Free Flip-Book Mathematics class 11th by Study Innovations. 254 Pages

Slope of the tangent (m2 ) = 2a1 / 3
b1/ 3

a1 / 3 2a1 / 3
m1 − m2 2b1 / 3 −
If θ is the angle between the two tangents, then ⇒ tan θ = 1 + m1m2 = a1 / 3 b1 / 3
2b1 / 3 2a1 / 3
1+ . b1 / 3

= 3 . 1 = 3 . 1 = 3; ∴θ = 60 o = π
2 2 3
 a 1 / 3  b 1 / 3 3
 b   a  2
+

Example: 24 The equation of the common tangent touching the circle (x − 3)2 + y 2 = 9 and the parabola y 2 = 4 x above the x-axis , is
Solution: (c)
[IIT Screening 2001]
Example: 25
Solution: (d) (a) 3y = 3x + 1 (b) 3y = −(x + 3) (c) 3y = x + 3 (d) 3y = −(3x + 1)

Any tangent to y2 = 4x is y mx 1 . It touches the circle if 3m + 1
m m
= + 3=
1 + m2
Y

or 9(1 + m2) =  3m + 1  2 or 1 =3 , ∴m = ± 1 (3,0) X
 m  m2 3 O3

For the common tangent to be above the x-axis, m = 1
3

∴ Common tangent is y = 1 x + 3 ⇒ 3y = x + 3
3

If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y 2 = 4ax and x 2 = 4ay

then [AIEEE 2004]

(a) d 2 + (3b − 2c)2 = 0 (b) d 2 + (3b + 2c)2 = 0 (c) d 2 + (2b − 3c)2 = 0 (d) d 2 + (2b + 3c)2 = 0

Given parabolas are y 2 = 4ax .....(i) and x 2 = 4ay .....(ii)

from (i) and (ii)  x2  2 = 4ax ⇒ x 4 − 64a 3 x = 0 ⇒ x=0, 4a ∴ y = 0, 4a
4a

So points of intersection are (0,0) and (4a,4a)

Given, the line 2bx + 3cy + 4d = 0 passes through (0,0) and (4a,4a)

∴ d = 0 ⇒ d 2 = 0 and (2b + 3c)2 = 0 ( a ≠ 0)

Therefore d 2 + (2b + 3c)2 = 0

Equations of Normal in Different forms a

(1) Point form : The equation of the normal to the parabola y 2 = 4ax at a point (x1, y1) is

y − y1 = − y1 (x − x1)
2a

P(x1,y1)

Tangent

Normal

Equation of normals of all other standard parabolas at (x1, y1)

Equation of parabolas Normal at (x1, y1)

y2 = −4ax y − y1 = y1 (x − x1)
2a

x 2 = 4ay y − y1 = − 2a (x − x1)
x1

x 2 = −4ay y − y1 = 2a (x − x1)
x1

(2) Parametric form: The equation of the normal to the parabola y 2 = 4ax at (at 2,2at) is y + tx = 2at + at 3

Equations of normal of all other standard parabola at 't'

Equations of parabolas Parametric co-ordinates Normals at 't'

y2 = −4ax (−at 2,2at) y − tx = 2at + at 3
x 2 = 4ay (2at,at 2) x + ty = 2at + at 3
x 2 = −4ay (2at,−at 2) x − ty = 2at + at 3

(3) Slope form: The equation of normal of slope m to the parabola y 2 = 4ax is y = mx − 2am − am3 at the
point (am2,−2am).

Equations of normal, point of contact, and condition of normality in terms of slope (m)

Equations of Point of contact in Equations of normal in Condition of normality
parabola terms of slope (m) terms of slope (m)

y2 = 4ax (am2,−2am) y = mx − 2am − am3 c = −2am − am3

y2 = −4ax (−am2,2am) y = mx + 2am + am3 c = 2am + am3
x 2 = 4ay
 − 2a , a  y = mx + 2a + a c = 2a + a
x 2 = −4ay  m m2  m2 m2

 2a ,− a  y = mx − 2a − a c = −2a − a
 m m2  m2 m2

Note :  The line lx + my + n = 0 is a normal to the parabola y 2 = 4ax if al(l 2 + 2m2 ) + m2n = 0

Point of intersection of normals at any two points on the Parabolaa

If R is the point of intersection then point of intersection of normals at any two points P(at12,2at1) and

Q(at 2 ,2at 2 ) on the parabola y2 = 4ax is R[2a + a(t12 + t 2 + t1t 2 ), − at1t 2 (t1 + t 2 )]
2 2

Y P(at12,2at1)
X′ A RX

Y′ Q(at 22 ,2at 2 )

Relation between 't1' and 't2' if Normal at 't1' meets the Parabola again at 't2' '

If the normal at the point P(at 2 ,2at 1 ) meets the parabola y2 = 4ax again at Q(at 2 ,2at 2 ) , then t2 = −t1 − 2
2 t1
1

Y P(at12,2at1)

X′ A X
Y′ Q(at 22 ,2at 2 )

Important Tips

• If the normals at points (at12,2at) and (at 2 ,2at 2 ) on the parabola y2 = 4ax meet on the parabola then t1t 2 =2
2

• If the normal at a point P(at 2 ,2at) to the parabola y 2 = 4ax subtends a right angle at the vertex of the parabola then t 2 = 2 .

• If the normal to a parabola y2 = 4ax , makes an angle φ with the axis, then it will cut the curve again at an angle tan −1  1 tan φ  .
 2 

• The normal chord to a parabola y 2 = 4ax at the point whose ordinate is equal to abscissa subtends a right angle at the focus.

• If the normal at two points P and Q of a parabola y 2 = 4ax intersect at a third point R on the curve. Then the product of the ordinate of P

and Q is 8a 2 .

Example: 26 If x + y = k is a normal to the parabola y 2 = 12x, then k is [IIT Screening 2000]
Solution: (b)
Example: 27 (a) 3 (b) 9 (c) –9 (d) –3
Solution: (b)
Example: 28 Any normal to the parabola y2 = 12x is y + tx = 6t + 3t 3 . It is identical with x + y = k if t = 1 = 6t + 3t 3
Solution: (b) 1 1 k

Example: 29 ∴ t = 1 and 1= 6 + 3 ⇒ k = 9
Solution: (a) k

The equation of normal at the point  a , a  to the parabola y2 = 4ax , is [Rajasthan PET 1984]
 4 
(d) 4 x − y + a = 0
(a) 4 x + 8y + 9a = 0 (b) 4 x + 8y − 9a = 0 (c) 4 x + y − a = 0

From y − y1 =− −y1 (x − x1)
2a

⇒ y − a = −a  x − a  ⇒ 2y + x = 2a + a = 9a ⇒ 2y + x− 9a =0 ⇒ 4x + 8y − 9a = 0
2a  4  4 4 4

The point on the parabola y 2 = 8x at which the normal is parallel to the line x − 2y + 5 = 0 is

(a) (−1 / 2, 2) (b) (1 / 2,−2) (c) (2, − 1 / 2) (d) (−2,1 / 2)

Let point be (h,k) . Normal is y−k = −k (x − h) or −kx − 4y + kh + 4k =0
4

Gradient = −K = 1 ⇒k = −2 . Substituting (h, k) and k = −2 in y2 = 8x , we get h= 1 . Hence point is  1 , − 2 
4 2 2  2 

Trick: Here only point  1 , − 2  will satisfy the parabola y2 = 8x .
 2 

The equations of the normal at the ends of the latus rectum of the parabola y 2 = 4ax are given by

(a) x 2 − y 2 − 6ax + 9a 2 = 0 (b) x 2 − y 2 − 6ax − 6ay + 9a 2 = 0

(c) x 2 − y 2 − 6ay + 9a 2 = 0 (d) None of these

The coordinates of the ends of the latus rectum of the parabola y 2 = 4ax are (a, 2a) and (a,−2a) respectively.

The equation of the normal at (a, 2a) to y2 = 4ax is y − 2a = −2a (x − a) using y − y1 = −y1 (x − x1 )
2a  2a 

Or x + y − 3a = 0 .....(i)

Similarly the equation of the normal at (a, –2a) is x − y − 3a = 0 .....(ii)

The combined equation of (i) and (ii) is x 2 − y 2 − 6ax + 9a 2 = 0 .

Example: 30 The locus of the point of intersection of two normals to the parabola x 2 = 8y, which are at right angles to each other, is

[Roorkee 1997]

(a) x 2 = 2(y − 6) (b) x 2 = 2(y + 6) (c) x 2 = −2(y − 6) (d) None of these

Solution: (a) Given parabola is x 2 = 8y .....(i)

Let (4t1 2t12 ) and Q(4 t 2 ,2t 2 ) be two points on the parabola (i)
2

Normal at P, Q are y − 2t12 1 (x − 4t1) ......(ii) and y − 2t 2 =− 1 (x − 4t2) ......(iii)
= − t1 2 t2

( )(ii)–(iii) gives2t2 − t12 = x 1 − 1  = x t1 − t2 , ∴ x = −2t1t 2 (t 2 + t1 ) .....(iv)
2 t2 t1 t1t2

From (ii), y = 2t12 − 1 (−2t1t 2 (t 2 + t1 ) − 4t1 ) = 2t12 + 2t2(t1 + t2) + 4 ⇒ y = 2t12 + 2t1t 2 + 2t 2 +4 .....(v)
t1 2

Since normals (ii) and (iii) are at right angles, ∴  − 1   − 1  = −1 ⇒ t1t 2 = −1
t1 t2

∴ From (iv), x = 2(t1 + t2) and from (v) y = 2t12 − 2 + 2t22 + 4 ⇒ y = 2[t12 + t 2 + 1] = 2[(t1 + t2)2 − 2t1t2 + 1]
2

⇒ y = 2[(t1 + t 2 )2 + 2 + 1] = 2[(t1 + t2)2 + 3 ⇒ y = 2 x2 + 3 = x2 +6⇒ x2 = 2(y − 6), which is the required locus.
 4  2

Co-normal Points
The points on the curve at which the normals pass through a common point are called co-normal points.

Q, R, S are co-normal points. The co- normal points are also called the YQ
feet of the normals.
P(x1,y1)
If the normal passes through point P(x1, y1) which is not on parabola, then X′ X
0
y1 = mx1 − 2am − am3 ⇒ am3 + (2a − x1 )m + y1 = 0 ......(i) R

Which gives three values of m. Let three values of m are m1,m2 and m3 , Y′ S

which are the slopes of the normals at Q, R and S respectively, then the

coordinates of Q, R and S are (am12 ,−2am1 ), (am 2 ,−2am 2 ) and (am32, − 2am3 ) respectively. These three points are
2

called the feet of the normals.

Now m1 + m2 + m3 = 0, m1m2 + m2m3 + m3m1 = (2a − x1 ) and m1 m 2 m 3 = − y1
a a

In general, three normals can be drawn from a point to a parabola.

(1) The algebraic sum of the slopes of three concurrent normals is zero.

(2) The sum of the ordinates of the co-normal points is zero.

(3) The centroid of the triangle formed by the co-normal points lies on the axis of the parabola.

(4) The centroid of a triangle formed by joining the foots of the normal of the parabola lies on its axis and is

given by  am12 + am 2 + am32 , 2am1 + 2am2 + 2am3  =  am12 + am22 + am32 ,0 
2 3 3

3

(5) If three normals drawn to any parabola y 2 = 4ax from a given point (h, k) be real, then h > 2a for a = 1 ,

normals drawn to the parabola y 2 = 4 x from any point (h, k) are real, if h > 2 .

(6) Out of these three at least one is real, as imaginary normals will always occur in pairs.

Circle through Co-normal points
Equation of the circle passing through the three (co-normal) points on the parabola y 2 = 4ax , normal at

which pass through a given point (α, β); is x2 + y2 − (2a + α)x − β y = 0
2

(1) The algebraic sum of the ordinates of the four points of intersection of a circle and a parabola is zero.

(2) The common chords of a circle and a parabola are in pairs, equally inclined to the axis of parabola.

(3) The circle through co-normal points passes through the vertex of the parabola.

(4) The centroid of four points; in which a circle intersects a parabola, lies on the axis of the parabola.

Example: 31 The normals at three points P, Q, R of the parabola y 2 = 4ax meet in (h, k), the centroid of triangle PQR lies on

Solution: (b) [MP PET 1999]
Example: 32
Solution: (c) (a) x = 0 (b) y = 0 (c) x = −a (d) y = a

Since the centroid of the triangle formed by the co-normal points lies on the axis of the parabola.

If two of the three feet of normals drawn from a point to the parabola y2 = 4 x be (1, 2) and (1, – 2)then the third foot is

(a) (2, 2 2) (b) (2, − 2 2) (c) (0, 0) (d) None of these

The sum of the ordinates of the foot = y1 + y2 + y3 = 0
∴ 2 + (−2) + y3 = 0 ⇒ y3 = 0

Equation of the Chord of contact of Tangents to a Parabolas

Let PQ and PR be tangents to the parabola y 2 = 4ax drawn from any external point P(x1, y1) then QR is

called the ‘Chord of contact’ of the parabola y2 = 4ax . Y

The chord of contact of tangents drawn from a point (x1, y1) to the X′ O Q Chord of
parabola y2 = 4ax is yy1 = 2a(x + x1) (x1,y1)P Y' contact X′

The equation is same as equation of the tangents at the point (x1, y1). R

Note :  The chord of contact joining the point of contact of two perpendicular tangents always passes

through focus. Q(at12,2at1)
 If tangents are drawn from the point (x1, y1) to the parabola

y 2 = 4ax, then the length of their chord of contact is P(x1,y1)

1 (y12 − 4ax1 )(y12 + 4a 2 )
|a|
R(at22,2at2)
 The area of the triangle formed by the tangents drawn from

(x1, y1) to y2 = 4ax and their chord of contact is (y12 − 4ax1 )3 / 2 .
2a

Equation of the Chord of the Parabola which is bisected at a given point s

The equation of the chord at the parabola y2 = 4ax bisected at the point (x1, y1) is given by T = S1, where

T = yy1 − 2a(x + x1) and S1 = y12 − 4ax1 . i.e., yy1 − 2a(x + x1) = y12 − 4ax1 Q(x2,y2)

P(x1,y1)
R(x3,y3)

Equation of the Chord joining any two points on the Parabola s

Let P(at 2 ,2at1 ), Q(at 2 ,2at 2 ) be any two points on the parabola y2 = 4ax . Then, the equation of the chord
1 2,

2at 2 − 2at1 2
( )joining − at12 + t2
these points is, y − 2at1 = at 2 (x − at12 ) or y − 2at1 = t1 x − at 2 or y(t1 + t 2 ) = 2x + 2at1t 2
2 1

(1) Condition for the chord joining points having parameters t1and t2 to be a focal chord: If the
(at12 ,2at1) and 2
chord joining points (at 2 ,2at 2 ) on the parabola passes through its focus, then (a,0) satisfies the

equation y(t1 + t2) = 2x + 2at1t2 ⇒ 0 = 2a + 2at1t 2 ⇒ t1t 2 = −1 or t2 = 1
− t1

(2) Length of the focal chord: The length of a focal chord having parameters t1 and t2 for its end points is a(t2 − t1)2 .

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

 a , − 2a  by virtue of relation t1t 2 = −1 .
t12 t1

 If one end of the focal chord of parabola is (at 2 ,2at) ,then other end will be  a , − 2at  and length
 t2 

a t 1  2
 t 
of chord = + .

 The length of the chord joining two point ' t1 ' and ' t2 ' on the parabola y 2 = 4ax is

a(t1 − t 2 ) (t1 + t 2 )2 + 4

 The length of intercept made by line y = mx + c between the parabola y 2 = 4ax is

4 a(1 + m2 )(a − mc) .
m2

Important Tips

• The focal chord of parabola y 2 = 4ax making an angle α with the x-axis is of length 4a cos ec 2α .

• The length of a focal chord of a parabola varies inversely as the square of its distance from the vertex.

• If l1 and l2 are the length of segments of a focal chord of a parabola, then its latus-rectum is 4l1l 2
l1 + l2

• The semi latus rectum of the parabola y2 = 4ax is the harmonic mean between the segments of any focal chord of the parabola.

Example: 33 If the points (au2,2au) and (av2,2av) are the extremities of a focal chord of the parabola y2 = 4ax, then [MP PET 1998,

93] (a) uv − 1 = 0 (b) uv + 1 = 0 (c) u + v = 0 (d) u − v = 0

Solution: (b) Equation of focal chord for the parabola y2 = 4ax passes through the point (au2,2au) and (av2,2av)

⇒ y − 2au = 2av − 2au (x − au2) ⇒ y − 2au = 2a(v − u) (x − au2) ⇒ y − 2au = 2 (x − au2)
av2 − au2 a(v − u)(v + u) v+u

It this is focal chord, so it would passes through focus (a, 0)

⇒ 0 − 2au = 2 (a − au2) ⇒ − uv − u2 = 1 − u2 , ∴ uv + 1 = 0
v+u

Trick : Given points (au2, 2au) and (av2, 2av) , then t1 = u and t2 = v , we know that t1t2 = −1 . Hence uv + 1 = 0 .

Example: 34 The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another
Solution: (c)
parabola with the directrix [IIT Screening 2002]
Example: 35
Solution: (c) (a) x = −a (b) x = − a (c) x = 0 (d) x = a
Example: 36 2 2
Solution: (a)
Let M(α, β ) be the mid point of PS.

