Coordinate Geometry

5.46 Coordinate Geometry Booster

So, OL = a2 - b2 Let q be the angle between the tangents
Then tan (q ) = m2 - m1
a2sec2q + b2cosec2q
1 + m1m2
= (a2 - b2 )sin q cos q
a2sin2q + b2cos2q = (m2 + m1)2 - 4m1m2
1 + m1m2

Now, NP = OL = 64 + 56 = 2 30 ¥ 9 = 6 6
fi NP = (a2 - b2 )sin q cos q 99 355
1 — 14
a2sin2q + b2cos2q 9

Therefore, Thus, q = tan -1 Ê 6 6ˆ
ar DOPN ÁË 5 ¯˜

= 1 ¥ ON ¥ NP 9. Normal at P(a cos q, b sin q) is
2 ax sec q – by cosec q = a2 – b2

= 1 ¥ ab(a2 - b2 )sin q cos q fi ax sec q – by cosec q = 14 – 5 = 9
2 a2sin2q + b2cos2q
which meets the ellipse again at Q. So
= 1 ¥ ab(a2 - b2 ) a2 cos 2q sec q – b2 sin 2q cosec q = 9
2 a2tan q + b2cot
q fi 14 cos 2q sec q – 5 sin 2q cosec q = 9
fi 28 cos2 q – 14 – 10 cos2 q = 9 cos q
£ 1 ¥ ab(a2 - b2 ) = (a2 - b2 ) fi 18 cos2 q – 9 cos q – 14 = 0
2 2ab 4

at tan q = b . fi (6 cos q – 7)(3 cos q + 2) = 0
a
fi cos q = - 2 , 7
Thus, the point P is 36

Ê a2 , b2 ˆ Thus, cos q = - 2
Á a2 + b2 ˜ 3
Ë a2 + b2 ¯
10. Given ellipse is x2 + y2 = 1
By symmetry, we have four such points. 25 16 …(i)

Ê a2 , ± b2 ˆ The equation of the chord bisected at (h, k) is
Thus, Á ± a2 + b2
a2 + b2 ˜ T = S1
Ë ¯
h2 k2
8. Given ellipse is x2 + y2 = 1 fi hx + ky = a2 + b2
(5/3) (5/2) a2 b2

The equation of any tangent to the given ellipse is fi hx + ky = h2 + k 2
25 16 25 16
y = mx + b2m2 + a2
fi (1/5) x + (2/5) y = (1/5)2 + (2/5)2
fi y = mx + 5 m2 + 5 25 16 25 16
23
Solving, we get …(ii)
which is passing through (1, 2). 4x + 5y = 4

So, 2 = m + 5 m2 + 5 Solving Eqs (i) and (ii), we get
23 x = 4, - 3 and y = - 12 , 16
55
fi (2 - m)2 = Ê 5 m2 + 5ˆ
ËÁ 2 3˜¯ Let the chord be AB, where

fi 6m2 – 24m + 24 = 15m2 + 10 A = ÊËÁ 4, - 12 ¯˜ˆ and B = ËÁÊ - 3, 16 ¯˜ˆ
fi 9m2 + 24m – 14 = 0 5 5

Let its roots are m1 and m2 Hence, the length the AB is

So, m1 + m2 = - 8 , m1m2 = - 14 (7)2 Ê 28ˆ 2 1 + 16 = 7 41
3 9 ÁË 5 ˜¯ 25 5
= + = 7

Ellipse 5.47

11. Any tangent to the ellipse is Ê b2 ˆ Ê 5ˆ
ËÁ ae, a ¯˜ ÁË 3˜¯
x cos q + y sin q = 1 L = = 2,
ab
…(i)

Its point of contact is P(a cos qc – b sin q) and its slope L¢ = Ê - b2 ˆ = ËÁÊ 2, - 53˜¯ˆ
ÁË ae, a ˜¯
is - b cot q . Also the focus is S(ae, 0).
a N = Ê — ae, b2 ˆ = Ê — 2, 5ˆ and
ËÁ a ¯˜ ËÁ 3¯˜
Any line through the focus S and perpendicular to tan-
gent (i) is

y - 0 = a tan q (x - ae) …(ii) N¢ = Ê — ae, - b2 ˆ = Ê — 2, - 5ˆ
b ËÁ a ¯˜ ËÁ 3˜¯

Also the equation of CP is Now, the tangent at L is

y - 0 = a tan q (x - 0) …(iii) xx1 + yy1 = 1
b 95

Eliminating q between Eqs (ii) and (iii), we get 5 y
3
Ê a2 ˆ x -x ae ˜ˆ¯ fi 2x + =1
ËÁ b2 ˜¯
ËÁÊ = 1 95

fi ÊÁË x - ae ˆ˜¯ = Ê b2 ˆ fi x + y =1
x ÁË a2 ˜¯ 9/2 3

Thus, P = Ê 9 , 0˜¯ˆ and S = (0, 3)
ÁË 2
ÁËÊ1 ae ¯˜ˆ Ê b2 ˆ
fi - x = ÁË a2 ˜¯ Therefore,
ar(quad PQRS) = 4 ¥ ar(DOPS)
fi Ê b2 ˆ = Ê ae ˆ = 4¥1¥9¥3
ÁË1 - a2 ¯˜ ËÁ x ¯˜ 22

= 27 s.u.

fi Ê a2 (1 - e2 ) ˆ = Ê ae ˆ 13. The equation of the tangent at (3 3 cos q, sin q ) to the
ËÁ1 - a2 ¯˜ ËÁ x ¯˜ given ellipse is

fi e2 = ÁËÊ ae ¯˜ˆ x ◊ 3 3 cos q + y ◊ sin q = 1
x 27

fi x=a fi x + y =1
e 3 3 sec q (cosec q )

Hence, the result. It is given that
12.
Y S = 3 3 sec q + cosec q

S fi dS = 3 3 sec q tan q - cosec q cot q
NL dq

For maximum or minimum, dS = 0
dq
X¢ R O PX gives

N¢ L¢ fi 3 3 sec q tan q - cosec q cot q = 0
Q
fi 3 3 sec q tan q = cosec q cot q
Y¢

Given ellipse is x2 + y2 = 1 fi cosec q cot q = 3 3
95 sec q tan q

b2 1- 5 = 2 fi cot3q = 3 3
a2 93
Now, e = 1- = fi cot q = 3

5.48 Coordinate Geometry Booster

fi q=p 25m2 + 4 = 4
6 m2 + 1

Hence, the value of q is p . fi 25m2 + 4 = 16(m2 + 1)
6 fi 9m2 = 12
fi m2 = 4
14. Let the mid-point be (h, k).
The equation of the ellipse is x2 + 2y2 = 2. 3
x2 + y2 = 1 fi m=± 2
2
and the equation of the tangent to the ellipse at 3

( 2 cos q, sin q ) is Hence, the equation of the common tangent is

x ◊ 2 cos q + y ◊ sin q = 1. y = - 2 x + 100 + 4
2 33

fi x + y = 1. y = - 2 x + 112
2 sec q cosec q 33

Let A = ( 2 sec q, 0) and B = (0, cosec q ) Ê 112 ˆ
7, 0) and B = ËÁ 0, 3 ˜¯
Thus, h = 2 sec q , k = cosec q Let A = (2
22
Thus, the length of the tangent
cos q = 1 and sin q = 1
fi 2 h 2k = 28 + 112 = 196 = 14
3 33
Squaring and adding, we get

1 1 17. Let (h, k) be the point of intersection of tangents at q
2h2 4k 2
+ =1 and j. Then

Hence, the locus of (h, k) is cos ÁËÊ q + j ¯ˆ˜ sin Ê q +jˆ
cos ÁËÊ q 2 ËÁ 2 ˜¯
1 1 h = and k = -jˆ
2x2 + 4y2 =1 a - j ¯˜ˆ b Ê q 2 ¯˜
2 cos ËÁ

15. The equation of any tangent to the given ellipse at (a Squaring and adding, we get
cos q, b sin q) is

x cos q + y sin q = 1 h2 + k2 = 1 - j ˜¯ˆ …(i)
ab a2 b2 cos2 ÊËÁ q 2 ( b = 3)
fi x + y =1
a sec q b cosec q It is given that,

Let A = (a sec q, 0) and B = (0, b cosec q) b(sin q + sin j) = 3

1 fi (sin q + sin j) = 1

D = ar (DOAB) = 2 ab sin q cosec q Ê q + jˆ Ê q -jˆ
ËÁ 2 ˜¯ ËÁ 2 ˜¯
fi 2 sin cos =1

=1 sin ËÁÊ q + j˜ˆ¯
ab sin 2q 2
Now, k = = 1 …(ii)
Hence, the minimum area is 1 sq. unit.
ab b cos Ê q - jˆ 2 cos2 ÁËÊ q - j ˜¯ˆ
ËÁ 2 ˜¯ 2
16. The equation of any tangent to the ellipse is

y = mx + a2m2 + b2 From Eqs (i) and (ii), we get

fi y = mx + 25m2 + 4 h2 + k2 = 2k
a2 b2 b

fi mx — y + 25m2 + 4 = 0 fi h2 + k 2 = 2k
25 9 3
which is also a tangent to the given circle. So, the
length of the perpendicular from the centre to the tan- Hence, the locus of (h, k) is
gent is equal to the radius of a circle
x2 + y2 = 2y
25 9 3

Ellipse 5.49

18. Given ellipse is x2 + y2 = 1 20. Any point on the ellipse be (a cos q + a sin q).
25 16 Clearly, the centre is (0, 0)

Y Now, distance, r = a2cos2q + a2sin2q = a
P
Also,
X¢ S O S¢Q(h, k)
a2 cos2 q + 2a2 sin q cos q + 2a2 sin2 q = 1

X fi a2 = 1 sin 2q
2q
1 + sin +

fi a= 1

Y¢ 1 + sin 2q + sin2q

Let the point Q be (h, k). Therefore, r = a = 1
Clearly, k is negative.
It is given that, 1 + sin 2q + sin2q

SP = NQ fi r= 1

fi –k = a + eh 3 + sin 2q — 1 cos 2q
22
fi -k =5+ 3h
5 r will be maximum when Dr will provide us minimum
value
Hence, the locus of Q is Hence, the maximum value of r
-y=5+ 3x
5 =2
3- 5
fi 3x + 5y + 25 = 0
19. The Equation of the tangent to the ellipse is 21. Let M(h, k) the mid-point of the chord PQ.
Since the length of PQ is 2c, so P and Q can be consid-
y = mx + a2m2 + b2 ered as (h + c cos q, k + c sin q) and (h – c cos q, k – c
sin q) respectively.
= 2x + 4a2 + b2
Thus, (h + c cos q )2 + (k + c sin q )2 =1
which is a normal to the given circle a2 b2
Clearly, it will pass through the centre of a circle, so
and (h - c cos q )2 + (k - c sin q )2 =1
0 = — 4 + 4a2 + b2 a2 b2

fi 4a2 + b2 = 4 Adding, we get

fi 4a2 + b2 = 16 b2h2 + a2k2 – a2b2 + c2(a2sin2q + a2cos2q) = 0
…(i)
Let P = ab = a 16 - 4a2
and subtracting, we get
fi P2 = a2(16 – 4a2)
fi Q = 16a2 – 4a4 4ch cos q - 4ch sin q = 0
a2 b2
fi dQ = 32a - 16a3
da sin q = cos q = 1 …(ii)
b2h -a2k h2b4 + k 2a4
d 2Q = 32 - 48a2
fi da2
From Eqs (i) and (ii), we get

For maximum or minimum, b2h2 + a2k2 – a2b2

dQ = 0 + c2 Ê b4 h2a 2 a4 + a4k 2b2 ˆ = 0
da ËÁ b4h2 +k 2 b4h2 + k 2a4 ¯˜

fi 32a – 16a3 = 0 Hence, the locus of M(h, k) is
fi 16a(a2 – 2) = 0

fi a = 0, ± 2 Ê b2x2 + a2 y2 ˆ = 1 Ê x2 - y2 ˆ
ÁË a4 y2 + b4x2 ¯˜ c2 ËÁ1 - a2 b2 ˜¯
when a = ± 2, then b = ± 2 2
Hence, the maximum value of ab is 4.

5.50 Coordinate Geometry Booster

22. Let the ellipse be x2 + y2 =1 and its pole be (h, k) Putting the values of sin a and cosa in Eq. (i), we get
a2 b2 x(x + y) + y(y - x) = a2( y + x)2 + b2( y - x)2

The equation of polar w.r.t. the given ellipse is fi (x2 + y2) = a2( y + x)2 + b2( y - x)2

hx + ky =1
a2 b2
fi (x2 + y2 ) = (a2 + b2 )(x2 + y2 ) + 2xy(a2 - b2 )
fi ky = - hx +1 fi (x2 + y2)2 = (a2 + b2)(x2 + y2) + 2xy(a2 – b2)
b2 a2 fi (x2 + y2)(x2 + y2 – a2 – b2) = 2xy(a2 – b2)

fi y = - b2h x + b2 …(i)
a2k k
which is the required locus.

which is a tangent to the parabola 24. Let the ellipse be x2 + y2 = 1 and the point P be (h, k).
ay2 = –2b2x a2 b2

fi y2 = Ê - 2b2 ˆ x The equation of any tangent to the ellipse be
ÁË a ˜¯

fi y2 = 4 Ê - b2 ˆ x …(ii) y = mx + a2m2 + b2
ËÁ 2a ¯˜
which is passing through P.
Since (i) is tangent to (ii), so

Ê b2 ˆ So, k = mh + a2m2 + b2
ËÁ 2a ¯˜
b2 - fi (k – mh)2 = (a2m2 + b2)
k fi k2 – 2hkm + m2h2 = (a2m2 + b2)
= Ê b2h ˆ fi (h2 – a2)m2 – 2hkm + (k2 – b2) = 0
ÁË a2k ˜¯
-

fi b2 = ak It has two roots, say m1 and m2.
k 2h

fi k 2 = 2b2 h Thus, m1 + m2 = 2hk
a (h2 - a2)

Hence, the locus of (h, k) is and m1m2 = k2 - b2
h2 - a2
y2 = Ê 2b2 ˆ x
ËÁ a ˜¯
It is given that, q1 + q2 = 2a
which represents a parabola. fi tan (q1 + q2) = tan (2a)

23. Let the equation of the ellipse be Ê x2 + y2 ˆ =1 fi tan q1 + tan q2 = tan (2a )
ÁË a2 b2 ¯˜ 1 - tan q1 ◊ tan q2

The equation of a tangent to the given ellipse is

x cos a + y sin a = a2cos2a + b2sin2a …(i) fi m1 + m2 = tan (2a )
1 - m1m2
After rotation, the equation of the tangent is

x cos (a + 90°) + y sin (a + 90°) fi h2 2hk b2 = tan (2a )
a2 - k2
= a2sin2a + b2cos2a - +

fi - x sin a + y cos a = a2sin2a + b2cos2a fi (h2 - k 2hk - a2) = tan (2a )
…(ii) 2 ) + (b2

On subtraction, we get Hence, the locus of P(h, k) is

(x + y) sin a + (x – y) cos a = 0 2xy = tan (2a )
y2 ) + (b2
fi sin a = cos a = 1 (x2 - - a2)

( y - x) (x + y) 2(x2 + y2 ) fi {(x2 – y2) + (b2 – a2)}tan (2a) = 2xy

Ellipse 5.51

25. Given ellipse is x2 + y2 = 1 So, h - 2 = k - 1 = - 2(- 4)
16 9 11 2

Y fi h - 2 = k -1= 4
B 11

Thus, Q = (6, 3)

X¢ C O AX As we know that,
SP + S ¢P = 2a

fi SP + QP = 2a

D Thus, S, P and Q are collinear.
Y¢ So, S¢Q : 4x – 5y = 9

Clearly, the vertices of the square lie on the director Therefore P is a point of intersection of
circle of the given ellipse x2 + y2 = 1 .
S¢Q : 4x – 5y = 9 and L : x + y = 5
16 9
Hence, the point P is Ê 34 , 11ˆ .
ÁË 9 9 ¯˜
The equation of the director circle is
x2 + y2 = 16 + 9 = 25 28. Clearly, the centre of the ellipse is (2, 2).

Thus, the length of AC is 10 which is diagonal of the Y
square.
S(3, 3) X
Thus, a 2 = 10
C(2, 2)
fi a=5 2 S(1, 1)
O
Hence, the length of the side of the square is 5 2 .
26. Let PQ be the double ordinate, where Since x-axis and y-axis are two perpendicular tangents
to the ellipse, so (0, 0) lies on the director circle and
P = (3 cos q, 2 sin q) and Q = (3 cos q, –2 sin q) (2, 2) is the centre of the director circle.

Let the point R(h, k) divides the double ordinate in the Thus, the radius = 22 + 22 = 8 = 2 2
ratio 2 : 1
Thus h = 3 cos q and k = 2 sin q Hence, the area of the director circle = 8p

3
Squaring and adding, we get

Ê h ˆ 2 Ê 3k ˆ 2 x2 y2
ÁË 3 ¯˜ ËÁ 2 ¯˜ a2 b2
+ = 1 29. The co-ordinates of any point on the ellipse + =1
be (a cos q, b sin q).
fi h2 + 9k 2 = 1
94 If it lies on the line bx = ay, we have

ba cos q = ab sin q

fi tan q = 1

Hence, the locus of (h, k) is fi q = p or 5p
44
x2 + 9y2 = 1 30. Let the equation of the ellipse be x2 + y2 = 1.
94 a2 b2

27. Let S = (2, –1) and S ¢ =(1, –1) and Q is the image of S Its polar form is 1 = cos2q + sin 2q
w.r.t. x + y = 5. r2 a2 b2

Q Let r1 and r2 be the lengths of the radius vectors CP and
P
CQ which are inclined at angles q and p + q.
A L=0 2
S¢ S A¢ cos2q sin 2q
So, 1 = a2 + b2
r12

and 1 = cos2q + sin 2q
r22 a2 b2

5.52 Coordinate Geometry Booster

Now, On solving the equations AB: x + 2y = 8 and AD: y = 2x
+ 14, and AB and DC, we get
1 + 1 = cos2q + sin 2q + sin 2q + cos2q
r12 r22 a2 b2 a2 b2 A = (–4, 6) and D = (–5, 4)
When c = –6, we get,
= 1 + 1
a2 b2 A = (3, 2) and D = (3, 0).

31. The given ellipse is LEVEL IV

4(x – 2y + 1)2 + 9(2x + y + 2)2 = 25 …(i) 1. Given ellipse is

Let X = x – 2y + 1, Y = 2x + y + 2

Equation (i) reduces to

4X2 + 9Y2 = 25 Y
P
fi X2 + Y2 =1
25/4 25/9

Thus, e = 1 - b2 X¢ O S(ae, 0) A X
a2
Q

= 1 - 25/9 = 1 - 4 = 5 Y¢
25/4 9 3

32. Let AD: y = 2x + c We have A = ar(DPQA)

So, BC: y = 2x + 4, AB: x + 2y = 8 = 1 (2b sin q )(a - a cos q )
2
and DC: x + 2y – 3 = 0

A y = 2x + c D = ab ËÊÁ sin q - 1 sin 2q ˜ˆ¯
2

dA = ab(cos q - cos 2q )
dq

O fi d2A = ab(— sin q + 2 sin 2q )
dq 2
B(0, 4)
For maximum or minimum
Let BC = 2a and AB = 2b.
Clearly, 2a = 5 fi a = 5 C(–1, 2) dA = 0
dq
2 gives
It is given that, ab(cos q – cos 2q) = 0

p ab = 5 p fi (cos q – cos 2q) = 0
2
fi cos 2q = cos q
fi ab = 5
2 fi cos 2q = cos (2p – q)

fi 5b= 5 fi 2q = 2p – q
22
fi 3q = 2p
fi b= 5
Also, b = c - 4 fi q = 2p = 120°
3
25
fi c-4 = 5 Clearly, d 2A is - ve
d 2q
25
fi c – 4 = ±10 Thus, A is maximum.
fi c = 4 ± 10 = 14, –6 Hence, the maximum value of A
when c = 14
= ab Ê sin 120° - 1 sin (240°)ˆ˜¯
ÁË 2

Ê 3+ 3 ˆ ab
= ÁË 2 4 ˜¯

= 3 3 ab s.u.
4

Ellipse 5.53

2. The equation of the tangent to the parabola y2 = 4x at But m1m2 = –1

(t2, 2t) is fi - 1 cos q ¥ 2 sin q = -1
2 sin q 4 cos q - 1
yt = x + t2

fi x – yt + t2 = 0 …(i)

and the equation of the normal fi cos q = 1
4x2 + 5y2 = 20 4 cos q - 1

i.e. x2 + y2 = 1 fi 4 cos q – 1 = cos q
54 fi 3 cos q = 1

at ( 5 cos j, 2 sin j) is fi cos q = 1
3
a2x - b2 y = a2 - b2
x1 y1 Thus, the radius of the circle,

fi 5x - 4y = 5 — 4 = 1 r = (a cos q - 1)2 + b2sin2q
5 cos j 2 sin j
Ê 4 1ˆ¯˜ 2 ÁÊË1 1 ˆ
ËÁ 3 9 ¯˜
fi ( 5 sec j)x - (2 cosec j) y = 1 …(ii) = - + 4 -

Since (i) and (ii) are the same line, so = 1 + 32 = 33
99 3
5 sec j = 2 cosec j = —1
1 t t2 Hence, the equation of the circle is

fi sec j = - 1 , cosec j = - 1 (x - 1)2 + y2 = 33 = 11
5t 2 2t 93

fi cos j = - t2 5, sin j = - 2t 4. Let the point P be (a cos q, b sin q) and the point Q be

Thus, 5t4 + 4t2 – 1 = 0 Ê a cos ËÁÊq + pˆ , b sin ËÁÊq + pˆˆ
fi 5t4 + 5t2 – t2 – 1 = 0 ËÁ 4 ¯˜ 4 ¯˜ ¯˜
fi 5t2(t2 + 1) – 1(t2 + 1) = 0
fi (5t2 – 1)(t2 + 1) = 0 The equation of the chord joining P and Q is
fi (5t2 – 1) = 0
x Ê pˆ y sin ËÊÁq p ¯˜ˆ Ê p ˆ
t=± 1 a cos ËÁ q + 8 ¯˜ + b + 8 = cos ËÁ 8 ˜¯
5
fi

Also, tan j = ±2 which is identical with
fi j = tan–1(±2) px + qy = r

3. The equation of the given ellipse is Comparing the co-efficients, we get
x2 + 4y2 = 16
cosÊËÁ q p ˆ˜¯ sin ËÁÊ q p ¯˜ˆ cosÊÁË p8 ¯˜ˆ
fi x2 + y2 = 1 + 8 + 8
16
Thus, = =
ap bq r

Given centre of the circle is C(1, 0). cos Ê q + pˆ = ap cos Ê pˆ
Let the equation of the circle be ËÁ 8 ˜¯ r ËÁ 8 ˜¯

(x – 1)2 + y2 = r2

Since the circle is the largest, so it will touch the ellipse and sin Ê q + pˆ = bq
at some point P(a cos q, b sin q). ËÁ 8 ˜¯ r
The equation of the tangent to the ellipse is
Squaring and adding, we get
x cos q + y sin q = 1
ab Ê ap ˆ 2 Ê bq ˆ 2
ËÁ r ¯˜ ËÁ r ˜¯
- b cos q - 1 cot q + = 1
a sin q 2
whose slope is m1 = =

and m2 = m(CP) = (a b sin q = 4 2 sin q a2p2 + b2q2 = r2
cos q - 1) cos q - 1 which is the required condition.

