Coordinate Geometry

6.28 Coordinate Geometry Booster

The equation of the given tangent is If the line (ii) be a tangent to the hyperbola (i), the co-
ordinates of the point of contact can be
y = mx + 9m2 - 4 …(ii)

Since Eqs (i) and (ii) are identical, so Ê a2m , ± b2 ˆ = Ê ± (-5), ± 4ˆ
ÁË ± c c ˜¯ ËÁ 3˜¯
a2 = 9, b2 = 4

Hence, the equation of the hyperbola is

x2 - y2 = 1 27. The equation of the given hyperbola is
94 4x2 – 9y2 = 36
fi 4x2 – 9y2 = 36
fi x2 - y2 = 1
24. The equation of any tangent to the parallel to 5 x – 4y + 94

7 = 0 is Here, a2 = 9, b2 = 4.
The equation of any tangent to the hyperbola
5x – 4y + l = 0

fi 4y = 5x + l x2 y2
a2 b2
fi y= 5x+l …(i) - =1 is
44
y = mx + a2m2 - b2
The equation of the given hyperbola is

4x2 – 9y2 = 36

fi x2 - y2 = 1 …(ii) fi y = mx + 9 m2 - 4
94
which is passing through (3, 2). So
Since, the line (i) is a tangent to the hyperbola (ii), so (2 – 3m)2 = 9m2 – 4

l2 = 9 Ê 25ˆ -4 fi 4 – 12m + 9m2 = 9m2 – 4
16 ËÁ 16 ¯˜ fi 4 – 12m = –4
fi m 2,m
fi l 2 = 225 - 64
16 16 3

fi l2 = 161 Hence, the equations of the tangents are

fi l = ± 161 x + 3 = 0 and y = 2 x .
Hence, the equations of the tangents are 3

5x - 4 y ± 161 = 0 28. We have,

25. The equation of any line perpendicular to 2 x12 - 3y12 -12 = 2.1 – 3. 4 – 12
3x + 4y + 10 = 0 is
4x – 3y + l = 0 = 2 – 12 – 12 = 2 – 24 = –22 < 0

So, the point (1, –2) lies outside of the hyperbola.

y = Ê 4ˆ x + l Thus, the number of tangents is 2.
ËÁ 3˜¯ 3
fi …(i) 29. The equation of the given hyperbola is

The equation of the given hyperbola is 3x2 – 4y2 = 12

9x2 – 16y2 = 144 fi x2 - y2 = 1
43
fi x2 - y2 = 1 …(ii) …(i)
16 9
Here, a2 = 4, b2 = 3 and m = 4
The line (i) will be a tangent to the hyperbola (ii), if The equation of any tangent to the hyperbola (i) is

ÊÁË l ¯˜ˆ 2 ÊËÁ 4 ˆ¯˜ 2
3 3
= 16 - 9 y = mx + a2m2 - b2

fi l2 = 256 – 81 = 175 fi y = 4x ± 64 - 3

fi l = ± 175 = ± 5 7 fi y = 4x ± 61

Hence, the equation of the tangents are Hence, the equations of tangents are
y = 4x + 61 and y = 4x - 61.
4x - 37 ± 5 7 = 0. …(i)
…(ii) 30. The equation of tangent to the curve
26. The given hyperbola is x2 – y2 – 8x + 2y + 11 = 0 is
xx1 – yy1 – 4(x – x1) + (y – y1) + 1 = 0
x2 – 9y2 = 9
fi 2x – y – 4(x + 2) + (y + 1) + 11 = 0
fi x2 - y2 = 1 fi –2x + 4 = 0
91 fi x–2=0

The given line is 5x + 12y – 9 = 0

fi y = Ê - 5 ˆ x + 3
ËÁ 12 ˜¯ 4

Hyperbola 6.29

31. When y = 2, then 4x2 = 36 34. The equation of the given curves are x2 - y2 = 1 and
fi x = ±3 94
Hence, the points are (3, 2) and (–3, 2). x2 y2
The equation of the tangent at (3, 2) is 9 + 4 =1.
12x – 6y = 24
fi 2x – y = 4 Y
Also, the equation of the tangent at (–3, 2) is
–12x – 6y = 24 X¢ c y=2
fi 2x + y + 4 = 0 X

32. The equation of the given hyperbola is y=2
9x2 – 16y2 = 144

fi x2 - y2 = 1 …(i)
16 9

The equation of any tangent to the hyperbola (i) can be x = –3 Y¢ x=3

considered as y = mx + 16m2 - 9 Here, the length of the major axis of the ellipse is equal
to the length of the transverse axis and also the length of
which is passing through (4, 3). So the minor axis is equal to the length of the conjugate axis.
(3 – 4m)2 = 16m2 – 9. Thus, the equations of the common tangents are x = ±3
and y = ±2.
fi 9 – 24 m + 16 m2 = 16 m2 – 9 35. The equation of any tangent to the hyperbola
fi 24m = 18 x2 - y2 = 1 is
fi m 3 and m 16 9

4
Let q be the angle between them. Then

3 3 1 y = mx + 16m2 - 9
3
tan(q ) = 4 3. = 4 = 4 fi m x - y + 16m2 - 9 = 0 …(i)
1 1 3
Y
4 4

fi q = tan -1 Ê 4ˆ
ÁË 3˜¯
M
Hence, the angle between the tangents is X¢ C X

tan -1 Ê 4 ˆ .
ÁË 3 ¯˜

33. The equation of any tangent to the hyperbola

x2 - y2 =1 is Y¢
a2 b2
If the tangent (i) is also the tangent to the circle x2 + y2
y = m1 x + a2m12 - b2 …(i) = 9, the length of the perpendicular from the centre of
the circle is equal to the radius of the circle. So

The equation of any tangent to the hyperbola 0 - 16m2 - 9 = 3
m2 +1
x2 - y2 =1 is
(-b2 ) (-a2)
fi 16m2 – 9 = 9(m2 + 1)
y = m2 x + (-b2 )m22 - (-a2 ) …(ii) fi 16m2 – 9m2 = 9 + 1 = 10
fi 7m2 = 10
Since the equations (i) and (ii) are identical, so

a2m12 - b2 = (-b2 )m22 - (-a2 ) fi m=± 7
10
fi m12 = 1 and m22 = 1
Hence, the equation of tangents are y = ± 7 x + 11.
Thus, m1 = ±1 = m2 10 5
Hence, the equations of the common tan gents are
36. The equation of any tangent to the parabola y2 = 8x is

y = ± x + a2 - b2 y = mx + 2 …(i)
m

6.30 Coordinate Geometry Booster

Y Then F1 = (ae, 0) and F2 = (–ae, 0).
The equation of any tangent to the hyperbola is

x secq - y tanq = 1 …(i)
ab

X¢ X Let p1 and p2 be two perpendiculars from foci upon the
C tangent (i).

fi p1 p2 = a2 (e2 -1)(e2sec2q -1)
((e2 -1) sec2q + tan2q )

Y¢ p1 p2 = a2 (e2 -1)(e2sec2q -1)
((e2 -1) sec2q + (sec2q -1))
Since the tangent to the parabola is also a tangent to the fi

hyperbola x2 - y2 = 1 , so fi p1 p2 = a2 (e2 -1)(e2sec2q - 1)
95 (e2sec2q - 1)

Ê 2ˆ 2
ÁË m ¯˜
5 x2 - 9 mx + = 45

fi 5 x2 - 9 Ê m2 x2 + 4 + 4mˆ¯˜ = 45 fi p1p2 = a2(e2 – 1)
ËÁ m2 fi p1p2 = b2
Hence, the product of the lengths of the perpendiculars
fi (5 - 9m2 )x2 - 36 x - 9 Ê 4 + 5ˆ¯˜ = 0 is b2.
ÁË m2 39. Let P(a, b) be the point of intersection of tangents at A
and B.
Now,
YB
D=0

fi (36)2 + 36(5 - 9m2 ) Ê 4 + 5ˆ˜¯ = 0
ËÁ m2

fi 36 + 20 + 25 - 36 - 45m2 = 0 P
m2 X¢

fi 4 + 5 - 9 m2 = 0 X
m2

fi 9m4 – 5m2 – 4 = 0

fi (m2 – 1)(9m2 + 4) = 0 A

fi (m2 – 1) = 0

fi m = ±1 Y¢

Hence, the equations of the common tangents are Clearly, the point of intersection of the tangents at A
and B is the chord of contact.
y = ± x ± 2. Therefore, the chord of contact AB is