α = at 2 + a ,β = 2at + 0 ⇒ 2α = at 2 + a, at = β Y (atS2(,2a,a0t))P
2 2
M(α,β)
β2
∴ 2α = a. a2 + a or 2aα = β2 + a2 X′ O X

∴ The locus is y2 = 4a (x − a ) = 4b(x − b), b = a
2 2   Y′
2 

∴ Directrix is (x − b) + b = 0 or x = 0 .

The length of chord of contact of the tangents drawn from the point (2, 5) to the parabola y2 = 8x, is [MNR 1976]

(a) 1 41 (b) 41 (c) 3 41 (d) 2 41
2 2

Equation of chord of contact of tangents drawn from a point (x1, y1) to parabola y2 = 4ax is yy1 = 2a(x + x1) . So that
5y = 2 × 2(x + 2) ⇒ 5y = 4 x + 8 .

Point of intersection of chord of contact with parabola y2 = 8x are  1 , 2 ,(8, 8), So the length of chord is 3 41 .
 2  2

If b, k are the intercept of a focal chord of the parabola y2 = 4ax , then K is equal to [Rajasthan PET 1999]

(a) ab (b) b (c) a (d) ab
b−a b−a b−a a−b

Let ' t1' ' t2' be the ends of focal chords
∴ . t1t2 = −1 . If S is the focus and P, Q are the ends of the focal chord, then

( )SP = (at12 − a)2 + (2at1 − 0)2 = a t12 + 1 = b (Given).... (i) ( at12,2at1 ) P

( )∴ SQ = a t22 + 1 = a 1 + 1 (Given)  t = − 1 ⇒ t 2 = 1 b
t12  t1 2  k S(a,0)
 2 t12  Q

= a(t12 + 1) = k ....(ii), ∴ b = t12 [Divide (i) by (ii)] (at22, 2at2)
t12 k

Putting in (1), we get a b + 1 =b⇒ ab +a = b ⇒ k = ab
 k  k b−a

Diameter of a Parabola .

The locus of the middle points of a system of parallel chords is called a diameter and in case of a parabola this

diameter is shown to be a straight line which is parallel to the axis of the Y P(x1,y1)
parabola. A
Y′ y=mx+c
The equation of the diameter bisecting chords of the parabola y 2 = 4ax
X′ P(h,k) X

of slope m is y= 2a Diameter
m
Q(x2,y2)
Note :  Every diameter of a parabola is parallel to its axis.

 The tangent at the end point of a diameter is parallel to corresponding system of parallel chords.

 The tangents at the ends of any chord meet on the diameter which bisects the chord.

Example: 37 Equation of diameter of parabola y2 = x corresponding to the chord x − y + 1 = 0 is [Rajasthan PET 2003]
Solution: (b)
(a) 2y = 3 (b) 2y = 1 (c) 2y = 5 (d) y =1

Equation of diameter of parabola is y= 2a , Here a= 1 , m 1 ⇒y= 2. 1 ⇒ 2y = 1
m 4 4
= 1

Length of Tangent, Subtangent ,Normal and Subnormal .

Let the parabola y2 = 4ax . Let the tangent and normal at P(x1, y1)meet the axis of parabola at T and G
respectively, and tangent at P(x1, y1)makes angle ψ with the positive direction of x-axis.

A(0, 0) is the vertex of the parabola and PN = y . Then,

(1) Length of tangent = PT = PN cosec ψ = y1 cosecψ Y (x1,y1)
(2) Length of normal = PG = PNcosec(90o −ψ ) = y1 secψ yy1=2a(x+x1) P
(3) Length of subtangent = TN = PN cotψ = y1 cotψ
TX(′–x1,0) ψ S(a,0) X
A (xN1,0)G(x1,2a,0)

(4) Length of subnormal = NG = PN cot(90o −ψ ) = y1 tanψ Y′ y2=4ax

where , tanψ = 2a = m, [slope of tangent at P(x, y)]
y1

Length of tangent, subtangent, normal and subnormal to y2 = 4ax at (at2, 2at)

(1) Length of tangent at (at 2,2at) = 2at cosecψ = 2at (1 + cot 2 ψ ) = 2at 1 + t 2

(2) Length of normal at (at 2 , 2at) = 2at secψ = 2at (1 + tan 2 ψ ) = 2a t 2 + t 2 tan 2 ψ = 2a (t 2 + 1)
(3) Length of subtangent at (at 2,2at) = 2at cotψ = 2at 2
(4) Length of subnormal at (at 2,2at) = 2at tanψ = 2a

Example: 38 The length of the subtangent to the parabola y2 = 16x at the point whose abscissa is 4, is
Solution: (c)
Example: 39 (a) 2 (b) 4 (c) 8 (d) None of these

Solution: (a) Since the length of the subtangent at a point to the parabola is twice the abscissa of the point. Therefore, the required

length is 8.

If P is a point on the parabola y 2 = 4ax such that the subtangent and subnormal at P are equal, then the coordinates of P
are

(a) (a, 2a) or (a,−2a) (b) (2a, 2 2a) or (2a,−2 2a)

(c) (4a, − 4a) or (4a, 4a) (d) None of these

Since the length of the subtangent at a point on the parabola is twice the abscissa of the point and the length of the
subnormal is equal to semi-latus-rectum. Therefore if P(x, y) is the required point, then 2x = 2a ⇒ x = a

Now (x, y) lies on the parabola y 2 = 4ax ⇒ 4a 2 = y 2 ⇒ y = ±2a
Thus the required points are (a, 2a) and (a,− 2a) .

Pole and Polar.
The locus of the point of intersection of the tangents to the parabola at the ends of a chord drawn from a fixed

point P is called the polar of point P and the point P is called the pole of the polar.

Equation of polar: Equation of polar of the point (x1, y1) with respect to parabola y 2 = 4ax is same as
chord of contact and is given by yy1 = 2a(x + x1)

(h,k) Q (h,k) T Q
T
Q′ Pole R
PoTla′ r P(x1,y1)
Pole Polar
R P
(x1,y1) R′ Q′
R′ T′

(1) Polar of the focus is directrix: Since the focus is (a, 0)

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

y 2 = 4ax .

(2) Any tangent is the polar of its point of contact: If the point P(x1y1) be on the parabola. Its polar and

tangent at P are identical. Hence the tangent is the polar of its own point of contact. (xP1,y1)
Coordinates of pole: The pole of the line lx + my + n = 0 with respect to the

parabola y2 = 4ax is  n , − 2am  . Q Q′
 l l 
R
R′

(i) Pole of the chord joining (x1, y1)and (x2, y2) is  y1 y 2 , y1 + y2  which is
 4a 2 

the same as the point of intersection of tangents at (x1, y1 ) and (x 2, y2 ) .

(ii) The point of intersection of the polar of two points Q and R is the pole of QR.

Characterstics of Pole and Polar.
(1) Conjugate points: If the polar of P(x1, y1)passes through Q(x 2 , y2 ) , then the polar of Q(x2, y2)goes

through P(x1, y1) and such points are said to be conjugate points.

Two points P(x1,y1) and Q(x 2 , y2 ) are conjugate points with respect to the parabola y2 = 4ax , if
y1y2 = 2a(x1 + x 2 ).

(2) Conjugate lines: If the pole of a line ax + by + c = 0 lies on the another line a1 x + b1y + c1 = 0 , then the
pole of the second line will lie on the first and such lines are said to be conjugate lines.

Two lines l1 x + m1y + n1 = 0 and l2 x + m2y + n2 = 0 are conjugate lines with respect to parabola y 2 = 4ax , if
(l1n2 + l 2n1 ) = 2am1m2

Note :  The chord of contact and polar of any point on the directrix always passes through focus.

 The pole of a focal chord lies on directrix and locus of poles of focal chord is the directrix.
 The polars of all points on directrix always pass through a fixed point and this fixed point is focus.

Example: 40 The pole of the line 2x = y with respect to the parabola y2 = 2x is
Solution: (a)
(a)  0, 1  (b)  1 , 0  (c)  0, − 1  (d) None of these
 2   2   2 

Let (x1, y1) be the pole of line 2x = y w.r.t. parabola y2 = 2x its polar is yy1 = x + x1

Also polar is y = 2x , ∴ y1 = 1 = x1 ,∴ x1 = 0, y1 = 1 . So Pole is  0, 1 
1 2 0 2  2 

Example: 41 If the polar of a point with respect to the circle x2 + y2 = r2 touches the parabola y2 = 4ax, the locus of the pole is
Solution: (a)
[EAMCET 1995]

(a) y2 = − r2 x (b) x2 = − r2 y (c) y2 = r2 x (d) x2 = r2 y
a a a a

Polar of a point (x1, y1) w.r.t. x 2 + y 2 = r 2 is xx1 + yy1 = r 2 i.e. yy1 = −xx1 + r 2

⇒y = − x1 x + r2 ⇒ y = mx + c, where m = − x1 ;c = r2
y1 y1 y1 y1

This touches the parabola y2 = 4ax , If c= a ⇒ r2 = a = − ay1
m y1 − x1 / y1 x1

∴Required locus of pole (x1, y1) is r2 = − ay i.e., y2 = − r2 x
y x a

Reflection property of a Parabola .
The tangent (PT) and normal (PN) of the parabola y2 = 4ax at. P are the internal and external bisectors of

∠SPM and BP is parallel to the axis of the parabola and ∠BPN = ∠SPN M Y P Light ray
Tangent α B
Note :  When the incident ray is parallel to the axis of the parabola, the β β α
X′ T Reflected ray
reflected ray will always pass through the focus. S
A N Light ray X

R Light ray
Y′ Light ray

Example: 42 A ray of light moving parallel to the x-axis gets reflected from a parabolic mirror
Solution: (a)
whose equation is (y − 2)2 = 4(x + 1). After reflection, the ray must pass through the point

(a) (0, 2) (b) (2, 0) (c) (0, –2) (d) (–1, 2)

The equation of the axis of the parabola is y − 2 = 0 , which is parallel to the x-axis. So, a ray parallel to x-axis is parallel

to the axis of the parabola. We know that any ray parallel to the axis of a parabola passes through the focus after
reflection. Here (0, 2) is the focus.

Conic sections

Definition.
An ellipse is the locus of a point which moves in such a way that its distance from a fixed point is in constant

ratio (<1) to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix
and the constant ratio is called the eccentricity of the ellipse, denoted by (e).

In other words, we can say an ellipse is the locus of a point which moves in a plane so that the sum of its
distances from two fixed points is constant and is more than the distance between the two fixed points.

Let S(α, β ) is the focus, ZZ ′ is the directrix and P is any point on the ellipse. Then by definition,

SP = e ⇒ SP = e.PM Z
PM
Ax+By+C=0
(x − α)2 + (y − β )2 = e Ax + By + C Directrix P(x,y)
A2 + B2 Z′ FocusS(α,β)

Squaring both sides, (A2 + B2 )[(x − α)2 + (y − β )2 ] = e 2 (Ax + By + C)2

Note :  The condition for second degree equation in x and y to represent an ellipse is that h2 − ab < 0 and

∆ = abc + 2 fgh − af 2 − bg 2 − ch2 ≠ 0

Example: 1 The equation of an ellipse whose focus is (–1, 1), whose directrix is x − y + 3 = 0 and whose eccentricity is 1 , is given by
Solution: (a) 2

[MP PET 1993]

(a) 7x2 + 2xy + 7y2 + 10x − 10y + 7 = 0 (b) 7x2 − 2xy + 7y2 − 10x + 10y + 7 = 0

(c) 7x2 − 2xy + 7y2 − 10x − 10y − 7 = 0 (d) 7x2 − 2xy + 7y2 + 10x + 10y − 7 = 0

Let any point on it be (x, y) then by definition,

(x + 1)2 + (y − 1)2 1 x−y+3
=2 12 + 12

Squaring and simplifying, we get

7x2 + 2xy + 7y2 + 10x − 10y + 7 = 0 , which is the required ellipse.

Standard equation of the Ellipse .
Let S be the focus, ZM be the directrix of the ellipse and P(x, y)is any point on the ellipse, then by definition

SP Y
PM M′ (0,b) p(x,y)
=e ⇒ (SP)2 = e 2 (PM)2 Directrix
Directrix
− x  2 ⇒ X′ Z′ C M Z X
(x − ae)2 + (y − 0)2 e 2  a  x2 y2 1 A′ S′(–ae,0) S(ae,0) A Axis
=  e a2 + a 2 (1 − = (–a,0) (0,–b) B′ (a,0)
e2)

x2 y2 x=–a/e Y′ x=a/e
a2 + b2
= 1 , where b2 = a 2 (1 − e 2 )

Since e < 1, therefore a2(1 − e2) < a2 ⇒ b2 < a 2 . Some terms related to the ellipse x2 y2 = 1, a > b :
a2 + b2

(1) Centre: The point which bisects each chord of the ellipse passing through it, is called centre (0, 0)

denoted by C. Y

M′ L1 B P L M

X' Z′ A′ S′ C A ZX
NS

L′1 B′ P′L′

Y′

(2) Major and minor axes: The diameter through the foci, is called the major axis and the diameter bisecting
it at right angles is called the minor axis. The major and minor axes are together called principal axes.

Length of the major axis AA′ = 2a , Length of the minor axis BB' = 2b

The ellipse x2 y2 = 1, is symmetrical about both the axes.
a2 + b2

(3) Vertices: The extremities of the major axis of an ellipse are called vertices.
The coordinates of vertices A and A′ are (a, 0) and (–a, 0) respectively.

(4) Foci: S and S′ are two foci of the ellipse and their coordinates are (ae, 0) and (–ae, 0) respectively.
Distance between foci SS′ = 2ae .

(5) Directrices: ZM and Z ′M ′ are two directrices of the ellipse and their equations are x = a and x = − a
e e
2a
respectively. Distance between directrices ZZ′ = e .

(6) Eccentricity of the ellipse: For the ellipse x2 y2 =1,
a2 + b2

b2 4b2  2b  2  Minor axis  2
a2 4a2  2a  Major axis
we have b2 = a 2 (1 − e)2 ⇒ e2 = 1 − = 1 − = 1 − ; e= 1 −

This formula gives the eccentricity of the ellipse.

(7) Ordinate and double ordinate: Let P be a point on the ellipse and let PN be perpendicular to the major
axis AA’ such that PN produced meets the ellipse at P ′ . Then PN is called the ordinate of P and PNP ′ the double

ordinate of P.

If abscissa of P is h, then ordinate of P, y2 = 1 − h2 ⇒y = b (a 2 − h2 ) (For first quadrant)
b2 a2 a (For fourth quadrant)

And ordinate of P′ is y = −b (a 2 − h2 )
a

Hence coordinates of P and P ′ are  h, b (a 2 − h2 )  and  h, −b (a 2 − h2 )  respectively.
 a  a 

(8) Latus-rectum: Chord through the focus and perpendicular to the major axis is called its latus rectum.
The double ordinates LL′ and L1 L1′ are latus rectum of the ellipse.

Length of latus rectum LL' = L1 L1′ = 2b 2 and end points of latus-rectum are L =  ae, b2  , L' =  ae, − b2 
a a a

and L1  ae, b2  ; L1 '  ae, − b 2 
a a
= − = −

(9) Focal chord: A chord of the ellipse passing through its focus is called a focal chord.

(10) Focal distances of a point: The distance of a point from the focus is its focal distance. The sum of the
focal distances of any point on an ellipse is constant and equal to the length of
the major axis of the ellipse. Y

M′ BP M
X′
Let P(x1, y1)be any point on the ellipse x2 y2 =1 CZ
a2 + b2 Z′ A′ S′ S A X

SP = ePM = e a − x1  = a − ex1 and S' P = ePM' = e a + x  = a+ ex1 B′
 e   e  Y′

1

∴ SP + S' P = (a − ex1) + (a + ex1) = 2a = AA' = major axis.

Example: 2 The length of the latus-rectum of the ellipse 5x2 + 9y2 = 45 is [MNR 1978, 80, 81; Kurukshetra CEE 1999]
Solution: (d)
Example: 3 (a) 5 / 4 (b) 5 / 2 (c) 5 / 3 (d) 10 / 3
Solution: (c)
Example: 4 Here the ellipse is x2 y2 =1
Solution: (a) 9+5

Example: 5 Here a2 = 9 and b2 = 5 . So, latus-rectum = 2b2 = 2(5) = 10 .
Solution: (c) a 3 3

Example: 6 In an ellipse the distance between its foci is 6 and its minor axis is 8. Then its eccentricity is [EAMCET 1994]

(a) 4 (b) 1 (c) 3 (d) 1
5 52 5 2

Distance between foci = 6 ⇒ 2ae = 6 ⇒ ae = 3 , Minor axis = 8 ⇒ 2b = 8 ⇒ b = 4 ⇒ b2 = 16

From b2 = a2(1 − e2), ⇒ 16 = a2 − a2e2 ⇒ a2 − 9 = 16 ⇒ a = 5

Hence, ae = 3⇒ e = 3
5

What is the equation of the ellipse with foci (±2,0) and eccentricity 1 [DCE 1999]
2

(a) 3x2 + 4y2 = 48 (b) 4 x2 + 3y2 = 48 (c) 3x2 + 4y2 = 0 (d) 4 x2 + 3y2 = 0

Here ae = ±2, e = 1 , ∴ a = ±4
2

Form b2 = a2(1 − e2) ⇒ b2 = 161 − 1  ⇒ b2 = 12
 4 

Hence, the equation of ellipse is x2 y2 =1 or 3x2 + 4y2 = 48
16 + 12

If P(x, y), F1 = (3,0) , F2 = (−3,0) and 16x2 + 25y2 = 400 , then PF1 + PF2 equals [IIT 1998]

(a) 8 (b) 6 (c) 10 (d) 12

We have 16 x 2 + 25y2 = 400 ⇒ x2 + y2 = 1 or x2 + y2 =1, where a2 = 25 and b2 = 16
25 16 a2 b2

This equation represents an ellipse with eccentricity given by e2 = 1 − b2 = 1 − 16 = 9 ⇒ e = 3/5
a2 25 25

So, the coordinates of the foci are (±ae,0) i.e. (3, 0) and (−3, 0) , Thus, F1 and F2 are the foci of the ellipse.