5.54 Coordinate Geometry Booster

5. Let R(h, k) be any point on the locus. x2 y2
a2 b2
Let OP: y = x tan a and OQ : y = –x tan a 7. Given ellipse is + =1

It is given that, Y
(h sin a – k cos a)2 + (h sin a + k cos a)2 = 2l2
P(x¢, y¢)
fi h2 sin2a + k2 cos2a = l2 L

fi h2 + k2 =1 X¢ A¢ S(–ae, 0) O NA X
l 2cosec2a l 2sec2a

which represents an ellipse.

6. Let the equation of the ellipse be x2 + y2 = 1.
a2 b2
Y¢

Let C be the centre and PQ be the chord whose equa- The equation of tangent at P is
tion is x cos a + y sin a = p.
xx¢ + yy¢ =1
Now, we make above two equations a homogeneous a2 b2
equation of 2nd degree.

Thus, x2 + y2 = Ê x cos a + y sin aˆ 2 The slope of the tangent is - b2 x¢
a2 b2 ÁË p ˜¯ a2 y¢

fi Ê p2 - cos2a¯˜ˆ x2 - 2xy sin a cos a and the slope of SP is y¢
ÁË a2 x¢ + ae

Ê p2 - sin2a¯ˆ˜ y¢ + b2x¢
ËÁ b2 + ae
+ y2 = 0 Now, tan q = x¢ a2 y¢
Ê y¢
Since the pair of diameters CP and CQ are at right an- 1 - ËÁ x¢ + ae ◊ b2 x¢ ˆ
gles, so a2 y¢ ¯˜

Ê p2 - cos2a ˆ + Ê p2 - sin 2a ˆ = 0 = a2b2 + b2 x¢ae
ËÁ a2 ¯˜ ËÁ b2 ¯˜ x¢y¢a2e2 + a2ey¢

fi p2 + p2 =1 = b2a(a + ex¢) = b2
a2 b2 a2ey¢(a + ex¢) aey¢

fi p 2 Ê 1 + 1ˆ =1 fi q = tan -1 Ê b2 ˆ
ÁË a2 b2 ˜¯ ÁË aey¢ ¯˜

fi p2 = Ê1 1 1ˆ = a2b2 8. Given ellipse is x2 + y2 =1
ËÁ a2 + b2 ˜¯ a2 + b2 a2 b2

fi p = ab and the circle is x2 + y2 = ab.
a2 + b2 On solving, we get,

Thus, PQ : x cos a + y sin a = ab x2 = a2b and y2 = ab2
a2 + b2 a+b a+b

Now, the length of the perpendicular draw a from the The equation of tangent to the ellipse at (x1, y1) is
centre to the line PQ
xx1 + yy1 =1 and the circle is xx1 + yy1 = ab
a2 b2
0 + 0 - ab
= a2 + b2 = ab The slope of the tangent to the ellipse is
a2 + b2
cos2a + sin2a m1 = - b2 x1
a2 y1
= constant
and the slope of tangent to the circle is
Hence, the line PQ touches a fixed circle whose centre

is (0, 0) and the radius is ab . m2 = - a
b
a2 + b2

Ellipse 5.55
…(i)
Now, tan q = m2 - m1 10. The equation of the tangent at P is …(ii)
1 + m1m2
x cos j + y sin j = 1
a Ê - b2 ˆ ab
b ÁË1 a2 ¯˜
= and the equation of the normal at P is
ab2 ax sec j – by cosec j = a2 – b2
1+ ba2
Then Q = (a sec j, 0)

= a2 - b2 a and R = Ê (a2 - b2 )cos j ˆ
a(a + b) b ËÁ a , 0¯˜

= a-b
ab

Thus, q = tan -1 Ê a - bˆ P
ÁË ab ¯˜ RaQ

9. The normal at P is
ax - by = a2 - b2

cos q sin q

and the normal at Q is

fi ax p ˜ˆ¯ - by p ¯˜ˆ = a2 - b2 Therefore, QR = a
cos ÁËÊq + 2 sin ÁÊËq + 2 fi a sec j - (a2 - b2 ) cos j = a

- ax - by = a2 - b2 a
sin q cos q
fi a2 sin2 j + b2 cos2 j = a2 cos j
fi ax + by = b2 - a2 fi a2 sin2 j + a2(1 – e2) cos2 j = a2 cos j)
sin q cos q fi a2(sin2 j + cos2 j) – a2e2 cos2 j = a2 cos j
fi a2 – a2e2 cos2 j = a2 cos j
The slope of the normal at P is fi e2 cos2 j + cos j – 1 = 0

m1 = a sin q = a tan q 11. Let the point P be (a cos j, b sin j)
b cos q b
and S = (ae, 0), S¢ = (–ae, 0)
and the slope of the normal at Q is
\ SP = a – ex = a – e cos j
m2 = - a cot q
b and S¢P = a + ae cos j

Also, SS¢ = 2ae

Let (a, b) be the incentre of DPSS¢. So

Now, tan a = m1 - m2 2ae◊a cosj + a(1- e cosj)(-ae)
1 + m1m2 a = + a(1- e cosj)(ae)

2ae + a(1- e cosj) + a(1+ e cosj)

a (tan q + cot q ) fi a = ae cos j

=b a2 Similarly, b = be sin j
b2 1+ e
1 -
Eliminating f, we get
ab 2
= (b2 - a2 ) sin 2q ËÁÊ a ˜¯ˆ 2 Ê b(1 + e) ˆ 2
ae ÁË be ¯˜
+ = 1

= 2ab ¥ 1 a2 + b 2 (1 + e)2 =1
a2e2 sin 2q a2e2 b2e2
fi

= 2a2 1 - e2 ¥ 1 Hence, the locus of incentre is
a2e2 sin 2q
x2 y2
2 1 - e2 a2e2 + b2e2 =1
e2 (sin 2q )
= (1 + e)2

5.56 Coordinate Geometry Booster

Let e1 be its eccentricity, then \ Circumcentre of DPQR be

b2e2 h = (a2 -b2)
4a
e1 = 1- (1 + e)2
a2e2 (cos a + cos b + cosg + cos (a+ b + g))

= 1 - a2 (1 - e2 ) and k = (a2 - b2 )
a2 (1 + e)2 4a

= 1 - 1 - e = 2e (sin a + sin b + sin g - sin (a+ b + g))
1+ e 1+ e
On simplification, we get

12. Let the point P be (a cos j, b sin j). cos (a + b + g) = h(a2 + 3b2 )
a(a2 - b2)
Y

BP and sin (a + b + g) = (a2 + 3b2 )k
b(a2 - b2)

q Squaring and adding, we get
CS
X¢ AX Ê (a2 + 3b2 )h ˆ 2 Ê (a2 + 3b2 )k ˆ 2
A¢ ËÁ a(a2 - b2) ¯˜ ËÁ b(a2 - b2) ¯˜
+ = 1

B¢

Hence, the locus of the centroid (h, k) is

Y¢ Ê (a2 + 3b2 )x ˆ 2 Ê (a2 + 3b2 ) y ˆ 2
ËÁ a(a2 - b2) ¯˜ ËÁ b(a2 - b2) ¯˜
tan q = b sin j = (b/a) sin j + = 1
a cos j — ae cos j - e
14. The equation of the tangent to the ellipse at P is
= 1 - e2 sin j
cos j - e x cos j + y sin j = 1 …(i)
ab

fi tan q cos j - e tan q = 1 - e2 sin j Let the equation of the auxilliary circle be
x2 + y2 = a2
2 tan (q /2) Ê1 - tan2 (j /2) ˆ …(ii)
1 - tan2 (q /2) ËÁ 1 + tan2 (j /2) e˜¯ Y
fi ¥ -

= 1 - e2 Ê 2 tan (q /2) ˆ Ê 1 - tan 2 (j /2) ˆ B
ËÁ 1+ tan 2 (q /2) ˜¯ ËÁ 1 + tan 2 (j /2) ¯˜
P

X¢ O A

On simplification, we get X

fi 1 + e tan ËÁÊ j2˜ˆ¯ - 1- e tan Ê qˆ = 0
ÁË 2¯˜

1 + e tan ÁÊË j2¯ˆ˜ = 1 - e tan Ê q ˆ Y¢
ÁË 2 ¯˜

fi tan Ê qˆ = 1+ e tan ÊËÁ j2ˆ¯˜ The equation of the pair of lines OA and OB are ob-
ÁË 2˜¯ 1- e tained by making homogeneous of (i) and (ii). So

Ê x y j˜¯ˆ 2
ÁË a b
13. Let the vertices of the equilateral triangle P, Q, R, x 2 + y 2 = a2 cos j + sin
whose eccentric angles are a, b, g respectively.
Let (h, k) be the centroid of DPQR. = a2 Ê x2 cos2j + y2 sin 2j + 2 xy sin jcos ˆ
ËÁ a2 b2 ab j¯˜
Then h = a (cos a + cos b + cos g)
and 3 fi (1 - cos2j) x 2 + Ê - a2 sin 2jˆ¯˜ y2 + (....) xy =0
ÁË1 b2
k = a (sin a + sin b + sin g)
3

Since DPQR is an equilateral triangle, so It is given that, –AOB = 90°. So
centroid = circumcentre. co-efficient of x2 + co-efficient of y2 = 0

Ellipse 5.57

fi 1 - cos2j + 1 - a2 sin 2j = 0 Since SF1 + S¢F1 = 2a = F1S¢ + F2S¢
b2
B

fi sin 2j + 1 - a2 e2 ) sin 2j = 0 F1
a2 (1 -

fi sin 2j + 1 - 1 sin 2j = 0 S¢ q A
- e2)
(1 S¢ C S

fi ÊÁË1 — 1 ˆ sin 2j = -1 2
— e2 ¯˜
1

Ê —e2 ˆ sin 2j B¢
ËÁ 1 — e2 ¯˜
fi = -1 Clearly, CS = ae and CF1 = ae¢
Let q be the angle between their axes.
fi e2 sin2 j = (1 – e2) Then, SF12 = a2e2 + a2e¢2 - 2a2ee¢ cos q
fi e2(1 + sin2 j) = 1 and S ¢F12 = a2e2 + a2e¢2 + 2a2ee¢ cos q

fi e2 = (1 + 1 Now, 2a = SF1 + S¢F1
sin 2j ) fi 4a2 = (SF1 + S¢F1)2

fi e= 1 fi 4a2 = SF12 + S ¢F12 + 2(SF1)(S ¢F1)

(1 + sin2j) fi 4a2 = 2a2(e2 + e¢2) +

15. Let two points on the ellipse be P and Q whose eccen- 2 (a2e2 + a2e¢2 ) - 4a4e2e¢2cos2q

tric angles are j1 and j2 where j1 – j2 = a. fi (2 – e2 – e¢2)2 = (e2 + e¢2)2 – 4e2e¢2 cos2 q
The equation of the tangent at P and Q are fi 4 – 4(e2 + e¢2) = –4e2e¢2 cos2 q
fi 1 – (e2 + e¢2) = –e2e¢2 cos2 q
x cos j1 + y sin j1 = 1
a b

and x cos j2 + y sin j2 =1 fi cos2q = Ê e2 + e¢2 - 1ˆ
a b ËÁ e2e¢2 ¯˜

Let R(h, k) be the point of intersection of the tangents. Ê e2 + e¢2 - 1ˆ
cos q = ÁË ee¢ ¯˜
a cos ÊËÁ j1 + j 2 ˜ˆ¯ b sin ËÊÁ j1 + j 2 ¯˜ˆ fi
2 2
Thus, h = and k =
ÁÊË j1 - j2 ¯ˆ˜ Ê j1 - j ˆ 17. Let the point of concurrency be (h, k).
cos 2 cos ËÁ 2 2 ˜¯ The equation of the normal to the given ellipse at (x¢,
y¢) is
cos Ê j1 + j2 ˆ sin Ê j1 + j 2 ˆ a2x - b2 y = a2 - b2
ËÁ 2 ¯˜ ÁË 2 ¯˜ x¢ y¢
h = and k =
a Ê aˆ b Ê a ˆ
cos ËÁ 2 ¯˜ cos ÁË 2 ˜¯
which is passing through (h, k). So

Squaring and adding, we get a2h - b2k = a2 - b2 …(i)
x¢ y¢ …(ii)

Ê hˆ 2 Ê kˆ 2 1 x2 y2
ËÁ a¯˜ ÁË b ¯˜ cos2 ÁËÊ a2 b2
+ = a ˆ˜¯ Also, the point (x¢ y¢) lies on + = 1. So
2
x¢2 + y¢2 =1
Hence, the locus of R(h, k) is a2 b2

x2 + y2 = sec2 ÊÁË a ˆ¯˜ On simplification, we get
a2 b2 2 –(a2 – b2) x¢4 + 2a2(a2 – b2)hx¢3 + (…)x¢2
–2a4(a2 – b2)hx¢ + a6h2 = 0
16. Let S and S¢ be the foci of one ellipse and F1 and F2 of
the other, where C being the common centre. which is a bi-quadratic equation. So, it has four roots,
say x1, x2, x3 and x4.
So, SF1S¢F2 will form a parallelogram.

5.58 Coordinate Geometry Booster

Then x1 + x2 + x3 + x4 = 2ha2 The equations of tangents at P and Q are
(a2 - b2) x cos a + y sin a = 1
ab
and x1x2x3x4 = a2
and x cos b + y sin b = 1
Now, Â1 + 1 + 1 + 1 = x1x2x3 ab

x1 x2 x3 x4 x1x2 x3x4 On solving, we get the point of intersection A, i.e.

2(a2 - b2 ) Ê a cos Ê a + b ˆ a sin ÊËÁ a + b ˆ˜¯ ˆ
a2h Á ÁË 2 ˜¯ 2 ˜
= A = Á , ˜
Á ˜
Hence, the value of Ë cos ÁËÊ a - b ¯ˆ˜ cos ÁËÊ a - b ˆ˜¯ ¯
2 2
Ê 1 1 1 1ˆ
(x1 + x2 + x3 + x4 ) ËÁ x1 + x2 + x3 + x4 ˜¯ Since the point A lies on x2 + y2 = 4 , so
a2 b2
2ha2 (a2 - b2)
= (a2 - b2) ¥ 2ha2 = 1 cos2 ÁËÊ a + b ˆ˜¯ sin 2 Ê a + b ˆ
cos2 ËÊÁ a 2 ÁË 2 ˜¯
+ = 4
x2 y2 - b ˜¯ˆ Ê a - b ˆ
a2 b2 2 cos2 ÁË 2 ¯˜
18. Let the ellipse be + =1.

Let P and Q be two points lie on the ellipse whose ec- fi cos2 Ê a - bˆ = 1
centric angles are a and b such that q = a – b. ËÁ 2 ˜¯ 4
Given that the tangents at P and Q are at right angles.
fi cos Ê a- bˆ = ± 1
Ê b ˆ Ê b ˆ ËÁ 2 ¯˜ 2
ÁË - a cot a ¯˜ ÁË - a cot b ¯˜ = -1
\ The equations of the normals PR and QR are
fi a2 sin a sin b + b2 cos a cos b = 0 ax sec a – by cosec a = a2 – b2
But the diameter parallel to the tangent at P will be
conjugate to the diameter CP, then its extremities will and ax sec b – by cosec b = a2 – b2
be (–a sin a, b cos a).
On simplification, we get

{ }ax a b ˆ¯˜

a2 - b2
Thus, d12 = a2sin2a + b2cos2a = ± cos ÊËÁ + cos (a + b) - 1
2 2

{ }by

a2 - b2
Similarly, d22 = a2sin2b + b2cos2b = ± sin ÁËÊ a + b ˆ¯˜ cos (a + b) + 1
Now, 2 2

d12d22 = (a2 sin2 a + b2 cos2 a)(a2 sin2 b + b2 cos2 b) Squaring and adding, we get
= (a2 sin a sin b + b2 cos a cos b)2
Ê ax ˆ 2 Ê by ˆ 2
ÁË a2 - b2 ˜¯ ËÁ a2 - b2 ˜¯
+

+ a2b2 (sin a cos b – cos a sin b)2 = cos2 (a + b) + 1 - cos2 (a + b)
4
= 0 + a2b2 sin2(a – b)
fi a2x2 + b2 y2 = 1 (a2 - b2)2
= a2b2 sin2q 4

fi d1d2 = ab sin q Hence, the result.

Hence, the result. 20. Let the circle

19. Given conics are x2 + y2 = 4 and x2 + y2 =1. x2 + y2 + 2gx + 2fy + c = 0 …(i)
a2 b2 a2 b2

Y x2 y2
a2 b2
intersect the ellipse + =1,

i.e. b2x2 + a2y2 = a2b2 …(ii)

X¢ A P X in four points (x1, y1), (x2, y2), (x3, y3) and (x4, y4), re-
spectively.
R(h, k)
Q It is given that, one point (x1, y1) = (h, k) is fixed and
other two points (x2, y2) and (x3, y3) are extremities of a
diameter of the ellipse.

Since j and p + j are the eccentric angles of the ex-

Y¢ tremities of diameters of ellipse,

Ellipse 5.59

So, (x2, y2) = (a cos j, b sin j) Integer Type Questions
and (x3, y3) = (–a cos j, –b sin j)

Thus, x1 + x2 + x3 = h and y1 + y2 + y3 = k 1. The equation of the tangent to the ellipse is
x + y =1
Now, from Eq. (i), we get
3 sec q 2 cosec q
x2 + y2 + 2gx + 2fy + c = 0

fi a2x2 + a2y2 + 2ga2x + 2fa2y + a2c = 0 Now,
D = ar DOAB
fi a2x2 + a2y2 + 2ga2x + a2c = –2fa2y = 1 ¥ 3 sec q ¥ 2 cosec q = 6
2 sin 2q
fi (a2x2 + a2y2 + 2ga2x + a2c)2 = 4f 2a4y2

fi (a2x2 + a2b2 – b2x2 + 2ga2x + a2c)2 Minimum area of the triangle = 6 s.u.

= 4f 2a2(a2b2 – b2x2) 2. We have F1 = (2, 0) and F2 = (–2, 0)
Now, A = ar(DPF1F2)
fi (a2 – b2)2x4 + 4a2g(a2 – b2)x3 + ax2 + bx + g = 0

where a, b and g are constants which is a bi-quadratic 5 cos q sin q 1
equation. =1 2
Let x1, x2, x3 and x4 are its roots. 01

4ga2 2 -2 0 1
(a2 - b2)
So, x1 + x2 + x3 + x4 = -

fi h + x4 = - 4ga2 = 1 (4 sin q ) = 2 sin q
(a2 - b2) 2

Hence, the maximum value of A is 2 s.u.