37. As we know that the locus of the perpendicular tan-

gents is the director circle. ax by
a2 b2
Hence, the equation of the director circle is - = 1

x2 + y2 = a2 – b2 = 16 – 9 = 7 b2xa b2
a2y b b
fi x2 + y2 = 7 x2 y2 fi y = -
a2 b2
38. Let F1 and F2 be two foci of the hyperbola - = 1.

which is a tangent to the parabola y2 = 4ax. So

Y - b2 a
b b2a ˆ
= Ê

N ÁË a2b ¯˜

X¢ F1 X fi b2 = Ê - b4 ˆ
F2 C ÁË a3 a ˜¯

M Hence, the locus of P(a, b) is
Y¢
y2 = Ê - b4 ˆ
ÁË a3 x¯˜

Hyperbola 6.31

esecq -1 fi ax + ay = a2 + b2
e

Then, p1 = sec2q tan 2q x y
a2 b2 Ê e(a2 + b2 )ˆ + b2 ˆ
+ fi + Ê a2 =1

ÁË a ¯˜ ÁË a ¯˜

esecq +1 Hence, the area of DOAB

and p2 = sec2q tan 2q 1 e(a2 + b2 ) Ê a2 + b2 ˆ
a2 b2 2 a ÁË a ˜¯
+ = ¥ ¥

fi p1 p2 = (esecq -1)(esecq + 1) = 1 ¥ e(a2 + b2 )2
2 a2
sec2q + tan 2q
a2 b2 = 1 ¥ a2e5
2
= a2b2 (e2sec2q -1)
(b2sec2q + a2tan2q ) 42. The equation of any normal at (a sec j, b tan j) to the

= a4 (e2 -1)(e2sec2q -1) hyperbola x2 - y2 =1 is
(a2 (e2 -1)sec2q + a2tan2q ) a2 b2

ax cos + by cot j = a2 + b2 …(i)

40. The equation of the normal to the curve x2 - y2 = 1 at The equation of any line perpendicular to (i) and pass-
16 9 ing through the origin is

(8, 3 3) is (b cot j)x – (a cos j)y = 0

a2x + b2 y = a2 + b2 fi bx – a sin jy = 0 …(ii)
x1 y1
fi 16x + 9y = 16 + 9 If we eliminate j between Eqs (i) and (ii), we get the
8 33 required locus of the foot of the perpendicular.
From (ii), we get,

fi 2 x + 3 y = 25 sin (j) = bx
Hence, the required equation of the normal is ay
2 x + 3 y = 25.
(a2 y2 - b2x2 )
41. Let, one end of the latus rectum of the given hyperbola fi cos (j) =

ay

is L Ê ae, b2 ˆ . (a2 y2 - b2x2 )
ÁË a ¯˜ and cot (j) =

Y bx
B From Eq. (i), we get

L ax ¥ (a2 y2 - b2x2 ) + by ¥ (a2 y2 - b2x2 )
ay bx
X¢ O S A C
= a2 + b2
L¢
Y¢ ( )fi (x2 + y2 ) (a2 y2 - b2x2 ) = (a2 + b2 ) xy

The equation of the normal to the given hyperbola at L fi (x2 + y2)2(a2y2 – b2x2) = (a2 + b2)2 x2y2
is
which is the required locus of the foot of the perpen-
a2x + b2 y = a2 + b2 dicular.
x1 y1
43. The equation of any normal to the hyperbola

x2 - y2 =1 is
a2 b2

ax cosj + by cotj = (a2 + b2) …(i)

6.32 Coordinate Geometry Booster

X¢ Y P Thus, the co-ordinates of G = Ê a2 + b2 ˆ
N MX ËÁ a cosj , 0˜¯

Q Clearly, the vertices, A = (a, 0) and A¢ = (–a, 0)

O Now, AG = Ê a2 + b2 ˆ
ÁË a cosj - a¯˜

and A¢G = Ê a2 + b2 + ˆ
ÁË a cosj a¯˜

Y¢ Therefore,

Since the normal (i) meets the x-axis at M and y-axis at = Ê a2 + b2 - ˆ Ê a2 + b2 + ˆ
ÁË a cosj a˜¯ ËÁ a cosj a˜¯
N respectively. Then, AG. A¢G

M= Ê a2 + b2 , ˆ and N = Ê 0, Ê a2 + b2 ˆ tan ˆ Ê 2 ˆ
ÁË a cosj 0¯˜ ÁË ËÁ b ¯˜ j ¯˜ Á ˜
ËÁ Ê a 2 + b2 ˆ sec2j a2 ˜¯
ËÁ a ¯˜
Let the co-ordinates of the point P be (a, b). = -

Since PM and PN are perpendiculars to the axes, so the = (a2e2 sec2j – a2)
= a2(e2 sec2j – 1)
co-ordinates of P are fi m = 2, n = 2, p = 2
Hence,
ÊÊ a2 + b2 ˆ sec j , Ê a2 + b2 ˆ ˆ (m + n + p)2 + 36 = 36 + 36 = 72.
ÁË ËÁ a ˜¯ ËÁ b ¯˜ tan j˜¯ 45. The equation of the normal to the hyperbola xy = c2 at

Therefore,

a = Ê a2 + b2 ˆ sec j and b = Ê a2 + b2 ˆ tan j Ê c ˆ
ÁË a ˜¯ ËÁ b ¯˜ ÁË t ¯˜
ct, is

fi a Ê a ˆ = secj and b Ê b ˆ = tanj xt3 – yt – ct4 + c = 0
ËÁ + ˜¯ ËÁ + ˜¯
a2 b2 a 2 b2 fi ct4 – xt3 + yt – c = 0

As we know that, which is passing through (a, b), so
sec2 j – tan2 j = 1
ct4 – at3 + bt – c = 0.

Ê a ˆ 2 Ê b ˆ 2 Let its four roots are t1, t2, t3, t4.
ÁË + ¯˜ ÁË + ¯˜
a2 - b 2 = 1

a2 b2 a2 b2 a
c
fi a2a2 – b2b2 = (a2 + b2)2 Therefore, t1 + t2 + t3 + t4 = ,
Hence, the locus of (a, b) is
0, S(t1t2t3) -b
a2x2 – b2y2 = (a2 + b2)2 S(t1t2) = = c

44. The equation of any normal to the hyperbola and S(t1t2t3t4) = 1.

x2 - y2 =1 at (f) is (i) x1 + x2 + x3 + x4 = c(t1 + t2 + t3 + t4 )
a2 b2
Ê a ˆ
ax cosj + by cotj = a2 + b2 = c ÁË c ¯˜ = a

Y

(ii) y1 + y2 + y3 + y4 = c Ê 1 + 1 + 1 + 1 ˆ
ÁË t1 t2 t3 t4 ¯˜

Q Ê (t1t2t3 ) ˆ
X¢ A¢ O A (t1t2t3t4 )¯˜
ÂÂ= c ËÁ

GX Ê Ê - b ˆ ˆ
Á ÁË c ˜¯ ˜
= cÁ ˜ = b
ÁÁË -1 ˜˜¯

Y¢

Hyperbola 6.33

(iii) x12 + x22 + x32 + x42 = c2 (t12 + t22 + t32 + t42 ) Â ÂTherefore, Ê 1ˆ = (x1x2 x3) = 2e2
ËÁ x1 ˜¯ x1x2 x3x4 b
{ }( )Â Â= c2 t1 2 - 2 (t1t2)
Ê 1 1 1 1 ˆ
ÌÏÓÔÔËÊÁ a ˆ 2 0Ô¸˝ (i) ( x1 + x2 + x3 + x4 ) ËÁ x1 + x2 + x3 + x4 ˜¯
c ¯˜ Ô˛
= c2 - = a 2

y1 2 - 2 ( y1y2 ) ( ) Â= Ê Ê 1 ˆˆ = 2b ¥ 2e2 = 4
( ) Â(iv) y12 + y22 + y32 + y42 = x1 ËÁ ÁË x1 ˜¯ ˜¯ e2 b