Since, the sum of the focal distance of a point on an ellipse is equal to its major axis, ∴ PF1 + PF2 = 2a = 10

An ellipse has OB as semi minor axis. F and F′ are its foci and the angle FBF′ is a right angle. Then the eccentricity of

the ellipse is [Pb. CET 2001, IIT 1997, DCE 2002]

(a) 1 (b) 1 (c) 2 (d) 1
2 2 3 3

Solution: (b) Since ∠FBF' = π Y
2 B
X’ F′ C F
∴ ∠FBC = ∠F' BC = π
4

∴ CB = CF ⇒ b = ae X

⇒ b2 = a2e2 ⇒ a2(1 − e2) = a2e2 Y′

⇒ 1 − e2 = e2 ⇒ 2e2 = 1 ⇒ e = 1 .
2

Example: 7 Let P be a variable point on the ellipse x2 + y2 = 1 with foci F1 and F2 . If A is the area of the triangle PF1F2, then the
Solution: (b) a2 b2

maximum value of A is [IIT 1994]

(a) 2abe (b) abe (c) 1 abe (d) None of these
2

Let P(a cos θ , b sin θ ) and F1(−ae,0), F2(ae,0)

A = Area of ∆PF1F2 = 1 a cos θ b sinθ 1 = 1 | 2aeb sin θ | = aeb | sinθ |
2 ae 0 1 2
− ae 0 1

∴ A is maximum, when | sinθ |= 1 .

Hence, maximum value of A = abe

Example: 8 The eccentricity of an ellipse, with its centre at the origin is 1 . If one of the directrices is x = 4 , then the equation of the
Solution: (b) 2

ellipse is [AIEEE 2004]

(a) 4 x2 + 3y2 = 1 (b) 3x2 + 4y2 = 12 (c) 4 x2 + 3y2 = 12 (d) 3x2 + 4y2 = 1

Given e = 1 , a = 4 . So, a = 2 ⇒ a2 = 4
2 e

From b2 = a2(1 − e2) ⇒ b2 = 41 − 1  = 4× 3 = 3
 4  4

Hence the equation of ellipse is x2 + y2 =1, i.e. 3x2 + 4y2 = 12
4 3

Equation of Ellipse in other form .

In the equation of the ellipse x2 y2 = 1, if a > b or a2 > b2 (denominator of Y
a2 + b2
y=b/e Z K
A(0,b)
x2 is greater than that of y 2 ), then the major and minor axis lie along x-axis and

y-axis respectively. But if a < b or a2 < b2 (denominator of x 2 is less than that of y 2 ), (0,be)

then the major axis of the ellipse lies along the y-axis and is of length 2b and the minor X′ B′ C (0,0) B X
axis along the x-axis and is of length 2a. (–a,0) (a,0)
y=–b/e (0,–be)
The coordinates of foci S and S’ are (0, be) and (0, – be) respectively. S′

The equation of the directrices ZK and Z' K' are y = ±b / e and eccentricity e is A′(0,– b)
Z′ K′
given by the formula a 2 = b 2 (1 − e 2 ) or e = 1 − a2
b2 Y′

Difference between both ellipse will be clear from the following table.

Ellipse x2 + y2 = 1
Basic fundamentals  b2 
 a 2
Centre
Vertices For a > b For b > a
Length of major axis
Length of minor axis (0, 0) (0, 0)
Foci (±a, 0) (0, ± b)
Equation of directrices
Relation in a, b and e 2a 2b
Length of latus rectum 2b 2a
(±ae, 0) (0,± be)
Ends of latus-rectum
x = ±a / e y = ±b / e

b2 = a2(1 − e2) a2 = b2(1 − e2)

2b 2 2a 2
a b

 ± ae,± b2   ± a2 ,±be 
a b

Parametric equations (a cosφ,b sinφ) (a cosφ,b sinφ) (0 ≤ φ < 2π )

Focal radii SP = a − ex1 and S' P = a + ex1 SP = b − ey1 and S' P = b + ey1
2a 2b
Sum of focal radii SP + S' P = 2ae 2be
Distance between foci 2a/e 2b/e
Distance between directrices
Tangents at the vertices x = –a, x = a y = b, y = –b

Example: 9 The equation of a directrix of the ellipse x2 y2 = 1 is
Solution: (a) 16 + 25

Example: 10 (a) y = 25 (b) x = 3 (c) x = −3 (d) x = 3
Solution: (c) 3 25

From the given equation of ellipse a2 = 16, b2 = 25 (since b > a )

So, a2 = b2(1 − e2) , ∴ 16 = 25(1 − e2) ⇒ 1− e2 = 16 ⇒ e2 = 9 ⇒ e = 3
25 25 5

∴ One directrix is y = b = 5 = 25
e 3/5 3

The distances from the foci of P(x1, y1) on the ellipse x2 + y2 = 1 are
9 25

(a) 4 ± 5 y1 (b) 5 ± 4 x1 (c) 5 ± 4 y1 (d) None of these
4 5 5

For the given ellipse b > a, so the two foci lie on y-axis and their coordinates are (0, ± be) ,

Where b = 5,a = 3 . So e = 1 − a2 = 1 − 9 = 4
b2 25 5

The focal distances of a point (x1, y1) on the ellipse x2 y2 = 1 , Where b2 > a2 are given by b ± ey1 . So, Required
a2 + b2

distances are b ± ey1 = 5± 4 y1 .
5

Parametric form of the Ellipse.

Let the equation of ellipse in standard form will be given by x2 y2 =1
a2 + b2

Then the equation of ellipse in the parametric form will be given by x = a cosφ, y = b sinφ , where φ is the

eccentric angle whose value vary from 0 ≤ φ < 2π . Therefore coordinate of any point P on the ellipse will be given

by (a cos φ , b sin φ)

Example: 11 The curve represented by x = 3(cos t + sin t), y = 4(cos t − sin t) is [EAMCET 1988; DCE 2000]
Solution: (a)
(a) Ellipse (b) Parabola (c) Hyperbola (d) Circle
Example: 12
Solution: (c) Given, x = 3(cos t + sin t), y = 4(cos t − sin t) ⇒ x = (cos t + sin t), y = (cos t − sin t)
3 4

Squaring and adding, we get x2 + y2 = (1 + sin 2t) + (1 − sin 2t) ⇒ x2 + y2 = 2, which represents ellipse.
9 16 9 16

The distance of the point 'θ ' on the ellipse x2 y2 = 1 from a focus is
a2 + b2

(a) a(e + cos θ ) (b) a(e − cos θ ) (c) a(1 + e cos θ ) (d) a(1 + 2e cos θ )

Focal distance of any point P(x, y) on the ellipse is equal to SP = a + ex . Here x = cos θ .

Hence, SP = a + ae cos θ = a(1 + e cos θ )

Special forms of an Ellipse.
(1) If the centre of the ellipse is at point (h,k) and the directions of the axes are parallel to the coordinate

axes, then its equation is (x − h)2 + (y − k)2 =1
a2 b
2

If we shift the origin at (h, k) without rotating the coordinate axes, then x = X + h and y = Y + k

(2) If the equation of the curve is (lx + my + n)2 + (mx − ly + p)2 =1 where lx + my + n = 0 and
a2 b2

mx − ly + P = 0 are perpendicular lines, then we substitute lx + my + n = X, mx − ly + p = Y , to put the equation
l2 + m2 l2 + m2

in the standard form.

Example: 13 The foci of the ellipse 25(x + 1)2 + 9(y + 2)2 = 225 are [MP PET 1998, UPSEAT 1991, 2000]
Solution: (d)
(a) (−1, 2),(6,1) (b) (−1, − 2),(1, 6) (c) (1,−2),(1,−6) (d) (−1, 2),(−1, − 6)

Given ellipse is (x + 1)2 + (y + 2)2 =1 i.e. X2 + Y2 =1, where X = x + 1 and Y = y + 2
9 25 9 25

Here a2 = 25, b2 = 9 [Type : X2 Y2 = 1]
b2 + a2

Eccentricity is given by e2 = a2 − b2 = 25 − 9 = 16 , ∴ e= 4
a2 25 25 5

Foci are given by Y = ±ae = ±5 4  = ±4
 5 

X = 0 ⇒ y + 2 = ±4 ⇒ y = −2 ± 4 = −6 or 2

x + 1 = 0 ⇒ x = −1 . Hence foci are (–1, –6) or (–1, 2).

Position of a point with respect to an Ellipse.

Let P(x1, y1)be any point and let x2 y2 = 1 is the equation of an ellipse. The point lies outside, on or inside
a2 + b2

the ellipse as if S1 = x 2 + y12 − 1 > 0, = 0, < 0
1 b2

a2 Y

P(outside)
P(on)

X′ P(inside) X
C

Y′

Example: 14 Let E be the ellipse x2 + y2 =1 and C be the circle x2 + y2 = 9 . Let P and Q be the points (1, 2) and (2, 1) respectively.
Solution: (d) 9 4

Then [IIT 1994]

(a) Q lies inside C but outside E (b) Q lies outside both C and E

(c) P lies inside both C and E (d) P lies inside C but outside E

The given ellipse is x2 + y2 =1. The value of the expression x2 + y2 −1 is positive for x = 1, y = 2 and negative for
9 4 9 4

x = 2, y = 1 . Therefore P lies outside E and Q lies inside E. The value of the expression x2 + y2 − 9 is negative for both

the points P and Q. Therefore P and Q both lie inside C. Hence P lies inside C but outside E.

Intersection of a Line and an Ellipse.

Let the ellipse be x2 y2 =1 ......(i) and the given line be y = mx + c ......(ii)
a2 + b2

Eliminating y from equation (i) and (ii), then x2 + (mx + c)2 =1
a2 b2

i.e., (a 2m2 + b 2 )x 2 + 2mca 2 x + a 2(c 2 − b 2 ) = 0

The above equation being a quadratic in x, its discriminant = 4m2c 2a 4 − 4a 2(a 2m2 + b 2 )(c 2 − b 2 )

{ }= b 2 (a 2m2 + b 2 ) − c 2

Hence the line intersects the ellipse in two distinct points if a 2m2 + b 2 > c 2 in one point if c 2 = a 2m2 + b 2 and
does not intersect if a 2m2 + b 2 < c 2 .

Equations of Tangent in Different formss

(1) Point form: The equation of the tangent to the ellipse x2 y2 = 1 at the point (x1, y1)is xx1 + yy1 =1
a2 + b2 a2 b2

(2) Slope form: If the line y = mx + c touches the ellipse x2 y2 = 1, then c 2 = a 2m2 + b 2 . Hence, the
a2 + b2

straight line y = mx ± a 2m2 + b 2 always represents the tangents to the ellipse.

Points of contact: Line y = mx ± a2m2 + b2 touches the ellipse x2 y2 = 1 at
a2 + b2

 ± a2m ,  b 2 
 a2m2 + b2 a 2m2 + b 2 

(3) Parametric form: The equation of tangent at any point φ(a cosφ,b sinφ) is x cos φ + y sin φ = 1
a b

Note :  The straight line lx + my + n = 0 touches the ellipse x2 y2 = 1 , if a2l 2 + b2m2 = n2.
a2 + b2

 The line x cosα + y sinα = p touches the ellipse x2 + y2 = 1 , if a 2 cos 2 α + b 2 sin 2 α = p 2 and that
a2 b2

point of contact is  a 2 cos α , b2 sinα  .
p p

 Two tangents can be drawn from a point to an ellipse. The two tangents are real and distinct or
coincident or imaginary according as the given point lies outside, on or inside the ellipse.

 The tangents at the extremities of latus-rectum of an ellipse intersect on the corresponding directrix.

Important Tips

• A circle of radius r is concentric with the ellipse x2 y2 =1, then the common tangent is inclined to the major axis at an angle
a2 + b2

tan−1  r2 − b2  .
a2 − r2

• The locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse x2 y2 = 1 is (x 2 + y2)2 = a2x 2 + b2y2 or
a2 + b2

r 2 = a2 cos2 θ + b2 sin2 θ (in polar coordinates)

• The locus of the mid points of the portion of the tangents to the ellipse x2 y2 = 1 intercepted between the axes is a2y2 + b2x2 = 4x2y2 .
a2 + b2

• The product of the perpendiculars from the foci to any tangent of an ellipse is equal to the square of the semi minor axis, and the feet of
these perpendiculars lie on the auxiliary circle.

Example: 15 The number of values of ‘c’ such that the straight line y = 4 x + c touches the curve x2 + y2 = 1 is [IIT 1998]
Solution: (c) 4 [IIT 1999]
Example: 16
(a) 0 (b) 1 (c) 2 (d) Infinite
Solution: (b,d)
We know that the line y = mx + c touches the curve x2 y2 = 1 iff c 2 = a2m2 + b2
a2 + b2

Here, a2 = 4, b2 = 1, m = 4 ∴ c2 = 64 + 1 ⇒ c = ± 65

On the ellipse 4 x2 + 9y2 = 1, the points at which the tangents are parallel to the line 8x = 9y are

(a)  2 , 1  (b)  −2 , 1  (c)  −2 , −1  (d)  2 , −1 
 5 5   5 5   5 5   5 5 

Ellipse is x2 y2 =1⇒ a2 = 1 , b2 = 1 . The equation of its tangent is 4 xx′ + 9yy′ = 1
1/4 +1/9 4 9

∴ m = − 4 x′ = 8 ⇒ x′ = −2y′ and 4 x′2 + 9y′2 = 1 ⇒ 4 x′2 + 9 x′2 =1⇒ x′ = ± 2
9y′ 9 4 5

When x′ = 2 , then y′ = −1 and when x′ = −2 , then y′ = 1 .
5 5 5 5

Hence points are  2 , −1  ,  −2 , 1 
 5 5   5 5 

Example: 17 If any tangent to the ellipse x2 + y2 = 1 intercepts equal lengths l on the axes, then l=
Solution: (b) a2 b2

Example: 18 (a) a2 + b2 (b) a2 + b2 (c) (a2 + b2)2 (d) None of these
Solution: (d)
The equation of any tangent to the given ellipse is x cos θ + y sin θ =1
a b

This line meets the coordinate axes at P a θ ,0  and Q 0, b 
 cos   sin θ 

∴ a = l = b ⇒ cos θ = a and sin θ b ⇒ cos2 θ + sin2 θ = a2 + b2 ⇒ l2 = a2 + b2 ⇒ l = a2 + b2 .
cos θ sin θ l =l l2 l2

The area of the quadrilateral formed by the tangents at the end points of latus- rectum to the ellipse x2 + y2 = 1 , is
9 5

IIT Screening 2003]

(a) 27/4 sq. units (b) 9 sq. units (c) 27/2 sq. units (d) 27sq. units

By symmetry the quadrilateral is a rhombus. So area is four times the area of the right angled triangle formed by the
tangents and axes in the 1st quadrant.

Now ae = a2 − b2 ⇒ ae = 2 ⇒ Tangent (in the first quadrant) at one end of latus rectum  2, 5  is 2 x + 5 . y =1
 3  9 3 5

i.e. x y = 1 . Therefore area = 4. 1 . 9 .3 = 27sq. units.
9/2+ 3 2 2

Equation of Pair of Tangents SS1 = T2 .

Pair of tangents: Let P(x1, y1)be any point lying outside the ellipse x2 y2 = 1 and let a pair of tangents
a2 + b2

PA, PB can be drawn to it from P.

Then the equation of pair of tangents PA and PB is SS1 = T 2 A
B
where S ≡ x2 + y2 −1 = 0 P(x1,y1)
a2 b2

S1 ≡ x 2 + y12 −1=0
1 b2

a2

T ≡ xx1 + yy1 −1 = 0
a2 b2

Director circle: The director circle is the locus of points from which perpendicular tangents are drawn to the ellipse.