4ga2 3. Given ellipse is 16x2 + 11y2 = 256.
(a2 - b2)
fi x4 = - - h x2 + y2 =1
16 (16/ 11)2
4fb2
Similarly, y4 = - (a2 - b2) - k The equation of the tangent to the given ellipse at

Since, (x4, y4) lies on the ellipse, so Ê 4 cos j, 16 sin j ˆ is
ÁË 11 ¯˜

x42 + y42 =1 x cos j + y sin j = 1 …(i)
a2 b2 4 (16/ 11)

Ê - 4ga2 - ˆ2 Ê - a 4fb2 - ˆ2 which is also a tangent to the circle
ËÁ a2 - b2 h¯˜ ËÁ 2 - b2 k ¯˜ x2 + y2 – 2x – 15 = 0
fi + =1
a2 b2 fi (x – 1)2 + y2 = 16

16a2g 2 + 8gh + 16b2 f 2 So, the length of the perpendicular from the centre to
(a2 - b2)2 a2 - b2 (a2 - b2)2
the circle is equal to the radius of a circle.

fi Ê cos j ˆ
ËÁ 4 ¯˜
h2 k2 -1
a2 b2
- 8fk + + =1 cos2j + sin2j =4
(a2 - b2)

2g2a2 2 f 2b2 16 256/11
(a2 - b2) (a2 - b2)
fi + gh + - fk =0 ÊÊ cos j ˆ - 1˜ˆ¯ 2 Ê cos2j + sin2j ˆ
ÁË ÁË 4 ˜¯ = 16ÁË 16 256/11˜¯
fi

fi 2(g 2a2 + f 2b2 ) + gh - fk = 0 fi Ê cos2j - 2 cos j ˆ = cos2j + 11 sin2j
(a2 - b2) ÁË 16 4 + 1˜¯ 16

fi 2(g2a2 + f2b2) = (gh – fk)(a2 – b2) fi 4 cos2j + 8 cos j – 5 = 0

Hence, the locus of the centre (–g, –f) is fi (2 cos j – 1)(2 cos j + 5) = 0
2(x2a2 + y2b2) = (hx – ky)(a2 – b2)
fi (2 cos j – 1) = 0
fi 2(a2x2 + b2y2) = (a2 – b2)(hx – ky)
fi cos j = 1
Hence, the result. 2

5.60 Coordinate Geometry Booster

fi j = p , 5p fi sin q = - 5
66 24

Hence, the number of values of f is 2. fi q = 2p - sin -1 Ê 5ˆ , p + sin -1 Ê 5ˆ
4. The required area of a parallelogram is made by the ËÁ 24˜¯ ÁË 24˜¯

tangents at the extremities of a pair of conjugate diam- Also, y-axis is one of the normals.
eter Hence, it has three normals.

= 4ab Previous Years’ JEE-Advanced Examinations
= 4¥3¥ 1
1. Given line is
4
= 3 s.u. x = -9
5. Given curve is x2 + y2 = 1 2

a - 10 4 - a fi 2x + 9 = 0
Let the point be P(h, k).
(4 – a)x2 + (a – 10)y2 – (a – 10)(4 – a) = 0
which represents an ellipse, if h2 – ab < 0 Given (x + 2)2 + y2 = 2 ¥ Ê 2x - 9ˆ
fi –(4 – a)(a – 10) < 0 3 ÁË 4 ˜¯
fi (a – 4)(a – 10) < 0
fi 4 < a < 10 fi (x + 2)2 + y2 = 4 ¥ Ê 2x - 9ˆ 2
Hence, the number of integral values of a is 5. 9 ËÁ 4 ¯˜
6. Given ellipse is
fi (x + 2)2 + y2 = 1 ¥ (2x - 9)2
x2 + 4y2 = 4 9
fi x2 + y2 = 4
fi 9((x + 2)2 + y2) = (2x – 9)2
41 fi 9(x2 + y2 + 4x + 4) = 4x2 – 36x + 81
fi 5x2 + 9y2 + 72x – 55 = 0
Let A be the area of the rectangle.
So, A = (4 cos q)(2 sin q) = 4 sin (2q) which represents an ellipse.
Thus, the greatest area of the rectangle is 4.
7. Given ellipse is 9x2 + 16y2 = 144 2. Clearly, 1 + 4 - 1 > 0
94
x2 + y2 = 1
16 9 fi P lies outside of E.
Hence, the value of Also, 1 + 4 – 9 = –4 < 0
PF1 + PF2 – 2 = 2a – 2 fi P lies inside C.
Thus, P lies inside C but outside E.
=8–2
=6 3. Given ellipse is x2 + y2 =1
8. Clearly, (S1F1) ◊ (S2F2) = b2 = 3 a2 b2
9. The minimum length of the intercept of the tangent to
the ellipse x2 + y2 = 1 between the axes = a + b = 7. So, F1 = (ae, 0) and F2 = (–ae, 0)
16 9 Let P be (a cos q, b sin q)

10. The equation of the normal to the ellipse is Then
a2x - b2 y = a2 - b2 ar(DPF1F2)
x1 y1
a cos q b sin q 1
fi 169x - 25y = 169 - 25 = 1 ae
13 cos q 5 sin q 2 01

fi 13x - 5y = 144 -ae 0 1
cos q sin q
= 1 ¥ 2ae ¥ b sin q
which is passing through (0, 6). So 2
0 - 30 = 144
sin q = abe ¥ sin q
Maximum value of A = abe
fi - 5 = 24
sin q = ab 1 - b2
a2

= b a2 - b2

Ellipse 5.61

4. Given ellipse is x2 + y2 = 1 Y X
16 9 A(h, k)

Ê 7 ˆ Q
Foci = (± ae, 0) = ËÁ ± 4 ◊ 4 0˜¯
, = (± 7, 0) P
X¢
O
Radius of a circle = 7 + 9 = 16 = 4

5. Given ellipse is x2 + y2 =1 Y¢
a2 b2

Y The equation of the tangent to the ellipse (i) is

x cos q + y sin q = 1 …(ii)
2
MP
X¢ F2(–ae, 0) F1(ae, 0) The equation of the 2nd ellipse can be written as

X x2 + y2 = 1 …(iii)
63

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

Y¢ So PQ is the chord of contact.

The equation of the chord of contact of the tangents

The equation of the tangent to the given ellipse at through A is
P(a cos q, b sin q) is
hx + ky = 1 …(iv)
63
x cos q + y sin q = 1
ab Since the equations (ii) and (iv) are identical, so

Now, hk
OM = d
6 =3 =1
cos q sin q

fi 0+ 0-1 = d 2

cos2q + sin 2q fi h = 3 cos q, k = 3 sin q
a2 b2
Now, squaring and adding, we get
1 cos2q sin 2q h2 + k2 = 9
d2 a2 b2
fi = + Therefore, the locus of A is

x2 + y2 = 9

Now, which is the director circle of x2 + y2 = 1 .
63
4a2 Ê - b2 ˆ = 4a2 - 4a2b2
ÁË1 d2 ˜¯ d2 Thus, the angle between the tangents at P and Q of the
ellipse x2 + 2y2 = 6 right angle.
Ê cos2q sin2q ˆ
= 4a2 - 4a 2b 2 ÁË a2 + b2 ˜¯ x2 y2
a2 b2
7. Let the ellipse be + =1.

= 4a2 – 4b2 cos2q – 4a2sin2q B = (0, b), F = (ae, 0) and F¢ = (–ae, 0)
= 4a2(1 – sin2q) – 4b2cos2q
= 4a2cos2q – 4b2cos2q It is given that, FBF¢ is a right angle. So
= 4 cos2q(a2 – b2) FB2 + F¢B2 = (FF¢)2
= 4 cos2q(a2e2)
= (2ae cos q)2 fi a2e2 + b2 + a2e2 + b2 = 4a2e2
= [(a + ae cos q) – (a – ae cos q)]2 fi 2a2d2 = 2b2
= (PF1 – PF2)2
fi a2e2 = b2 = a2(1 – e2)
6. Given ellipse is
x2 + 4y2 = 4 fi a2d2 = a2 – a2e2
fi 2a2e2 = a2
fi 2e2 = 1

fi x2 + y2 = 1 …(i) fi e2 = 1
41 2

fi e= 1
2

5.62 Coordinate Geometry Booster

8. Ans. (c) Therefore, ar DOPN = 1 ¥ ON ¥ NP
Given ellipse is 2
16x2 + 25y2 = 400
= 1 ¥ ab(a2 - b2 ) sin q cos q
fi x2 + y2 = 1 2 a2sin2q + b2cos2q
25 16
= 1 ¥ ab(a2 - b2 ) q
Now, 2 a2tan q + b2cot
PF1 + PF2 = 2a
= 10 £ 1 ¥ ab(a2 - b2 ) = (a2 - b2 )
2 2ab 4

9. Given ellipse is x2 + y2 =1 at tan q = b
a2 b2 a

Y Thus, the point P is

N Ê a2 , b2 ˆ
P Á a2 + b2 ˜
Ë a2 + b2 ¯
X¢ O
L X By symmetry, we have four such points.

Ê a2 , ± b2 ˆ
Thus, Á ± a2 + b2 ˜
Y¢ a2 + b2 ¯
Ë

Let P lies in the first quadrant. 10. Given ellipse is 4x2 + 9y2 = 1 …(i)
So P = (a cos q, b sin q) Differentiating w.r.t. x, we get,
The equation of the tangent at P is
8x + 18y ◊ dy = 0
x cos q + y sin q = 1 dx
ab
fi dy = — 8x = - 4x
-1 dx 18y 9y
Now, ON =
= cos2q + sin 2q Since the tangent is parallel to 8x = 9y, so
a2 b2 — 4x = 8
9y 9
ab

a2sin2q + b2cos2q fi —x=2
y1
The equation of ON is
x sin q - y cos q = 0 fi x = –2y.
ba Put x = –2y in Eq. (i), we get
fi 4(4y2) + 9y2 = 1
and the equation of the normal at P is fi 25y2 = 1

ax sec q – by cosec q = a2 – b2 fi y2 = 1
25
So, OL = a2 - b2

a2sec2q + b2cosec2q fi y=±1
5
= (a2 - b2 ) sin q cos q
a2sin2q + b2cos2q when y = ± 1 , then x = 2
55
Now, NP = OL
(a2 - b2 ) sin q cos q Hence, the points are Ê - 2 , 1ˆ and Ê 2 , - 1ˆ .
ÁË 5 5¯˜ ÁË 5 5¯˜
fi NP =
a2sin2q + b2cos2q

Ellipse 5.63

11. Let the co-ordinates of P be (a cos q, b sin q) and of Q fi ËÁÊ1 - ae ˆ = Ê b2 ˆ
be (a cos q, a sin q) respectively. x ¯˜ ËÁ a2 ˜¯

Y fi Ê b2 ˆ = Ê ae ˆ
Q ËÁ1 - a2 ˜¯ ËÁ x ˜¯

R fi Ê a2 (1 - e2 ) ˆ = Ê ae ˆ
P ËÁ1 - a2 ¯˜ ËÁ x ¯˜

X¢ O X

fi e2 = Ê ae ˆ
ËÁ x ¯˜

Y¢ fi x=a
e

Let R(h, k) divides PQ in the ratio r : s. Hence, the result.
Then h = s(a cos q ) + r(a cos q ) = a cos q
13. Given ellipse is x2 + y2 = 1
r+s 95

and k = s(b sin q ) + r(a sin q ) = (ar + bs) sin q Y
r+s r+s
S
NL

Thus, h = cos q, k(r + s) = sin q X¢ R O PX
a (ar + bs)

Squaring and adding, we get N¢ L¢
Q

h2 + k2(r + s)2 =1 Y¢
a2 (ar + bs)2
b2 1- 5 = 2
Hence, the locus of R(h, k) is Now, e= 1- a2 = 93

x2 + (r + s)2 y2 =1 L = Ê b2 ˆ = Ê 2, 5ˆ
a2 (ar + bs)2 ÁË ae, a ˜¯ ÁË 3¯˜

which represents an ellipse. L¢ = Ê - b2 ˆ = Ê 2, - 5ˆ
12. Any tangent to the ellipse is ËÁ ae, a ¯˜ ÁË 3¯˜

x cos q + y sin q = 1 …(i) N = Ê b2 ˆ = ÊËÁ —2, 53ˆ¯˜
ab ËÁ — ae, a ˜¯

Its point of contact is P(a cos qc b sin q) and its slope and N¢ = Ê - b2 ˆ = Ê — 2, - 5ˆ
is - b cot q . Also the focus is S(ae, 0). ËÁ — ae, a ¯˜ ËÁ 3¯˜

a Now, the the tangent at L is

Any line through the focus S and the perpendicular to xx1 + yy1 = 1
tangent (i) is 95

y - 0 = a tan q (x - ae) …(ii) 5 y
b 3
fi 2x + =1
Also the equation of CP is
95
x + y =1
y - 0 = a tan q (x - 0) …(iii) fi 9/2 3
b

Eliminating q between Eqs (ii) and (iii), we get Thus, P = Ê 9 , 0¯ˆ˜ and S = (0, 3)
ÁË 2
Ê a2 ˆ x -x ae ¯ˆ˜
ÁË b2 ¯˜ ÁÊË = 1 Therefore,

ar(quad PQRS) = 4 ¥ ar(DOPS)

fi Ê x - aeˆ = Ê b2 ˆ = 4¥1¥9¥3
ÁË x ˜¯ ÁË a2 ¯˜ 22

= 27 s.u.

5.64 Coordinate Geometry Booster

14. Any tangent to the ellipse Thus, the equation of the common tangent is
x cos q + y sin q = 1
ab y=- 2 x+4 7
33
Let it meet the x-axis at A and y-axis at B. Then the co-
ordinates of A and B are This tangent meets the co-ordinate axes in P and Q re-
spectively.

A = Ê a q , 0˜¯ˆ and B = Ê 0, b q ˆ So, P = (2 Ê 4 7ˆ
ÁË cos ËÁ sin ¯˜ 7, 0) and Q = ÁË 0, 3 ˜¯

Let M(h, k) be the mid-point of AB.

Then 2h = a and 2k = b Lengthof PQ = 28 + 112 = 84 + 112 = 196 = 14
cos q sin q 3 3 33

fi h = a and k = b x2 y2
2 cos q 2 sin q a2 b2
16. Given ellipse is + =1
fi cos q = a and sin q = b
2h 2k The equation of any tangent to the given ellipse at P(a
cos q, b sin q) is
Squaring and adding, we get
fi x cos q + y sin q = 1
a2 + b2 =1 ab
4h2 4k 2
fi x + y =1
fi a2 + b2 = 4 a sec q b cosec q
h2 k2

Hence, the locus of M(h, k) is Y
B
a2 + b2 = 4
x2 y2 P

15. The equation of any tangent to the given ellipse is X¢ O A X

y = mx + 25m2 + 4 …(i)

Y Y¢
Q
M Here, A = (a sec q, 0) and B = (0, cosec q)
B(0, 2) Thus,

X¢ A¢(–5, 0) O P X ar(DOAB) = 1 ¥ a sec q ¥ b cosec q
B¢(1, –2) A(5, 0) 2

= ab
|sin 2q|

Y¢ ≥ ab, Ê |sin 2q| £1 fi 1 ≥ 1¯˜ˆ
ËÁ |sin 2q|
Hence OM = 4

fi m.0 - 0 + 25m2 + 4 = 4 17. Given ellipse is
m2 + 1 x2 + 4y2 = 4

fi 25m2 + 4 = 4 fi x2 + y2 = 1
m2 + 1 41

fi (25m2 + 4) = 16(m2 + 1) L¢ L
fi 9m2 = 12
fi 3m2 = 4 X¢ S¢ S X
fi m=± 2 V

3 Q P
=- 2 Y¢
( m < 0)
3

Ellipse 5.65

Thus, e = 1- b2 = 1- 1 = 3 Y
a2 4 2
M
Foci: B(0, 1)

S = (ae, 0) = ( 3, 0) X¢ A N O X
A(3, 0)

and S ¢ = (—ae, 0) = (- 3, 0)

End-points of locus recta:

L = Ê b2 ˆ = ÁËÊ 3, 1 ˜ˆ¯ Y¢
ÁË ae, a ˜¯ 2
Put x = 3(1 – y) in Eq. (i), we get
and L¢ = Ê — ae, b2 ˆ = Ê - 3, 1ˆ 9(1 – y)2 + y2 = 9
ËÁ a ¯˜ ËÁ 2¯˜
fi 9(y2 – 2y + 1) + y2 = 9
Thus, P = ÁÊË 3, - 1 ¯˜ˆ and Q = ËÊÁ - 3, - 1 ˜ˆ¯ fi 10y2 – 18y = 0
2 2 fi 5y2 – 9y = 0

As we know that, the focus is the mid-point of P and Q. fi y = 0, 9
Thus, the focus of a parabola is ÁÊË 0, - 12ˆ¯˜ . 5

and the length of PQ = 2 3 Thus, the y co-ordinate of M is 9 .
5

Now,

ar(DOAM) = 1 ¥ OA ¥ MN
2
Now, 4a = 2 3 fi a = 3
2 = 1 ¥3¥ 9
25

Thus, the vertices of a desired parabola = 27
10
= Ê 0, - 1 ± a¯ˆ˜ = Ê - 1 ± 3ˆ
ÁË 2 ÁË 0, 2 2 ¯˜ 19. Given ellipse is x2 + 4y2 = 16

Therefore, two desired parabolas are fi x2 + y2 = 1 …(i)
16 4
fi x2 = ± 4a Ê y - Ê - 1 ± 3ˆˆ X
ÁË ÁË 2 2 ˜¯ ˜¯ Y

fi x2 = 2 Ê y + 1 + 3ˆ P
3 ÁË 2 2 ¯˜ M
X¢ C Q
or

x2 = -2 Ê y + 1 - 3ˆ
3 ÁË 2 2 ˜¯

fi x2 = 2 3 y + (3 + 3) Y¢

or Let the co-ordinates of P be (4 cos q, 2 sin q).
x2 = - 2 3 y + (3 — 3) The equation of the normal to the given ellipse at P(4
cos q, 2 sin q) is
18. Given ellipse is …(i)
x2 + 9y2 = 9 4x sec q – 2y cosec q = 42 – 22
fi 4x sec q – 2y cosec q = 12
fi x2 + y2 = 1 fi 2x sec q – y cosec q = 6
91 So, the point Q is (3 cos q, 0).
Let M(h, k) be the mid-point of PQ. So
The equation of the auxiliary circle is
x2 + y2 = 9 h = 7 cos q and k = sin q
2
and the equation of the line AB is

fi x + y =1
31

x = 3(1 – y)

5.66 Coordinate Geometry Booster

\ 4h2 + k 2 = 1 Solving Eqs (i) and (ii), we get
49
x = 11, y = 8
Hence, the locus of M(h, k) is 55
4x2 + y2 = 1
49 …(ii) Hence, the orthocentre is Ê 11 , 8ˆ .
ËÁ 5 5¯˜
The equation of the latus rectum of (i) is
22. Let the point be M(x, y).
x = ± ae = ± 4 ¥ 3 = ± 2 3 It is given that PM = Length of perpendicular from Q to
2 AB

Put x = ± 2 3 in Eq. (ii), we get x + y -1
fi (x — 3)2 + ( y - 4)2 = 3
2
y2 = 1 - 48 = 1 Ê 1ˆ
49 49 ËÁ 3¯˜ + 1

fi y=±1 fi 10((x – 3)2 + (y – 4)2) = (x + 3y – 3)2
7 fi 10(x2 + y2 – 6x – 8y + 25) = (x + 3y – 3)2
fi 10(x2 + y2 – 6x – 8y + 25)
Hence, the required points are ËÁÊ ± 2 3, ± 71˜¯ˆ .
= x2 + 9y2 + 9 + 6xy – 6x – 18y
Y P(3, 4) fi 9x2 + y2 – 6xy – 54x – 62y + 241 = 0
A
23. Equation of ellipse is
(y + 2)(y – 2) + l(x + 3)(x – 3) = 0

X¢ O B X Y
B(0, 4) Y

S(–3, 2) P(3, 2)

Y¢

20. Given ellipse is x2 + y2 = 1 X¢ C X
94 R(–3, –2) Q(3, –2)
…(i)
Hence AB is the chord of contact. So
…(ii)
xx1 + yy1 = 1 …(i) Y¢
94 …(ii)
fi 3x + 4y = 1 which is passing through B(0, 4). So
94 6 ¥ 2 + l(–9) = 0

fi x + y =1 fi 9l = 12
3
fi l=4
Solving Eqs (i) and (ii), we get 3

A = Ê - 9 , 8ˆ , B = (3, 0) Thus, the required ellipse is
ÁË 5 5¯˜ ( y2 - 4) + 4 (x2 - 9) = 0
3
21. The equation of the altitude through A is
fi 3(y2 – 4) + 4(x2 – 9) = 0
y=8 fi 4x2 + 3y2 = 48
5
fi x2 + y2 = 1
and the slope of AB is - 1 12 16
3
Thus, e = 1 - 12 = 1
The equation of the altitude through P is 16 2

y – 4 = 3(x – 3)

Ellipse 5.67

24. Given ellipse is x2 + y2 = 1 Therefore,
43
8 D1 - 8D 2 = 8 ¥ 45 5 - 8 ¥ 9
Y 5 58 2
P
= 45 – 36

=9

25. Clearly, the point P(h, 1) lies on the ellipse. So

X¢ O M(h, 0) R(x, 0) X h2 + 1 = 1
63

Q fi h2 = 1 - 1 = 2
6 33

Y¢ fi h2 = 2
2
when x = h, then
fi h2 = 4
y2 = 1 - h2 = 4 - h2 fi h = ±2
3 44 Now, the tangent at (2, 1) is
fi y = ± 3 4 - h2
fi 2x + y = 1
2 63
x + y =1
Thus, the points P and Q are 33

Ê 3 h2 ˆ Ê 3 h2 ˆ fi x+y=3
ËÁ h, 2 ˜¯ ËÁ 2 ¯˜
4 - and h, - 4 - Hence, the value of h is 2.