Â= (b)2 - 2c2 Ê 1ˆ (ii) Similarly we can have
ËÁ t1t2 ¯˜
( y1 + y2 + y3 + Ê 1 + 1 + 1 + 1 ˆ
Â= y4 ) ËÁ y1 y2 y3 y4 ¯˜
(b )2 2c2 Ê t1 t2 ˆ
- ËÁ t1t2t3t4 ¯˜ Ê Ê 1ˆ ˆ
ËÁ ÁË y1 ¯˜ ¯˜
( )=
= (b)2 Â y1 Â

(v) x1 ◊ x2 ◊ x3 ◊ x4 = c4(t1t2t3t4) = –c4 = 4.
47. The equation of the chord of contact of tangents drawn
(vi) y1 ◊ y2 ◊ y3 ◊ y4 = c4 Ê 1 ˆ
ËÁ t1t2t3t4 ˜¯ from the point (2, 3) is

= c4 Ê 1ˆ = -c4 2x - y = 1
ÁË -1˜¯ 92

46. the equation of any normal to the given hyperbola at fi 4x – 9y = 18
48. Any tangent to the hyperbola x2 - y2 = 1 is
(x, y) is
94
x (k - y) = y (x - b) …(i)
a2 b2 x secq - y tanq = 1
32

Ê 1 1ˆ y kx Y
ËÁ a2 b2 ˜¯ b a2
fi + xy - - = 0

fi y = b2kx …(i) X¢ C
a2 (e2x - b)
X
M(h, k)
The equation of the hyperbola is
B

x2 - y2 =1 …(ii)
a2 b2

Solving Eqs (i) and (ii), we get, Y¢

x2 - b4k 2 x2 =1 Let the tangent intersects the x-axis at A and y-axis at B,
a2 b2a4 (e2x - b)2 respectively.
Then A = (3 cos q, 0) and B = (0, –2 cot q)
fi a2e4x4 – 2ba2e2x3 – (a2b2 + b2k2 + a4e4)x2 + 2ba4e2x Let (h, k) be the mid-point of AB.
Therefore,
+ a4b2 – 0 …(iii)
2h = 3 cos q and 2k = –2 cot q
Let x1, x2, x3, x4 are the roots of Eq. (iii).

Then, x1 + x2 + x3 + x4 = 2b , fi secq = 3 and tanq = - 1
e2 2h k

 (x1x2 ) = - a2b2 + b2k 2 + a4e2 As we know that,
a2e4 sec2 q – tan2 q = 1

 (x1x2 x3 ) = - 2ba2 fi 9 - 1 =1
e2 4h2 k2

Âand (x1x2 x3 x4 ) = a2b2 Hence, the locus of (h, k) is 9 - 1 =1.
e4 4 x2 y2

Hyperbola

6.34 Coordinate Geometry Booster

49. Any point on the circle x2 + y2 = a2 be (a cos q, a sin q). x2 - y2 = Ê hx - ky ˆ 2
a2 b2 ËÁ a2 - b2 ˜¯
Y Ê h2 k2 ˆ 2
Q ËÁ a2 b2 ˜¯

P M(h, k) 1 Ê h2 k2 ˆ 2 1 Ê h2 k2 ˆ2
a2 ËÁ a2 b2 ¯˜ b2 ÁË a2 b2 ¯˜
X¢ O X fi - x2 - - y2

R h2 k2
a4 b4
= x2 + y2 - 2hk xy …(iii)
a2b2

Y¢ Equation (iii) will be a right angle, if co-efficient of
x2 + co-efficient of y2 = 0
The equation of the chord of contact from the point
k2 ˆ 2 k2 ˆ 2
(a cos q, a sin q) to the hyperbola x2 – y2 = a2 is fi 1 Ê h2 - b2 ˜¯ - h2 - 1 Ê h2 - b2 ¯˜ - k2 = 0
a2 ÁË a2 a4 b2 ËÁ a2 b4
(a cos q)x – (a sin q)y = a2

fi x cos q – y sin q = a …(i)

Let the mid-point be (h, k). Ê h2 k2 ˆ 2 Ê 1 1 ˆ Ê h2 k2 ˆ
ËÁ a2 b2 ˜¯ ÁË a2 b2 ˜¯ ËÁ a4 b4 ˜¯
The equation of the chord bisected at (h, k) to the hy- fi - - = +

perbola x2 – y2 = a2 is

hx – ky = h2 – k2 …(ii) Hence, the locus of (h, k) is

Since the Eqs (i) and (ii) are identical, so Ê x2 y2 ˆ 2 Ê 1 1ˆ Ê x2 y2 ˆ
ÁË a2 b2 ˜¯ ÁË a2 b2 ˜¯ ÁË a4 b4 ˜¯
h = -k = h2 - k2 - - = + .
cosq sinq a
51. Let the point on the parabola be (h, k).
ah -ak
fi cosq = h2 - k2 and sin q = h2 - k2 Y

Squaring and adding, we get

Ê ah ˆ 2 Ê -ak ˆ 2 Q
ËÁ h2 - k ¯˜ ËÁ h2 - k2 ˜¯ P(h, k)
2 + = 1
Q
fi a2(h2 + k2) = (h2 – k2)2 X¢ X
Hence, the locus of (h, k) is a2(x2 + y2) = (x2 – y2)2.
50. Let the mid-point be (h, k). O
R

Y
A

M(h, k) Y¢

X¢ O X The equation of the chord of contact of the parabola is

yk = 2a(x + h) …(i)

B fi y = 2 a x + 2 ah
kk
Y¢ …(ii)

The equation of the chord bisected at (h, k) to the given

hyperbola is Since, the line (ii) is a tangent to the hyperbola

T = S1 x2 - y2 = 1, so,
a2 b2
fi hx - ky = h2 - k2 …(i)
a2 b2 a2 b2 c2 = a2m2 – b2

The equation of the hyperbola is Ê 2ah ˆ 2 Ê 2a ˆ 2
ËÁ k ˜¯ ËÁ k ¯˜
x2 y2 fi = a2 - b2
a2 b2
- =1 …(ii)

Since the chord (i) subtends right angle at the centre, so fi 4a2h2 = 4a4 – k2b2
we can write Hence, the locus of (h, k) is

4a2x2 = 4a4 – y2b2

Hyperbola 6.35
…(i)
52. Any tangent to the hyperbola x2 - y2 =1 at R(q) is fi hx – ky = h2 – k2
a2 b2 fi ky = hx + (k2 – h2) X

x secq - y tanq = 1 Ê h ˆ Ê k 2 - h2 ˆ
ab ÁË k ¯˜ ÁË k ¯˜
…(i) fi y = x +

Y Y

R

P M(h, k)
X¢ O
X¢ C M(h, k) X
B
Q

Y¢ Y¢

Let the mid-point of PQ be (h, k). If the line (i) touches the parabola y2 = 4 ax, so

Then the equation of the chord bisected at (h, k) to the

ellipse x2 + y2 =1 is c= a
a2 b2 m

hx + ky = h2 + k2 …(ii) fi Ê h2 - k2 ˆ = a = ak
a2 b2 a2 b2 ËÁ k ¯˜ (h /k ) h

Therefore, the Eqs (i) and (ii) are identical. So fi h(h2 – k2) = ak2

secq /a = -tanq /b = 1 Hence, the locus of (h, k) is
h k +
a2 b2 Ê h2 k2 ˆ x(x2 – y2) = ay2
ÁË a2 b2 ¯˜
54. If (h, k) be the mid-point of the chord of the hyperbola

fi secq = -tanq = 1 x2 - y2 =1, then
h k + a2 b2
a b Ê h2 k2 ˆ
ÁË a2 b2 ˜¯ T = S1

hk fi hx - ky = h2 - k2
a2 b2 a2 b2
fi secq = a and tan q = b
Ê h2 + k2 ˆ Ê h2 + k2 ˆ Ê h2 k2 ˆ
ËÁ a2 b2 ¯˜ ËÁ a2 b2 ¯˜ fi ky = hx - ËÁ a2 - b2 ¯˜
b2 a2

We know that, fi y Ê b2h ˆ x - b2 Ê h2 - k2 ˆ …(i)
sec2 q – tan2 q = 1 = ËÁ a2k ¯˜ k ÁË a2 b2 ˜¯