Let P(x1, y1) be any point on the locus. Equation of tangents through P(x1, y1)is given by SS1 = T 2

i.e.,  x 2 + y2 − 1  x 2 + y12 − 1 =  xx1 + yy1 − 1 2 P(x1,y1)
a 2 b2 a 1 b2  a 2 b2
2 B 90°

They are perpendicular, So coeff. of x 2 + coeff. of y 2 = 0 A′ C A
B′
∴  1 1   x 2 y12 1  x 2 y12  = 0 or x12 + y12 = a2 + b2
 a2 + b2  1 + b2 − − 1 + b4

a2 a4

Hence locus of P(x1, y1)i.e., equation of director circle is x 2 + y 2 = a 2 + b 2

Example: 19 The angle between the pair of tangents drawn from the point (1, 2) to the ellipse 3x2 + 2y2 = 5 is [UPSEAT 2001]

(a) tan−1(12 / 5) (b) tan−1(6 / 5) (c) tan−1(12 / 5) (d) tan−1(6 / 5)

Solution: (c) The combined equation of the pair of tangents drawn from (1,2) to the ellipse 3x2 + 2y2 = 5 is

(3x2 + 2y2 − 5)(3 + 8 − 5) = (3x + 4y − 5)2 [using SS1 = T 2 ]

⇒ 9x2 − 24 xy − 4y2 + ....... = 0

The angle between the lines given by this equation is tan θ = 2 h2 − ab
a+b

Where a = 9 , h = −12, b = −4 ⇒ tanθ = 12 / 5 ⇒θ = tan−1(12 / 5)

Example: 20 The locus of the point of intersection of the perpendicular tangents to the ellipse x2 y2 = 1 is [Karnataka CET 2003]
Solution: (c) 9+4

Example: 21 (a) x2 + y2 = 9 (b) x2 + y2 = 4 (c) x2 + y2 = 13 (d) x2 + y2 = 5

The locus of point of intersection of two perpendicular tangents drawn on the ellipse is x2 + y2 = a2 + b2, which is called
“director circle”.

Given ellipse is x2 y2 = 1 . ∴Locus is x2 + y2 = 9 + 4 , i.e. x2 + y2 = 13 .
9+4

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse

x2 + 2y2 = 2 between the coordinate axes, is [IIT Screening 2004]

(a) 1 + 1 =1 (b) 1 + 1 =1 (c) 1 + 1 =1 (d) 1 + 1 =1
x2 2y2 4x2 2y2 2x2 4y2 2x2 y2

Solution: (c) Let the point of contact be R ≡ ( 2 cos θ , sinθ )

Equation of tangent AB is x cos θ + y sinθ = 1 Y
2
B Q(h,k)

⇒ A ≡ ( 2 sec θ ,0); B ≡ (0, cosesθ ) R

Let the middle point Q of AB be (h, k). X’ O A X
P
h sec θ ,k cosecθ cos θ 1 , sinθ 1 1 1 =1
⇒ = 2 = 2 ⇒ = h2 = 2k ⇒ 2h2 + 4k2 Y’

Thus required locus is 11 =1
2x2 + 4y2

Equations of Normal in Different forms .

(1) Point form: The equation of the normal at (x1, y1)to the ellipse x2 y2 = 1 is a2x b2y = a2 − b2.
a2 + b2 x1 − y1

Tangent P(x1,y1)

Normal

Q(x2,y2)

(2) Parametric form: The equation of the normal to the ellipse x2 y2 = 1 at (a cos φ, b sin φ) is
ax sec φ − by cosecφ = a 2 − b 2 . a2 + b2

(3) Slope form: If m is the slope of the normal to the ellipse x2 y2 = 1 , then the equation of normal is
a2 + b2

y = mx ± m(a 2 − b 2 )
a2 + b2m2

The coordinates of the point of contact are  ± a2 , ± mb 2 
 a2 + b2m2 a 2 + b 2m2 

Note :  If y = mx + c is the normal of x2 y2 = 1 , then condition of normality is c2 = m2 (a 2 − b 2 )2 .
a2 + b2 (a 2 + b 2m2 )

 The straight line lx + my + n = 0 is a normal to the ellipse x2 y2 = 1, if a2 b2 =  a2 − b2  2 .
a2 + b2 l2 + m2 n2

 Four normals can be drawn from a point to an ellipse.

Important Tips

• If S be the focus and G be the point where the normal at P meets the axis of an ellipse, then SG = e.SP , and the tangent and normal
at P bisect the external and internal angles between the focal distances of P.

Y Normal
P(x1,y1)

X′ S′ C G S T X

Y′

• Any point P of an ellipse is joined to the extremities of the major axis then the portion of a directrix intercepted by them subtends a
right angle at the corresponding focus.

• With a given point and line as focus and directrix, a series of ellipse can be described. The locus of the extermities of their minor axis is
a parabola.

• The equations to the normals at the end of the latera recta and that each passes through an end of the minor axis, if e 4 + e 2 + 1 = 0
• If two concentric ellipse be such that the foci of one be on the other and if e and e’ be their eccentricities. Then the angle between their

axes is cos−1 e2 + e′2 −1 .
ee′

Example: 22 The equation of normal at the point (0, 3) of the ellipse 9x2 + 5y2 = 45 is [MP PET 1998]

(a) y − 3 = 0 (b) y + 3 = 0 (c) x-axis (d) y-axis

Solution: (d) For x2 y2 = 1 , equation of normal at point (x1, y1) , is (x − x1)a2 = (y − y1)b2
Example: 23 a2 + b2 x1 y1

Here, (x1, y1) = (0, 3) and a2 =5, b2 = 9 , Therefore (x − 0) .5 = (y − 3) .9 or x=0 i.e., y-axis.
0 3

If the normal at any point P on the ellipse cuts the major and minor axes in G and g respectively and C be the centre of the

ellipse, then [Kurukshetra CEE 1998]

(a) a2(CG)2 + b2(Cg)2 = (a2 − b2)2 (b) a2(CG)2 − b2(Cg)2 = (a2 − b2)2

(c) a2(CG)2 − b2(Cg)2 = (a2 + b2)2 (d) None of these

Solution: (a) Let at point (x1, y1) normal will be (x − x1) a 2 = (y − y1)b2
x1 y1
Example: 24
Solution: (b) At G, y=0 ⇒ x = CG = x1(a2 − b2) and at g, x = 0 ⇒ y = Cg = y1(b2 − a2)
a2 b2

x12 + y12 =1 ⇒ a2(CG)2 + b(Cg)2 = (a2 − b2)2 .
a2 b2

The equation of the normal to the ellipse x2 y2 = 1 at the positive end of the latus-rectum is
a2 + b2

(a) x + ey + e3a = 0 (b) x − ey − e3a = 0 (c) x − ey − e2a = 0 (d) None of these

The equation of the normal at (x1, y1) to the given ellipse is a2x − b2y = a2 − b2 . Here, x1 = ae and y1 = b2
x1 y1 a

So, the equation of the normal at positive end of the latus- rectum is

a2x − b2y = a 2e 2 [ b2 = a2(1 − e2) ] ⇒ ax − ay = a 2e 2 ⇒ x − ey − e3a = 0
ae b2 / a e

Auxiliary Circle .
The circle described on the major axis of an ellipse as diameter is called an auxiliary circle of the ellipse.

If x2 y2 = 1 is an ellipse, then its auxiliary circle is x2 + y2 = a2 Y
a2 + b2
P(x,y)

Eccentric angle of a point: Let P be any point on the X′ φ X
CM
ellipse x2 y2 = 1. Draw PM perpendicular from P on the major axis of the x2 + y2 =1
a2 + b2 Y′ a2 b2

ellipse and produce MP to meet the auxiliary circle in Q. Join CQ. The angle

∠XCQ = φ is called the eccentric angle of the point P on the ellipse.

Note that the angle ∠XCP is not the eccentric angle of point P.

Properties of Eccentric angles of the Co-normal points .

(1) The sum of the eccentric angles of the co-normal points on the ellipse x2 y2 = 1 is equal to odd multiple of π .
a2 + b2

(2) If α, β ,γ are the eccentric angles of three points on the ellipse, the normals at which are concurrent, then
sin(α + β ) + sin(β + γ ) + sin(γ + α) = 0 .

(3)Co-normal points lie on a fixed curve: Let P(x1, y1 ), Q(x 2 , y2 ) , R(x3 , y3 ) and S(x 4 , y4 ) be co-normal

points, then PQRS lie on the curve (a 2 − b 2 )xy + b 2kx − a 2hy = 0

This curve is called Apollonian rectangular hyperbola. Q(x2,y2) P(x1,y1)

T(h,k)

R(x3,y3) S(x4,y4)

Note :  The feet of the normals from any fixed point to the ellipse lie at the intersections of the apollonian

rectangular hyperbola with the ellipse.

Important Tips

• The area of the triangle formed by the three points, on the ellipse x2 y2 =1, whose eccentric angles are θ ,φ and ψ is
a2 + b2

2ab sin φ −ψ  sinψ −θ  sin θ −φ  .
 2  2   2 

• The eccentricity of the ellipse x2 y2 = 1 is given by 2cot w = e 2 sin 2θ , where w is one of the angles between the normals at the
a2 + b2 (1 − e 2 )

points whose eccentric angles are θ and π +θ .
2

Example: 25 The eccentric angle of a point on the ellipse x2 y2 = 1 , whose distance from the centre of the ellipse is 2, is
Solution: (a) 6 +2

Example: 26 [WB JEE 1990]
Solution: (a)
(a) π / 4 (b) 3π / 2 (c) 5π / 3 (d) 7π / 6

Let θ be the eccentric angle of the point P. Then the coordinates of P are ( 6 cosθ , 2 sinθ )
The centre of the ellipse is at the origin, It is given that OP = 2

⇒ 6 cos 2 θ + 2 sin 2 θ = 2 ⇒ 6 cos 2 θ + 2 sin 2 θ = 4 ⇒ 3 cos 2 θ + sin 2 θ = 2 ⇒ 2 sin 2 θ = 1

⇒ sin 2 θ = 1 ⇒ sinθ = ± 1 ⇒ θ = ±π / 4
2 2

The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the

point-whose eccentric angle is π / 4 , is

(a)  a2 − b2  ab (b)  a2 + b2  ab (c) 1  a2 − b2  ab (d) 1  a2 + b2  ab
a2 + b2 a2 − b2 ab a2 + b2 ab a2 − b2

The given point is ( a cos π / 4, b sinπ / 4) i.e.  a, b  .
2 2

So, the equation of the tangent at this point is xy 2 ......(i)
a +b =

∴ p1 = length of the perpendicular form (0, 0) on (i) = 0 + 0 − 2 2ab
a b a2 + b2
=
1/a2 +1/b2

Equation of the normal at  a, b  is a2x b2y = a2 − b2⇒ 2ax − 2by = a 2 − b 2 .....(ii)
2 2 −

a/ 2 b/ 2

Therefore, p2 = length of the perpendicular form (0, 0) on (ii) = a 2b 2 = a2 − b2
2(a 2 + b 2 )
( 2a )2 + (− 2b )2

So, area of the rectangle = p1p2 = 2ab a2 − b2 =  a2 − b2  ab
× 2(a2 + b2) a2 + b2

a2 + b2

Chord of Contact .

If PQ and PR be the tangents through point P(x1, y1) to the ellipse Y
C
x2 y2 = 1, then the equation of the chord of contact QR is xx1 + yy1 = 1 or P(x1,y1) Q
a2 + b2 a2 b2 X’ R
X
T = 0 at ( x1, y1 )

Y′

Equation of Chord with Mid point (x1, y1) .

The equation of the chord of the ellipse x2 y2 = 1, whose mid point be (x1 , y1 ) is T = S1 , where
a2 + b2

T = xx1 + yy1 − 1 = 0 , S1 = x12 + y12 −1 = 0 Q(x2,y2)
a2 b2 a2 b2 P (x1,y1)

R(x3,y3)

Equation of the Chord joining two points on an Ellipse .

Let P(a cosθ ,b sinθ ) ; Q(a cosφ,b sinφ)be any two points of the ellipse x2 y2 = 1 . Then ,the equation of
a2 + b2

the chord joining these two points is y − b sinθ = b sinφ − b sinθ (x − a cosθ )
a cosφ − a cosθ

Thus, the equation of the chord joining two points having eccentric angles θ and φ on the ellipse

x2 y2 =1 is x cos θ + φ  + y sin θ + φ  = cos θ −φ 
a2 + b2 a  2  b  2   2 

Note :  If the chord joining two points whose eccentric angles are α and β cut the major axis of an ellipse

at a distance ‘c’ from the centre, then tan α tan β = c − a .
2 2 c + a

 If α and β be the eccentric angles of the extremities of a focal chord of an ellipse of eccentricity e,

then tan α tan β + 1 e = 0.
2 2 1± e

Example: 27 What will be the equation of that chord of ellipse x2 y2 = 1 which passes from the point (2,1) and bisected on the point
Solution: (d) 36 + 9

[UPSEAT 1999]

(a) x + y = 2 (b) x + y = 3 (c) x + 2y = 1 (d) x + 2y = 4

Let required chord meets to ellipse on the points P and Q whose coordinates are (x1, y1) and (x 2 , y2 ) respectively

 Point (2,1) is mid point of chord PQ

∴ 2= 1 (x1 + x 2 ) or x1 + x2 = 4 and 1 = 1 (y1 + y2 ) or y1 + y2 =2
2 2

Again points (x1, y1 ) and (x2, y2) are situated on ellipse; ∴ x12 + y12 = 1 and x 2 + y 2 =1
36 9 2 2

36 9

On subtracting x 2 − x 2 + y 2 − y12 =0 or y2 − y1 = − (x2 + x1) = −4 = −1
2 1 2 9 x2 − x1 4(y2 + y1) 4×2 2

36

∴ Gradient of chord PQ = y2 − y1 = −1
x2 − x1 2

Therefore, required equation of chord PQ is as follows, y − 1 = − 1 (x − 2) or x + 2y =4
2

Alternative: S1 = T (If mid point of chord is known)

∴ 22 12 −1 = 2x + 1y −1 ⇒ x + 2y = 4
36 + 9 36 9

Example: 28 What will be the equation of the chord of contact of tangents drawn from (3, 2) to the ellipse x 2 + 4y 2 = 9
Solution: (a)
Example: 29 (a) 3x + 8y = 9 (b) 3x + 8y = 25 (c) 3x + 4y = 9 (d) 3x + 8y + 9 = 0
Solution: (a)
The required equation is T = 0 i.e. , 3x + 4(2y) − 9 = 0 or 3x + 8y = 9 .
Example: 30
Solution: (a) A tangent to the ellipse x 2 + 4y 2 = 4 meets the ellipse x 2 + 2y 2 = 6 at P and Q. The angle between the tangents at P and
Example: 31
Solution: (d) Q of the ellipse x 2 + 2y 2 = 6 is [IIT 1997]

(a) π (b) π (c) π (d) π
2 3 4 6

The given ellipse x 2 + 4y 2 = 4 can be written as x2 y2 = 1 .....(i)
4 +1
A(h,k) Q

Any tangent to ellipse (i) is x cos θ + y sinθ =1 .....(ii)
2

Second ellipse is x 2 + 2y 2 =6 , i.e. x2 y2 =1 .....(iii) P
6 +3

Let the tangents at P, Q meet at (h, k) .

∴Equation of PQ, i.e. chord of contact is hx + ky = 1 .....(iv)
6 3

Since (ii) and (iv) represent the same line, ∴ h/6 2 = k/3 = 1 ⇒ h = 3 cosθ and k = 3 sinθ
(cosθ ) / sin θ 1

So, h2 + k2 = 9 or x 2 + y 2 = 9 is the locus of (h, k) which is the director circle of the ellipse x2 y2 =1
6 +3

∴ The angle between the tangents at P and Q will be π / 2 .

The locus of mid-points of a focal chord of the ellipse x2 y2 = 1 is [EAMCET 1995]
a2 + b2

(a) x 2 y 2 ex (b) x 2 y 2 ex (c) x 2 + y 2 = a 2 + b 2 (d) None of these
a2 + b2 = a a2 − b2 = a

Let (h, k) be the mid point of a focal chord. Then its equation is S1 = T or hx ky h2 k2 . This passes through
a2 + b2 = a2 + b2

(ae, 0) , ∴ hae = h2 + k2 . So, locus of (h, k) is xe x 2 y 2
a2 a2 b2 a = a2 + b2

If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

(a) cosα + cos β (b) sinα − sin β (c) cosα − cos β (d) sinα + sin β
cos(α − β ) sin(α − β ) cos(α − β ) sin(α + β )

The equation of a chord joining points having eccentric angles α and β is given by

x cos α + β  + y sin α + β  = cos α − β 
a  2  b  2   2 

If it passes through (ae,0) then e cos α +β  = cos α −β 
 2   2 

cos α − β  2 sin α + β  cos α − β 
 2   2   2 
⇒ e = ⇒e = ⇒ e = sinα + sin β
cos α + β  sin α + β  cos α + β  sin(α + β )
 2  2  2   2 

Pole and Polar .

Let P(x1,y1)be any point inside or outside the ellipse. A chord through P intersects the ellipse at A and B
respectively. If tangents to the ellipse at A and B meet at Q(h,k) then locus of Q is called polar of P with respect to
ellipse and point P is called pole.

Q(h,k) A Q(h,k) A
Polar
Polar A′ B B′ A′
P(x1,y1) Pole Q′
P(x1,y1)
B

T B′
Pole

Equation of polar: Equation of polar of the point (x1, y1) with respect to ellipse x2 y2 =1 is given by
a2 + b2

xx1 + yy1 =1 (i.e. T = 0 )
a2 b2

Coordinates of pole: The pole of the line lx + my + n = 0 with respect to ellipse x2 y2 =1 is
a2 + b2

P  − a 2l , − b2m  P
n n
Q′
Q R′
R

Note :  The polar of any point on the directrix, passes through the focus.

 Any tangent is the polar of its own point of contact.
Properties of pole and polar

(1) If the polar of P(x1, y1)passes through Q(x 2, y2 ) , then the polar of Q(x 2, y2 ) goes through P(x1, y1) and
such points are said to be conjugate points.