Let the tangents at P and Q meet at R(x1, 0). 26. The equation of P1 is y2 – 8x = 0 and P2 is y2 + 16x = 0
Therefore PQ is a chord of contact. So Tangent to y2 = 8x passes through (–4, 0)

xx1 = 1 0 = m1(-4) + 2
4 m1

x= 4 1 = 2
x1 m12
fi

which is an equation of PQ at x = h. So Also, tangent to y2 + 16x = 0 passes through (2, 0)

h= 4 0 = m2 ¥ 2 - 4
x1 m2

m22 = 2

fi x1 = 4 Hence, the value of Ê1 + m22 ˆ
h ÁË m12 ¯˜

Now, D(h) = area of DPQR =2+2

= 1 ¥ PQ ¥ RT = 4.
2
27. For the given line, point of contact for

=1¥2 3¥ 4 - h2 ¥ (x1 - h) E1: x2 + y2 = 1 is Ê a2 , b2 ˆ
22 a2 b2 ËÁ 3 3 ˜¯

fi D¢(h) = 3(4 + 2h2 ) ¥ 4 - h2 h and for E2: x2 + y2 = 1 is Ê B2 , A2 ˆ
2 B2 A2 ËÁ 3 3 ˜¯

which is always decreasing. Point of contact of x + y = 3 and the circle is (1, 2)

Also, the general point on x + y = 3 can be taken as

D1 = Maximum of D(h) = 45 5 at h = 1 ÁËÊ1 r ,2 rˆ
8 2 2 2 ˜¯

D2 = Minimum of D(h) = 9 at h =1 where r = 2 2
2 3

5.68 Coordinate Geometry Booster

ÊÁË 1 83¯˜ˆ ËÊÁ 5 34˜ˆ¯ For orthocentre: One altitude is y = 0 (MN is per-
3 3
So, required points are , and , pendicular)

Comparing with points of contact of ellipse Other altitude is

a2 = 5, B2 = 8 and b2 = 4, A2 = 1 (y - 6) = 5 ËÊÁ x - 32˜¯ˆ
26

e1e2 = 2 7 and e12 + e22 = 43 Hence, the orthocentre is ÁÊË - 9 , 0˜¯ˆ
10 40 10

28. (i) (ii) Equation of tangent at M and N are

x ± y 6 =1
68

Thus, R is (6, 0)
Equation of normal at M is

(y - 6) = - 6 ÁÊË x - 3 ˜ˆ¯
2 2

Thus, Q is ÁÊË 7 , 0˜ˆ¯
2

ar(DMQR) = 1 ¥ 6 ¥ 5 = 5 6
2 24

Here, e = 1, F1 = (-1, 0), F2 = (1, 0) ar(MF1NF2 ) = 6+3 6=4 6
3 22 2

The given parabola is y2 = 4x Ratio = 5 6 : 4 6 = 5
428
Thus, M = ÁËÊ 3 , 6¯˜ˆ and N = ÁËÊ 3, - 6¯˜ˆ
2 2

Hyperbola 6.1

CHAPTER Hyperbola

6

CONCEPT BOOSTER i.e. S¢P – SP = 2a P
S
1. INTRODUCTION S¢

The word ‘hyperbola’ has ben derived from the Greek lan- Definition 3
guage meaning ‘over-thrown’ or ‘excessive’, from which the
English term hyperbole is also derived. The term hyperbola is A conic section is said to be a hyperbola, if its eccentricity is
believed to have been coined by Apollonius of Perga (c.262– more than 1, i.e. e > 1.
c.190 BC), who was a Greek geometer and astronomer noted
for his writings on conic sections. His innovative method- Definition 4
ology and terminology, especially in the field of the conic
sections, the conics. According to him, hyperbola, the incli- A conic
nation of the plane to the base of the cone exceeds that of the ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
side of the cone
represents a hyperbola if
2. MATHEMATICAL DEFINITIONS (i) D π 0
(ii) h2 – ab > 0,
Definition 1 ahg
It is the locus of a point which moves in a plane in such a way where D = h b f
that its distance from a fixed point (focus) to a fixed straight gf c
line (directrix) is constant (> 1), i.e. = abc + 2fgh – af2 – bg2 – ch2

SP = e (> 1) Definition 5
PM
The section of a double right circular cone and a plane is said
MP to be a hyperbola if the plane is parallel to the axis of a double
right circular cone.
S (Focus)

Z (Directrix)

Definition 2

A hyperbola is a conic section defined as the locus of all
points P in the plane, the difference of whose distances from
two fixed points (the foci S and S¢) is a constant,

6.2 Coordinate Geometry Booster

Definition 6 fi x2 (e2 - 1) - a2 y2 - 1) = 1
a2 (e2 - 1) (e2
Let z, z1 and z2 be three complex numbers such that
|z – z1| – |z – z2| = k, fi x2 - y2 = 1,
a2 b2
where k < |z1 – z2|
Then the locus of z is known as a hyperbola.

P(z) where, b2 = a2(e2 – 1)
which is the standard equation of a hyperbola.

S¢(z2) A¢ 2a A S(z1) 4. DEFINITION AND BASIC TERMINOLOGY OF THE

HYPERBOLA x2 - y 2 =1
a2 b 2

3. STANDARD EQUATION OF A HYPERBOLA (i) Centre: C(0, 0). All chords passing through C and are

Y bisected at C.
(ii) Vertices: A(a, 0) and A¢(–a, 0)
BM P (iii) Co-vertices: B(0, b) and B¢(0, –b)
(iv) Transverse axis: AA¢ = 2a
(v) Conjugate axis: BB¢ = 2b

(vi) Eccentricity: e = 1 + b2
a2
X¢ A¢ Z¢ C Z A N S X
(vii) Latus rectum: Any chord passing through the focus

and perpendicular to the axis is known as latus rectum.

B¢ 2b2
The length of the latus rectum = .
Y¢
a

Let ZM be the directrix, S be the focus and SZ be the per- (viii) End-points of a latus recta: Let LL¢ and L1L¢1 be
two latus recta pass through the focus S(ae, 0) and
pendicular to the directrix.
S¢(–ae, 0).

From the definition of hyperbola, we can write Then, Ê b2 ˆ ; L¢ Ê ae, - b2 ˆ ;
LÁË ae, a ˜¯ ËÁ a ¯˜
SA = e.AZ …(i)

and SA¢ = eA¢Z …(ii) Ê b2 ˆ L1¢ Ê b2 ˆ
ËÁ a ˜¯ ÁË a ˜¯
Let the length of AA¢ = 2a and C be the mid-point of AA¢. L1 -ae, ; -ae, -

Adding Eqs (i) and (ii), we get

SA + SA¢ = e(AZ + A¢Z) Y
L1
fi (CS – CA) + (CS + CA) = e(AZ + AA¢ – AZ) L

fi 2CS = e(AA¢) = 2e

fi CS = ae …(iii) X¢ S¢ SX

Subtracting Eq. (i) from Eq. (ii), we get O

(SA¢ – SA) = e(A¢Z – AZ)

fi AA¢ = e((CA¢ + CZ) – (CA – CZ)) L¢1
Y¢
fi 2a = e ◊ 2CZ L¢

fi CZ = a …(iv) (ix) Equation of the latus recta: x = ±ae
e (x) Co-ordinates of Foci: S(ae, 0) and S¢(–ae, 0)
(xi) Distances between two foci (Focal length): 2ae
Let P(x, y) be any point on the curve and PM be the (xii) Equation of directrices: x = ± a

perpendicular to the directrix. e

Now from the definition of hyperbola, we get,

SP2 = e2 ◊ PM2 2

fi (x - ae)2 + y2 = e2 Ê ex - aˆ = (ex - a)2 2a
ÁË e2 ˜¯ (xiii) Distance between the directrices:

e

fi x2 – 2aex + a2e2 + y2 = (e2x2 – 2aex + a2) (xiv) Focal distances
SP = ex – a and S¢P = ex + a
fi x2(e2 – 1) – y2 = a2(e2 – 1)
(xv) |S¢P – SP| = 2a

Hyperbola 6.3
S¢
P Join NU.
S Let –UCN = j
Here P and U are the corresponding points of the hy-
perbola and the auxiliary circle and f is the eccentric
angle of P, where 0 £ f < 2p.
Now, U = (a cos j, a sin j)

(xvi) Diameter: Any chord passing through the centre of the Also, x = CN = CN ◊CU = a.secj
hyperbola is known as the diameter. CU

(xvii) Focal chord: Any chord passing through the focus is Thus the co-ordinates of P be (a sec f, y).
called the focal chord. Since the point P lies on the hyperbola, so

YP a2sec2j - y2 =1
a2 b2
X¢ C S X
fi sec2j - y2 =1
Q b2

Y¢ fi y2 = sec2j -1 = tan 2j
b2
(xviii) Auxiliary circle: Any circle is drawn with centre C
and the transverse axis as a diameter is called the aux- fi y2 = b2 tan2j
iliary circle. fi y = ±b tan j
The equation of the auxiliary circle is given by fi y = b tan j
x2 + y2 = a2 (since P lies in the first quadrant)
Hence, the parametric equations of the hyperbola are
Y x = a sec j and y = b tan j.
(xx) Any point on the hyperbola can be considered as (a sec
X¢ A¢ O AX j, b tan j).
(xxi) Eccentric angle: If two points P(a sec j, b tan j) and
U(a cos j, a sin j) are the corresponding points on the
hyperbola and the auxiliary circle, then f is called the
eccentric angle of the point P on the hyperbola, where
0 £ j < 2p.

Y¢ Y

(xix) Parametrics equations: Let the auxiliary circle be P
x2 + y2 = a2 and U (a cos j, b sin j) be any point on the U
auxiliary circle.
X¢ f NX
O
Y

P Y¢
U

X¢ f NX (xxii) Conjugate hyperbola: Corresponding to every hyper-
O
bola, there exists a hyperbola in which the conjugate

and the transverse axes of one is equal to the transverse

and the conjugate axes of the other. Such types of hy-

Y¢ perbolas are known as the conjugate hyperbola.

x2 y2 The equation of the conjugate hyperbola to the
a2 b2
Let P(x, y) be any point on the hyperbola - =1. hyperbola x2 - y2 =1 is
a2 b2
Draw PN perpendicular to x-axis
x2 y2
Let NU ba tangent to the auxiliary circle. a2 - b2 = -1

6.4 Coordinate Geometry Booster

(xxvii) Condition of a focal chord: The equation of
the chord joining the points P(a sec j1, b tan j1)
and Q(a sec j2, b tan j2) is

x cos Ê j1 - j2 ˆ - y sin Ê j1 + j2 ˆ = cos Ê j1 + j2 ˆ
a ÁË 2 ˜¯ b ËÁ 2 ¯˜ ËÁ 2 ˜¯

which is passing through the focus S(ae, 0). So

e cos Ê j1 - j2 ˆ = cos Ê j1 + j2 ˆ
ËÁ 2 ˜¯ ÁË 2 ¯˜
It is also known as vertical hyperbola.
(xxiii) Rectangular or equilateral hyperbola: If the semi- cos Ê j1 - j2 ˆ
ÁË 2 ¯˜
transverse axis is equal to the semi-conjugate axis of fi = 1
a hyperbola, i.e. a = b, then it is known as the rectan-
gular hyperbola or the equilateral hyperbola. cos Ê j1 + j2 ˆ e
ÁË 2 ¯˜
Y
cos Ê j1 - j2 ˆ - cos Ê j1 + j 2 ˆ
B ËÁ 2 ˜¯ ÁË 2 ˜¯
fi = 1- e

cos Ê j1 - j2 ˆ + cos Ê j1 + j2 ˆ 1+ e
ÁË 2 ¯˜ ËÁ 2 ¯˜
X¢ A¢ C AX
j1 j2 1- e
fi tan Ê 2 ˆ tan Ê 2 ˆ = 1+ e
ËÁ ˜¯ ÁË ¯˜
B¢
(xxviii) Condition of a focal chord with respect to
Y¢ eccentricity (e): As we know that if the chord
passing through the focus, then
The equation of a rectangular hyperbola is
x2 – y2 = a2. e cos Ê j1 - j2 ˆ = cos Ê j1 + j2 ˆ
ËÁ 2 ˜¯ ËÁ 2 ˜¯
Clearly its eccentricity (e)

= 1 + b2 = 1 + a2 = 1+1 = 2 fi e ¥ 2 sin Ê j1 + j 2 ˆ cos Ê j1 - j2 ˆ
a2 a2 ÁË 2 ˜¯ ÁË 2 ¯˜

(xxiv) Equation of a hyperbola whose axes are paral- = 2 sin Ê j1 + j2 ˆ cos Ê j1 + j2 ˆ
ÁË 2 ¯˜ ËÁ 2 ¯˜
lel to the co-ordinate axes and the centre (h, k) is

(x - h)2 - (y - k)2 =1, fi e ¥ (sin j1 + sin j2) = sin (j1 + j2)
a2 b2
fi e = sin(j1 + j2 )
where the foci are (h ± ae, k) and the directrix is (sinj1 + sinj2 )

x = h± a which is the required condition.
e

Y (xxix) Rule to find out the centre of the hyperbola
P(x, y) If f(x, y) = 0 is the equation of a hyperbola, the

centre of the hyperbola is obtained by the rela-

C(h, k) tion df =0 and df = 0.
dx dy
X¢ O X
Let

f(x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

be the equation of a hyperbola.

Y¢ Now, d f = 0
dx
(xxvi) Equation of the chord joining the points P(j1) …(i)
and Q(j2): The equation of the chord joining the fi 2ax + 2hy + 2g = 0
points fi ax + y + g = 0

P(a sec j1, b tan j1) and Q(a sec j2, b tan j2) is and d f = 0
dy

x cos Ê j1 - j2 ˆ - y sin Ê j1 + j2 ˆ = cos Ê j1 + j2 ˆ fi 2hx + 2by + 2f = 0
a ÁË 2 ˜¯ b ÁË 2 ˜¯ ËÁ 2 ¯˜ fi hx + by + f = 0
…(ii)

Hyperbola 6.5

Solving Eqs (i) and (ii), we get the required 6. INTERSECTION OF A LINE AND A HYPERBOLA
centre.
(xxx) Polar form of a hyperbola Y
The equation of hyperbola is

x2 - y2 =1 X¢ C
a2 b2
X

Putting x = r cos q and y = r sin q, we get y = mx + c
Y¢
1 = cos2q - sin 2q
r2 a2 b2

fi r2 = a2b2 = a2 (e2 -1) Let the hyperbola be x2 - y2 =1 …(i)
b2cos2q - a2sin2q (e2cos2q -1) a2 b2 …(ii)

(xxxi) Polar form of a hyperbola if centred at focus and the line be y = mx + c

The equation of a hyperbola is From Eqs (i) and (ii), we get

x2 - y2 =1 x2 - (mx + c)2 =1
a2 b2 a2 b2

If centred C(0, 0) at the focus (ae, 0), then the fi b2x2 – a2(mx + c)2 = a2b2

polar form of the hyperbola becomes fi (a2m2 – b2)x2 + 2mca2x + a2(b2 + c2) = 0
Now, D = 4m2c2a4 – 4a2(a2m2 – b2)(b2 + c2)
r = a(e2 -1)
1- e cosq

= 4(m2c2a4 – a4m2b2 – a4m2c2 + a2b4 + a2b2c2)

= 4(–a4m2b2 + a2b4 + a2b2c2)

r = 4a2b2(b2 + c2 – a2m2)
q
(i) The line y = mx + c will never intersect the hyper-
a (0, 0) bola, if
D<0
(xxxii) Polar form of a rectangular hyperbola fi c2 < a2m2 – b2
The equation of rectangular hyperbola is
x2 – y2 = a2 (ii) The line y = mx + c will be a tangent to the hyper-
Putting x = r cos q and y = r sin q, we get, bola if D = 0
r2 cos 2q = a2 fi c2 = a2m2 – b2
This is known as the condition of tangency.
5. POSITION OF A POINT WITH RESPECT TO A HYPERBOLA
(iii) The line y = mx + c will intersect the hyperbola in
two real and distinct points, if
D>0
fi c2 > a2m2 – b2

(iv) Any tangent to the hyperbola can be considered as

y = mx + a2m2 - b2 .

(v) Co-ordinates of the point of contact.
If the line y = mx + c be a tangent to the hyperbola

x2 - y2 =1, the co-ordinates of the point of contact is
a2 b2

Ê a2m , ± b2 ˆ .
ÁË ± c c ¯˜

which is also known as m-point on the hyperbola.

The point (x1, y1) lies outside, on or inside the hyperbola (vi) Number of tangents: If a point lies outside, on

x2 y2 and inside of a hyperbola, the number of tangents
a2 b2
- =1 according as are 2, 1 and 0 respectively.

x12 y12 (vii) If the line lx + my + n = 0 be a tangent to the
a2 b2
- -1< 0, = 0, > 0 hyperbola x2 - y2 = 1, then
a2 b2

a2l2 – b2m2 = n2

6.6 Coordinate Geometry Booster
Y
7. DIFFERENT FORMS OF TANGENTS

(i) Point form: The equation of the tangent to the hy-

perbola x2 - y2 =1 at the point (x1, y1) is P(h, k)
a2 b2

xx1 - yy1 =1 X¢ O X
a2 b2

(ii) Parametric form: Equation of tangent to the hyper-

bola x2 - y2 =1 at (a sec js b tan j) is
a2 b2
Y¢

x secj - y cosecj = 1 Let it passes through the point (h, k).
ab Then,

(iii) The point of intersection of the tangents at P(q) and k = mh + a2m2 - b2

Q(j) on the hyperbola x2 - y2 =1 is fi (k - mh)2 = ( a2m2 - b2 )2
a2 b2

Ê a cos Ê q -j ˆ b sin Ê q + j ˆ ˆ fi k2 + m2h2 – 2kmh = a2m2 – b2
Á ËÁ 2 ˜¯ ÁË 2 ˜¯ ˜
Á , ˜ fi (h2 – a2)m2 – 2khm + (k2 + b2) = 0
ÁËÁ ˜¯˜
cos Ê q + j ˆ cos Ê q + j ˆ It has two roots, say m1 and m2. Then
ËÁ 2 ˜¯ ËÁ 2 ˜¯
m1 ◊m2 = k2 + b2
h2 - a2
(iv) Slope form: The equation of the tangent to the hyper-

bola x2 - y2 =1 in terms of the slope m is fi k2 + b2 = -1
a2 b2 h2 - a2

y = mx + a2m2 - b2 fi k2 + b2 = –h2 + a2
fi h2 + k2 = a2 – b2
The co-ordinates of the point of contact are Hence, the locus of (h, k) is

Ê a2m , ± b2 ˆ . x2 + y2 = a2 – b2
ËÁ ± a2m2 - b2 a2m2 ¯˜
- b2 Notes

(v) The equation of the tangent to the conjugate hy- x2 - y2 = 1 exists
a2 b2
x2 y2 1. The director circle of a hyperbola
a2 b2
perbola - = -1 is only when a > b.
2. If a < b, the equation of the director circle x2 + y2 =
y = mx + b2 - a2m2
a2 – b2 does not exist.
The co-ordinates of the point of contact are
3. The equation of the director circle to the conjugate

Ê a2m , ± b2 ˆ hyperbola x2 - y2 = -1 is
ËÁ ± b2 - a2m2 b2 - a2m2 ¯˜ a2 b2

(vi) The line y = mx + c be a tangent to the hyperbola x2 + y2 = b2 – a2

x2 y2 It exists only when b > a.
a2 b2
- =1, if 4. If b < a, the equation of the director circle does not

c2 = a2m2 – b2 exist.

(vii) Director circle: The locus of the point of intersec- (viii) Pair of tangents

tion of two perpendicular tangents to a hyperbola Y

is known as the director circle of the hyperbola.

The equation of the director circle to the hyper-

bola x2 - y2 =1 is
a2 b2
x2 + y2 = a2 – b2 X¢ O X

The equation of any tangent to a hyperbola is P(x1, y1)
y = mx + a2m2 - b2 .