Ê hˆ2 Êkˆ2
ËÁ a˜¯ ÁË b ˜¯
fi Ê h2 - =1
ÁË a2 k2 2 h2 k2 2
b2 ˆ Ê a2 b2 ˆ M(h, k) P
+ ¯˜ ÁË + ¯˜
Q
fi Ê h2 - k2 ˆ = Ê h2 + k2 ˆ2 X¢ O X
ÁË a2 b2 ¯˜ ÁË a2 b2 ¯˜

Hence, the locus of (h, k) is Ê x2 - y2 ˆ = Ê x2 + y2 ˆ 2
ËÁ a2 b2 ˜¯ ÁË a2 b2 ˜¯ .
Y¢
53. If (h, k) be the mid-point of the chord of the hyperbola
x2 – y2 = a2, then If the line (i) be a tangent to the circle x2 + y2 = c2, then
T = S1 C2 = A2(1 + m2)

6.36 Coordinate Geometry Booster

fi b4 Ê h2 - k2 ˆ 2 = c2 Ê b4h2 ˆ Thus, k = -h = -c2
k2 ÁË a2 b2 ¯˜ ÁË1 + a4k 2 ¯˜ 1 t at 2

fi (b2h2 – a2k2)2 = c2(a4k2 + b2h2) fi t = -h , t2 = -c2
Hence, the locus of (h, k) is k ak

(b2x2 – a2y2)2 = c2(a4y2 + b4x2) Eliminating t, we get,
h2a = –kc2
fi Ê x2 - y2 ˆ2 = c2 Ê x2 + y2 ˆ
ÁË a2 b2 ¯˜ ËÁ a4 b4 ¯˜ Hence the locus of (h, k) is

yc2 + x2a – 0

55. If (h, k) be the mid-point of the chord of the circle fi y = Ê - a ˆ x2.
x2 + y2 = a2, then ÁË c2 ˜¯

T = S1 57. Let the point P be (h, k).
fi hx + ky = h3 + k2
Then the equation of the chord of contact of the circle
hˆ Ê h2 + k 2 ˆ
Ê k ¯˜ ËÁ k ˜¯ x2 + y2 = a2 is
fi y = ËÁ - x + (i)
hx + ky = a2 …(i)

Y Y
P

R

R

X¢ O X Q

O M(h, k) X¢ O X
Q

P

Y¢ Y¢

If the line (i) be a tangent to the hyperbola x2 - y2 =1, The equation of the normal chord of the hyperbola
a2 b2
so, c2 = a2m2 – b2 x2 - y2 =1 at (f) is
a2 b2
h2 2 2 h2
Ê + k ˆ Ê k2 ˆ ax cos j – by cot j = a2 + b2 …(ii)
fi ÁË k ¯˜ = a2 ËÁ ¯˜ - b2
Equations (i) and (ii) are identical. Therefore,

fi (h2 + k2)2 = (a2h2 – b2k2) a cosj = - b cotj = (a2 + b2)
h k a2
Hence, the locus of (h, k) is

(x2 + y2)2 = (a2x2 – b2y2) sec j = a3 b2 ) and tan j = - a2b
56. Any tangent to the parabola y2 = 4ax at (at2, 2at) is h(a2 + k(a2 + b2)

yt = x + at2 …(i)

Y We have,
P sec2 j – tan2 j = 1

Ê a3 ˆ 2 Ê a2b ˆ 2
ÁË h(a2 + b2 )˜¯ ÁË k(a2 + b2 )˜¯
fi - - = 1

X¢ M(h, k) X fi Ê a6 - a4b2 ˆ = (a2 + b2 )2
Q O ÁË h2 k 2 ¯˜

fi Ê a2 - b2 ˆ = Ê a2 + b2 ˆ2
ÁË h2 k2 ˜¯ ËÁ a2 ˜¯
Y¢

If (h, k) be the mid-point of the chord of the hyperbola Hence, the locus of (h, k) is

xy = c2, then Ê a2 b2 ˆ Ê a2 + b2 ˆ2
ÁË x2 y2 ¯˜ ËÁ a2 ˜¯
xk + yh = c2 …(ii) - =

Therefore, the Eqs (i) and (ii) are identical.

Hyperbola 6.37

58. If (h, k) be the mid-point of the chord of the hyperbola x1 x2 a4
y1 y2 b4
x2 - y2 =1, then fi = -
a2 b2
Thus, m = 4 and n = 4
T = S1
Ê m + n ˆ 10 Ê 4 + 4 ˆ 10
ÁË 4 ˜¯ ÁË 4 ¯˜
fi hx - ky = h2 - k2 …(i) Hence, the value of = = 210 =1024.
x2 b2 a2 b2
60. The equation of the polar with respect to the hyperbola
Y
x2 y2
P a2 - b2 =1 is

M(h, k) xx1 - yy1 =1
X¢ O S a2 b2

Q fi x(-ae) - y◊0 =1
a2 b2

Y¢ fi x=-a
e
which is passing through the focus (ae, 0) of the given
61. The equation of the polars from points (x1, y1) and (x2,
hyperbola.
y2) to the hyperbola
Therefore,
x2 y2
aeh = h2 - k2 a2 - b2 = 1 are
a2 a2 b2
xx1 yy1
fi h2 - k2 = eh a2 - b2 =1 …(i)
a2 b2 a
xx2 yy2
Hence, the locus of (h, k) is and a2 - b2 =1 …(ii)

x2 - y2 = Ê eˆ x Now, slopes of (i) and (ii) are
a2 b2 ËÁ a˜¯

59. The equations of the chord of contact of the tangents to m1 = b2 x1 and m2 = b2 x2
a2 y1 a2 y2
the given hyperbola at (x1, y1) and (x2, y2) are

xx1 - yy1 =1 …(i) Since the polars of the given points are perpendicular,
a2 b2 …(ii)
xx2 yy2 so
a2 b2
and - =1 m1 ¥ m2 = –1

Y fi Ê b2 x1 ˆ ¥ Ê b2 x2 ˆ = -1
ÁË a2 y1 ¯˜ ÁË a2 y2 ˜¯

Q(x 2, y 2) ) fi x1x2 = - a4
y1 y2 b4
y

1

,

X¢ Q(x X x1x2 a4
O y1 y2 b4
1

fi + = 0

62. Let (x1, y1) be the pole of the hyperbola. Then the equa-

tion of the polar from a point (x1, y1) w.r.t. the hyperbola
x2 –3y2 = 3 is

Y¢ xx1 – 3yy1 = 3 …(i)
The equation of the given polar is
The slopes of the lines (i) and (ii) are
x–y=3 …(ii)
b2 b2
m1 = Ê a2 x1 ˆ and m2 = Ê a2 x1 ˆ Therefore, the Eqs (i) and (ii) are identical. So
ËÁ y1 ˜¯ ÁË y1 ˜¯
x1 = -3y1 = 3
Since, (i) and (ii) meet at right angles, so 1 -1 3

m1m2 = –1 fi x1 = 1, y1 = 1
3
Ê b2 x1 ˆ Ê b2 x1 ˆ
fi ÁË a2 y1 ˜¯ ¥ ÁË a2 y1 ¯˜ = -1 Hence, the pole is ÊËÁ1, 1ˆ .
3¯˜

6.38 Coordinate Geometry Booster

63. Let the pole be (h, k). Hence, the locus of (h, k) is

The equation of the polar from the point (h, k) w.r.t. the x2 + y2 = a2
4
hyperbola x2 - y2 =1 is
a2 b2
fi 4x2 + y2 = 4a2
hx ky
a2 - b2 =1 …(i) 65. Let (h, k) be the pole.