(2) If the pole of a line l1x + m1y + n1 = 0 lies on the another line l2 x + m2y + n2 = 0 , then the pole of the
second line will lie on the first and such lines are said to be conjugate lines.

(3) Pole of a given line is same as point of intersection of tangents at its extremities.

Example: 32 The pole of the straight line x + 4y = 4 with respect to ellipse x 2 + 4y 2 = 4 is [EAMCET 2002]
[EAMCET 1995]
(a) (1, 4) (b) (1, 1) (c) (4, 1) (d) (4, 4)

Solution: (b) Equation of polar of (x1, y1 ) w.r.t the ellipse is xx1 + 4yy1 = 4 .....(i)

Comparing with x + 4y = 4 .....(ii)

x1 = 4y1 =1 ⇒ x1 = 1, y1 = 1 . ∴ Coordinates of pole (x1, y1) = (1, 1)
1 4

Example: 33 If the polar with respect to y2 = 4ax touches the ellipse x2 + y2 = 1, the locus of its pole is
α2 β2

(a) x2 − y2 β 2) = 1 (b) x2 + β 2y2 =1
α2 (4a2α 2 / α2 4a2

(c) α 2 x 2 + β 2y 2 = 1 (d) None of these

Solution: (a) Let P(h, k) be the pole. Then the equation of the polar is ky = 2a(x + h) or y = 2a x+ 2ah .
k k

x2 y2  2ah  2 2  2a  2 + β 2, (using c 2 = a 2m2 + b 2 )
α2 β2  k   k 
This touches + = 1 , So = α

⇒ 4a 2h2 = 4a 2α 2 + k2β 2. So, locus of (h, k) is 4a2 x 2 = 4a 2α 2 + β 2y 2 or x2 − y2 =1
α2
 4a 2α 2 
β2

Diameter of the Ellipse.

Definition : The locus of the mid- point of a system of parallel chords of an ellipse is called a diameter and the

chords are called its double ordinates i.e. A line through the centre of an ellipse is called a diameter of the ellipse.

The point where the diameter intersects the ellipse is called the vertex of the diameter.

Equation of a diameter to the ellipse x2 y2 = 1 : Let y = mx + c be Y
a2 + b2
y=mx+c

a system of parallel chords of the ellipse x2 + y2 = 1, where m is a constant and X′ X
a2 b2
x2 y2 =1 y = − b2 x
c is a variable. a2 + b2 a 2m
Y′

The equation of the diameter bisecting the chords of slope m of the ellipse x2 y2 = 1 is y = − b2 x, which
a2 + b2 a 2m

is passing through (0, 0).

Conjugate diameter: Two diameters of an ellipse are said to be conjugate diameter if each bisects all chords

parallel to the other. Y

Conjugate diameter of circle i.e. AA′ and BB′ are perpendicular to each QB 90° A P
other. Hence, conjugate diameter of ellipse are PP′ and QQ′ . Hence, angle X′ θ
C X
between conjugate diameters of ellipse > 90o . P′ A′
B′ Q′
Now the coordinates of the four extremities of two conjugate diameters are

P(a cosφ,b sinφ); P′(−a cosφ,−b sinφ) ; Q(−a sinφ,b cosφ);Q′(a sinφ,−b cosφ) Y′

If y = m1x and y = m2 x be two conjugate diameters of an ellipse, then m1m2 = − b2
a2

(1) Properties of diameters
(i) The tangent at the extremity of any diameter is parallel to the chords it bisects or parallel to the conjugate diameter.

(ii) The tangent at the ends of any chord meet on the diameter which bisects the chord.

(2) Properties of conjugate diameters (a cosφ',b sinφ')
(i) The eccentric angles of the ends of a pair of conjugate diameters of an D P(a cosφ,b sinφ)
ellipse differ by a right angle,

π A′ A
2 C
i.e. φ − φ′ =
P′ D′

(ii) The sum of the squares of any two conjugate semi-diameters of an ellipse is constant and equal to the sum
of the squares of the semi axes of the ellipse, i.e. CP 2 + CD2 = a2 + b2

(iii) The product of the focal distances of a point on an ellipse is equal to the square of the semi-diameter
which is conjugate to the diameter through the point,

i.e., SP.S′P = CD2 D P(a cosφ,b sinφ)

S′ C S
P′ D′

(iv) The tangents at the extremities of a pair of conjugate diameters form a parallelogram whose area is

constant and equal to product of the axes, i.e. D M Q Y′ P X
X′ R′ C R
Area of parallelogram = (2a)(2b) = Area of rectangle contained under
major and minor axes. P′ D′

Q′ Y

(v) The polar of any point with respect to ellipse is parallel to the diameter to the one on which the point lies.
Hence obtain the equation of the chord whose mid point is (x1,y1) , i.e. chord is T = S1 .

(3) Equi-conjugate diameters: Two conjugate diameters are called equi-conjugate, if their lengths are
equal i.e. (CP)2 = (CD)2

∴ a 2 cos 2 φ + b2 sin2 φ = a 2 sin 2 φ + b2 cos 2 φ

⇒ a 2(cos2 φ − sin2 φ) − b2(cos2 φ − sin2 φ) = 0 ⇒ (a 2 − b2)(cos2 φ − sin2 φ) = 0

 (a 2 − b 2 ) ≠ 0 , ∴ cos 2φ = 0 . So, φ = π or 3π
4 4

∴ (CP) = (CD) = (a 2 + b2) for equi-conjugate diameters.
2

Important Tips

• If the point of intersection of the ellipses x2 y2 = 1 and x2 + y2 = 1 be at the extremities of the conjugate diameters of the
a2 + b2 α2 β2

former, then a2 + b2 = 2
α2 β2

• The sum of the squares of the reciprocal of two perpendicular diameters of an ellipse is constant.

• In an ellipse, the major axis bisects all chords parallel to the minor axis and vice-versa, therefore major and minor axes of an ellipse
are conjugate diameters of the ellipse but they do not satisfy the condition m1. m2 = −b2 / a2 and are the only perpendicular
conjugate diameters.

Example: 34 If one end of a diameter of the ellipse 4 x 2 + y 2 = 16 is ( 3,2) , then the other end is

(a) (− 3, 2) (b) ( 3,−2) (c) (− 3,−2) (d) (0, 0)

Solution: (c) Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other
Example: 35 end are (− 3,−2) .

Solution: (a) If θ and φ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse x2 y2 = 1 ,then θ − φ is equal to
a2 + b2

(a) ± π (b) ±π (c) 0 (d) None of these
2

Let y = m1 x and y = m2 x be a pair of conjugate diameter of an ellipse x2 y2 = 1 and let P(a cos θ , b sinθ ) and
a2 + b2

Q (a cos φ, b sin φ) be ends of these two diameters. Then m1m2 = − b2
a2

⇒ b sinθ −0 × b sinφ −0 = − b2 ⇒ sinθ sinφ = − cosθ cos φ ⇒ cos(θ − φ) = 0 ⇒ θ − φ = ±π / 2 .
a cosθ −0 a cos φ −0 a2

Subtangent and Subnormal .

Let the tangent and normal at P(x1,y1)meet the x-axis at A and B respectively.

Length of subtangent at P(x1,y1)to the ellipse x2 y2 =1 is DA = CA − CD = a2 − x1
a2 + b2 x1

Length of sub-normal at P(x1,y1) to the ellipse x2 y2 =1 is Y P(x1,y1)
a2 + b2 A′ DA
CB
−  x1 b2 x1  b2 X
a2 a2
BD = CD − CB = x1 − = x1 = (1 − e 2 )x1 . X′

Y′

Note :  The tangent and normal to any point of an ellipse bisects respectively the internal and external angles

between the focal radii of that point.

Example: 36 Length of subtangent and subnormal at the point  − 5 3 ,2 of the ellipse x2 + y2 = 1 are
 2 25 16

(a)  5 3 − 10  , 83 (b)  5 3 + 10  , 83 (c)  5 3 + 12  , 16 3 (d) None of thee
 2 3  5  2 3  10  2 3  5

Solution: (a) Here a2 25, b2 = 16, x1 −5 3 . Length of subtangent = a2 − x1 25 5 3 = 5 3 10 .
= = 2 x1 = + 2−3
−5 3/2 2

Length of subnormal = b2 x1 = 16  − 5 3  = 83
a2 25  2  5

Concyclic points .

Any circle intersects an ellipse in two or four points. They are called concyclic points and the sum of their
eccentric angles is an even multiple of π .

Q(β) P(α)
R(γ) S(δ)

If α, β.γ ,δ be the eccentric angles of the four concyclic points on an ellipse, then α + β + γ + δ = 2nπ , where
n is any integer.

Note :  The common chords of a circle and an ellipse are equally inclined to the axes of the ellipse.

Important Tips

• The centre of a circle x 2 + y 2 + 2gx + 2 fy + c = 0 passing through the three points, on an ellipse x2 y2 = 1 (whose eccentric angles
a2 + b2

are a, β,γ ) is g  a 2 −b 2  {cos α + cos β + cos γ cos(α + γ )}and − f  b 2 −a 2  {sin α + sin β + sin γ − sin(α + γ )}
4a 4a
− = + + β = + β

• P ′CP and D′CD are conjugate diameters of an ellipse and α is the eccentric angles of P. Then the eccentric angles of the point where
the circle through P, P ′, D again cuts the ellipse is π / 2 − 3α .

Reflection property of an Ellipse .

Let S and S′ be the foci and PN the normal at the point P of the ellipse, then ∠SPS′ = ∠SQS′ . Hence if an

incoming light ray aimed towards one focus strike the concave side of the mirror Q B Tangent Light
in the shape of an ellipse then it will be reflected towards the other focus. P ray

αα X
A
X′ A′ S′ C NS
Normal

Reflected ray B′

Example: 37 A ray emanating from the point (−3,0) is incident on the ellipse 16x 2 + 25y2 = 400 at the point P with ordinate 4. Then
Solution: (a)
the equation of the reflected ray after first reflection is

(a) 4 x + 3y = 12 (b) 3x + 4y = 12 (c) 4 x − 3y = 12 (d) 3x − 4y = 12

For point P y-coordinate =4 Y
Given ellipse is 16x 2 + 25y 2 = 400 (0,4) P

16x 2 + 25(4)2 = 400 , ∴ x = 0 X

∴co-ordinate of P is (0, 4) X′ (–3,0) S’ C S(3,0)

e2 =1− 16 = 9
25 25
Y′
3
∴ e = 5

∴ Foci (±ae,0) , i.e. (±3,0)

∴ Equation of reflected ray (i.e.PS) is x + y = 1 or 4x + 3y = 12 .
3 4

Conic sections

Definition.

A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its

distance from a fixed point in the same plane to its distance from a fixed line is Z

always constant which is always greater than unity.

Fixed point is called focus, fixed straight line is called directrix and the M Directrix P
constant ratio is called eccentricity of the hyperbola. Eccentricity is denoted by e Z′ S(Focus)
and e > 1.

A hyperbola is the particular case of the conic

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

When , abc + 2 fgh − af 2 − bg 2 − ch2 ≠ 0 i.e., ∆ ≠ 0 and h2 > ab .

Let S(h,k) is the focus, directrix is the line ax + by + c = 0 and the eccentricity is e. Let P(x1, y1) be a point
which moves such that SP = e.PM

⇒ (x1 − h)2 + (y1 − k)2 = e. ax1 + by1 + c
a2 + b2

⇒ (a 2 + b2 )[(x1 − h)2 + (y1 − k)2 ] = e 2(ax1 + by1 + c)2

Hence, locus of (x1,y1) is given by (a2 + b2)[(x − h)2 + (y − k)2] = e 2(ax + by + c)2
Which is a second degree equation to represent a hyperbola (e > 1).

Example: 1 The equation of the conic with focus at (1, – 1), directrix along x − y + 1 = 0 and with eccentricity 2 is
Solution: (c)
[EAMCET 1994; DCE 1998]
Example: 2
Solution: (a) (a) x 2 − y 2 = 1 (b) xy = 1 (c) 2xy − 4 x + 4y + 1 = 0 (d) 2xy + 4 x − 4y − 1 = 0

Here, focus (S) = (1, –1), eccentricity (e)= 2
From definition , SP = e PM

(x − 1)2 + (y + 1)2 = 2.(x − y + 1)
12 + 12

⇒ (x − 1)2 + (y + 1)2 = (x − y + 1)2 ⇒ 2xy − 4 x + 4y + 1 = 0 , which is the required equation of conic (Rectangular hyperbola)

The centre of the hyperbola 9x2 − 36x − 16y2 + 96y − 252 = 0 is [Karnataka CET 1993]

(a) (2, 3) (b) (– 2, – 3) (c) (–2, 3) (d) (2, – 3)

Here a = 9, b = −16, h = 0 , g = −18 , f = 48 , c = −252

Centre of hyperbola =  hf − bg , gh − af  =  (0) (48) − (−16)(−18) , (−18)(0) − (9)(48)  = (2, 3)
 ab − h2 ab − h2  (9)(−16) − 0 (9)(−16) − 0

Standard equation of the Hyperbola .

Let S be the focus, ZM be the directrix and e be the eccentricity of the hyperbola, then by definition,

⇒ SP =e ⇒ (SP)2 = e 2 (PM)2
PM

e 2  x a  2 Y PLQ
 e  M′ M′
⇒ (x − a.e)2 + (y − 0)2 = − L1

x2 y2 x2 y2 (x,y) N
a2 a 2 (e 2 a2 − b2
X′ axis
Z A (a,0) S (ae,0) X
⇒ − = 1 ⇒ = 1 , where b2 = a 2 (e 2 − 1) (–ae,0)S′ (–a,0) A′ Z′ C

− 1) Directrix
Directrix
L1′ L′ Q′
This is the standard equation of the hyperbola. x=–a/e Y′ x=a/e

Some terms related to hyperbola : Let the equation of hyperbola is x2 y2 =1
a2 − b2

(1) Centre : All chords passing through C are bisected at C. Here C (0,0)

(2) Vertex: The point A and A′ where the curve meets the line joining the foci S and S ′ are called vertices of
hyperbola. The co-ordinates of A and A′ are (a, 0) and (– a, 0) respectively.

(3) Transverse and conjugate axes : The straight line joining the vertices A and A′ is called transverse axis
of the hyperbola. The straight line perpendicular to the transverse axis and passing through the centre is called
conjugate axis.

Here, transverse axis = AA′ = 2a

Conjugate axis = BB′ = 2b

(4) Eccentricity : For the hyperbola x2 y2 =1
a2 − b2

 2b  2  Conjugate axis  2
 2a   Transverse axis 
We have b 2 = a 2 (e 2 − 1), e= 1 + = 1 +

(5) Double ordinates : If Q be a point on the hyperbola, QN perpendicular to the axis of the hyperbola and
produced to meet the curve again at Q′ . Then QQ′ is called a double ordinate at Q.

If abscissa of Q is h, then co-ordinates of Q and Q′ are  h, b h2 − a2  and  h, − b h2 − a 2  respectively.
 a   a 

(6) Latus-rectum : The chord of the hyperbola which passes through the focus and is perpendicular to its
transverse axis is called latus-rectum.

Length of latus-rectum LL′ = L1L1′ = 2b 2 = 2a(e 2 − 1) and end points of latus-rectum L ae, b2  ;
a a

L′ ae, − b2 ; L1  − ae , b2  ; L1′  − ae, − b2  respectively.
a a a

(7) Foci and directrices: The points S(ae, 0) and S′(−ae, 0) are the foci of the hyperbola and ZM and

Z′M ′ are two directrices of the hyperbola and their equations are x= a and x = − a respectively.
e e

Distance between foci SS′ = 2ae and distance between directrices ZZ′ = 2a / e .

(8) Focal chord : A chord of the hyperbola passing through its focus is called a focal chord.
(9) Focal distance : The difference of any point on the hyperbola from the focus is called the focal distance of the point.

