1

c2 = a2 m2 + b2

2 2

2( )b a bk

= a2 2 2

2 2b ha k

+ b2 .

h2 + k2 = (a + b)2 . . . (ii)From (i) and (ii) tangents at P and Q are at rightangle.

Prove that the portion of the normal at any point of the ellipse 2 2

2 2 1x ya b

intercepted

between the curve and major axis is ( )b rra

where r and r are the focal distances of the point at which the

normal is drawn.

Illustration - 88

SOLUTION :Equation of Norrmal at

P cosax

– sinby

= a2 – b2 ,

G 2 2

cos , 0a ba

P

G

Y

X

PG =

22 22cos cos ( sin )a b a b

a

PG = ba 2 2 2 2cos sinb a . . . (i)

rr = SP . SP = (a – ex) (a + ex) = a2 – e2 x2

= a2 – e2 a cos2 = a2 + (b2 – a2) cos2 = a2 sin2 + b2 cos2 . . . (ii)

From (i) and (ii)

PG = ba rr .

Section 9 75

Conic Section

Find the locus of a point from which the two tangents to the ellipse are inclined at an angle.Illustration - 89

SOLUTION :Equation of tangent of slope m is

y = mx + 2 2 2a m b . . . (i)

Point P (x1, y1) lies on (i)

y1 = mx1 + 2 2 2a m b m2 (x1

2 – a2) – 2x1y1 m + (y12 – b2) = 0

Let roots be m1 and m2

m1 + m2 = 1 12 21

2x yx a

Px (x , y)1

Y

X

and m1 m2 = 2 212 21

y bx a

tan = 1 2

1 21m m

m m

tan2 (1 + m1 m2)2

= (m1 + m2)2 – 4 m1 m2 . . . (ii)

Using (ii),

tan2

22 212 21

1 y bx a

=

2 2 2 21 1 1

2 2 22 2 11

4 4x y y bx ax a

Locus is tan2 (x2 + y2 – a2 – b2)2

= 4 [x2b2 + a2y2 – a2b2]

The tangent at a point P on an ellipse intersects the major axis is T, and N is the foot of theperpendicular from P to same axis. Show that the circle drawn on NT as diameter intersects the auxiliarycircle orthogonally.

Illustration - 90

SOLUTION :Let P (a cos , b sin ) ; N (a cos , 0),

T , 0cos

a

Circle on NT as diameter is

(x – a cos ) cosax

+ y2 = 0 . . . (i)

Auxillary circle is x2 + y2 = a2 . . . (ii)For circles (i) and (ii) to intersect orthogonally,

2 (g)1(g2) + 2f1 f2 = c1 + c2

i.e. 0 = a2 – a2 = 0Hence two circles intersect orthogonally

P

N A T

Y

X

Section 976

Conic Section

If P is any point on the ellipse 2 2

2 2 1x ya b

where ordinate is y, prove that the angle

between the tangent at P and the focal distance of P is tan–1 (b2/aey).

SOLUTION :

Tangent at P (x, y) is 2 2xx yya b = 1 where

2 2

2 2x ya b = 1. . . . (i)

Slope of tangent is = – 2

2,x by a

Slope of SP is = y

x ae

where S (–ae, 0)

tan =

2

2

2

21

y x bx ae y a

y x bx ae y a

= 2 2 2 2 2

2 2 3( )a y b x b x aex y a b a ey

2 2 2Using b = a (1-e )

= 2 2 2

2 2 3a b b x ae

x y a e a ey

=

2

2( )( )

b a a exa ey a ex

= 2b

aey [using (i)]

Prove that in general four normals can be drawn to an ellipse from any point and the sumof the eccentric angles of the feet of these normal is equal to an odd multiple of two right angles.

SOLUTION :

Equation of Normal cos sinax by

= a2 – b2

As it passes through (h, k) cos sinah bk

= a2 – b2

Replace cos = 2

211

tt

, sin = 2

21

tt , where t = tan 2

4 2 2 3 2 22( ) 2( ) 0bk t ah a b t ah a b t bk t – bk = 0

Its roots are tan 2r

, r = 1, 2, 3, 4

tan 31 2 42 2 2 2

= 1 3

2 41S S

S S

= = tan 2

2 4as 0, 1bkS Sbk

Illustration - 91

Illustration - 92

Section 9 77

Conic Section

1 2 3 42

= n + 2

1 + 2 + 3 + 4 = (2n + 1)

Let P be a point on the ellipse 2 2

2 2 1x ya b

, a > b > 0. Let the line parallel to y-axis

passing through P meet the circle x2 + y2 = a2 at the point Q such that P and Q are on the same side of the x-axis. For two positive real numbers r and s, find the locus of the point R on PQ such that PR : RQ = r : s andP varies over the ellipse.

SOLUTION :

x1 = ( cos ) ( cos )r a s a

r s

x1 = a cos 1cos xa

. . . (i)

y1 = ( sin ) ( sin )r a s b

r s

(r + s) y1 = (ar + bs) sin

1( )sin r s y

ar bs

. . . (ii)

From (i) and (ii), 22

1 1 ( )x y r sa ar bs

= 1.

Hence required locus is 22

22 1x r sy

ar bsa

.

Illustration - 93

Q (a cos , a sin )

R (x , y ) 1 1

P (a cos , b sin )

S

r

Section 978