The equation of the normal chord of the given hyper- Then the equation of the polar w.r.t. the hyperbola

bola at (f) is x2 - y2 =1 is
a2 b2
ax cos j – by cot j = (a2 + b2) …(ii)

Equations (i) and (ii) are identical. Therefore, hx - ky =1
a2 b2
a cosj = b cotj = (a2 + b2)
(h/a2 ) (k /b2 ) 1
fi y = Ê b2h ˆ x + Ê - b2 ˆ …(i)
h k ÁË a2k ¯˜ ÁË k ¯˜
fi cosj = (a2 + b2 ) a3 , cot j = (a2 + b2 ) b3

The foci of the given hyperbola are (ae, 0) and (–ae, 0)

fi secj = a3 , tan j = b3 The equation of the circle is
+ b2 + b2)k
(a2 )h (a2 (x – ae)(x + ae) + y2 = 0

We know that, fi x2 + y2 = (ae)2 …(ii)
sec2 j – tan2 j = 1
If the line (i) be a tangent to the circle (ii), then

c2 = a2(1 + m2)

Ê a3 ˆ 2 Ê b3 ˆ 2 b4 Ê b4k2 ˆ
ÁË + b2 )h ¯˜ ÁË + b2)k ˜¯ k2 ËÁ1 a4h2 ˜¯
fi (a2 - (a2 = 1 fi = (ae)2 +

Ê a3 ˆ 2 Ê b3 ˆ 2 b4 Ê 4h2 + b4k 2 ˆ
ÁË h ¯˜ ËÁ k ˜¯ k2 ËÁ a 2k 2 ¯˜
fi - = (a2 + b2 )2 fi = e2 a

Hence, the locus of (h, k) is fi a2b4 = (a4h2 + b4k2)
(a6y2 – b6x2) = (a2 + b2)2(x2y2) e2

64. Let (h, k) be the pole. fi (a4h2 + b4k 2 ) = a4b4
The equation of the polar from the point (h, k) w.r.t. the (a2 + b2)
parabola y2 = 4ax is

yk = 2a(x + h) = 2ax + 2ah

fi y = Ê 2aˆ x + Ê 2 ahˆ …(i) fi Ê h2 + k2 ˆ = (a2 1 b2 )
ÁË k ¯˜ ËÁ k ˜¯ ËÁ a4 b4 ˜¯ +

Y Hence, the locus of (h, k) is Ê x2 + y2 ˆ = 1
ËÁ a4 b4 ˜¯ + b2)
Q (a2

66. The equation of the chord is

P 7x + y = 2y = –7x + 2 …(i)
X¢
X Hence, the equation of the diameter is

R y = b2x = 7x = - x
a2m 3 ¥ (-7) 3

Y¢ x + 3y = 0
67. The given line is 3x + 4y + 10 = 0
If the line (i) be a tangent to the hyperbola x2 – y2 = a2,
y = Ê - 3ˆ x + Ê - 5ˆ …(i)
then ÁË 4˜¯ ÁË 2 ˜¯

c2a2m2 – a2

Ê 2ah ˆ 2 Ê 2a ˆ 2 Hence, the equation of the diameter corresponds to the
ÁË k ¯˜ ËÁ k ¯˜ line (i) is
fi = a2 - a2

fi 4h2 = 4a2 – k2 y = b2x = 4x = - 16 x
a2m 27
fi h2 + k2 = a2 9 Ê - 3ˆ
4 ËÁ 4¯˜

fi 16x + 27y = 0

Hyperbola 6.39

68. The slope of the given chord is m= Ê - 2 ˆ . 73. The combined equation of the asymptotes is
ÁË 3 ¯˜ (2x – y – 3)(3x + y – 7) = 0

Hence, the equation of the diameter parallel to the fi 6x2 – xy – y2 – 23x + 4y + 21 = 0
Let the equation of the hyperbola be
given chord is
6x2 – xy – y2 – 23x + 4y + l = 0,
y = - b2x = - 4x = -2 x where l is any constant such that it represents two
a2m 3 straight lines which passes through (1, 1), so l = 15.
9 Ê - 2ˆ Hence, the equation of the hyperbola becomes
ÁË 3¯˜
6x2 – xy – y2 –23x + 4y + 15 = 0.
fi 2x + 3y = 0 74. The combined equation of the asymptotes parallel to

69. Let the equation of the diameter, which is conjugate to the lines 2x + 3y = 0 and 3x + 2y = 0 is
(2x + 3y + l)(3x + 2y + m) = 0
x = 2y is
which is passing through (1, 2).
y = m1 x. y = Ê 1ˆ x and y = m1x Therefore,
As we know that two diameters ËÁ 2˜¯
(2 ◊ 1 + 3 ◊ 2 + l)(3 ◊ 1 + 2 ◊ 2 + m) = 0
are conjugates, if fi (8 + l)(7 + m) = 0
fi l = –8, m = –7
m1 m2 = b2 Thus, the combined equation of the assymptotes is
a2
(2x + 3y – 8)(3x + 2y – 7) = 0
fi m1 ¥ 1 = 16 Let the equation of the hyperbola be
2 9
2x + 3y = 0 and 2x – 3y = 0
fi m1 = 32 which is passing through (5, 3). So
9
(2 ◊ 5 + 3 ◊ 3 – 8)(3 ◊ 5 + 2 ◊ 3 – 7) + l = 0
Hence, the equation of the conjugate diameters is fi 11 ¥ 14 + l = 0
fi l = –154
y = 32 x Hence, the equation of the hyperbola is
9
(2x + 3y – 8)(3x + 2y – 7) – 154 = 0.
fi 32x = 9y 75. The equation of the given hyperbola is

70. Equations of the asymptotes to the hyperbola x2 – 2y2 = 2.
So, the equations of its asymptotes are
xy – 2x – 3y = 0 is
x - 2 y = 0 and x + 2 y = 0.
xy – 2x – 3y + l = 0,
Let any point on the hyperbola be P ( 2 sec j, tanj) .
where l is any constant such that it represents two Let PM and PN are two perpendiculars from the point
P to the asymptotes.
straight lines. Then, PM ◊ PN

Therefore,

abc + 2 fgh – af 2 – bg 2 – ch 2 = 0

Ê -3ˆ Ê 1ˆ Ê 1ˆ 2
ÁË 2 ˜¯ ÁË 2¯˜ ÁË 2¯˜
fi 0 + 2 ¥ ¥ (-1) ¥ - 0 - 0 - l = 0

fi l=6 = 2 secj - 2 tanj ¥ 2 secj + 2 tanj
Hence, the required asymptotes are 1+ 2 1+ 2

xy – 2x – 3y + 6 = 0 = 2 ¥ (secj - tanj) ¥ (secj + tanj)
fi (x – 2)(y – 3) = 0 3
fi x = 2 and y = 3
71. The equations of the asymptotes of the given curve is = 2 ¥ (sec2 j - tan2 j)
3
3x2 + 10xy + 8y2 + 14x + 22y + l = 0,
where l is any constant such that it represents two = 2.
straight lines. 3

Therefore, 76. The equation of the given hyperbola is x2 - y2 = 1 .
abc + 2fgh – af2 – bg2 – ch 2 = 0 94

fi 3 ◊ 8 ◊ l – +2 ◊ 7 ◊ 11 ◊ 5 – 3 ◊ 121 – 8 ◊ 49 – l ◊ 25 = 0 Thus, the equations of the asymptotes are
fi 24l – + 770 – 363 – 392 – 25l = 0
fi l = 15 Ê x + yˆ Ê x - yˆ = 0
Hence, the combined equation of the given asymptotes ÁË 3 2 ¯˜ ÁË 3 2 ¯˜

is fi Ê x + yˆ = 0 and Ê x - yˆ = 0
3x2 + 10xy + 8y2 + 14x + 22y + 15 = 0 ÁË 3 2 ¯˜ ÁË 3 2 ¯˜

72. Since the asymptotes are perpendicular to each other,

so the hyperbola is rectangular. Hence, its eccentricity

is 2 . fi 2x + 3y = 0 and 2x – 3y = 0

6.40 Coordinate Geometry Booster

The equation of any tangent to the hyperbola The equation of the tangent to the hyperbola (i) at P is

x2 - y2 = 1 is x secj - y tanj = 1 …(ii)
94 ab

The equations of the asymptotes of the hyperbola (i)

x sec j - y tan j = 1 are
32
bx – ay = 0 and bx + ay = 0

Let the points of intersection of 2x + 3y = 0, 2x – 3y = 0 Let the points of the intersection of the asymptotes and
and x secj - y tanj = 1 are O, P and Q respectively.
the tangent are O, Q, R respectively.
32
Therefore, O = (0, 0), Then, O = (0, 0),

Q = [a(sec j + tan j), b(sec j + tan j)] and

R = [a(sec j – tan j), b(sec j – tan j)]

P = Ê 3 , - 2 ˆ Clearly, mid-point of QR is (a sec j, b tan j), which is
ËÁ + + ¯˜
secj tanj secj tanj co-ordinates of P.