From the figure, SP = ePM = e x 1 − a  = ex1 − a, S′P = ePM ′ = e x 1 + a  = ex1 +a
 e   e 

The difference of the focal distance of a point on the hyperbola is constant and is equal to the length of
transverse axis.

| S′P − SP |= 2a = AA′ = Transverse axis

Example: 3 The eccentricity of the hyperbola which passes through (3, 0) and (3 2, 2) is [UPSEAT 2000]
Solution: (b)
(a) (13) (b) 13 (c) 13 (d) None of these
Example: 4 3 4
Solution: (c)
Example: 5 Let equation of hyperbola is x 2 / a 2 − y 2 / b 2 = 1 . Point (3, 0) lies on hyperbola
Solution: (c)
So, (3)2 0 =1 or 9 =1 or a2 =9 and point (3 2,2) also lies on hyperbola. So, 3( 2)2 (2)2 =1
Example: 6 a2 − b2 a2 a2 − b2
Solution: (d)
Put a2 = 9 we get, 18 4 =1 or 2 − 4 =1 or 4 = 1− 2 or 4 =1 or b2 = 4
9 − b2 b2 − b2 b2

We know that b 2 = a 2(e 2 − 1) . Putting values of a 2 and b 2

4 = 9(e 2 − 1) or e2 −1 = 4 or e2 =1+ 4 or e = (1 + 4 / 9) or e = (13) / 9 = 13 .
9 9 (c) (± 5,0) 3

The foci of the hyperbola 9x 2 − 16y 2 = 144 are [MP PET 2001]

(a) (± 4,0) (b) (0, ± 4) (d) (0, ± 5)

The equation of hyperbola is x2 y2 =1.
16 −9

Now, b2 = a 2(e 2 − 1) ⇒ 9 = 16(e 2 − 1) ⇒ e= 5 . Hence foci are (± ae,0) =  ± 4 . 5 , 0  i.e., (± 5,0)
4  4 

If the foci of the ellipse x2 + y2 = 1 and the hyperbola x2 − y2 = 1 coincide, then the value of b2 is
16 b2 144 81 25

[MNR 1992; UPSEAT 2001; AIEEE 2003]

(a) 1 (b) 5 (c) 7 (d) 9

For hyperbola, x2 − y2 = 1
144 81 25

A= 144 , B = 81 , e1 = 1+ B2 = 1 + 81 = 225 5
25 25 A2 144 144 = 4

Therefore foci = (±ae1 , 0) =  ± 12 . 5 , 0  = (± 3, 0) . Therefore foci of ellipse i.e., (±4 e,0) = (±3,0) (For ellipse a = 4 )
 5 4 

⇒ e= 3 , Hence b2 = 16 1 − 9  = 7 .
4  16 

If PQ is a double ordinate of hyperbola x2 y2 =1 such that CPQ is an equilateral triangle, C being the centre of the
a2 − b2
hyperbola. Then the eccentricity e of the hyperbola satisfies
[EAMCET 1999]

(a) 1 < e < 2 / 3 (b) e = 2 / 3 (c) e = 3 / 2 (d) e > 2 / 3

Let P(a sec θ , b tanθ ) ; Q(a sec θ , − b tanθ ) be end points of double ordinates and C(0, 0) is the centre of the hyperbola

Now PQ = 2b tanθ ; CQ = CP = a 2 sec 2 θ + b 2 tan 2 θ P (a sec θ, b tan θ)
Q(a sec θ,–b tanθ)
Since CQ = CP = PQ , ∴ 4b 2 tan 2 θ = a 2 sec 2 θ + b 2 tan 2 θ

⇒ 3b 2 tan 2 θ = a 2 sec 2 θ ⇒ 3b 2 sin 2 θ = a 2 C(0, 0)
⇒ 3a 2 (e 2 − 1) sin 2 θ = a 2 ⇒ 3 (e 2 − 1) sin 2 θ = 1

⇒ 1 = sin 2 θ <1 (sin 2 θ < 1)
3(e 2 − 1)

⇒ e 1 1 < 3 ⇒ e2 − 1 > 1 ⇒ e2 > 4 ⇒ e> 2
2− 3 3 3

Conjugate Hyperbola .
The hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of a

given hyperbola is called conjugate hyperbola of the given hyperbola.

Hyperbola x2 − y2 =1 − x2 + y2 =1 or x2 − y2 = −1 Y
Fundamentals a2 b2 a2 b2 a2 b2
Centre
Length of transverse axis (0, 0) (0, 0) (0S,–b)
Length of conjugate axis 2b
Foci 2a 2a Z B(0,b) y= b/e
Equation of directrices
Eccentricity 2b (0, ± be) X′ C

Length of latus rectum (± ae, 0) X

Parametric co-ordinates x = ±a / e y = ±b / e Z′ y= –b/e
Focal radii B (0,–b)
Difference of focal radii
(S′P − SP) e=  a 2 +b 2   a2 + b2  S′
Tangents at the vertices a2 b2 (0,–b)
Equation of the transverse e=
axis
Equation of the conjugate axis 2b2 2a2 Y′
a b
(b sec φ, a tan φ), 0 ≤ φ < 2π
(a sec φ, b tan φ) , 0 ≤ φ < 2π
SP = ey1 − b & S′P = ey1 + b
SP = ex1 − a & S′P = ex1 + a 2b
2a

x = −a, x = a y = −b, y = b
y=0 x=0

x=0 y=0

Note :  If e and e′ are the eccentricities of a hyperbola and its conjugate, then 1 1 =1.
e2 + e′2

 The foci of a hyperbola and its conjugate are concyclic.

Example: 7 The eccentricity of the conjugate hyperbola of the hyperbola x 2 − 3y 2 = 1 , is [MP PET 1999]
Solution: (a)
(a) 2 (b) 2 (c) 4 (d) 4
3 3

The given hyperbola is x2 − y2 = 1 . Here a 2 = 1 and b2 = 1
1 1/ 3 3

Since b 2 = a 2 (e 2 − 1) ⇒ 1 = 1(e 2 − 1) ⇒ e2 = 4 ⇒ e= 2
3 3 3

If e′ is the eccentricity of the conjugate hyperbola, then 1 + 1 =1 ⇒ 1 =1− 1 =1− 3 = 1 ⇒ e′ = 2.
e2 e′2 e′2 e2 4 4

Special form of Hyperbola .

If the centre of hyperbola is (h, k) and axes are parallel to the co-ordinate axes, then its equation is

(x − h)2 + (y − k)2 =1. By shifting the origin at (h, k) without rotating the co-ordinate axes, the above equation
a2 b
2

reduces to X2 Y2 = 1, where x = X + h, y = Y + k .
a2 − b2

Example: 8 The equation of the hyperbola whose foci are (6, 4) and (– 4, 4) and eccentricity 2 is given by [MP PET 1993]

(a) 12x 2 − 4y 2 − 24 x + 32y − 127 = 0 (b) 12x 2 + 4y 2 + 24 x − 32y − 127 = 0

(c) 12x 2 − 4y 2 − 24 x − 32y + 127 = 0 (d) 12x 2 − 4y 2 + 24 x + 32y + 127 = 0

Solution: (a) Foci are (6, 4) and (– 4, 4) and e = 2 .

∴ Centre is  6 − 4 , 4 + 4  = (1, 4)
 2 2 

So, ae + 1 = 6 ⇒ ae = 5 ⇒ a = 5 and b = 5 3
2 2

Hence, the required equation is (x − 1)2 − (y − 4)2 =1 or 12x 2 − 4y2 − 24 x + 32y − 127 = 0
25 / 4 (75 / 4)

Example: 9 The equations of the directrices of the conic x 2 + 2x − y 2 + 5 = 0 are

(a) x = ± 1 (b) y = ±2 (c) y = ± 2 (d) x = ± 3

Solution: (c) (x + 1)2 − y2 −1+ 5 = 0 ⇒ − (x + 1)2 + y2 =1
4 4

Equation of directrices of y2 − x2 =1 are y = ± b
b2 a2 e

Here b = 2 , e = 1 + 1 = 2 . Hence, y = ± 2 ⇒ y = ± 2 .
2

Auxiliary circle of Hyperbola .

Let x2 y2 =1 be the hyperbola with centre C and transverse axis A′A . Y (x,y)
a2 − b2 Q P

Therefore circle drawn with centre C and segment A′A as a diameter is called X′ φ90o N X

auxiliary circle of the hyperbola x2 y2 =1 (– a,0)A′ (0,0)C A(a,0)
a2 − b2

∴ Equation of the auxiliary circle is x 2 + y 2 = a 2 Y′

Let ∠QCN = φ

Here P and Q are the corresponding points on the hyperbola and the auxiliary circle (0 ≤ φ < 2π )

(1) Parametric equations of hyperbola : The equations x = a secφ and y = b tanφ are known as the

parametric equations of the hyperbola x2 y2 = 1 . This ( a secφ,b tanφ ) lies on the hyperbola for all values of φ.
a2 − b2

Position of points Q on auxiliary circle and the corresponding

point P which describes the hyperbola and 0 ≤ φ < 2π

φ varies from Q(a cos φ, a sinφ) P(a sec φ, b tan φ)

0 to π I I
2

π to π II III
2 III II
IV IV
π to 3π
2

3π to 2π
2

Note :  The equations x = a coshθ and y = b sin hθ are also known as the parametric equations of the

Example: 10 hyperbola and the co-ordinates of any point on the hyperbola x2 − y2 =1 are expressible as
Solution: (c) a2 b2

(a cos hθ , b sin hθ ), where cos hθ = eθ + e −θ and sin hθ = eθ − e −θ .
2 2

The distance between the directrices of the hyperbola x = 8 secθ , y = 8 tanθ is [Karnataka CET 2003]

(a) 16 2 (b) 2 (c) 8 2 (d) 4, 2

Equation of hyperbola is x = 8 secθ , y = 8 tanθ ⇒ x = sec θ , y = tanθ
8 8

∴sec 2 θ − tan 2 θ = 1 ⇒ x2 y2 =1
82 − 82

Here a = 8,b = 8 . Now e = 1 + b2 = 1 + 82 = 2
a2 82

∴ Distance between directrices = 2a = 2×8 = 8 2.
e 2

Position of a point with respect to a Hyperbola .

Let the hyperbola be x2 y2 = 1. Y P (outside)
a2 − b2 A′ C A
P (on)
Then P(x1, y1) will lie inside, on or outside the hyperbola x2 y2 =1 X′ P(inside)
a2 − b2 X

according as x12 − y12 −1 is positive, zero or negative.
a2 b2
Y′

Example: 11 The number of tangents to the hyperbola x2 y2 = 1 through (4, 1) is [AMU 1998]
4−3

(a) 1 (b) 2 (c) 0 (d) 3

Solution: (c) Since the point (4, 1) lies inside the hyperbola  16 − 1 − 1 > 0 ; ∴ Number of tangents through (4, 1) is 0.
4 3

Intersection of a Line and a Hyperbola.

The straight line y = mx + c will cut the hyperbola x2 y2 =1 in two points may be real, coincident or
a2 − b2

imaginary according as c 2 >, =, < a2m2 − b2 .

Condition of tangency : If straight line y = mx + c touches the hyperbola x2 y2 = 1 , then c2 = a2m2 − b2 .
a2 − b2

Equations of Tangent in Different forms.

(1) Point form : The equation of the tangent to the hyperbola x2 y2 = 1 at (x1, y1) is xx1 − yy1 = 1.
a2 − b2 a2 b2

(2) Parametric form : The equation of tangent to the hyperbola x2 y2 =1 at (a secφ,b tanφ) is
a2 − b2

x sec φ − y tanφ = 1
a b

(3) Slope form : The equations of tangents of slope m to the hyperbola x2 y2 =1 are
a2 − b2

y = mx ± a2m2 − b2 and the co-ordinates of points of contacts are  ± a2m ,± b2  .
 a2m2 − b2 a2m2 − b2 

Note :  If the straight line lx + my + n = 0 touches the hyperbola x2 y2 = 1 , then a2l 2 − b2m2 = n2 .
a2 − b2

 If the straight line x cosα + y sinα = p touches the hyperbola x2 y2 = 1,
a2 − b2

then a2 cos2 α − b2 sin2 α = p2
 Two tangents can be drawn from an outside point to a hyperbola.

Important Tips

• For hyperbola x2 − y2 =1 and y2 − x2 =1, the equation of common tangent is y = ±x ± a2 − b2 , points of contacts are
a2 b2 a2 b2

 ± a2 ;± b2  and length of common tangent is 2 . (a2 + b2) .
 a2 − b2 a2 − b2  a2 − b2

• If the line y = mx ± a2m2 − b2 touches the hyperbola x2 y2 =1 at the point (a secθ ,b tanθ ) , then θ = sin −1  b  .
a2 − b2  am 

Example: 12 The value of m for which y = mx + 6 is a tangent to the hyperbola x2 − y2 =1, is [Karnataka CET 1993]
100 49

(a) 17 (b) 20 (c) 3 (d) 20
20 17 20 3

Solution: (a) For condition of tangency, c 2 = a 2m2 − b 2 . Here c = 6 , a = 10, b = 7

Then, (6)2 = (10)2 . m2 − (7)2

36 = 100m2 − 49 ⇒ 100m2 = 85 ⇒ m2 = 17 ⇒ m= 17
20 20

Example: 13 If m1 and m2 are the slopes of the tangents to the hyperbola x2 y2 = 1 which pass through the point (6, 2), then
Solution: (a, b) 25 − 16

Example: 14 (a) m1 + m2 = 24 (b) m1m2 = 20 (c) m1 + m2 = 48 (d) m1m2 = 11
11 11 11 20

The line through (6, 2) is y − 2 = m (x − 6) ⇒ y = mx + 2 − 6m

Now, from condition of tangency (2 − 6m)2 = 25m2 − 16

⇒ 36m2 + 4 − 24m − 25m2 + 16 = 0 ⇒ 11m2 − 24m + 20 = 0

Obviously, its roots are m1 and m2 , therefore m1 + m2 = 24 and m1m2 = 20
11 11

The points of contact of the line y = x − 1 with 3x2 − 4y2 = 12 is [BIT Ranchi 1996]

(a) (4, 3) (b) (3, 4) (c) (4, – 3) (d) None of these

Solution: (a) The equation of line and hyperbola are y = x − 1 .....(i) and 3x2 − 4y2 = 12 .....(ii)
From (i) and (ii), we get 3x2 − 4(x − 1)2 = 12
Example: 15 ⇒ 3x2 − 4(x2 − 2x + 1) = 12 or x2 − 8x + 16 = 0 ⇒ x = 4
Solution: (b) From (i), y = 3 so points of contact is (4, 3)
Example: 16
Solution: (c) Trick : Points of contact are  ± a2m , ± b2  .
 a2m2 − b2 a2m2 − b2 
Example: 17
Solution: (b) Here a2 = 4 , b2 = 3 and m = 1 . So the required points of contact is (4, 3).

P is a point on the hyperbola x2 y2 = 1, N is the foot of the perpendicular from P on the transverse axis. The tangent to
a2 − b2

the hyperbola at P meets the transverse axis at T. If O is the centre of the hyperbola, then OT.ON is equal to

(a) e 2 (b) a 2 (c) b2 (d) b2
a2

Let P(x1, y1) be a point on the hyperbola. Then the co-ordinates of N are (x1,0) . Y (x1P,y1
OT )
The equation of the tangent at (x1, y1) is xx1 − yy1 =1
a2 b2 (xN1,0) X
X′

This meets x-axis at T a2 , 0  ; ∴ OT.ON = a2 × x1 = a2
x1 x1
Y′

If the tangent at the point (2 sec φ,3 tanφ) on the hyperbola x2 y2 = 1 is parallel to 3x − y + 4 = 0 , then the value of φ
4−9

is [MP PET 1998]

(a) 45o (b) 60o (c) 30o (d) 75o

Here x = 2 secφ and y = 3 tanφ

Differentiating w.r.t. φ

dx = 2 sec φ tanφ and dy = 3 sec 2 φ
dφ dφ

∴ Gradient of tangent dy = dy / dφ = 3 sec 2 φ ; ∴ dy = 3 cosecφ .....(i)
dx dx / dφ 2 sec φ tan φ dx 2

But tangent is parallel to 3x − y + 4 = 0 ; ∴ Gradient m = 3

From (i) and (ii), 3 cosecφ =3 ⇒ cosecφ = 2 , ∴ φ = 30o
2

The slopes of the common tangents to the hyperbola x2 − y2 =1 and y2 − x2 =1 are [Roorkee 1997]
9 16 9 16

(a) – 2, 2 (b) – 1, 1 (c) 1, 2 (d) 2, 1

Given hyperbola are x2 − y2 =1 .....(i) and y2 − x2 =1 .....(ii)
9 16 9 16

Any tangent to (i) having slope m is y = mx ± 9m2 − 16 .....(iii)

Putting in (ii), we get, 16[mx ± 9m2 − 16]2 − 9x 2 = 144

⇒ (16m2 − 9)x 2 ± 32m( 9m2 − 16) x + 144m2 − 256 − 144 = 0

⇒ (16m2 − 9)x 2 ± 32m( 9m2 − 16) x + (144m2 − 400) = 0 .....(iv)

If (iii) is a tangent to (ii), then the roots of (iv) are real and equal.
∴ Discriminant = 0; 32 × 32m2(9m2 − 16) = 4(16m2 − 9)(144m2 − 400) = 64 (16m2 − 9)(9m2 − 25)

⇒ 16m2(9m2 − 16) = (16m2 − 9)(9m2 − 25) ⇒ 144m4 − 256m2 = 144m4 − 481m2 + 225

⇒ 225m2 = 225 ⇒ m2 = 1 ⇒ m = ±1

Equation of Pair of Tangents.

If P(x1, y1) be any point outside the hyperbola x2 y2 =1 then a pair of tangents PQ, PR can be drawn to it from P.
a2 − b2
Y
The equation of pair of tangents PQ and PR is SS1 = T 2 T (h,k)Q

where, S = x2 y2 −1, S1 = x12 − y12 −1, T = xx1 − yy1 −1 X′ A′ CP A X
a2 − b2 a2 b2 a2 b2 (x1,y1)
R

Y′

Director circle : The director circle is the locus of points from which perpendicular tangents are drawn to

the given hyperbola. The equation of the director circle of the hyperbola x2 y2 =1 is x2 + y2 = a2 − b2
a2 − b2
Y
P (h, k)

90o

X′ A′ C AX

Y′

Example: 18 The locus of the point of intersection of tangents to the hyperbola 4 x2 − 9y2 = 36 which meet at a constant angle π / 4 , is
Solution: (a)
(a) (x2 + y2 − 5)2 = 4(9y2 − 4 x2 + 36) (b) (x 2 + y2 − 5) = 4(9y2 − 4 x 2 + 36)

(c) 4 (x 2 + y2 − 5)2 = (9y2 − 4 x 2 + 36) (d) None of these

Let the point of intersection of tangents be P(x1, y1) . Then the equation of pair of tangents from P(x1, y1) to the given

hyperbola is (4 x 2 − 9y2 − 36)(4 x12 − 9y12 − 36) = [4 x1x − 9y1y − 36]2 ......(i)

From SS1 = T 2 or x 2(y12 + 4) + 2x1y1xy + y2(x12 − 9) + ..... = 0 .....(ii)

Since angle between the tangents is π / 4 .