Y¢

Hyperbola 6.7

The combined equation of the pair of tangents 9. CHORD OF CONTACT

drawn from a point P(x1, y1) lies outside of the hy- If a tangent is drawn from a point P(x1, y1) to the hyperbola

perbola x2 - y2 =1 to the hyperbola x2 - y2 = 1 is x2 - y2 =1 touching the hyperbola at Q and R, the equation
a2 b2 a2 b2 a2 b2

SS1 = T2 of the chord of contact QR is

Ê x2 y2 ˆ Ê x12 y12 ˆ Ê xx1 yy1 -1˜¯ˆ 2 xx1 - yy1 =1
ÁË a2 b2 1˜¯ ÁË a2 b2 - 1¯˜ ËÁ a2 b2 a2 b2
fi - - - = -

8. DIFFERENT FORMS OF NORMALS

P

F1 F2

10. EQUATION OF THE CHORD BISECTED AT A POINT

The equation of the chord of the hyperbola x2 - y2 =1
a2 b2
(i) Point form: The equation of the normal to the hyper- bisected at the point (x1, y1) is

bola x2 - y2 =1 at (x1, y1) is SS1 = T2
a2 b2
Ê x2 y2 ˆÊ x12 y12 ˆ Ê xx1 yy1 -1˜ˆ¯ 2
a2x + b2 y = a2 + b2 i.e. ËÁ a2 - b2 -1¯˜ ËÁ a2 - b2 - 1˜¯ = ÁË a2 - b2
x1 y1
Y

(ii) Parametric form: The equation of the normal to the B

hyperbola x2 - y2 =1 at (a sec j, b tan j) is
a2 b2

ax cos j + by cot j = a2 + b2 X¢ C M(x1, y1)
X
(iii) Slope form: The equation of the normal to the hyper- Y¢
A
bola x2 - y2 =1 in terms of the slope m is 11. POLE AND POLAR
a2 b2 B
Y
y = mx m(a2 + b2 ) Q
a2 - m2b2

The co-ordinates of the point of contact are

Ê a2 , mb2 ˆ
ÁË ± a2 - b2m2 a2 - b2m2 ¯˜

X¢ A X
O
(iv) The line y = mx + c will be a normal to the hyperbola if
P(x1, y1) Polar
A¢
Ê m2 (a2 + b2)2 ˆ
c2 = ËÁ (a 2- m2b2 ) ˜¯

which is also known as the condition of the normalcy Y¢ R B¢
to a hyperbola.
Let P be any point inside or outside of the hyperbola. Sup-
pose any straight line through P intersects the hyperbola at

6.8 Coordinate Geometry Booster

A and B. Then the locus of the point of intersection of the Hence, the locus of the mid-point is y= b2x .
tangents to the hyperbola at A and B is called the polar of the a2m
given point P with respect to the hyperbola and the point P is
called the pole of the polar. 13. CONJUGATE DIAMETERS

The equation of the polar from a point (x1, y1) to the Two diameters are said to be conjugate when each bisects all
the chords parallel to the others.
hyperbola x2 - y2 =1 is xx1 - yy1 = 1
a2 b2 a2 b2 Y

Properties related to pole and polar
(i) The polar of the focus is the directrix.
(ii) Any tangent is the polar of the point of contact.

(iii) The pole of a line lx + my + n = 0 with respect to the

x2 y2 X¢ X
a2 b2
ellipse - =1 is

Ê - a2l , - b2m ˆ Y¢
ÁË n n ¯˜

(iv) The pole of a given line is same as point of intersection If y = m1x and y = m2x be two conjugate diameters,
then,
of tangents at its extremities.
m1 ◊m2 = b2
(v) If the polar of P(x1, y1) passes through Q(x2, y2), the a2
polar of Q(x2, y2) goes through P(x1, y1) and such points
are said to be conjugate points. Properties of diameters

(vi) If the pole of a line lx + my + n = 0 lies on the another Property-I
line l¢x + m¢y + n¢ = 0, the pole of the second line will If a pair of the diameters be conjugate with respect to a
lie on the first and such lines are said to be conjugate hyperbola, they are conjugate with respect to its conjugate
lines. hyperbola also.

12. DIAMETER Property-II
The parallelogram formed by the tangents at the extremities
The locus of the mid-points of a system of parallel chords of of the conjugate diameters of a hyperbola has its vertices ly-
a hyperbola is called a diameter of the hyperbola. ing on the asymptotes and is of conjugate area.

Y Property-III
P(x1, y1) If the normal at P meets the transverse axis in G, then
SG = e.SP
X¢ C R(h, k)
AA Also, the tangent and the normal bisects the angle
between the focal distances of P.

Q(x1, y1) Property-IV
If a pair of conjugate diameters meet the hyperbola in P and
Y¢ P¢ and its conjugate in D and D, then the asymptotes bisect

PD, P¢D ◊ PD¢ and P¢D¢.

The equation of a diameter to a hyperbola x2 - y2 =1 is 14. ASYMPTOTES
a2 b2
b2
y = a2m x Asymptote Asymptote

y= b x Y y = b x
a a
Let (h, k) be the mid-point of the chord y = mx + c of

the hyperbola x2 - y2 =1. b x2 – y2 –1
a2 b2 a2 b2
–a a
Then, T = S1 –b a x

fi xh - yk -1= h2 - k2 -1
a2 b2 a2 b2

b2h
Slope = a2k

Hyperbola 6.9

An asymptotes of any hyperbola or a curve is a straight line be the equations of the hyperbola, its asymptote and its
conjugate, respectively, we can write from the above
which touches it in two points at infinity. equations,

The equation of the asymptotes of the hyperbola C + H = 2A

x2 - y2 =1 are
a2 b2
15. RECTANGULAR HYPERBOLA
y=± bx
a A hyperbola is said to be rectangular, if the angle between its
asymptote is 90°.
As we know that the difference between the 2nd de-
Ê bˆ
gree curve and pair of asymptotes is constant. Thus, 2 tan -1 ËÁ a ˜¯ = 90°

Given hyperbola is x2 - y2 =1 Ê bˆ
a2 b2 ËÁ a ¯˜
fi tan -1 = 45°

The equation of the pair of asymptotes are

x2 - y2 + l = 0 …(i) fi b = tan 45°
a2 b2 a

Equation (i) represents a pair of straight lines, then fi b=a
Hence, the equation of the rectangular hyperbola is
D=0
x2 – y2 = a2
fi 1 Ê - 1 ˆ ◊l + 0 - 0 - 0 - l ◊0 = 0 The eccentricity (e) of the rectangular hyperbola is
a2 ËÁ b2 ˜¯

fi l=0 x2 y2 e= 1+ a2 = 1+1 = 2
a2 b2 a2
From Eq. (i) we get the equation of asymptotes as - = 0.

fi y=±bx 16. RECTANGULAR HYPERBOLA XY = C2
a

14.1 Important Points Related to Asymptotes The equation of a rectangular hyperbola is x2 – y2 = a2 and its
asymptote are
(i) The asymptotes pass through the centre of the hyper-
x – y = 0 and x + y = 0,
bola. where the asymptotes are inclined at 45° and 135°,
respectively
(ii) A hyperbola and its conjugate have the same asymp-
If we rotate the axes through an angle of 45° in clockwise
totes. direction without changing the origin, then we replace x by
[x cos (–45°) – y sin (–45°)] and y by [x sin (–45°) + y cos
(iii) The equation of the hyperbola and its asymptotes differ (–45°)],

by a constant only.

(iv) The equation of the asymptotes of a rectangular hyper-

bola x2 – y2 = a2 are y = ±x i.e. x by Ê x+ yˆ and y by Ê -x + yˆ
ÁË 2 ˜¯ ÁË 2 ¯˜
(v) The angle between the asymptotes of the hyperbola

x2 - y2 =1 is Then the equation, x2 – y2 = a2 reduces to
a2 b2
Ê x + yˆ 2 Ê -x + yˆ 2
ËÁ 2 ¯˜ ËÁ 2 ¯˜
2 tan -1 Ê b ˆ - = a2
ÁË a ¯˜
1 (2xy + 2xy) = a2
(vi) The bisectors of the angles between the asymptotes are fi 2

the co-ordinate axes. fi xy = a2 = c2 (say)
2
(vii) No tangent to the hyperbola can be drawn from its cen-
fi xy = c2
tre.

(viii) Only one tangent to the hyperbola can be drawn from a

point lying on its asymptotes other than its centre. y
y=x
(ix) Two tangents can be drawn to the hyperbola from any

of its external points which does not lie at its asymp-

totes.

(x) Let H: x2 - y2 -1= 0 , A: x2 - y2 = 0 and x
a2 b2 a2 b2 v = –x

C: x2 - y2 +1= 0
a2 b2

6.10 Coordinate Geometry Booster

x2 = y1 = 1 xy = 1 Ê cˆ
O ÁË t ¯˜
(vi) The equation of the tangent at ct, to the hyperbola

F1 q xy = c2 is x + yt = 2c .
F2 t

(vii) The equation of the normal at (x1, y1) to the hyperbola
xy = c2 is

xx1 - yy1 = x12 - y12

Rectangular hyperbola (viii) The equation of the normal at t to the hyperbola xy = c2

Properties of rectangular hyperbola is
xt3 – yt – ct4 + c = 0

(i) The asymptotes of the rectangular hyperbola xy = c2 are 17. REFLECTION PROPERTY OF A HYPERBOLA

x = 0 and y = 0 If an incoming light ray passing through one focus, after
striking the convex side of the hyperbola, it will get reflected
(ii) The parametric equation of the rectangular hyperbola towards other focus.

xy = c2 are x = ct and y = c . Y
t

(iii) Any point on the rectangular hyperbola xy = c2 can be

considered as Ê ct, cˆ .
ËÁ t ¯˜

(iv) The equation of the chord joining the points t1 and t2 is X¢ S¢ C SX
x + yt1t2 – c(t1 + t2) = 0 Y¢

(v) The equation of the tangent to the rectangular hyperbola

xy = c2 at (x1, y1) is
xy1 + x1y = c2

EXERCISES

LEVEL I 5. Find the equation of the hyperbola, whose one focus is
(Problems Based on Fundamentals)
(2, 1), the directrix is x + 2y = 1 and the eccentricity is
ABC OF HYPERBOLA
2.
1. Find the centre, the vertices, the co-vertices, the length
of transverse axis, the conjugate axis and the latus rec- 6. Find the equation of the hyperbola, whose distance be-
tum, the eccentricity, the foci and the equation of direc-
trices of each of the following hyperbolas. tween foci is 16 and the eccentricity is 2 .

(i) x2 - y2 = 1 7. Find the equation of the hyperbola,whose foci are (6, 4)
94
and (–6, 4) and the eccentricity is 2.
(ii) x2 - y2 = -1
16 9 8. Find the eccentricity of the hyperbola whose latus rec-

(iii) 9x2 – 16y2 – 36x + 96y – 252 = 0 tum is half of its transverse axis.
2. Find the equation of the hyperbola, whose centre is
9. If e1 and e2 be the eccentricities of a hyperbola and its
(1, 0), one focus is (6, 0) and the length of transverse
axis is 6. conjugate, prove that 1 + 1 =1.
3. Find the equation of the hyperbola, whose centre is e12 e22
(3, 2), one focus is (5, 2) and one vertex is (4, 2).
4. Find the equation of the hyperbola, whose centre is 10. An ellipse and a hyperbola are confocal (have the same
(–3, 2), one vertex is (–3, 4) and eccentricity is 5/2.
focus) and the conjugate axis of the hyperbola is equal

to the minor axis of the ellipse. If e1 and e2 are the ec-
centricities of the ellipse and the hyperbola, prove that
e12 + e22 = 2 .

11. Find the centre of the hyperbola

4(2y – x – 3)2 – 9(2x + y – 1)2 = 80

12. Find the centre of the hyperbola

3x2 – 5y2 – 6x + 20y – 32 = 0

Hyperbola 6.11

13. Prove that the straight lines TANGENT AND TANGENCY

x- y = 2013 and x+ y =1 30. Find the equation of the tangent to the curve x2 – y2 – 8x
ab ab 2013
+ 2y + 11 = 0 at (2, 1).

always meet on a hyperbola. Ê 2 ˆ 31. Find the equation of the tangent to the curve 4x2 – 3y2 =
ÁË 2 ¯˜
14. Prove that the locus represented by x = 3 1 + t and 24 at y = 2.
1 - t
32. Find the angle between the tangents to the curve

y = 4t is a hyperbola. 9x2 – 16y2 = 144 drawn from the point (4, 3).
t2 -1
15. Prove that the locus represented by x = 1 (et + e-t ) 33. Find the equations of the common tangent to the curves

2 x2 - y2 =1 and x2 - y2 = -1 .
a2 b2 b2 a2
and y = 1 (et - e-t ) is a hyperbola.
2 34. Find the equations of the common tangents to the

16. If the equation x2 + y2 = 1 represents a curves x2 - y2 = 1 and x2 + y2 =1.
2014 - l 2013 - l 94 9 4

hyperbola, find l. 35. Find the equation of the common tangents to the curves

17. If the foci of the ellipse x2 + y2 =1 and the hyperbola x2 - y2 = 1 and x2 + y2 = 9.
16 b2 16 9

x2 - y2 = 1 coincide, find the value of b2. 36. Find the equation of the common tangents to the curves
144 81 25
y2 = 8x and x2 - y2 = 1 .
18. If the latus rectum subtends right angle at the centre of 95

x2 y2 37. Find the locus of the point of intersection of the perpen-
a2 b2
the hyperbola - = 1, find its eccentricity. dicular tangents to the curve x2 - y2 = 1 .
16 9
19. Find the location of point (1, 4) w.r.t the hyperbola
2x2 – 3y2 = 6. 38. Find the product of the perpendiculars from foci upon

20. If (l, –1) is an exterior point of the curve 4x2 – 3y2 = any tangent to the hyperbola x2 - y2 =1.
1 such that the length of the interval where l lies is m, a2 b2
find the value of m + 10.
39. If the tangent to the parabola y2 = 4ax intersects the

INTERSECTIONS OF A LINE AND A HYPERBOLA hyperbola x2 - y2 =1 at A and B respectively, find the
a2 b2
21. Find the points common to the hyperbola 25x2 – 9y2
= 225 and the straight line 25x + 12y – 45 = 0. locus of the points of intersection of tangents at A and B.

22. For what value of l, does the line y = 3x + l touch the NORMAL AND NORMALCY
hyperbola 9x2 – 5y2 = 45?
40. Find the equation of the normal to the curve x2 - y2 = 1
23. For all real values of m, the straight line 16 9

y = mx + 9m2 - 4 is a tangent to a hyperbola, find at (8, 3 3) .

the equation of the hyperbola. 41. A normal is drawn at one end of the latus rectum of the
24. Find the equations of tangents to the curve 4x2 – 9y2 =
hyperbola x2 - y2 =1 which meets the axes at points
36, which is parallel to 5x – 4y + 7 = 0. a2 b2
25. Find the equations of tangents to the curve 9x2 – 16y2 =
A and B respectively. Find the area of the DOAB.
144, which is perpendicular to the straight line 3x + 4y
+ 10 = 0. 42. Prove that the locus of the foot of the perpendicular
26. If the line 5x + 12y – 9 = 0 touches the hyperbola
x2 – 9y2 = 9, then find its point of contact. from the centre upon any normal to the hyperbola
27. Find the equation of tangents to the curve 4x2 – 9y2 = 36
from the point (3, 2). x2 - y2 =1 is
28. Find the number of tangents from the point (1, –2) to a2 b2
the curve 2x2 – 3y2 = 12.
29. Find the equation of the tangent to the curve 3x2 – 4y2 = (x2 + y2)(a2y2 – b2x2) = (a2 + b2)2x2y2
12 having slope 4.
43. A normal to the hyperbola x2 - y2 =1 meets the axes
a2 b2

in M and N and the lines MP and NP are drawn perpen-

diculars to the axes meeting at P. Prove that the locus

of P is the hyperbola a2x2 – b2y2 = (a2 + b2)2.

6.12 Coordinate Geometry Booster

44. If the normal at j on the hyperbola x2 - y2 =1 meets 52. A tangent to the hyperbola x2 - y2 =1 cuts the ellipse
a2 b2 in a2 Q. b2 of the
x2 y2
the transverse axis at G such that a2 + b2 =1 points P and Find the locus

AG ◊ A¢G = am(en secp q – 1), mid-point PQ.

where A, A¢ are the vertices of the hyperbola and m, n 53. Chords of the hyperbola x2 – y2 = a2 touch the parabola

and p are positive integers, find the value of (m + n + y2 = 4ax. Prove that the locus of their mid-points is the

p)2 + 36. curve y2(x – a) = x3.

45. If the normals at (xi, yi), i = 1, 2, 3, 4 on the rectangular 54. A variable chord of the hyperbola x2 - y2 =1 is a tan-
hyperbola xy = c2 meet at the point (a, b), prove that a2 b2

(i) x1 + x2 + x3 + x4 = a gent to the circle x2 + y2 = c2. Prove that the locus of its
(ii) y1 + y2 + y3 + y4 = b
(iii) x12 + x22 + x32 + x42 = a 2 mid-points is

Ê x2 y2 ˆ 2 Ê x2 y2 ˆ
ËÁ a2 b2 ˜¯ ÁË a4 b4 ˜¯
(iv) y12 + y22 + y32 + y42 = b 2 - = c2 +
(v) x1 ◊ x2 ◊ x3 ◊ x4 = –c4
55. A variable chord of the circle x2 + y2 = a2 touches the
(vi) y1 ◊ y2 ◊ y3 ◊ y4 = –c4.
x2 y2
46. If the normals at (xi, yi), i = 1, 2, 3, 4 on the hyperbola hyperbola a2 - b2 = 1. Prove that the locus of its mid-

x2 - y2 =1 are concurrent, prove that points is (x2 + y2)2 = a2x2 – b2y2.
a2 b2
56. A tangent to the parabola y2 = 4ax meets the hyperbola
Ê 1 1 1 1 ˆ
(i) ( x1 + x2 + x3 + x4 ) ÁË x1 + x2 + x3 + x4 ¯˜ = 4 xy = c2 in two points P and Q. Prove that the locus of

the mid-point of PQ lies on a parabola.

( y1 y4 )ÁËÊ 1 1 1 1ˆ 57. From a point P, tangents are drawn to the circle x2 +
y1 y2 y3 y4 ˜¯
(ii) + y2 + y3 + + + + = 4 y2 = a2. If the chord of contact of the circle is a normal

chord of a hyperbola x2 - y2 = 1, prove that the locus
a2 b2
CHORD OF CONTACT/CHORD BISECTED AT A POINT
Ê a2 b2 ˆ Ê a2 + b2 ˆ 2
47. Find the equation of the chord of contact of tangents of the point P is ÁË x2 - y2 ¯˜ = ÁË a2 ¯˜ .

from the point (2, 3) to the hyperbola x2 - y2 = 1 . 58. Prove that the locus of the mid-points of the focal chord
94
x2 y2 x2 y2
48. Find the locus of the mid-points of the portions of the of the hyperbola a2 - b2 =1 is a2 - b2 = ex .
a
tangents to the hyperbola x2 - y2 = 1 included be-
94 59. If the chords of contact of tangents from two points

tween the axes. (x1, y1) and (x2, y2) to the hyperbola x2 - y2 =1 are
a2 b2
49. From the points on the circle x2 + y2 = a2, tangents are

drawn to the hyperbola x2 – y2 = a2. Prove that the locus x1x2 am
y1 y2 bn
of the mid-points of the chord of contact is at right angles such that = - , where m, n are

(x2 – y2)2 = a2(x2 + y2)

50. Prove that the locus of the mid-points of the hyperbola Ê m + nˆ10
ËÁ 4 ˜¯
x2 y2 positive integers, find the value of .
a2 b2
- =1 which subtend right angle at the centre is

Ê x2 y2 ˆ2 Ê 1 1 ˆ Ê x2 y2 ˆ POLE AND POLAR
ËÁ a2 b2 ˜¯ ÁË a2 b2 ¯˜ ËÁ a4 b4 ¯˜
- - = + 60. Find the polar of the focus (–ae, 0) with respect to the

51. Tangents are drawn from a point P to the parabola hyperbola x2 - y2 =1.
a2 b2
y2 = 4ax. If the chord of contact of the parabola be a
61. If the polars of (x1, y1) and (x2, y2) with respect to the
x2 y2
tangent to the hyperbola a2 - b2 =1, find the locus x2 y2
a2 b2
of the point P. hyperbola - =1 are at right angles, prove that

x1x2 + a4 = 0.
y1 y2 b4

Hyperbola 6.13

62. Find the pole of the line x – y = 3 w.r.t. the hyperbola x2 y2
a2 b2
x2 – 3y2 = 3. 77. Let P be a variable point on the hyperbola - =1

63. Prove that the locus of the poles of the normal chords such that its distance from the transverse axis is equal

with respect to the hyperbola x2 - y2 =1 is the curve to its distance from an asymptote to the given hyper-
a2 b2
bola. Prove that the locus of P is (x2 – y2)2 = 4x2(x2 – a2).
y2a6 – x2b6 = (a2 + b2)2x2y2.
78. Show that the tangent at any point of a hyperbola cuts
64. Prove that the locus of the poles with respect to the
off a triangle of constant area from the asymptotes and
parabola y2 = 4ax of the tangent to the hyperbola x2 – y2
that the portion of it intercepted between the asymp-
= a2 is the ellipse 4x2 + y2 = 4a2.
totes is bisected at the point of contact.
65. Prove that the locus of the pole with respect to the
79. If p1 and p2 are the perpendiculars from any point on
x2 y2
hyperbola a2 - b2 =1 of any tangent to the circle, the hyperbola x2 - y2 =1 on its asymptotes, prove
a2 b2
whose diameter is the line joining the foci, is the ellipse
1 1 1
x2 + y2 = 1 . that p1 p2 = a2 + b2 .
a4 b4 + b2
a2 80. If the normal at t1 to the hyperbola xy = c2 meets it again
at t2, prove that t13t2 = -1 .
DIAMETER
81. A triangle has its vertices on a rectangular hyperbola.
66. Prove that the equation of the diameter which bisects
Prove that the orthocentre of the triangle also lies on
the chord 7x + y – 2 = 0 of the hyperbola x2 - y2 = 1
the same hyperbola.

is x + 3y = 0. 37 82. Find the locus of the poles of the normal chords of the

rectangular hyperbola xy = c2

67. Find the equation of the diameter of the hyperbola 83. If the angle between the asymptote is 2a, prove that the
x2 - y2 = 1 , which corresponds the line 3x + 4y + 10
94 eccentricity of the hyperbola is sec a.
= 0.
84. A circle cuts the rectangular hyperbola xy = 1 in points
68. Find the equation of the diameter to the hyperbola
x2 - y2 = 1 parallel to the chord 2x + 3y + 5 = 0. (xr, yr), r = 1, 2, 3, 4, prove that x1x2x3x4 = 1 and y1y2y3y4
94 =1.