Ê 3 2 ˆ Thus, the area of DOQR
ÁË - - ¯˜
and Q = secj tanj , secj tanj 00

Hence, the area of DOPQ = 1 a(secj + tanj) b(secj + tanj)
2 a(secj - tanj) -b(secj - tanj)

0 0 00
3 -2
= 1 secj + tan j = 1 -ab - ab
23 secj + tan j 2
secj - tan j 2
0 = ab.
secj - tan j
0 79. The equations of the asymptotes of the given hyperbola

x2 - y2 =1 are
a2 b2
= 1 (6 + 6) = 6 s. u.
2 bx – ay = 0 and bx + ay = 0

Let P be any point on the given hyperbola be

77. Let the point P be (a sec j, b tan j). (a sec j, b tan j).

The equations of the asymptotes of the given hyperbola It is given that, p1 and p2 be the lengths of perpendicu-
lars from the point P to the asymptotes
x2 - y2 =1 are
a2 b2 bx – ay = 0 and bx + ay = 0

bx – ay = 0 and bx + ay = 0 p1 = absecj - ab tanj
b2 + a2
Let PM and PN be two perpendiculars from the point P Thus,

to the transverse axis and the asymptote

bx – ay = 0 and p2 = absecj + ab tanj
Thus, PM = b tan j b2 + a2

and PN = absecj - ab tanj Therefore, 1
b2 + a2 p1 p2

It is given that, PM = PN Ê a2 + b2 ˆ
ÁË tan j )(ab sec j j ) ˜¯
= (ab secj - ab + ab tan

fi b tanj = absecj - ab tanj = Ê a2 + b2 ˆ = 1 + 1
b2 + a2 ËÁ a2b2 ¯˜ a2 b2

fi tanj = a secj - tanj Hence, the result. Ê c ˆ
b2 + a2 ÁË t1 ˜¯
80. The equation of the normal at ct1, is

78. Let the equation of the hyperbola be t13x - t1y - ct14 + c = 0

x2 y2 …(i)
a2 b2
- =1 …(i) Also it meets the hyperbola again at

and any point on the hyperbola (i) be Ê c ˆ
ÁË ct2, t2 ˜¯
P(a sec j, b tan j)

Hyperbola 6.41

Therefore, Eliminating t, we get
(h2 – k2)2 + 4c2hk = 0
ct2t13 - c t1 - ct14 + c = 0
t2 Hence, the locus of (h, k) is
(x2 – y2)2 + 4c2xy = 0
fi t2t13 - t1 - t14 +1= 0
t2 83. The equations of the asymptotes of the rectangular
hyperbola xy = c2 are x = 0 and y = 0.
fi t22t13 - t1 - t14t2 + t2 = 0 Clearly, the angle between the asymptotes

fi t2t13(t2 - t1) + (t2 - t1) = 0 =p
2
fi (t2t13 + 1)(t2 - t1) = 0
fi 2a = p
fi (t2t13 + 1) = 0 ( t1 π t2) 2

fi t2t13 = -1 fi a=p
4
Hence, the result.
Thus, the eccentricity,
81. Let P, Q and R are the vertices of a triangle such that

P = Ê ct1, c ˆ , Q = Ê ct2 , c ˆ , R = Ê ct3 , c ˆ Ê pˆ
ÁË t1 ˜¯ ËÁ t2 ˜¯ ËÁ t3 ¯˜ ÁË 4 ¯˜
e= 2 = sec = seca
c-c
t3 t2 =- 1
Now, slope of QR = ct3 - ct2 t2t3 84. Let the equation of the circle be …(i)
x2 + y2 + 2gx + 2fy + c + 0 …(ii)
Therefore, slope of PM is t2t3.
The equation of the perpendicular PM on QR is and the equation of the given hyperbola be
xy = 1
y - c = t2t3 ( x - ct1) …(i)
t1 Solving, we get

Similarly, the equation of the perpendicular BN on PR is x2 + 1 + 2 gx + 2f +c=0
x2 x

y - c = t1t3 ( x - ct2 ) …(ii) fi x4 + 2gx3 + cx2 + 2fx + 1 = 0
t2
Let its roots are x1, x2, x3 and x4.
Solving Eq. (i) and (ii), we get, Then, x1x2x3x4 = 1
Similarly, we can easily prove that
x = - c and y = -ct1t2t3.
t1t2t3 y1y2y3y4 = 1
85. Any point on the rectangular hyperbola

Thus, the point Ê - c , - ct1t2t3 ˆ lies on the rectan- xy = c2 is P Ê ct, cˆ
ËÁ t1t2t3 ¯˜ ËÁ t ¯˜

gular hyperbola xy = c2. The equation of any tangent to the rectangular hyperbola

Hence, the result. xy = c2 at t is

82. The equation of the normal to the rectangular hyperbola x + yt = 2c …(i)
t
xy = c2 at t is

xt2 - y = ct3 - c …(i) The equation of any normal to the rectangular hyperbola
t
xy = c2 at t is

Let the pole be (h, k). xt3 – yt – xt4 + c = 0 …(ii)

Then the equation of the polar from the point (h, k) to Therefore, a1 = 2ct, a2 = 2c
t
the rectangular hyperbola

xy = c2 is Ê 1 ˆ Ê 1 ˆ
ËÁ t3 ˜¯ ÁË t ¯˜
xk + yh = 2c2 (ii) and b1 = c t - , b2 = c - t 3

Therefore, the Eqs (i) and (ii) are identical. So

k h= 2c2 Now,
t2 -1 ct3 - c
= Ê 1ˆ 2c2 Ê1 ˆ
ËÁ t3 ˜¯ t ÁË t ¯˜
a1a2 + b1b2 = 2c 2t t - + - t 3

t

fi t2 = - k and h = - 2ct = 2c2 Ê t 2 - 1 + 1 - t 2 ˆ = 0
h t4 -1 ÁË t2 t2 ˜¯

6.42 Coordinate Geometry Booster

86. We have, e1 = 2 and e2 = 2 Q = Ê 2ct1t3 , 2c ˆ
ËÁ t1 + t3 t1 + t3 ¯˜
Now, (e1 + e2 )2 = ( 2 + 2)2 = (2 2)2 = 8
87. Any point on the given hyperbola and R = Ê 2ct2t3 , 2c ˆ respectively
ÁË t2 + t3 t2 + t3 ˜¯
x2 - y2 = 1 be P ( 2 secj, tanj) .
2 Hence, the area of the DPQR
The equations of the asymptotes of the given hyperbola
are 2ct1t2 2c 1
t1 + t2 t1 + t2
x - 2 y = 0 and x + 2 y = 0. = 1 2ct1t3 2c 1
2 t1 + t3 t1 + t3
Let PM and PN be the lengths of perpendiculars from 2ct2t3 2c 1
the point P on the asymptotes. t2 + t3 t2 + t3
Thus, = 2c2 (t1 - t2 )(t2 - t3)(t3 - t1)
(t1 + t2 )(t1 + t2 )(t1 + t2 )
PM.PN

= 2secj - 2 tanj ¥ 2secj + 2 tanj
1+ 2 1+ 2

=2 89. The given rectangular hyperbola is
3
xy = 18 …(i)
= (1± 2 2, 2)
Replacing x by x cos (45°) + y sin (45°) and y by –x sin

88. Let the points A, B, C be (45°) + y cos (45°) in (i), we get

Ê cˆ , Ê c ˆ , Ê cˆ respectively. Ê x+ y ˆ Ê - x+ yˆ = 18
ÁË ct1, t1 ˜¯ ÁË ct2, t2 ¯˜ ËÁ ct3, t3 ˜¯ ËÁ 2 2 ˜¯ ÁË 2 2 ¯˜