∴ tan(π / 4) = 2 [x12y12 − (y12 + 4)(x12 − 9)] . Hence locus of P(x1, y1) is (x 2 + y2 − 5)2 = 4(9y2 − 4 x 2 + 36) .
y12 + 4 + x12 − 9

Equations of Normal in Different forms .

(1) Point form : The equation of normal to the hyperbola x2 y2 =1 at (x1,y1) is a2x b2y = a2 + b2 .
a2 − b2 x1 + y1

(2) Parametric form: The equation of normal at (a secθ ,b tanθ ) to the hyperbola x2 y2 =1 is
a2 − b2

ax cosθ + by cotθ = a2 + b2 Y Tangent
P(x1,y1) Normal
(3) Slope form: The equation of the normal to the hyperbola
C AX
x2 y2 = 1 in terms of the slope m of the normal is y = mx  m(a 2 + b2 ) X′ A′
a2 − b2 a2 − b2m2 Y′

(4) Condition for normality : If y = mx + c is the normal of x2 y2 =1
a2 − b2

then c =  m(a2 + b2) or c2 = m2(a 2 + b 2 )2 , which is condition of normality.
a2 − m2b2 (a 2 − m2b2 )

(5) Points of contact : Co-ordinates of points of contact are  ± a2 , mb2 
 a2 − b2m2 a 2 − b2m2 

Note :  If the line lx + my + n = 0 will be normal to the hyperbola x2 y2 = 1 ,then a2 b2 = (a 2 + b2)2 .
a2 − b2 l2 − m2 n2

Important Tip

• In general, four normals can be drawn to a hyperbola from any point and if α, β,γ ,δ be the eccentric angles of these four co-normal

points, then α + β + γ + δ is an odd multiple of π .

• If α, β,γ are the eccentric angles of three points on the hyperbola. x2 y2 =1, the normals at which are concurrent, then,
a2 − b2

sin(α + β ) + sin(β + γ ) + sin(γ + α) = 0

• If the normal at P meets the transverse axis in G, then SG = e.SP . Also the tangent and normal bisect the angle between the focal

distances of P.

• The feet of the normals to x2 y2 =1 from (h,k) lie on a2y(x − h) + b2x(y − k) = 0 .
a2 − b2

Example: 19 The equation of the normal to the hyperbola x2 − y2 =1 at the point (8, 3 3) is [MP PET 1996]
Solution: (d) 16 9

Example: 20 (a) 3x + 2y = 25 (b) x + y = 25 (c) y + 2x = 25 (d) 2x + 3y = 25
Solution: (a)
From a2x + b2y = a2 + b2
x1 y1

Here a2 = 16 , b2 = 9 and (x1, y1) = (8, 3 3)

⇒ 16x 9y = 16 + 9 i.e., 2x + 3y = 25 .
+
8 33

If the normal at 'φ' on the hyperbola x2 y2 = 1 meets transverse axis at G, then AG.A′G =
a2 − b2

(Where A and A′ are the vertices of the hyperbola)

(a) a2(e4 sec 2 φ − 1) (b) (a2e4 sec 2 φ − 1) (c) a2(1 − e4 sec 2 φ) (d) None of these

The equation of normal at (a sec φ,b tanφ) to the given hyperbola is ax cosφ + by cotφ = (a2 + b2)

This meets the transverse axis i.e., x-axis at G. So the co-ordinates of G are   a2 + b2  sec φ, 0  and the co-ordinates of
a 

the vertices A and A′ are A(a,0) and A′(−a,0) respectively.

∴ AG.A′G =  − a +  a2 + b2  sec φ   a +  a2 + b2  sec φ  =  a2 + b2 2 sec 2 φ − a2 = (ae2)2 sec 2 φ − a2 = a 2 (e 4 sec 2 φ − 1)
 a   a  a

Example: 21 The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axis at L and M respectively, then
Solution: (a) the locus of the middle point of LM is a hyperbola whose eccentricity is

(a) e (b) e (c) e (d) None of these
e2 −1 e4 −1 a2e2 − 1

The equation of the normal at P(a sec φ,b tanφ) to the hyperbola is ax cosφ + by cotφ = a2 + b2 = a2e2

It meets the transverse and conjugate axes at L and M, then L(ae2 sec φ,0) ; M  0, a2e2 tanφ 
b

Let the middle point of LM is (α, β ) ; then α = ae2 sec φ ⇒ sec φ = 2α .....(i)
2 ae 2

and β = a2e2 tanφ ⇒ tanφ = 2bβ ......(ii)
2b a 2e 2

 1 = sec 2 φ − tan2 φ ; 1= 4α 2 − 4b2β 2 , ∴ Locus of (α, β ) is x2 − y2 =1
a2e4 a4e4
 a 2e 4   a4e4 
4 4b2

 a 2e 4 a4e4 
4 + 4b 2
It is a hyperbola, let its eccentricity e1 = = 1 + a2 = a2 + b2 = a 2e 2 ; ∴ e1 = e.
 a2e4  b2 b2 a 2 (e 2 − 1) e2 −1
4

Equation of Chord of Contact of Tangents drawn from a Point to a Hyperbola .

Let PQ and PR be tangents to the hyperbola x2 y2 =1 drawn from any external point P (x1, y1) .
a2 − b2
Y
Then equation of chord of contact QR is
Q
or xx 1 yy1 =1
a2 − b2 X′ A′ CP A X
(x1,y1) R
or T = 0 (At x1 , y1 )

Y′

Equation of the Chord of the Hyperbola whose Mid point (x1, y1) is given h

Equation of the chord of the hyperbola x2 y2 = 1, bisected at the given Y
a2 − b2 C
Q(x2,y2)
xx1 yy1 x 2 y12 Y′
point (x1, y1 ) is a2 − b2 −1 = 1 − b2 −1 P
X′ (x1,y1) X
a2 A
R (x3,y3)
i.e., T = S1

Note :  The length of chord cut off by hyperbola x2 y2 =1 from the line y = mx + c is
a2 − b2

2ab [c 2 − (a 2m2 − b 2 )](1 + m2 )
(b 2 − a 2m2 )

Equation of the Chord joining Two points on the Hyperbola .

The equation of the chord joining the points P(a secφ1, b tanφ1) and Q(a secφ2,b tanφ2 ) is

y − b tan φ1 = b tan φ2 − b tan φ1 (x − a sec φ1)
a sec φ2 − a sec φ1

x cos φ1 −φ2  − y sin φ1 +φ2  = cos φ1 +φ2 
a  2  b  2   2 

Note :  If the chord joining two points (a secθ1,b tanθ1) and (a secθ 2,b tanθ 2) passes through the focus

of the hyperbola x2 y2 = 1 , then tan θ1 tan θ2 = 1− e .
a2 − b2 2 2 1+ e

Example: 22 The equation of the chord of contact of tangents drawn from a point (2, –1) to the hyperbola 16x 2 − 9y2 = 144 is

(a) 32x + 9y = 144 (b) 32x + 9y = 55 (c) 32x + 9y + 144 = 0 (d) 32x + 9y + 55 = 0

Solution: (a) From T = 0 i.e., xx1 − yy1 = 1 . Here, 16x 2 − 9y2 = 144 i.e., x2 − y2 =1
Example: 23 a2 b2 9 16
Solution: (a)
So, the equation of chord of contact of tangents drawn from a point (2, –1) to the hyperbola is 2x − (−1) y =1
Example: 24 9 16
Solution: (b)
Example: 25 i.e., 32x + 9y = 144
Solution: (a)
The point of intersection of tangents drawn to the hyperbola x2 y2 = 1 at the points where it is intersected by the line
a2 − b2

lx + my + n = 0 is

(a)  − a 2l , b 2m  (b)  a 2l , − b2m  (c)  − a 2n , b2n  (d)  a 2n , − b2n 
n n n n l m l m

Let (x1, y1) be the required point. Then the equation of the chord of contact of tangents drawn from (x1, y1) to the given

hyperbola is xx1 − yy1 =1 ......(i)
a2 b2

The given line is lx + my + n = 0 .....(ii)

Equation (i) and (ii) represent the same line

∴ x1 = − y1 = 1 ⇒ x1 = − a2l , y1 = b2m ; Hence the required point is  − a 2l , b2m  .
a 2l b2m −h n n n n

What will be equation of that chord of hyperbola 25x 2 − 16y 2 = 400 , whose mid point is (5, 3) [UPSEAT 1999]

(a) 115x − 117y = 17 (b) 125x − 48y = 481 (c) 127x + 33y = 341 (d) 15x + 121y = 105

According to question, S = 25x 2 − 16y 2 − 400 = 0 ......(i)
Equation of required chord is S1 = T

Here S1 = 25(5)2 − 16(3)2 − 400 = 625 − 144 − 400 = 81 and T = 25xx1 − 16yy1 − 400 , where x1 = 5 , y1 = 3
⇒ 25x(5) − 16y(3) − 400 = 125x − 48y − 400

So, from (i) required chord is 125x − 48y − 400 = 81 ⇒ 125x − 48y = 481 .

The locus of the mid-points of the chords of the circle x 2 + y 2 = 16 which are tangent to the hyperbola

9x 2 − 16y 2 = 144 is [Roorkee 1997]

(a) (x 2 + y 2 )2 = 16x 2 − 9y 2 (b) (x 2 + y 2 )2 = 9x 2 − 16y 2

(c) (x 2 − y 2 )2 = 16x 2 − 9y 2 (d) None of these

The given hyperbola is x2 y2 =1 ……(i)
16 −9

Any tangent to (i) is y = mx + 16m2 − 9 ……(ii)
……(iii)
Let (x1, y1 ) be the mid point of the chord of the circle x 2 + y 2 = 16
Then equation of the chord is T = S1 i.e., xx1 + yy1 − (x12 + y12 ) = 0
Since (ii) and (iii) represent the same line.

∴ m = −1 = 16m2 − 9
x1 y1 − (x12 + y12 )

⇒ m = − x1 and (x 2 + y12 )2 = y12 (16m2 − 9) ⇒ (x12 + y12 )2 = 16 . x12 y12 − 9y12 = 16 x 2 − 9y12
y1 1 y12 1

∴ Locus of (x1, y1 ) is (x 2 + y 2 )2 = 16x 2 − 9y 2 .

Pole and Polar g

Let P be any point inside or outside the hyperbola. If any straight line drawn through P interesects the
hyperbola at A and B. Then the locus of the point of intersection of the tangents to the hyperbola at A and B is
called the polar of the given point P with respect to the hyperbola and the point P is called the pole of the polar.

The equation of the required polar with (x1, y1) as its pole is

xx1 yy1 A
a2 b2
− =1 (h,k) Q A′ Pole
Polar P(x1,y1) X
X′
Q′ B
Note :  Polar of the focus is the directrix.
B′
 Any tangent is the polar of its point of contact.

(1) Pole of a given line : The pole of a given line lx + my + n = 0 with respect to the hyperbola

x2 y2 =1 is (x1, y1) =  − a 2l , b2m  Q (h, k) B′
a2 − b2 n n
A′

Polar
X
X′ Pole
P
(x1,y1)
A B
(2) Properties of pole and polar Q ′

(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 lx + my + n = 0 lies on the another line l′x + m′y + n′ = 0 then the pole of the second
line will lie on the first and such lines are said to be conjugate lines.

(iii) Pole of a given line is same as point of intersection of tangents as its extremities.

Important Tips

• If the polars of (x1, y1) and (x2, y2) with respect to the hyperbola x2 y2 = 1 are at right angles, then x1 x 2 a4 =0
a2 − b2 y1 y 2 + b4

Example: 26 If the polar of a point w.r.t. x2 y2 = 1 touches the hyperbola x2 y2 = 1 , then the locus of the point is [Pb. CET 1999]
a2 + b2 a2 − b2

(a) Given hyperbola (b) Ellipse

(c) Circle (d) None of these

Solution: (a) Let (x1, y1 ) be the given point.
Example: 27
Its polar w.r.t. x2 + y2 =1 is xx1 + yy1 =1 i.e., y= b2 1 − xx1  = − b2 x1 x+ b2
a2 b2 a2 b2 y1 a2 a 2y1 y1

This touches x2 y2 =1 if  b2  2 = a2 .  b2 x1  − b2 ⇒ b4 = a2b4 x12 − b2 ⇒ b2 = b 2 x12 −1 ⇒ x12 − y12 =1
a2 − b2 y1  a 2y1  y12 a 4 y12 y12 a 2 y12 a2 b2

∴ Locus of (x1, y1 ) is x2 y2 = 1 . Which is the same hyperbola.
a2 − b2

The locus of the poles of the chords of the hyperbola x2 y2 = 1 , which subtend a right angle at the centre is
a2 − b2

(a) x2 y2 1 1 (b) x2 y2 1 1 (c) x2 y2 1 1 (d) x2 y2 1 1
a4 + b4 = a2 − b2 a2 − b2 = a2 − b2 a4 − b4 = a2 + b2 a4 − b4 = a2 − b2

Solution: (a) Let (x1, y1 ) be the pole w.r.t. x2 y2 =1 ......(i)
a2 − b2

Then equation of polar is hx ky =1 .....(ii)
a2 − b2

The equation of lines joining the origin to the points of intersection of (i) and (ii) is obtained by making homogeneous (i)

with the help of (ii), then  x 2 − y2  =  hx − ky  2 ⇒ x 2  1 − h2  − y 2  1 k2  + 2hk xy = 0
a 2 b2  a2 b2  a2 a4 b2 + b4 a 2b 2

Since the lines are perpendicular, then coefficient of x 2 + coefficient of y 2 = 0

1 h2 1 k2 =0 or h2 k2 1 1 . Hence required locus is x2 y2 1 1
a2 − a4 − b2 − b4 a4 + b4 = a2 − b2 a4 + b4 = a2 − b2

iameter of the Hyperbola .

The locus of the middle points of a system of parallel chords of a hyperbola is called a diameter and the point

where the diameter intersects the hyperbola is called the vertex of the diameter. Y (x1,y1)
P
x2 y2
Let y = mx + c a system of parallel chords to a2 − b2 =1 for different

b2x X′ CX
a 2m
chords then the equation of diameter of the hyperbola is y = , which is Q R(h,k)
Y′ (x2,y2)
passing through (0, 0)

Conjugate diameter : Two diameters are said to be conjugate when each bisects all chords parallel to the others.

If y = m1x , y = m2 x be conjugate diameters, then m1m2 = b2 .
a2

Note :  If a pair of diameters be conjugate with respect to a hyperbola, they are conjugate with respect to its

conjugate hyperbola also.

 In a pair of conjugate diameters of a hyperbola. Only one meets the curve in real points.

 The condition for the lines AX 2 + 2HXY + BY 2 = 0 to be conjugate diameters of x2 y2 = 1 is
a2 − b2

a2A = b2B.

Important Tips

 If CD is the conjugate diameter of a diameter CP of the hyperbola x2 y2 = 1 , where P is (a sec φ, b tan φ) then coordinates of D is
a2 − b2

(a tan φ, b sec φ) , where C is (0, 0).

Example: 28 If a pair of conjugate diameters meet the hyperbola and its conjugate in P and D respectively, then CP 2 − CD2 =
Solution: (b)
Example: 29 (a) a 2 + b 2 (b) a 2 − b 2 (c) a2 (d) None of these
b2

Coordinates of P and D are (a sec φ, b tan φ) and (a tan φ, b sec φ) respectively.

Then (CP)2 − (CD)2 = a 2 sec 2 φ + b 2 tan 2 φ − a 2 tan 2 φ − b 2 sec 2 φ

= a 2 (sec 2 φ − tan 2 φ) − b 2 (sec 2 φ − tan 2 φ) = a 2 (1) − b 2 (1) = a 2 − b 2 .

If the line lx + my + n = 0 passes through the extremities of a pair of conjugate diameters of the hyperbola x2 y2 = 1 then
a2 − b2

(a) a 2l 2 − b 2m2 = 0 (b) a 2l 2 + b 2m2 = 0 (c) a 2l 2 + b 2m2 = n2 (d) None of these

Solution: (a) The extremities of a pair of conjugate diameters of x2 y2 =1 are (a sec φ, b tan φ) and (a tan φ, b sec φ) respectively.
a2 − b2

According to the question, since extremities of a pair of conjugate diameters lie on lx + my + n = 0

∴ l(a sec φ) + m(b tan φ) + n = 0 ⇒ l (a tanφ) + m(b sec φ) + n = 0 ……(i)

Then from (i), al sec φ + bm tan φ = −n or a 2l 2 sec 2 φ + b 2m2 tan 2 φ + 2ablm sec φ tan φ = n2 ……(ii)

And from (ii), al tan φ + bm sec φ = −n or a 2l 2 tan 2 φ + b 2m2 sec 2 φ + 2ablm sec φ tan φ = n2 ……(iii)

Then subtracting (ii) from (iii)

∴ a 2l 2 (sec 2 φ − tan 2 φ) + b 2m2 (tan 2 φ − sec 2 φ) = 0 or a 2l 2 − b 2m2 = 0 .