69. In the hyperbola 16x2 – 9y2 = 144, find the equation of 85. If the tangent and the normal to a rectangular hyperbola
the diameter which is conjugate to the diameter whose
equation is x = 2y. xy = c2 at a point cuts off intercepts a1 and a2 on one
axis and b1, b2 on the other axis, prove that a1a2 + b1b2
ASYMPTOTES = 0.

70. Find the asymptotes of the curve xy – 3y – 2x = 0. 86. If e1 and e2 be the eccentricities of the hyperbola
71. Find the equations of the asymptotes of the curve xy = c2 and x2 – y2 = a2, find the value of (e1 + e2)2.

(a sec j, a tan j). 87. Find the product of the lengths of the perpendiculars
72. Find the eccentricity of the hyperbola whose asymp-
drawn from any point on the hyperbola x2 - y2 = 1 to
totes are 3x + 4y = 10 and 4x – 3y = 5. 2
73. Find the equation of a hyperbola whose asymptotes are
its asymptote.
2x – y = 3 and 3x + y = 7 and which pass through the
point (1, 1). 88. If A, B and C be three points on the rectangular hyper-
74. The asymptotes of a hyperbola having centre at the
point (1, 2) are parallel to the lines 2x + 3y = 0 and bola xy = c2, find
3x + 2y = 0. If the hyperbola passes through the point
(5, 3), prove that its equation is (2x + 3y – 8)(3x + 2y (i) the area of the DABC
– 7) – 154 = 0.
75. Find the product of the lengths of the perpendiculars (ii) the area of the triangle formed by the tangents at
from any point on the hyperbola x2 – 2y2 = 2 to its as-
ymptotes. A, B and C.
76. Find the area of the triangle formed by any tangent to
the hyperbola x2 - y2 = 1 and its asymptotes. 89. Find the length of the transverse axis of the rectangular

94 hyperbola xy = 18.

90. Prove that the locus of a point whose chord of contact

with respect to the circle x2 + y2 = 4 is a tangent to the

hyperbola xy = 1 is a hyperbola.

91. Find the asymptotes of the hyperbola xy = hx + ky.

92. If e be the eccentricity of the hyperbola x2 - y2 =1
a2 b2

and q is the angle between the asymptotes, find cos (q/2).

93. A ray is emanating from the point (5, 0) is incident on

the hyperbola 9x2 – 16y2 = 144 at a point P with ab-

scissa 8. Find the equation of the reflected ray after first

reflection and point P lies in first quadrant.

6.14 Coordinate Geometry Booster

94. A ray is coming along the line 2x – y + 3 = 0 (not 7. If the eccentricity of the hyperbola x2 – y2 sec2a = 5 is
through the focus) to the hyperbola x2 - y2 = 1 .
43 3 times the eccentricity of the ellipse x2 sec2a + y2 =
After striking the hyperbolic mirror, it is reflected 5, the value of a is
(not through the other focus). (a) p/6 (b) p/4 (c) p/3 (d) p/2
Find the equation of the line containing the reflected 8. For all real values of m, the straight line
ray.
y = mx + 9m2 - 4 is a tangent to the curve

LEVEL II (a) 9x2 + 4y2 = 36 (b) 4x2 + 9y2 = 36
(Mixed Problems) (c) 9x2 – 4y2 = 36 (d) 4x2 – 9y2 = 36

9. The foci of the ellipse x2 + y2 =1 and the hyperbola
16 b2

1. The magnitude of the gradient of the tangent at extrem- x2 - y2 = 1 coincide. The value of b2 is
144 81 25
x2 y2
ity latus rectum of the hyperbola a2 - b2 =1 is equal to (a) 5 (b) 7 (c) 9 (d) 4

(a) be (b) e (c) ab (d) ae 10. The locus of the mid-points of the parallel chords with

2. The eccentricity of the hyperbola conjugate to the hy- gradient m of the rectangular hyperbola xy = c2 is

perbola x2 - y2 = 1 is (a) mx + y = 0 (b) y – mx = 0
4 12
(c) my – x = 0 (d) my + x = 0

(a) 2 (d) 4 11. The locus of the foot of the perpendicular from the cen-
3 3
(b) 2 (c) 3 tre of the hyperbola xy = c2 on a variable tangent is

(a) (x2 – y2)2 = 4c2xy (b) (x2 + y2)2 = 2c2xy

3. The asymptote of the hyperbola x2 - y2 =1 form with (c) (x2 + y2) = 4c2xy (d) (x2 + y2)2 = 4c2xy
a2 b2
12. P is a point on the hyperbola x2 - y2 = 1, N is the foot
any tangent to the hyperbola a triangle whose area is a2 b2

a2 tan l in magnitude, its eccentricity is of the perpendicular from P on the transverse axis. The

(a) sec l (b) cosec l tangent to the hyperbola at P meets transverse axis at T.

(c) sec2 l (d) cosec2 l If O be the centre of the hyperbola, OT.ON is

4. The equation x2 + y2 = 1 ( p π 4, 9) represents (a) e2 (b) a2 (c) b2 (d) b2/a2
29 - p 4 - p
13. If PN be the perpendicular from a point on the rectan-

(a) an ellipse if p is any constant greater than 4 gular hyperbola (x2 – y2) = a2 on any on its asymptotes,

(b) a hyperbola if p is any constant between 4 and 29 the locus of the mid-point of PN is a/an

(a) circle (b) parabola

(c) a rectangular hyperbola if p is any constant greater (c) ellipse (d) hyperbola

than 29 14. The equation to the chord joining two points (x1, y1) and
(x2, y2) on the rectangular hyperbola xy = c2 is
(d) no real curve if p is less than 29.

5. Locus of the feet of the perpendiculars drawn from (a) x + y = 1 (b) x + y = 1

either foci on a variable tangent to the hyperbola x1 + x2 y1 + y2 x1 - x2 y1 - y2

16y2 – 9x2 = 1

(a) x2 + y2 = 9 (b) x2 + y2 = 1 (c) x + y =1 (d) x + y =1
9 y1 + y2 x1 + x2 y1 - y2 x1 - x2

(c) x2 + y2 = 7 (d) x2 + y2 = 1 15. If P(x1, y1), Q(x2, y2), R (x3, y3) and S(x4, y4) are 4 con-
144 16 cyclic points on the rectangular hyperbola xy = c2, the

6. The locus of the point of intersection of the lines co-ordinates of the orthocentre of DPQR is

3x - y = 4 3t = 0 (a) (x4 – y4) (b) (x4, y4)

(c) (–x4, –y4) (d) (–x4, y4)

and 3 tx + ty - 4 3 = 0 16. The chord PQ of the rectangular hyperbola xy = a2

(where t is a parameter) is a hyperbola, whose eccen- meets the axis of x at A, C is the mid-point of PQ and O
tricity is
the origin. The DACO is a/an

(a) equilateral (b) isosceles triangle

(a) 3 (b) 2 (c) 2/ 3 (d) 4/3 (c) right angled D (d) right isosceles triangle.

17. A conic passes through the point (2, 4) and is such that

the segment of any of its tangents at any point con-

Hyperbola 6.15

tained between the co-ordinates is bisected at the point the equation of the circle through the points of intersec-
of tangency. The foci of the conic are tion of two conics is
(a) x2 + y2 = 5
(a) (2 2, 0) and (— 2 2, 0)
(b) 5(x2 + y2 ) - 3x - 4y = 0
(b) (2 2, 2 2) and (—2 2, - 2 2)
(c) 5(x2 + y2 ) + 3x + 4y = 0
(c) (4, 4) and (– 4, – 4)
(d) x2 + y2 = 25

(d) (4 2, 4 2) and (—4 2, —4 2) 26. At the point of intersection of the rectangular hyper-
bola xy = c2 and the parabola y2 = 4ax, tangents to the
18. The latus rectum of the conic satisfying the differential rectangular hyperbola and the parabola make angles
equation, xdy + ydx = 0 and passing through the point q and j, respectively with the axis of x, then
(2, 8) is (a) q = tan–1(–2 tan j) (b) j = tan–1(–2 tan j)

(a) 4 2 (b) 8 (c) 8 2 (d) 16 (c) q = 1 tan-1(- tanj) (d) j = 1 tan-1(- tanq )
22
19. If the normal to the rectangular hyperbola xy = c2 at the

point t meets the curve again at t1, the value of t3t1 is 27. The area of the quadrilateral formed with the foci of the

(a) 1 (b) –1 (c) 0 (d) None x2 y2 x2 y2
a2 b2 a2 b2
20. With one focus of the hyperbola x2 - y2 = 1 as the hyperbola - =1 and - = -1 is
9 16
(a) 4a2 + b2) (b) 2(a2 + b3)
centre, a circle is drawn which is the tangent to the
(d) 1 (a2 + b2 )
hyperbola with no part of the circle being outside the (c) (a2 + b2) 2
hyperbola. The radius of the circle is
28 The eccentricity of the hyperbola whose latus rectum
(a) < 2 (b) 2 (c) 11/3 (d) None
is 8 and the conjugate axis is equal to half the distance
x2 y2
21. AB is a double ordinate of the hyperbola a2 - b2 =1 between the foci is

such that DAOB (where O is the origin) is an equilateral (a) 4 (b) 4 (c) 2 (d) None.
triangle, the eccentricity of the hyperbola satisfies 3 3 3

(b) 1 < e < 2 29. If P( 2 secq , 2 tanq ) is a point on the hyperbola
3
(a) e > 3 whose distance from the origin is 6 , where P is in

the first quadrant, then q is equal to

(c) e = 2 (d) e> 2 (a) p (b) p (c) p (d) None
3 3 4 3 6

22. If the product of the perpendicular distances from any 30. An ellipse and a hyperbola have same centre origin, the

point on the hyperbola of the eccentricity from its as- same foci and the minor axis of the one is the same as

ymptotes is equal to 6, the length of the transverse axis the conjugate axis of the other. If e1 and e2 be their ec-

of the hyperbola is centricities, respectively, then 1 + 1 is equal to
e12 e22
(a) 3 (b) 6 (c) 8 (d) 12

23. If x + iy = j + iy where i = -1 and j and y are (a) 1 (b) 2 (c) 3 (d) None
non-zero real parameters, then j = constant and y =
31. The number of possible tangents can be drawn to the

constant, represents two systems of rectangular hyper- curve 4x2 9y2 = 36, which are perpendicular to the

bola which intersect at an angle of straight line 5x + 2y – 10 = 0 is

(a) p/6 (b) p/3 (c) p/4 (d) p/2 (a) 0 (b) 1 (c) 2 (d) 4

24. The tangent to the hyperbola xy = c2 at the point P in- 32. The equation of a tangent passing through (2, 8) to the
tersects the x-axis at T and the y-axis at T ¢. The normal
hyperbola 5x2 – y2 = 5 is

to the hyperbola at P intersects the x-axis at N and the (a) 3x – y + 2 = 0 (b) 3x + y – 14 = 0

y-axis at N¢. The area of DPNT and PN¢T ¢ are D and D¢ (c) 23x – 3y – 2 2 = 0 (d) 3x – 23y + 178 = 0.

respectively, then 1 + 1 is 33. If P(x1, y1), Q(x2, y2), R(x3, y3 and (Sx4, y4) are four con-
D D¢ cyclic points on the rectangular hyperbola xy = c2, the

(a) equal to 1 (b) depends on t co-ordinates of the orthocentre of DPQR are

(c) depends on c (d) equal to 2 (a) (x4, y4) (b) (x4, – y4)
(c) (–x4, – y4) (d) (–x4, y4).
25. The ellipse 4x2 + y = 5 and the hyperbola 4x2 – y2 = 4

have the same foci and they intersect at right angles,

6.16 Coordinate Geometry Booster

34. If the hyperbolas x2 + 3xy + 2y2 + 2x + 3y + 2 = 0 and 45. If two distinct tangents can be drawn from the point
(a, 2) on different branches of the hyperbola
x2 + 3xy + 2y2 + 2x + 3y + c = 0 are conjugate of each

other, the value of c is x2 - y2 = 1 , then
9 16
(a) –2 (b) 0 (c) 4 (d) 1

35. A rectangular hyperbola circumscribes a DABC, it will (a) a < 3 (b) a > 2
2 3
always pass through its

(a) orthocentre (b) circumcentre (c) |a| > 3 (d) |a| > 5

(c) centroid (d) incentre x2 y2
a2 b2
36. The co-ordinates of a point on the hyperbola 46. From any point on the hyperbola - = 1, tangents

x2 - y2 = 1 , which is nearest to the line are drawn to the hyperbola x2 - y2 = 2 . The area cut
24 18 a2 b2

3x + 2y + 1 = 0 are

(a) (6, 3) (b) (–6, –3) off by the chord of contact on the asymptotes is

(c) (6, –3) (d) (–6, 3) (a) a/2 (b) ab (c) 2ab (d) 4ab

37. The latus rectum of the hyperbola 9x2 – 16y2 – 18x – 47. A hyperbola passes through (2, 3) and has asymptotes

32y – 151 = 0 is 3x – 4y + 5 = 0 and 12x + 5y = 40, the equation of its

(a) 9/4 (b) 9 (c) 3/2 (d) 9/2 transverse axis is

38. If the eccentricity of the hyperbola x2 – y` sec2 a = 5 is (a) 77x – 21y – 265 = 0 (b) 21x – 77y + 265 = 0

3 times the eccentricity of the ellipse x2 sec2 a + y2 = (c) 21x + 77y – 326 = 0 (d) 21x + 77y – 273 = 0
25, the value of a can be
48. The centre of a rectangular hyperbola lies on the line

(a) p (b) p (c) p (d) p y = 2x. If one of the asymptotes is x + y + c = 0, the
6 4 3 2
other asymptote is

(a) x – y – 3c = 0 (b) 2x – y + c = 0

39. If values of m for which the line y = mx + 2 5 touches (c) x – y – c = 0 (d) None

the hyperbola 16x2 – 9y2 = 144 are the roots of x2 – 49. The equation of a rectangular hyperbola, whose asymp-

(a + b)x – 4 = 0, then the value of (a + b) is totes are x = 3 and y = 5 and passing through (7, 8) is
(a) xy – 3y + 5x + 3 = 0 (b) xy + 3y + 4x + 3 = 0
(a) 2 (b) 4 (c) 0 (d) none
(c) xy – 3y + 5x – 3 = 0 (d) xy – 3y + 5x + 5 = 0
40. The locus of the feet of the perpendiculars drawn from 50. The equation of the conjugate axis of the hyperbola

either focus on a variable tangent to the hyperbola xy – 3y – 4x + 7 = 0 is

16x2 – 9y = 1 is (a) x + y = 3 (b) x + y = 7

(a) x2 + y2 = 9 (b) x2 + y2 = 1/9 (c) y – x = 3 (d) none

(c) x2 + y2 = 7/144 (d) x2 + y2 = 1/16 51. The curve xy = c (c > 0) and the circle x2 + y2 = 1 touch

41. The locus of the foot of the perpendicular from the cen- at two points, the distance between the points of con-

tre of the hyperbola xy = 1 on a variable tangent is tact is

(a) (x2 – y2)2 = 4xy (b) (x2 – y2)2 = 2xy (a) 2 (b) 3 (c) 4 (d) 2 2

(c) (x2 + y2)2 = 2xy (d) (x2 + y2) = 4xy 52. Let the curves (x – 1)(y – 2) = 5 and (x – 1)2 + (y – 2)2

42. The tangent at a point P on the hyperbola x2 - y2 =1 = r2 intersect at four points P, Q, R, S. If the centroid of
a2 b2
DPQR lies on the line y = 3x – 4, the locus of S is

meets one of the directrix in F. If PF subtends an angle (a) y = 3x (b) x2 + y2 + 3x + 1 = 0

q at the corresponding focus, the value of q is (c) 3y = x + 1 (d) y = 3x + 1

(a) p (b) p (c) 3p (d) None 53. The ellipse 4x2 + 9y2 = 36 and the hyperbola a2x2 – y2
4 2 4
= 4 intersect at right angles, the equation of the circle

x2 y2 through the point of intersection of the two conics is
a2 b2
43. The number of points on the hyperbola - = 3 (a) x2 + y2 = 5

, from which mutually perpendicular tangents can be (b) 5(x2 + y2 ) = 3x + 4y

drawn to the circle x2 + y2 = a2, is (c) 5(x2 + y2 ) + 3x + 4y = 0

(a) 0 (b) 2 (c) 3 (d) 4 (d) x2 + y2 = 25

44. If the sum of the slopes of the normal from a point P to

the hyperbola xy = c2 is equal to l, where l ΠR+, the 54. The angle between the lines joining the origin to the
points of intersection of the line 3 x + y = 2 and the
locus of the point P is curve 3 x + y = 2 is

(a) x2 = lc2 (b) y2 = lc2

(c) xy = lc2 (d) x/y = lc2.

Hyperbola 6.17

-1 Ê 2ˆ (b) p 7. The perpendicular from the centre upon the normal at
ËÁ 3 ¯˜ 6
(a) tan x2 y2
a2 b2
any point of the hyperbola - =1 meets at Q.

(c) tan -1 Ê 3ˆ (d) p Prove that the locus of Q is
ËÁ 2 ˜¯ 2
(x2 + y2)(a2y2 – b2x2) = (a2 + b2)x2y2

8. From the points on the circle x2 + y2 = a2, tangents are

55. If S = 0 be the equation of the hyperbola x2 + 4xy + drawn to the hyperbola x2 – y2 = a2. Prove that the locus

3y2 – 4x + 2y + 1 = 0, the value of k for which S + k = 0 of the mid-points of the chord of contact is

represents its asymptotes is (x2 – y2)2 = a2(x2 + y2)

(a) 20 (b) –16 (c) –22 (d) 18 9. Prove that the locus of the mid-points of the hyperbola

LEVEL III x2 - y2 =1 which subtend right angle at the centre is
(Problems for JEE Advanced) a2 b2

Ê x2 - y2 ˆ2 Ê 1 - 1 ˆ = Ê x2 + y2 ˆ .
ÁË a2 b2 ˜¯ ËÁ a2 b2 ¯˜ ÁË a4 b4 ˜¯
1. For all real values of m, the straight line

y = mx + 9m2 - 4 is a tangent to a hyperbola, find 10. Tangents are drawn from a point P to the parabola

the equation of the hyperbola. y2 = 4ax. If the chord of contact of the parabola be a

2. Find the equations of the common tangent to the curves tangent to the hyperbola x2 - y2 =1, find the locus
of the point P. a2 b2
x2 - y2 =1 and x2 - y2 = -1 and also find its length.
a2 b2 b2 a2
11. Chords of the hyperbola x2 – y2 = a2 touch the parabola
3. Find the equation of the common tangents to the curves
y2 = 4ax. Prove that the locus of their mid-points is the
x2 - y2 = 1 and x2 + y2 = 9.
curve y2(x – a) = x3.