(i) Then the area of the DABC fi x2 – y2 = –18
Hence, the length of the transverse axis
ct1 c 1
t1 = 2a = 2.6 = 12 units.
90. Let (h, k) be any point.
= 1 ct2 c 1
2 t2 The equation of the chord of contact of the tangents
from (h, k) to the circle x2 + y2 = 4 is
ct3 c 1
t3 hx + ky = 4.
Also, the given hyperbola is

xy = 1

1 c2 t12 1 t1 fi x Ê 4 - hxˆ =1
2 t1t2t3 t22 1 t2 ËÁ k ¯˜
= t32 1 t3

= c2 ¥ (t1 - t2 )(t2 - t3)(t3 - t1) fi 4x – hx2 = k
2 t1t2t3 fi hx2 – 4x + k = 0
Thus, its roots are equal. So
(ii) The equations of the tangents at A, B and C are
D=0
x + yt1 = 2c …(i) fi 16 – 4hk = 0
t1 fi hk = 4
Hence, the locus of (h, k) is xy = 4.
x + yt2 = 2c …(ii) 91. The combined equation of the asymptotes of the given
t2 hyperbola is

and x + yt3 = 2c …(iii) xy – hx – ky + l = 0
t3 where l is any constant such that it represents two
straight lines.
Thus, the points of intersections of (i) and (ii), (i) Therefore,

abc + 2fgh – af 2 – bg 2 – ch2 = 0.

and (iii), (ii) and (iii) meet at P, Q, R respectively. fi 0+ 2ËÊÁ - kˆ Ê - hˆ Ê 1ˆ - l = 0
2 ˜¯ ÁË 2 ¯˜ ËÁ 2˜¯ 4
Ê 2ct1t2 2c ˆ
Thus, P = ÁË t1 + t2 , t1 + t2 ¯˜ , fi l = hk

Hyperbola 6.43

Hence, the asymptotes are If (i) and (ii) are the same, then
a2m2 – b2 = –b2m2 + a2
xy – hx – ky + hk = 0
fi (x – h)(y – k) = 0 fi (a2 + b2) m2 = (a2 + b2)
fi (x – h) = 0 and (y – k) = 0 fi m2 = 1
fi m = ±1
92. We have, q = 2 tan -1 Ê bˆ Hence, the equation of the common tangent be
ËÁ a ˜¯ y = ± x ± a2 - b2 .

fi tan Ê q ˆ = b 3. The equation of any tangent to the hyperbola
ËÁ 2 ¯˜ a
x2 - y2 = 1 is
Also, 16 9
y = mx + 16m2 - 9
e= 1+ b2 = 1 + tan 2 Ê q ˆ = sec Ê q ˆ
a2 ÁË 2 ¯˜ ÁË 2 ¯˜

fi cos Ê q ˆ = 1
ËÁ 2 ˜¯ e

93. The equation of the given hyperbola is fi mx - y + 16m2 - 9 = 0
9x2 – 16y2 = 144 which is also a tangent of x2 + y2 = 9. So

fi x2 - y2 = 1 …(i) 16m2 - 9 = 3
16 9 m2 +1

Let any point on the given hyperbola be P(8, k).

Since the point P lies on (i), so fi 16m2 - 9 = 3 m2 + 1
fi 16m2 – 9 = 9m2 + 9
64 - k 2 = 1 fi 16m2 – 9m2 = 9 + 9
16 9 fi 7m2 = 18

fi k2 = 27

fi k=3 3 fi m=±3 2
7
Hence, the co-ordinates of P be (8, 3 3) .
Hence, the equation of the common tangent be
Thus, the equation of the reflected ray is

y - 3 3 = 0 - 3 3 (x - 8) y = ± Ê 3 2ˆ x+ 288 - 9
-5 - 8 ÁË 7 ¯˜ 7

fi 3 3 x -13y +15 3 = 0 fi 7 y = ± (3 2)x +15

LEVEL III 4. The equation of any normal to the hyperbola

x2 - y2 =1 at (f) is
a2 b2
1. As we know that y = mx + c will be the tangent to the
ax cos j + by cot j = a2 + b2
x2 y2
hyperbola a2 - b2 =1 if Y

c2 = a2m2 – b2 = 9m2 – 4

Hence, the equation of the hyperbola is

x2 - y2 = 1 Q
94 X¢ A¢ O A

2. The equation of any tangent to the hyperbola GX

x2 - y2 =1 is y = mx + a2m2 - b2 …(i)
a2 b2

and the equation of any tangent to the hyperbola Y¢

x2 - y2 = -1 , i.e. ( x2 ) - y2 ) =1 is Thus, the co-ordinates of G = Ê a2 + b2 ˆ
b2 a2 -b2 (-a2 ÁË a cosj , 0˜¯

y = mx + (-b2 )m2 + a2 …(ii) Clearly, the vertices, A = (a, 0) and A¢ = (–a, 0)

6.44 Coordinate Geometry Booster

Ê a2 + b2 ˆ (v) x1x2x3x4 = c4(t1t2t3t4) = –c4
ÁË a cosj - a˜¯
Now, AG = Ê 1 ˆ Ê 1ˆ
ÁË t1t2t3t4 ˜¯ ÁË -1¯˜
(vi) y1y2y3y4 = c4 = c4 = -c4

and A¢G = Ê a2 + b2 + ˆ 6. The equation of any normal to the given hyperbola at
ÁË a cosj a˜¯
(x, y) is
Therefore,
x y
Ê a2 + b2 ˆÊ a2 + b2 ˆ a2 (k - y) = b2 (x - b)
ËÁ a cosj - a˜¯ ÁË a cosj + a˜¯
AG ◊ A¢G =

Ê 2 ˆ fi Ê 1 + 1 ˆ xy - y - kx = 0
Á ˜ ËÁ a2 b2 ˜¯ b a2
ÁË Ê a2 + b2 ˆ ¯˜
= ËÁ a ˜¯ sec2j - a2
b2kx
fi y = a2 (e2x - b) …(i)
…(ii)
= (a2c2 sec2 j – a2)

= a2 (e2 sec2 j – 1) The equation of the hyperbola is

fi m = 2, n = 2, p = 2 x2 - y2 =1
a2 b2
Hence, (m+ n + p)2 + 36 = 36 + 36 = 72.

5. The equation of the normal to the hyperbola xy = c2 at Solving Eqs (i) and (ii), we get

Ê ct, cˆ is x2 b4k 2 x2
ÁË t ¯˜ a2 b2a4 (e2x - b)2
- =1
xt3 – yt – ct4 + c = 0

fi ct4 – xt3 + yt – c = 0 fi a2e4x4 – 2ba2e2x3 – (a2b2 + b2k2 + a4e4) x2
+ 2ba4e2x + a4b2 = 0
which is passing through (a, b), so ct4 – at3 + bt – c = 0 …(iii)
Let x1, x2, x3, x4 are the roots of Eq. (iii).
Let its four roots are t1t2, t3, t3.
Therefore,

t1 + t2 + t3 + t4 = a , S(t1t2) = 0, Then, x1 + x2 + x3 + x4 = 2b ,
c e2

S(t1t2t3) = -b and S(t1t2t3t4) = 0. Â (x1x2 ) = - a2b2 + b2k 2 + a4e2
c a2e4

(i) x1 + x2 + x3 + x4 = c(t1 + t2 + t3 + t4 ) 2ba2
e2
= c Ê a ˆ = a  (x1x2 x3 ) = -
ÁË c ˜¯
a2b2
Ê 1 1 1 1 ˆ Âand (x1x2 x3x4 ) = e4
ÁË t1 t2 t3 t4 ˜¯
(ii) y1 + y2 + y3 + y4 = c + + +

Ê Ê b ˆ ˆ Â ÂTherefore, Ê 1ˆ = (x1x2 x3) = 2e2
Á ËÁ c ¯˜ ˜ ÁË x1 ˜¯ x1x2 x3x4 b
Ê (t1t2t3) ˆ cÁ - ˜
(t1t2t3t4 )¯˜ ÁÁË ˜¯˜
ÂÂ= cÁË = -1 = b Ê 1 1 1 1 ˆ
x4 ) ËÁ x1 x2 x3 x4 ¯˜
(i) ( x1 + x2 + x3 + + + +

(iii) x12 + x22 + x32 + x42 = c2 (t12 + t22 + t32 + t42 ) ( ) Â= Ê Ê 1 ˆˆ = 2b ¥ 2e2 = 4
x1 ÁË ËÁ x1 ˜¯ ¯˜ e2 b
{ }( )Â Â= c2 t1 2 - 2 (t1t2)
(ii) Similarly, we can have

ÏÔÓÔÌËÊÁ a ˆ 2 0Ô¸˝ Ê 1 1 1 1ˆ
c ¯˜ Ô˛ y4 ) ÁË y1 y2 y3 y4 ¯˜
= c2 - = a 2 ( y1 + y2 + y3 + + + +

y1 2 - 2 ( y1y2 )
( ) Â(iv) y12 + y22 + y32 + y42 = ( )= Ê Â Ê 1 ˆˆ
 y1 ÁË ËÁ y1 ˜¯ ¯˜
Ê 1ˆ
Â= (b)2 - 2c2 ËÁ t1t2 ˜¯ = 4.