Subtangent and Subnormal of the Hyperbolaa Y
Let the tangent and normal at P(x1, y1 ) meet the x-axis at A and B respectively. (x1, y1)
P
Length of subtangent AN = CN − CA = x1 − a2
x1
X′ N BX
(a 2 + b 2 ) = b2 CA
Length of subnormal BN = CB − CN a2 x1 x1 a2 = (e 2
= − x1 − 1)x1

Y′

Reflection property of the Hyperbola

If an incoming light ray passing through one focus (S) strike convex side of the hyperbola then it will get

reflected towards other focus (S′) Y M Light ray
∠TPS′ = ∠LPM = α αL
Reflected ray
α P
Normal

X′ (–ae,0)S′ A′ C T A S (ae,0) N X

Tangent

Y′

Example: 30 A ray emanating from the point (5, 0) is incident on the hyperbola 9x2 − 16y2 = 144 at the point P with abscissa 8; then
Solution: (a) the equation of reflected ray after first reflection is (Point P lies in first quadrant)

(a) 3 3x − 13y + 15 3 = 0 (b) 3x − 13y + 15 = 0 (c) 3 3x + 13y − 15 3 = 0 (d) None of these

Given hyperbola is 9x 2 − 26y 2 = 144 . This equation can be rewritten as x2 y2 =1 .....(i)
16 −9

Since x coordinate of P is 8. Let y-coordinate of P is α Y M light ray
∴ (8,α) lies on (i)
Reflected ray αL
64 α2 =1; ( P lies in first quadrant) Tangent Normal
16 9
∴ − ∴ α = 27 X′ S′(–ae,0) A′ C T A S(ae,0) N X

α=3 3

Hence coordinate of point P is (8, 3 3) Y′

 Equation of reflected ray passing through P(8, 3 3) and S′(−5, 0 ); ∴ Its equation is y−3 3 = 0−3 3 (x − 8)
−5− 8

or 13y − 39 3 = 3 3x − 24 3 or 3 3 x − 13y + 15 3 = 0

Asymptotes of a Hyperbola
An asymptote to a curve is a straight line, at a finite distance from the origin, to which the tangent to a curve

tends as the point of contact goes to infinity.

The equations of two asymptotes of the hyperbola x2 y2 =1 are y = ± b x or x ± y = 0.
a2 − b2 a a b

Note :  The combined equation of the asymptotes of the hyperbola x2 y2 =1 is x2 y2 = 0.
a2 − b2 a2 − b2

 When b = a i.e. the asymptotes of rectangular hyperbola x2 − y2 = a2 are y = ±x , which are at right angles.

 A hyperbola and its conjugate hyperbola have the same asymptotes.

 The equation of the pair of asymptotes differ the hyperbola and Y
the conjugate hyperbola by the same constant only i.e. Hyperbola
– Asymptotes = Asymptotes – Conjugated hyperbola or, Asymptotes
B

 x2 − y2 − 1 −  x2 − y2  =  x2 − y2  −  x2 − y2 + 1 . X′ A′ C AX
a2 b2 a2 b2 a2 b2 a2 b2
B′ x2 y2 =1
a2 − b2

 The asymptotes pass through the centre of the hyperbola. Y′ x2 y2
a2 b2
 The bisectors of the angles between the asymptotes are the − = −1
coordinate axes.

 The angle between the asymptotes of the hyperbola S=0 i.e., x2 y2 =1 is 2 tan−1 b or 2 sec−1 e .
a2 − b2 a

 Asymptotes are equally inclined to the axes of the hyperbola.

Important Tips

 The parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes

and is of constant area. Y

Area of parallelogram QRQ′R′ = 4(Area of parallelogram QDCP) = 4ab = Constant DQ
P
 The product of length of perpendiculars drawn from any point on the hyperbola x2 y2 =1 to the X′ RM X
a2 − b2 C
P′ R′
asymptotes is a 2b 2 . Q′ D′
a2 + b2 Y′

Example: 31 From any point on the hyperbola, x2 y2 =1 tangents are drawn to the hyperbola x2 − y2 =2. The area cut-off by
Solution: (d) a2 − b2 a2 b2

the chord of contact on the asymptotes is equal to

(a) ab (b) ab (c) 2ab (d) 4ab
2

Let P(x1, y1) be a point on the hyperbola x2 − y2 = 1 , then x12 − y12 =1
a2 b2 a2 b2

The chord of contact of tangent from P to the hyperbola x2 y2 =2 is xx1 − yy1 =2 .....(i)
a2 − b2 a2 b2

The equation of asymptotes are x − y =0 ......(ii)
a b ......(iii)

And xy =0
a +b

The point of intersection of the asymptotes and chord are  x1 2a /b, x1 2b / b ; x1 2a /b, x1 − 2b / b  , (0, 0)
/ a − y1 / a − y1 / a + y1 / a + y1

∴ Area of triangle = 1 | (x1y2 − x2y1)| = 1  x12 / − 8a b / b2  = 4ab .
2 2 a2 − y12

Example: 32 The combined equation of the asymptotes of the hyperbola 2x 2 + 5xy + 2y2 + 4 x + 5y = 0 [Karnataka CET 2002]
Solution: (d)
(a) 2x 2 + 5xy + 2y2 = 0 (b) 2x 2 + 5xy + 2y2 − 4 x + 5y + 2 = 0 = 0

(c) 2x 2 + 5xy + 2y2 + 4 x + 5y − 2 = 0 (d) 2x 2 + 5xy + 2y2 + 4 x + 5y + 2 = 0

Given, equation of hyperbola 2x 2 + 5xy + 2y2 + 4 x + 5y = 0 and equation of asymptotes

2x 2 + 5xy + 2y2 + 4 x + 5y + λ = 0 ......(i) which is the equation of a pair of straight lines. We know that the standard
equation of a pair of straight lines is ax 2 + 2hxy + by2 + 2gx + 2 fy + c = 0

Comparing equation (i) with standard equation, we get a = 2, b = 2 , h= 5 , g = 2, f = 5 and c =λ .
2 2

We also know that the condition for a pair of straight lines is abc + 2 fgh − af 2 − bg 2 − ch2 = 0 .

Therefore, 4λ + 25 − 25 − 8 − 25 λ = 0 or −9λ + 9 =0 or λ=2
2 4 4 2

Substituting value of λ in equation (i), we get 2x 2 + 5xy + 2y2 + 4 x + 5y + 2 = 0 .

Rectangular or Equilateral Hyperbola
(1) Definition : A hyperbola whose asymptotes are at right angles to each other is called a rectangular

hyperbola. The eccentricity of rectangular hyperbola is always 2 .

The general equation of second degree represents a rectangular hyperbola if ∆ ≠ 0, h2 > ab
and coefficient of x 2 + coefficient of y 2 = 0

The equation of the asymptotes of the hyperbola x2 y2 =1 are given by y = ± b x .
a2 − b2 a

The angle between these two asymptotes is given by tanθ = b −  − b  = 2b / a = 2ab .
a  a  1 − b2 / a2 a2 − b2

1 + b  −b 
a  a 

If the asymptotes are at right angles, then θ = π / 2 ⇒ tan θ = tan π ⇒ 2ab = tan π ⇒ a2 − b2 =0
2 a2 − b2 2

⇒ a = b ⇒ 2a = 2b . Thus the transverse and conjugate axis of a rectangular hyperbola are equal and the equation

is x2 − y2 = a2 . The equations of the asymptotes of the rectangular hyperbola are y = ±x i.e., y = x and y = −x .

Clearly, each of these two asymptotes is inclined at 45 to the transverse axis.

(2) Equation of the rectangular hyperbola referred to its asymptotes as the axes of coordinates :
Referred to the transverse and conjugate axis as the axes of coordinates, the equation of the rectangular hyperbola is

x2 − y2 = a2 …..(i)

The asymptotes of (i) are y = x and y = – x. Each of these two asymptotes is inclined at an angle of 45 with

the transverse axis, So, if we rotate the coordinate axes through an angle of − π / 4 keeping the origin fixed, then

the axes coincide with the asymptotes of the hyperbola and x = X cos(−π / 4) − Y sin(−π / 4) = X +Y and
2

y = X sin(− π / 4)+ Y cos(−π / 4) = Y −X . Y
2 XY=c2

Substituting the values of x and y in (i),

We obtain the  X +Y 2 −  Y −X 2 = a2 ⇒ XY = a2 ⇒ XY = c2 X′ O X
2 2 2

where c2 = a2 . Y′
2

This is transformed equation of the rectangular hyperbola (i).

(3) Parametric co-ordinates of a point on the hyperbola XY = c2 : If t is non–zero variable, the
coordinates of any point on the rectangular hyperbola xy = c2 can be written as (ct,c / t) . The point (ct,c / t) on the

hyperbola xy = c2 is generally referred as the point ‘t’.

For rectangular hyperbola the coordinates of foci are (±a 2, 0) and directrices are x = ±a 2 .

For rectangular hyperbola xy = c 2 , the coordinates of foci are (±c 2, ± c 2) and directrices are

x + y = ±c 2 .

(4) Equation of the chord joining points t1 and t2 : The equation of the chord joining two points
cc
−
 ct1, c  and  ct2 , c  on the hyperbola xy = c2 is y − c = t2 − t1 (x − ct1) ⇒ x + y t1t2 = c (t1 + t2) .
t1 t2 t1 ct2 ct1

(5) Equation of tangent in different forms

(i) Point form : The equation of tangent at (x1, y1) to the hyperbola xy = c2 is xy1 + yx1 = 2c2 or x y =2
x1 +y

1

(ii) Parametric form : The equation of the tangent at  ct, c  to the hyperbola xy = c2 is x + yt = 2c .On
 t  t

replacing x1 by ct and y1 by c on the equation of the tangent at (x1, y1) i.e. xy1 + yx1 = 2c2 we get x + yt = 2c .
t t

Note :  Point of intersection of tangents at ' t1' and ' t2' is  2ct1 t 2 , t1 2c 
t1 + t 2 +t
2

(6) Equation of the normal in different forms : (i) Point form : The equation of the normal at (x1, y1) to the

hyperbola xy = c2 is xx1 − yy1 = x12 − y12 . As discussed in the equation of the tangent, we have  dy  = − y1
 dx ( x1
x1 , y1)

So, the equation of the normal at (x1, y1) is y − y1 = −1 (x − x1) ⇒ y − y1 = x1 (x − x1)
y1
 dy 
 dx ( x1 ,y1 )

⇒ yy1 − y12 = xx1 − x12 ⇒ xx1 − yy1 = x12 − y12
This is the required equation of the normal at (x1, y1) .

(ii) Parametric form: The equation of the normal at  ct, c  to the hyperbola xy = c2 is
 t 

xt3 − yt − ct4 + c = 0 . On replacing x1 by ct and y1 by c / t in the equation.

We obtain xx1 − yy1 = x12 − y12, xct − yc = c 2t 2 − c2 ⇒ xt 3 − yt − ct4 +c = 0
t t2

Note :  The equation of the normal at  ct, c  is a fourth degree in t. So, in general, four normals can be
 t 

drawn from a point to the hyperbola xy = c2

 If the normal at  ct, c  on the curve xy = c2 meets the curve again in ' t ′ ' then; t′ = −1 .
 t  t3

 Point of intersection of normals at ' t1' and 't2' is  c {t1t 2 (t12 + t1t2 + t 2 ) − 1} , c {t13 t 3 + (t12 + t1t 2 + t 2 )} 
t1t 2 (t1 + t2 2 2 2 (t1 + t 2 ) 2

) t1t

Important Tips

• A triangle has its vertices on a rectangular hyperbola; then the orthocentre of the triangle also lies on the same hyperbola.
• All conics passing through the intersection of two rectangular hyperbolas are themselves rectangular hyperbolas.

• An infinite number of triangles can be inscribed in the rectangular hyperbola xy = c2 whose all sides touch the parabola y2 = 4ax .

Example: 33 If 5x 2 + λy 2 = 20 represents a rectangular hyperbola, then λ equals
Solution: (c)
Example: 34 (a) 5 (b) 4 (c) – 5 (d) None of these
Solution: (d)
Since the general equation of second degree represents a rectagular hyperbola if ∆ ≠ 0, h2 > ab and coefficient of
Example: 35
Solution: (a) x 2 + coefficient of y 2 = 0 . Therefore the given equation represents a rectangular hyperbola if λ + 5 = 0 i.e., λ = −5

If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the mid-point of PN is

(a) Circle (b) Parabola (c) Ellipse (d) Hyperbola

Let xy = c 2 be the rectangular hyperbola, and let P(x1, y1) be a point on it. Let Q (h, k) be the mid-point of PN. Then the

coordinates of Q are  x1 , y1  . Y
2 xy=c2

∴ x1 =h and y1 =k⇒ x1 =h and y1 = 2k Q (h, k) P(x1, y1)
2
O
But (x1, y1 ) lies on xy = c2. X′ N X

∴ h.(2k) = c 2 ⇒ hk ⇒ c 2 / 2

Hence, the locus of (h, k) is xy = c 2 / 2 , which is a hyperbola.

If the normal at  ct, c  on the curve xy = c 2 meets the curve again in t′, then Y′
 t 

(a) t′ = − 1 (b) t′ = − 1 (c) t′ = 1 (d) t′2 = − 1
t3 t t2 t2

The equation of the tangent at  ct, c  is ty = t 3 x − c t 4 +c
 t 

If it passes through  c t ′, c  then
 t′ 

⇒ tc = t 3ct′ − ct 4 +c ⇒ t = t3t′2 − t4t′ + t′ ⇒ t − t′ = t 3t′(t′ − t) ⇒ t′ = − 1
t′ t3

Example: 36 If the tangent and normal to a rectangular hyperbola cut off intercepts a1 and a2 on one axis and b1 and b2 on the
Solution: (c) other axis, then

Example: 37 (a) a1b1 + a2b2 = 0 (b) a1b2 + b2a1 = 0 (c) a1a2 + b1b2 = 0 (d) None of these
Solution: (a)
Let the hyperbola be xy = c 2 . Tangent at any point t is x + yt 2 − 2ct = 0
Example: 38
Solution: (b) Putting y=0 and then x=0 intercepts on the axes are a1 = 2ct and b1 = 2c
t

Normal is xt 3 − yt − ct 4 + c = 0 .

Intercepts as above are a2 = c (t 4 − 1) , b2 = − c(t 4 − 1)
t
t 3

∴ a1a2 + b1b2 = 2ct × c (t 4 − 1) + 2c × − c(t 4 − 1) = 2c 2 (t 4 − 1) − 2c 2 (t 4 − 1) = 0 ; ∴ a1a2 + b1b2 = 0 .
t3 t t t2 t2

A variable straight line of slope 4 intersects the hyperbola xy = 1 at two points. The locus of the point which divides the

line segment between these two points in the ratio 1 : 2 is [IIT 1997]

(a) 16x 2 + 10xy + y 2 = 2 (b) 16x 2 − 10xy + y 2 = 2 (c) 16x 2 + 10xy + y 2 = 4 (d) None of these

Let P(h, k) be any point on the locus. Equation of the line through P and having slope 4 is y − k = 4(x − h) .....(i)

Suppose this meets xy = 1 ......(ii) in A(x1, y1 ) and B(x 2 , y2 )

Eliminating y between (i) and (ii), we get 1 − k = 4(x − h)
x

⇒ 1 − xk = 4 x 2 − 4hx ⇒ 4 x 2 − (4h − k) x − 1 = 0 ......(iii)

This has two roots say x1, x2 ; x1 + x2 = 4h − k ......(iv) and x1 x 2 = − 1 ......(v)
4 4

Also, 2x1 + x2 =h [ P divides AB in the ratio 1 : 2]
3

i.e., 2x1 + x 2 = 3h ......(vi)

(vi) – (iv) gives, x1 = 3h − 4h − k = 8h + k and x2 = 3h − 2 . 8h + k = − 2h + k
4 4 4 2

Putting in (v), we get 8h + k  − 2h + k  = − 1
4  2  4

⇒ (8h + k)(2h + k) = 2 ⇒ 16h2 + 10hk + k 2 = 2

∴ Required locus of P(h, k) is 16x 2 + 10xy + y 2 = 2 .

PQ and RS are two perpendicular chords of the rectangular hyperbola xy = c 2 . If C is the centre of the rectangular

hyperbola, then the product of the slopes of CP, CQ, CR and CS is equal to

(a) – 1 (b) 1 (c) 0 (d) None of these

Let t1, t2, t3, t4 be the parameters of the points P, Q, R and S respectively. Then, the coordinates of P, Q, R and S are

 ct1, c  ,  ct 2 , c  ,  ct3 , c  and  ct4 , c  respectively.
t1 t2 t3 t4

Now, PQ⊥RS ⇒ c − c × c − c = −1 ⇒ 11 = −1 ⇒ t1t2t3t4 = −1 .....(i)
− − − t1t2 × − t3t4
t2 t1 t4 t3
ct2 ct1 ct4 ct3

∴ Product of the slopes of CP,CQ,CR and CS

1 × 1 × 1 × 1 = 1 =1 [Using (i)]
t12
t 2 t 2 t 2 t12 t 2 t 2 t 2
2 3 4 2 3 4