16 9 x2 y2 12. Prove that the locus of the mid-points of the rectangu-
If the normal at f on the hyperbola a2 b2
4. - =1 meets lar hyperbola xy = c2 of constant length 2d is

(x2 + y2)(xy – c2) = d2xy

the transverse axis at G such that x2 y2
a2 b2
AG ◊ A¢G = am(en secp q – 1), 13. A variable chord of the hyperbola - =1 is a tan-

where A, A¢ are the vertices of the hyperbola and m, n

and p are positive integers, find the value of gent to the circle x2 + y2 = c2. Prove that the locus of its

(m + n + p)2 + 36 mid-points is

5. If the normals at (xi, yi), i = 1, 2, 3, 4 on the rectangular Ê x2 y2 ˆ 2 Ê x2 y2 ˆ
hyperbola xy = c2 meet at the point (a, b), prove that ÁË a2 b2 ˜¯ ËÁ a4 b4 ¯˜
- = c2 +

(i) x1 + x2 + x3 + x4 = a 14. A variable chord of the circle x2 + y2 = a2 touch the
(ii) y1 + y2 + y3 + y4 = a

(iii) x12 + x22 + x32 + x42 = b 2 hyperbola x2 - y2 = 1. Prove that the locus of its mid-
a2 b2
(iv) y12 + y22 + y32 + y42 = b 2
points is (x2 + y2)2 = a2x2 – b2y2.
(v) x1 ◊ x2 ◊ x3 ◊ x4 = –c4
(vi) y1 ◊ y2 ◊ y3 ◊ y4 = –c4 15. A tangent to the parabola y2 = 4ax meets the hyperbola

6. If the normals at (xi, yi), i = 1, 2, 3, 4 on the hyperbola xy = c2 in two points P and Q. Prove that the locus of

x2 y2 the mid-point of PQ lies on a parabola.
a2 b2
- =1 are concurrent, prove that 16. From a point P, tangents are drawn to the circle x2 +

y2 = a2. If the chord of contact of the circle is a normal

Ê 1 1 1 1 ˆ chord of a hyperbola x2 - y2 = 1, prove that the locus
x4 ) ÁË x1 x2 x3 x4 ¯˜ a2 b2
(i) ( x1 + x2 + x3 + + + + = 4 .

Ê 1 1 1 1ˆ of the point P is Ê a2 - b2 ˆ = Ê a2 + b2 ˆ 2 .
y4 ) ËÁ y1 y2 y3 y4 ¯˜ ÁË x2 y2 ˜¯ ËÁ a2 ¯˜
(ii) ( y1 + y2 + y3 + + + + = 4.

6.18 Coordinate Geometry Booster

17. The normal to the hyperbola x2 - y2 =1 meets the 4. A tangent to the hyperbola x2 - y2 = 1 cuts the ellipse
a2 b2 a2 b2

axes in M and N, and lines MP and NP are drawn at x2 + y2 =1 in P and Q. Prove that the locus of the
a2 b2
right angles to the axes. Prove that the locus of P is the

hyperbola a2x2 – b2y2 = (a2 + b2)2. Ê x2 y2 ˆ 2 Ê x2 y2 ˆ
ÁË a2 b2 ¯˜ ÁË a2 b2 ˜¯
18. Prove that the locus of the point of intersection of tan- mid-point of PQ is + = - .

gents to a hyperbola which meet at a constant angle b

is the curve 5. If a triangle is inscribed in a rectangular hyperbola,

(x2 + y2 + b2 – a2)2 = 4 cot2b(a2b2 – b2x2 + a2b2) prove that its orthocentre lies on the curve.

19. Prove that the chords of a hyperbola, which touch the 6. Prove that the locus of the poles of the normal chords

conjugate hyperbola, are bisected at the point of contact. of the rectangular hyperbola xy = c2 is the curve
(x2 – y2)2 + 4c2xy = 0.
20. A straight line is drawn parallel to the conjugate axis of 7. If a circle cuts a rectangular hyperbola xy = c2 in A, B,

a hyperbola to meet it and the conjugate hyperbola in C and D and the parameters of these four points be t1, t2,
t3 and t4 respectively. Prove that the centre of the circle
the points P and Q. Show that the tangents at P and Q through A, B and C is

meet the curve y4 Ê y2 + x2 ˆ = 4x2 and the normals
b4 ËÁ b2 a2 ¯˜ a2

meet on the axis of x. Ê c Ê + t2 + t3 + 1ˆ , c Ê 1 + 1 + 1 + ˆˆ
21. Prove that the locus of the mid-points of the chords of ËÁ 2 ËÁ t1 t1t2t3 ¯˜ 2 ËÁ t1 t2 t3 t1t2t3¯˜ ¯˜

the circle x2 + y2 = 16, which are tangent to the hyper- 8. If a triangle is inscribed in a rectangular hyperbola,
bola 9x2 – 16y2 = 144 is (x2 + y2)2 = 16x2 – 9y2.
22. Find the asymptotes of the curve 2x2 + 5xy + 2y2 + 4x + prove that the orthocentre of the triangle lies on the
5y = 0. Also, find the general equation of all hyperbolas
having the same asymptotes. curve.
23. Find the equation of the hyperbola whose asymptotes 9. If a circle cuts a rectangular hyperbola xy = c2 in A, B,
are the straight lines x + 2y + 3 = 0 and 3x + 4y + 5 = 0
and pass through the point (1, –1). C, D and the parameters of the four points be t1, t2, t3,
and t4 respectively, prove that the centre of the mean
24. Let C be the centre of the hyperbola x2 - y2 =1 and position of the four points bisects the distance between
a2 b2
the centres of the two curves.
the tangent at any point P meets the asymptotes at the
10. A circle of variable radius cuts the rectangular hyper-
bola x2 – y2 = 9a2 in points P, Q, R and S. Find the equa-
tion of the locus of the centroid of DPQR.

points Q and R. Prove that the equation to the locus

of the centre of the circle circumscribing, the DCQR is Integer Type Questions

4(a2x2 – b2y2) = (a2 + b2)2. 1. Find the eccentricity of the hyperbola conjugate to the

25. If P, Q, R be three points on the rectangular hyperbola hyperbola x2 - y2 = 1 .
4 12
xy = c2, whose abscissae are x1, x2, x3, prove that the
area of DPQR is

c2 ¥ (x1 - x2 )(x2 - x3)(x3 - x1) 2. If the foci of the ellipse x2 + y2 = 1 and the hyperbola
2 x1 x2 x3 16 b2

LEVEL IV x2 - y2 = 1 coincide, find the value of (b2 + 1).
(Tougher Problems for JEE 144 81 25
Advanced)
3. If e1 and e2 be the eccentricities of a hyperbola and its
1. A tangent to the parabola x2 = 4ay meets the hyperbola
xy = k2 in two points P and Q. Prove that the locus of conjugate, find the value of Ê1 + 1 ˆ .
the mid-point of PQ lies on the parabola. ÁË e12 e22 + 3˜¯

2. Find the equation of the chord of the hyperbola 25x2 – 4. Find the number of tangents to the hyperbola
16y2 = 400 which is bisected at the point (6, 2).
x2 - y2 = 1 from the point (4, 3).
3. Find the locus of the mid-points of the chords of the 43
circle x2 + y2 = 16 which are tangents to the hyperbola x2 y2
9x2 – 16y2 = 144. 5. Find the number of points on the hyperbola a2 - b2 = 3

from which mutually perpendicular tangents can be

drawn to the circle x2 + y2 = a2.

Hyperbola 6.19

6. If the latus rectum of the hyperbola 9x2 – 16y2 – 18x – 1. The equation of the asymptotes are

32y – 151 = 0 is m, find the value of (2m – 3). (a) 5xy + 2x2 + 3y2 = 0 (b) 3x2 + 4y2 + 6xy = 0

7. If the number of possible tangents can be drawn to the (c) 2x3 + 2y2 – 5xy = 0 (d) 2x2 + y2 – 5xy = 0
curve 4x2 – 9y2 = 36, which are perpendicular to the
2. The angle subtended by AB at the centre of the

straight line 5x + 2y = 10 is m, find the value of (m + 4). hyperbola is

8. If values of m for which the line y = mx + 2 5 touches (a) sin -1 Ê 4 ˆ (b) sin -1 Ê 2 ˆ
ÁË 5 ˜¯ ËÁ 5 ¯˜
the hyperbola 16x2 – 9y2 = 144 are the roots of x2 –

(a + b)x – 4 = 0, find the value of (a + b + 3). -1 Ê 3ˆ -1 Ê 1ˆ
ÁË 5˜¯ ÁË 5˜¯
9. The curve xy = c, (c > 0) and the circle x2 + y2 = 1 touch (c) sin (d) sin
at two points, find the distance between the points of
Ê 7ˆ
contact. 3. The equation of the tangent to the hyperbola at ÁË -1, 2˜¯
is
10. If the product of the perpendicular distances from any

point on the hyperbola x2 - y2 =1 of eccentricity (a) 5x + 2y = 2 (b) 3x + 2y = 4
a2 b2 (c) 3x + 4y =11 (d) 3x – 4y = 10
e = 3 from its asymptote is equal to 6, find the length

of the transverse axis of the hyperbola. Passage III
11. The tangent to the hyperbola xy = c2 at the point P in-
A point P moves such that the sum of the slopes of the nor-
tersects the x-axis at T and the y-axis at T ¢. The normal
to the hyperbola at P intersects the x-axis at N and N ¢, mals drawn from it to the hyperbola xy = 16 is equal to the
respectively. The area of Ds PNT and PN ¢T ¢ are D and
sum of the ordinates of the feet of normals. The locus of the

P is a curve C.

Ê c2 c2 ˆ 1. The equation of the curve C is
ËÁ D D¢ + 4˜¯
D¢ respectively, find the value of + . (a) x2 = 4y (b) x2 = 16y (c) x2 = 12y (d) y2 = 8x

12. Find the area of the triangle formed by any tangent 2. If the tangent to the curve C cuts the co-ordinate axes at

to the hyperbola x2 - y2 = 1 and its asymptotes. A and B, the locus of the mid-point of AB is
94
(a) x2 = 4y (b) x2 = 2y

(c) x2 + 2y = 0 (d) x2 + 4y = 0

3. The area of the equilateral triangle, inscribed in the

Comprehension Link Passage curve C, having one vertex as the vertex of the curve C

is

Passage I (a) 772 3 s. u. (b) 776 3 s. u.
The locus of the foot of the perpendicular from any focus of
a hyperbola upon any tangent to the hyperbola is an auxiliary (c) 760 3 s. u. (d) 768 3 s. u.
circle of the hyperbola. Consider the foci of a hyperbola as
(–3, –2) and (5, 6) and the foot of the perpendicular from the Passage IV
focus (5, 6) upon a tangent to the hyperbola as (2, 5).
The vertices of DABC lie on a rectangular hyperbola such that
1. The conjugate axis of the hyperbola is
the orthocentre of the triangle is (3, 2) and the asymptotes of

(a) 4 11 (b) 2 11 (c) 4 22 (d) 2 22 the rectangular hyperbola are parallel to the co-ordinate axes.

2. The directrix of the hyperbola corresponding to The two perpendicular tangents of the hyperbola intersect at

the focus (5, 6) is the point (1, 1).

(a) 2x + 2y – 1 = 0 (b) 2x + 2y – 11 = 0 1. The equation of the asymptotes is

(c) 2x + 2y – 7 = 0 (d) 2x + 2y – 9 = 0 (a) xy – x + y – 1 = 0 (b) xy – x – y + 1 = 0

3. The point of contact of the tangent with the hyper- (c) 2xy + x + y (d) 2xy = x + y + 1

bola is 2. The equation of the rectangular hyperbola is

Ê 2 31ˆ Ê 7 23ˆ (a) xy = 2x + y – 2 (b) xy = 2x + y + 5
ËÁ 9 3 ¯˜ ÁË 4 4 ¯˜
(a) , (b) , (c) xy = x + y + 1 (d) xy = x + y + 10

Ê 2 9˜ˆ¯ Ê 7 7˜ˆ¯ 3. The number of real tangents that can be drawn from the
ËÁ 3 ÁË 9
(c) , (d) , point (1, 1) to the rectangular hyperbola is

(a) 4 (b) 0 (c) 3 (d) 2

Passage II Passage V
The portion of the tangent intercepted between the asymp- A line is drawn through the point P(–1, 2) meets the hyper-
totes of the hyperbola is bisected at the point of contact. bola xy = c2 at the (points A and B lie on the same side of P)
and Q is a point on the line segment AB.
Consider a hyperbola whose centre is at the origin. A line
x + y = 2 touches this hyperbola at P(1, 1) and intersects the 1. If the point Q be chosen such that PA, PQ, and PB are
in AP, the locus of the point Q is
asymptotes at A and B such that AB = 6 2 .

6.20 Coordinate Geometry Booster

(a) x = y + 2xy (b) x = y + xy (C) 2 (R) a circle
+ e-iq (S) a parabola
(c) 2x = y + 2xy (d) 2x = y + xy x = (eiq

2. If the point Q be chosen such that PA, PQ and PB are in )

GP, the locus of the point Q is and

(a) xy – y + 2x – c2 = 0 (b) xy + y – 2x + c2 = 0 Ê eiq - e-iq ˆ
ËÁ eiq + e-iq ˜¯
(c) xy + y + 2x + c2 = 0 (d) xy – y – 2x – c2 = 0 y = i ¥ is

3. If the point Q be chosen such that PA, PQ, and PB are

in HP, the locus of the point Q is (D) x = i (e-iq - eiq )
2
(a) 2x – y = 2c2 (b) x – 2y = 2c2

(c) 2x + y + 2c2 = 0 (d) x + 2y = 2c2

Passage VI and
The graph of the conic x2 – (y – 1)2 = 1 has one tangent line
with positive slope that passes through the origin, the point of y = Ê eiq + e-iq ˆ is
tangency being (a, b). ÁË sin 2q ˜¯

1. The value of sin -1 Ê aˆ is 3. Match the following columns
ËÁ b ˜¯ The locus of the point of intersection of two perpen-
dicular tangents to a conic is a director circle to the
(a) 5p/12 (b) p/6 (c) p/3 (d) p/4 given conic.

2. The length of the latus rectum of the conic is

(a) 1 (b) 2 (c) 2 (d) none Column I Column II
(P) x2 + y2 = 45
3. The eccentricity of the conic is (A) The director circle of
x2 - y2 = 1 is (Q) x2 + y2 = 7
(a) 4/3 (b) 3 (c) 2 (d) none 16 9

Matrix Match (B) The director circle of
(For JEE-Advanced Examination Only) x2 + y2 = 1 is
20 25
1. Match the following columns
(C) The director circle of (R) x2 + y2 = 0
Let z, z1 and z2 be three complex numbers and b, c ΠR+. x2 Рy2 = 16 is
Then the locus of z

Column I Column II

(A) is an ellipse, if (P) |z – c| = b (D) The director circle of (S) x + 2 = 0
x2 + y2 = 25 is (T) x2 + y2 = 50
(B) is a hyperbola, (Q) |z – z1| + |z – z2| = 2b,
if where 2b > |z1 – z2| (E) The director circle of
y2 = 8x is
(C) is a straight (R) |z – z1| – |z – z2| = 2b
line, if Where 2b < |z1 – z2|

(D) is a circle, if (S) |z – z1| + |z – z2| = |z1 – z2| (F) The director circle of (U) x + 4 = 0
xy = 1 is

2. Match the following columns: 4. Match the following columns:
The locus of a variable point P, whose co-ordinates are The equation of the common tangent between the given
given by curves

Column I Column II
(P) an ellipse
(A) Ê 1 - t 2 ˆ Column I Column II
3ÁË 1 + t 2 ¯˜ (P) y = x + 7
x = (A) x2 + y2 = 9
and
and x2 - y2 = 1 is
16 9
y = Ê 8t ˆ is
ËÁ 1 - t2 ˜¯
(B) x2 - y2 = 1 (Q) x2 - y2 = 1
(B) x = 1 (et + e-t ) (Q) a hyperbola 25 9 9 25
2
and
and
y = 1 (et - e-t ) is y = 3 2 x + 16 is
77
2

Hyperbola 6.21

(C) y2 = 8x (R) y = x + 2 (B) Tangents are drawn from a (Q) p
and (S) y = 3 x + 6 2 point (4, 3) to the hyperbola
xy = –1 is x2 - y2 = 1 , the angle be- 2
53 16 9
(D) x2 – y2 = 9 tween their tangents is
and
x2 + y2 = 4 (C) The angle between the as- (R) Ê 24 ˆ
ymptotes to the hyperbola ÁË 25 ¯˜
x2 - y2 = 1 is sin -1

5. Match the following columns:

Column I Column II 16 9

(A) The product of the perpendicu- (P) 16 (D) The angle between the as- (S) p
lars from the foci of any tangent ymptotes to the rectangular
to the hyperbola x2 - y2 = 1 is hyperbola x2 – y2 = 2013 is 3
16 9

(B) The product of the perpendicu- (Q) 9 Questions asked in Previous Years’
lars from the foci of any tan- JEE-Advanced Examinations
gent to the ellipse x2 + y2 = 1
25 16 1. The equation x2 - y2 = 1, r > 1 represents a/an
is 1- r 1+ r

(C) The product of the perpendicu- (R) 144/25 (a) ellipse (b) hyperbola

lars from any point on the on (c) circle (d) None [IIT-JEE, 1981]

x2 - y2 = 1 2. Each of the four inequalities given below defines a re-
9 16
the hyperbola to gion in the xy-plane. One of these four regions does not

its asymptotes is have the following property. For any two points (x1, y1)

(D) The product of the perpen- (S) 2/3 and (x2, y2) in the region, the point Ê x1 + x2 , y1 + y2 ˆ
ÁË 2 2 ˜¯
diculars from any point on the
is also in the region. The inequality defining the region
hyperbola x2 - y2 = 1 to its
2 is

asymptotes is (a) x2 + 2y2 £ 1 (b) max.{|x|, |y|} £ 1

(c) x2 – y2 £ 1 (d) y2 – x £ 0

6. Match the following columns: [IIT-JEE, 1981]

Column I Column II No questions asked from 1982 to 1993.
(P) have a common
(A) Two intersecting 3. The equation 2x2 + 3y2 – 8x – 18y + 35 = k represents
circles tangent
(Q) have a common (a) no locus if k > 0
(B) Two mutually
external circles normal (b) an ellipse if k < 0
(R) do not have a com-
(C) Two circles, one (c) a point if k = 0
strictly inside the mon tangent
other (d) a hyperbola if k > 0 [IIT-JEE, 1994]
(S) do not have a com-
(D) Two branches of a mon normal. No questions asked in 1995.
hyperbola
4. An ellipse has eccentricity 1/2 and one focus at S(1/2, 1).
7. Match the following columns: Its one directrix is the common tangent (nearer to S) to
the circle x2 + y2 = 1 and x2 – y2 = 1. The equation of the
ellipse in the standard form is … [IIT-JEE, 1996]

5. A variable straight line of slope 4 intersects the hyper-

Column I Column II bola xy = 1 at two points. Find the locus of the point

which divides the line segment between these two

(A) Tangents are drawn from a (P) -1 Ê 3ˆ points in the ratio 1 : 2. [IIT-JEE, 1997]
point on the circle x2 + y2 = 11 ÁË 5¯˜
to the hyperbola x2 - y2 = 1 , sin 6. If the circle x2 + y2 = a2 intersects the hyperbola xy = c2

25 14 in four points P(x1, y1), Q(x2, y2), R(x3, y3) and S(x4, y4),
then

the angle between the tangent (a) x1 + x2 + x3 + x4 = 0 (b) y1 + y2 + y3 + y4 = 0
is (c) x1 ◊ x2 ◊ x3 ◊ x4 = c4 (d) y1 ◊ y2 ◊ y3 ◊ y4 = c4

[IIT-JEE, 1998]

6.22 Coordinate Geometry Booster

7. The angle between a pair of tangents drawn from a (c) focus of the hyperbola is (5, 0)

point P to the parabola y2 = 4ax is 45°. Show that the (d) vertex of the hyperbola is (5 3, 0)

locus of the point P is a hyperbola. [IIT-JEE, 2006]
14. A hyperbola, having the transverse axis of length 2 sin q
8. If x = 9 is the chord of contact of the hyperbola x2 – y2

= 9, the equation of the corresponding pair of tangents is confocal with the ellipse 3x2 + 4y2 = 12, then its equa-

is tion is

(a) 9x2 – 8y2 + 18x – 9 = 0 (a) x2 cosec2 q – y2 sec2 q = 1
(b) x2 sec2 q – y2 cosec2 q = 1
(b) 9x2 – 8y2 – 18x – 9 = 0 (c) x2 sin2 q – y2 cos2 q = 1
(d) x2 cos2 q – y2 sin2 q = 1
(c) 9x2 – 8y2 – 18x + 9 = 0

(d) 9x2 – 8y2 + 18x + 9 = 0 [IIT-JEE, 1999] [IIT-JEE, 2007]

9. Let P(a sec q, b tan q) and Q(a sec j, b tan j), where 15. Consider a branch of the hyperbola
x2 - 2y2 - 2 2x - 4 2 y - 6 = 0 with the vertex at the
q +j = p , be two points on the hyperbola x2 - y2 =1.
2 a2 b2
point A. Let B be one of the end points of its latus rec-

If (h, k) is the point of the intersection of the normals at tum. If C be the focus of the hyperbola nearest to the
P and Q, then k is point A, the area of DABC is

(a) Ê a2 + b2 ˆ (b) - Ê a2 + b2 ˆ Ê 2ˆ (b) Ê 3 ˆ
ËÁ a ¯˜ ËÁ a ¯˜ (a) ËÁ1- 3 ˜¯ ËÁ 2 - 1¯˜

(c) Ê a2 + b2 ˆ (d) - Ê a2 + b2 ˆ Ê 2ˆ (d) Ê 3 ˆ
ÁË b ˜¯ ÁË b ˜¯ (c) ÁË1+ 3 ¯˜ ËÁ 2 + 1¯˜

[IIT-JEE, 1999] [IIT-JEE, 2008]
No questions asked in 2000 to 2002. 16. An ellipse intersects the hyperbola 2x2 – 2y2 = 1 or-

10. For a hyperbola x2 - y2 =1, which of the fol- thogonally. The eccentricity of the ellipse is reciprocal
cos2a sin 2a of that of the hyperbola. If the axes of the ellipse are
along the co-ordinate axes, then the
lowing remains constant with the change of a? (a) ellipse is x2 + 2y2 = 2
(b) ellipse is x2 + 2y2 = 4
(a) abscissae of vertices (b) abscissae of foci (c) foci of the ellipse are (±1, 0)

(c) eccentricity (d) directrix

[IIT-JEE, 2003] (d) foci of the ellipse are (± 2, 0) .
[IIT-JEE, 2009]
11. The point of contact of the line 2x + 6 y = 2 and the
hyperbola x2 – 2y2 = 4 is 17. Match the following columns:

(a) (4, - 6) (b) ( 6, 1) Column I Column II
(A) Circle
(c) (1/2, 1/ 6) (d) (1/6, 3/2) (P) The locus of the point (h, k)
(B) Parabola for which the line
[IIT-JEE, 2004] (C) Ellipse hx + ky = 1
12. Tangents are drawn to the circle x2 + y2 = 9 from a touches the circle
x2 + y2 = 4.
points on the hyperbola x2 - y2 = 1 . Find the locus of
94 (Q) A point z in the complex
plane satisfying
the mid-point of the chord of contact. [IIT-JEE, 2005] |z + 2| – |z – 2| = ±3
13. Let a hyperbola passes through the focus of an ellipse
(R) The points of the conic have
x2 + y2 = 1 . The transverse and conjugate axis of this parametric representations
25 16
hyperbola coincide with the major and minor axes of x= 3 Ê 1 - t 2 ˆ and
the given ellipse, also the product of eccentricities of ËÁ 1 + t 2 ¯˜
the given ellipse and hyperbola is 1, then the

(a) hyperbola is x2 - y2 = 1 y = Ê 1 2t 2 ˆ
9 16 ËÁ +t ˜¯

(b) hyperbola is x2 - y2 = 1
9 25

Hyperbola 6.23

(D) Hyperbola (S) The eccentricity of the If the hyperbola passes through a focus of the
conic lies in the interval ellipse, then
1£x< . (a) the hyperbola is x2 - y2 = 1

(T) The points z in the complex 32
plane satisfying
Re(z + 1)2 = |z|2 + 1 (b) a focus of hyperbola is (2, 0)
(c) the eccentricity of hyperbola is 2
[IIT-JEE, 2009]
3
Comprehension
(d) the hyperbola is x2 – 3y2 = 3 [IIT-JEE, 2011]
The circle x2 + y2 – 8x = 0 and the hyperbola x2 - y2 = 1
94 23. Tangents are drawn to the hyperbola x2 - y2 = 1,
94
intersect at the points A and B.
18. The equation of the common tangent with positive parallel to the straight line 2x – y = 1. The points

slope to the circle as well as to the hyperbola is of contact of the tangents on the hyperbola are

(a) 2x - 5y - 20 = 0 (b) 2x - 5y + 4 = 0 (a) Ê 2 9 , 1ˆ (b) Ê - 2 9 , - 1ˆ
ÁË 2 2 ˜¯ ÁË 2 2 ¯˜

(c) 3x – 4y + 8 = 0 (d) 4x – 3y + 4 = 0 (c) (3 3, - 2 2) (d) (-3 3, 2 2)

19. The equation of a circle with AB as its diameter is

(a) x2 + y2 – 12x + 24 = 0 [IIT-JEE, 2012]

(b) x2 + y2 + 12x + 24 = 0 No questions asked in between 2013-2014.
24. Consider the hyperbola H : x2 – y2 = 1 and a circle S
(c) x2 + y2 + 24x – 12 = 0
with center N(x2, 0). Suppose that H and S touch each
(d) x2 + y2 – 24x – 12 = 0 [IIT-JEE, 2010] other at a point P(x1, y1) with x1 > 1 and y1 > 0. The
common tangent to H and S at P intersects the x-axis at
20. The line 2x + y = 1 is the tangent to the hyperbola point M. If (l, m) is the centroid of the triangle DPMN,

x2 - y2 =1. If this line passes through the point of then the correct expression(s) is (are)
a2 b2

intersection of the nearest directrix and the x-axis, dI 1
dx1 3x12
the eccentricity of the hyperbola is … (a) =1- for x1 > 1

[IIT-JEE, 2010]

21. Let P(6, 3) be a point on the hyperbola x2 - y2 = 1. If (b) dm = x1 for x1 > 1
a2 b2 dx1 3( x12
- 1)

the normal at the point P intersects the x-axis at (9, 0), dI 1
dx1 3x12
the eccentricity of the hyperbola is (c) =1+ for x1 > 1

(a) 5 (b) 3 (c) 2 (d) 3 (d) dm = 1 for y1 > 0 [IIT-JEE-2015]
2 2 dy1 3

22. Let the eccentricity of the hyperbola x2 - y2 =1 No questions asked in 2016.
a2 b2

be reciprocal to that of the ellipse x2 + 4y2 = 4.

ANSWERS

LEVEL II 26. (a) 27. (b) 28. (c) 29. (a) 30. (b)
31. (a) 32. () 33. (c) 34. (b) 35. (a)
1. (b) 2. (a) 3. (a) 4. (b) 5. (d) 36. (c) 37. (d) 38. (b) 39. (c) 40. (d)
6. (b) 7. (b) 8. (d) 9. (b) 10. (a) 41. (d) 42. (b) 43. (a) 44. (a) 45. (a)
11. (d) 12. (b) 13. (d) 14. (a) 15. (c) 46. (a) 47. (d) 48. (c) 49. (d) 50. (b)
16. (b) 17. (c) 18. (c) 19. (b) 20. (b) 51. (a) 52. (a) 53. (a) 54. (c) 55. (c)
21. (d) 22. (b) 23. (d) 24. (c) 25. (a)

6.24 Coordinate Geometry Booster

LEVEL III INTEGER TYPE QUESTIONS

1. x2 - y2 = 1 1. 2 2. 8 3. 4 4. 1 5. 0
94 6. 6 7. 4 8. 3 9. 2 10. 6
11. 6 12. 6
2. y = ± x ± a2 - b2
COMPREHENSIVE LINK PASSAGES
3. y = ± 7 x + 11
10 5 Passage I: 1. (d) 2. (b) 3. (c)
Passage II: 1. (a) 2. (c) 3. (b)
4. 72 Passage III: 1. (b) 2. (c) 3. (d)
10. 4a2x2 + b2y2 = 4a4 Passage IV: 1. (b) 2. (c) 3. (d)
22. 2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0 Passage V: 1. (c) 2. (b) 3. (a)
23. 3x2 + 10xy + 8y2 + 4x + 6y + 1 = 0 Passage VI: 1. (d) 2. (c) 3. (d)

LEVEL IV MATRIX MATCH

2. 75x – 16y = 418 1. (A) Æ Q; (B) Æ R; (C) Æ S; (D) Æ P
2. (A) Æ P; (B) Æ Q; (C) Æ Q; (D) Æ Q
3. (x2 + y2)2 = 16x2 – 9y2 3. (A) Æ Q; (B) Æ P; (C) Æ R; (D) Æ T; (E) Æ S;

Ê 2hˆ 2 Ê 2k ˆ 2 (F) Æ R
ÁË 3 ˜¯ ÁË 3 ¯˜ 4. (A) Æ Q; (B) Æ P; (C) Æ R; (D) Æ S
10. x - + y - = a2 5. (A) Æ Q; (B) Æ P; (C) Æ R; (D) Æ S
6. (A) Æ P, Q; (B) Æ P, Q; (C) Æ Q, R; (D) Æ Q, R
7. (A) Æ Q; (B) Æ P; (C) Æ R; (D) Æ S

HINTS AND SOLUTIONS

LEVEL I (c) Co-vertices: B(a, 0) = B(4, 0)
and B¢(–a, 0) = B¢(–4, 0)
1. (i) The equation of the given hyperbola is x2 - y2 = 1
94 (d) The length of the transverse axis = 2a = 6
(e) The length of the conjugate axis = 2b = 8
(a) Centre: (0, 0)
(b) Vertices: A(a, 0) = A(3, 0) (f) The length of the latus rectum = 2a2 = 16
b3
and A (–a, 0) = A(–3, 0)
(c) Co-vertices: B(0, b) = B(0, 2) (g) Eccentricity = e = 1+ a2 = 1+ 9 = 5
b2 16 4
and B¢(0, –b) = B¢(0, –2)
(d) The length of the transverse axis = 2a = 6 (h) Equation of the directrices :
(e) The length of the conjugate axis = 2b = 4
y = ± b = ± 3 = ± 12 .
(f) The length of the latus rectum = 2b2 = 8 e 5/4 5
a3
(iii) The equation of the hyperbola is
(g) Eccentricity = e = 1+ b2 = 1+ 4 = 13 9x2 – 16y2 – 36x + 96y – 252 = 0
a2 9 3
fi 9(x2 – 4x) – 16(y2 – 6y) = 252
fi 9(x – 2)2 – 16(y – 3)2 = 252 + 36 – 144 = 144

(h) Equation of the directrices: fi 9(x - 2)2 - 16(y - 3)2 = 1
144 144
x=±a=± 2 =± 6
e 13/3 13 fi (x - 2)2 - ( y - 3)2 = 1
16 9
(ii) The equation of the given hyperbola is
x2 - y2 = -1 (a) Centre : (0, 0)
16 9 fi X = 0, Y = 0
fi x – 2 = 0, y – 3 = 0
(a) Centre: (0, 0) fi x = 2 and y = 3
(b) Vertices: A(0, b) = A(0, 3) Hence, the centre is (2, 3).

and A(0, –b) = A(0, –3)

Hyperbola 6.25

(b) Vertices : (±a, 0) fi b= 4
fi X = ±a, Y = 0 5
fi x – 2 = 4, y – 3 = 0
fi x = 2 ± 4, y = 3 Also, a2 = b2 (e2 -1) = 16 Ê 25 - 1¯˜ˆ = 84
Hence, the vertices are (6, 3) and (–2, 3). 25 ÁË 4 25

(c) Co-vertices: (0, ±b) Hence, the equation of the hyperbola is
fi X = 0, Y = ±b
fi x –2 = 0, y – 3 = ±3 (x + 3)2 - ( y - 2)2 =1
fi x = 2, y = 3 ± 3 a2
Hence, the co-vertices are (2, 6) and (2, 0) b2

(d) The length of the transverse axis = 2a = 8 fi (x + 3)2 - ( y - 2)2 = 1
(e) The length of the conjugate axis = 2b = 6 84 16 25

5. From the definition of the hyperbola, we can write

(f) Eccentricity = e = 1+ 9 = 5 SP = e ,
16 4 PM

(g) Co-ordinates of Foci : (±ae, 0) where S = focus, P = (x, y)

X = ±5, Y = 0 fi SP2 = e2 PM2

fi X – 2 = ±5, y – 3 = 0 fi (x - 2)2 + (y - 1)2 = 4 ÔÏÊ x + 2y -1ˆ 2 ¸Ô
ÔÓÌÁË 1+ 4 ¯˜ ˝
fi X = 2 ± 5, y = 3 Ô˛

Hence, the co-ordinates of the foci are (7, 3) fi 5{(x – 2)2 + (y – 1)2} = 4(x + 2y – 1)2

and (–3, 3) 6. Given relation is

2. The equation of the hyperbola with centre (1, 0) is 2 ae = 16

(x -1)2 - y2 =1 fi ae = 8
a2 b2
fi a = 8/e fi a = 2

Here, 2a = 6 fi a = 3 Now, b2 – a2(e2 – 1) – a2e2 – a2 = 64 – 4 = 60
Also, one focus = (6, 0)
fi 1 + ae = 6 Hence, the equation of the hyperbola is
fi ae = 5
fi 3e = 5 x2 - y2 =1
fi e = 5/3 a2 b2
Therefore,
fi x2 - y2 = 1
4 60

b2 = a2 (e2 - 1) = 9 ÁÊË 25 - 1ˆ¯˜ = 16 7. Since foci are (6, 4) and (–6, 4), so the axis of the
9 hyperbola is parallel to x-axis.
It is given that the distance between two foci is 12, so
Hence, the equation of the hyperbola is 2 ae = 12
fi ae = 6
(x -1)2 - y2 = 1 fi 2a=6
9 16 fi a=3
Also,
3. Since the focus is (5, 2) and the vertex is (4, 2), so the b2 – a2(e2 – 1) – a2e2 –a2 – 36 – 9 = 27
axis of the hyperbola is parallel to x-axis and a = 1, Hence, the equation of the hyperbola is
ae = 2
Let the equation of the hyperbola be x2 y2
a2 b2
(x - 3)2 - ( y - 2)2 =1 …(i) - =1
a2 b2
fi x2 - y2 = 1
As we know that, the relation in a, b and e with respect 9 27

to a hyperbola is 8. Given relation is
2b2 = 1 ¥ (2a) = a
b2 – a2(e2 – 1) – a2e2 – a2 – 4 – 1 – 3 a2
Now from Eq. (i), we get
fi 2b2 = a
(x - 3)2 - ( y - 2)2 = 1 a
13

4. Since the focus is (–3, 2) and the vertex is (–3,4), so the fi 2b2 = a2
axis of the hyperbola is parallel to y-axis and be = 2 Also, relation in a, b and e is

fi b¥5 =2 b2 = a2(c2 – 1)
2

6.26 Coordinate Geometry Booster

fi a2 = a2 (e2 -1) fi 3(x2 –2x) – 5(y2 – 4y) = 32
2 fi 3(x – 1)2 – 5(y – 2)2 = 32 + 3 – 20 = 15

fi (e2 -1) = 1 fi 3(x -1)2 - 5( y - 2)2 = 1
2 15 15

fi e= 3 fi (x -1)2 - ( y - 2)2 = 1
2 53
x2 y2
9. Let the equation of the hyperbola is a2 - b2 =1 and 13. The given straight lines are …(i)
x - y = 2013 …(ii)
its conjugate is x2 - y2 = -1 . ab
a2 b2
Therefore, and
x+ y= 1
b2 a2 a b 2013
a2 b2
e1 = 1 + and e2 = 1 + Multiplying Eqs (i) and (ii), we get

Ê a2 + b2 ˆ Ê a2 + b2 ˆ Ê x - yˆ Ê x + yˆ = 2013 ¥ 1 =1
ËÁ a2 ˜¯ ËÁ b2 ¯˜ ÁË a b ¯˜ ËÁ a b ¯˜ 2013
e12 = and e22 =
fi x2 y2
a2 b2
fi - =1

fi 1 + 1 = Ê a2 ˆ + Ê b2 b2 ˆ which represents a hyperbola.
e12 e22 ËÁ a2 + b2 ˜¯ ÁË a2 + ¯˜
14. We have,

= Ê a2 + b2 ˆ =1 x = Ê 1 + t2 ˆ and y = 4t
ÁË a2 + b2 ˜¯ 3ËÁ 1 - t2 ¯˜ t2 -1

x2 y2 x2 y2 Ê1+ t2 ˆ 2 Ê 2t ˆ 2
a2 b2 9 4 ËÁ 1 - t2 ¯˜ ÁË -t ˜¯
10. Let the equation of the ellipse be + =1 and the fi - = - 2

1

equation of the hyperbola is is x2 - y2 =1 fi x2 - y2 = (1+ t2 )2 - 4t2 = Ê1- t2 ˆ 2 =1
a2 b2 9 4 (1 - t2 )2 ËÁ 1 - t2 ¯˜

Also, it is given that,

2b = 2a fi x2 - y2 = 1
fi b=a 94

Let e1 and e2 be the eccentricities of the ellipse and the which represents a hyperbola.
hyperbola.
15. We have,
Then e1 = 1 - b2 = 1 - a2 =0
a2 a2 x = 1 (et + e-t ) and y = 1 (et - e-t )
22
and e2 = 1+ b2 = 1+ a2 = 2
a2 a2 fi 2x = (et + e–t) and 2y = (et – e–t)
fi 4x2 – 4y2 = (et + e–t)2 – (et – e–t)2
Thus, e12 + e22 = 2 fi 4x2 – 4y2 = 2 + 2
fi x2 – y2 = 1
11. The equation of the hyperbola is Which represents a rectangular hyperbola.
4(2y – x – 3)2 – 9(2x + y – 1)2 = 80
16. The given equation of the hyperbola is
The centre of the hyperbola is obtained from the equa- x2 + y2 = 1

tions 2014 - l 2013 - l

2y – x – 3 = 0 and 2x + y – 1 = 0 fi (2013 – l)x2 + (2014 – l)y2
Solving, we get, –(2013 – l)(2014 – l) = 0

x = –2/5 fi (2013 – l)x2 + (l – 2014)y2
and y = 13/10 –(2013 – l)(2014 – l) = 0

Hence, the centre is Ê - 2 , 13ˆ . The given equation represents a hyperbola, if
ÁË 5 10¯˜ h2 – ab > 0

12. The equation of the given hyperbola is fi 0 – (2013 – l)(l – 2014) > 0
3x2 – 5y2 – 6x + 20y – 32 = 0 fi (2013 – l)(l – 2014) < 0
fi 2013 < l < 2014
fi l Π(2013, 2014)

Hyperbola 6.27

17. The given equation of the hyperbola is 19. We have,

x2 - y2 = 1 2 x12 - 3 y12 -1 = 2 – 48 – 1 = – 47 < 0
144 81 25
Thus, the point (1, 4) lies outside of the hyperbola.
We have, 20. Since the point (l, –1) is an exterior point of the curve

e= 1+ b2 = 1+ 81 = 225 = 15 = 5 4x2 – 3y2 = 1, so
a2 144 144 12 4 4l2 – 3 – 1 < 0
fi 4l2 – 4 < 0
Also, a2 = 144 fi l2 – 1 < 0
25 fi (l + 1)(l – 1) < 0
fi –1 < l < 1
Thus, the foci are fi l Œ (–1, 1)
Thus, the length of the interval, where l lies is 2.
(± ae, 0) = Ê ± 12 ¥ 5 , 0ˆ¯˜ = (± 3, 0). Therefore, m = 2
ËÁ 5 4 Hence, the value of m + 10 = 2 + 10 = 12.

Now, for the ellipse,

ae = 3 21. As we know that if the line y = mx + c be a tangent

fi a2e2 = 9 x2 y2
a2 b2
Thus, b2 = a2(1 – e2) to the hyperbola - =1, the co-ordinates of the

= a2 – a2e2

= 16 – 9 = 7 Ê a2m , b2 ˆ
ÁË ± c c ¯˜
Hence, the value of b2 is 7. point of contact is ± .

18. Let LL¢ be the latus rectum of the given hyperbola.

Ê b2 ˆ Ê b2 ˆ The equation of the given hyperbola is
L ËÁ ae, a ¯˜ ËÁ a ˜¯
Therefore, and L¢ ae, - and the cen- 25x2 – 9y2 = 225

tre of the hyperbola is C(0, 0) fi x2 - y2 = 1 …(i)
9 25 …(ii)
(b2 /a) b2
Now, slope of CL = m1 = ae = a2e Also, the given line is

25x + 12y = 45

(-b2 /a) b2 fi 12y = –25x + 45
ae a2e
and the slope of m2 = = - fi y = - 25 x + 45
12 12
Since, the latus rectum subtends right angle at the cen-
Here, a = 3, b = 5 and m = – 25/12.
tre, so
Thus, the common point is
m1 ¥ m2 = –1
Ê a2m , b2 ˆ 20ˆ
fi Ê b2 ˆ ¥ Ê b2 ˆ = -1 ËÁ ± c ± c ¯˜ = Ê 5, ± 3 ¯˜ .
ÁË a 2e ¯˜ ËÁ - a 2e ˜¯ ËÁ

Ê b4 ˆ 22. As we know that, the line y = mx + c will be a tangent
ËÁ a4e2 ¯˜
fi =1 to the hyperbola x2 - y2 = 1, if
a2 b2

fi b4 = a4e2 c2 = a2m2 – b2 …(i)
fi a4(e2 – 1)2 = a4e2
fi (e2 – 1)2 = e2 The equation of the hyperbola is
fi e4 – 3e2 + 1 = 0 9x2 – 5y2 = 45

fi e2 = 3 ± 5 fi x2 - y2 = 1
2 59

Here, a2 = 5, b2 = 9, m = 3, c = l

fi e2 = 3 + 5 Therefore, from Eq. (i), we get
2
l2 = a2m2 – b2 = 15 – 9 = 6

fi l=± 6

e2 = 3 + 5 = 6 + 2 5 Ê 5 + 1ˆ 2 23. The equation of any tangent to the hyperbola
24 ÁË 2 ¯˜
fi = x2 y2
a2 b2
- =1 is

fi Ê 5 +1ˆ y = mx + a2m2 - b2 …(i)
e = ËÁ 2 ¯˜