Â= Ê t1 t2 ˆ 7. The equation of the normal to the hyperbola
(b )2 - 2c2 ÁË t1t2t3t4 ¯˜
x2 y2
a2 - b2 =1 at (a sec q, b tan q) is

= (b)2

Hyperbola 6.45

ax cos q + by cot q = a2 + b2 …(i) 9. Let the mid-point be M(h, k)
Let the point Q be (a, b). The equation of the chord of the hyperbola
The equation of any line perpendicular to (i) is
x2 - y2 =1 is
(b cot q)x – (a cos q)y + k = 0 a2 b2
which is passing through the centre. So
T = S1
k=0
Thus, (b cot q)x – (a cos q) y = 0 fi hx - ky = h2 - k2
which meets at Q. a2 b2 a2 b2
Thus,
First we make it a homogeneous equation of 2nd
(b cot q)a – (a cos q)b = 0
degree.

fi sinq = ba Ê hx ky ˆ 2
ab Á ˜
Á a2 - b2 ˜ = Ê x2 - y2 ˆ
ÁÁË h2 - k2 ¯˜˜ ÁË a2 b2 ¯˜
a2b 2 - b2a 2
fi cosq = a2 b2
ab
Ê h2x2 k 2 y2 ˆ Ê x2 y2 ˆ Ê h2 k2 ˆ
a2b 2 - b2a 2 fi ÁË a4 + b4 - 2(…) xy˜¯ = ÁË a2 - b2 ¯˜ ÁË a2 - b2 ¯˜
and cotq =
which subtends right angle at the centre, i.e.
ba co-efficient of x2 + co-efficient of y2 = 0
Putting the values of cos q and cot q, in Eq. (i) we get

Ê a2b 2- b2a 2 ˆ Ê a2b 2 - b2a 2 ˆ a2 b2 Ê h2 1 Ê h2 k2 ˆ k2 1 Ê h2 k2 ˆ ˆ
ax ÁÁË ab ˜¯˜ ÁËÁ ba ˜˜¯ ÁË a4 a2 ÁË a2 b2 ¯˜ b4 b2 ËÁ a2 b2 ˜¯ ˜¯
+ by = + - - + + - = 0

Hence, the locus of Q is fi Ê h2 + k2 ˆ = Ê1 + 1ˆ Ê h2 - k2 ˆ
ÁË a4 b4 ˜¯ ËÁ a2 b2 ¯˜ ÁË a2 b2 ¯˜
( ) ( )x a2 y2 - b2 x2 + y a2 y2 - b2 x2 = a2 + b2
Hence, the locus of M(h, k) is
yx

( )fi Ê x yˆ Ê x2 y2 ˆ Ê 1 1 ˆ Ê x2 y2 ˆ
ÁË y + x ¯˜ a2 y2 - b2x2 = a2 + b2 ÁË a4 + b4 ˜¯ = ËÁ a2 + b2 ˜¯ ÁË a2 - b2 ¯˜

fi Ê x + yˆ2 (a2 y2 - b2 x2 ) = (a2 + b2 )2 10. Let the point P be (h, k).
ËÁ y x ¯˜ The equation of the tangent to the parabola
y2 = 4ax at P is
fi (x2 + y2)2(a2y2 – b2x2) = (a2 + b2)2 yk = 2a(a + h)

8. Any point on the circle x2 + y2 = a2 is (a cos q, a sin q). fi y = 2ax + 2ah
kk
The chord of contact of this point with respect to the

hyperbola x2 – y2 = a2 is which is a tangent to the hyperbola. So

x cos q – y sin q = a …(i) c2 = a2m2 – b2

If its mid-point be (h, k), then it is same as Ê 2ahˆ 2 Ê 2a ˆ 2
ÁË k ˜¯ ËÁ k ¯˜
T = S1 fi = a2 - b2
i.e. hx – ky = h2 – k2

Comparing Eqs (i) and (ii), we get fi 4a2h2 = 4a4 - b2
k2 k2
cosq = sinq = a
h k h2 - k2 fi 4a2h2 + k2b2 = 4a4
Hence, the locus of (h, k) is
We know that
sin2 q + cos2 q = 1 4a2x2 + b2y2 = 4a2
11. Let M(h, k) be the mid-point of the hyperbola
Ê ah ˆ 2 Ê ak ˆ 2
ÁË h2 - k ¯˜ ÁË h2 - k2 ˜¯ x2 – y2 = a2
fi 2 + = 1 The equation of the chord at M is

fi a2(h2 + k2) = (h2 – k2)2 T = S1
Hence, the locus of (h, k) is fi hx – ky = h2 – k2
fi ky = hx – (h2 – k2)
a2(x2 + y2) = (x2 – y2)2

6.46 Coordinate Geometry Booster

Ê hˆ Ê h2 - k 2 ˆ Hence, the locus of the mid-point M(h, k) is
ÁË k ¯˜ ÁË k ¯˜
fi y = x - Ê x2 y2 ˆ 2 Ê x2 y2 ˆ
ËÁ a2 b2 ¯˜ ÁË a4 b4 ¯˜
- = c2 +

which is a tangent to the parabola y2 = 4ax. So

c= a 14. Let the mid-point be M(h, k).
m The equation of the chord of the circle
x2 + y2 = a2 is
fi Ê k2 - h2 ˆ = ak T = S1
ÁË k ˜¯ h fi hx + ky = h2 + k2
fi ky = –hx + (h2 + k2)
fi ak2 = k2h – h3
Ê h ˆ Ê h2 + k 2 ˆ
fi k2(h – a) = h3 ËÁ k ¯˜ ÁË k ¯˜
fi y = - x +
Hence, the locus of M(h, k) is

y2(x – a) = x3 which is a tangent to the hyperbola. So

12. Let M(h, k) be the mid-point of the chord of length 2d c2 = a2m2 – b2

inclined at an angle q with the x-axis. Then its extremi- Ê h2 + k2 ˆ 2 Ê hˆ 2
ËÁ k ˜¯ ËÁ k ˜¯
ties are fi = a2 - b2

(h + d cos q ck + d sin q)

and (h – d cos q ck – d sin q) fi (h2 + k2)2 = (a2h2 – k2b2)
Hence, the locus of M(h, k) is
These extremities lie on the hyperbola xy = c2
(x2 + y2)2 = (a2x2 – b2y2)
So, (h + d cos q) (k + d sin q) = c2 …(i) 15. Let the point P be (at2, 2at).

and (h – d cos q)(k – d sin q) = c2 …(ii)

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

hk + d2 sin q cos q = c2 …(iii) Y

and h sin q + k cos q = 0 P

i.e. tanq = - k …(iv) Q
h M(h, k)
X¢ R O
Eliminating q between Eqs (iii) and (iv), we get X

(hk - c 2 ) Ê h2 + k2 ˆ = d 2k
ÁË h2 ˜¯ h

fi (hk – c2)(h2 + k2) = d2hk Y¢

Hence, the locus of M(h, k) is The equation of the tangent to the parabola at P is

(xy – c2)(x2 + y2) = d2xy yt = x + at2 …(i)

13. Let the mid-point be M(h, k). x2 y2 Given hyperbola is
a2 b2
The equation of the chord of the hyperbola - =1 xy = c2 …(ii)
is
The equation of the chord of the hyperbola xy = c2 at

T = S1 M(h, k) is

h2 k2 xk + yh = 2hk …(ii)
a2 b2
fi hx - ky = - So, Eqs (i) and (ii) are the same line. So
a2 b2
1 = - t = - at2
fi ky = hx Ê h2 - k2 ˆ k h 2 hk
b2 a2 - ÁË a2 b2 ¯˜
fi t = - h and t2 = - 2h
Ê b2h ˆ b2 Ê h2 k2 ˆ ka
ËÁ a2k ¯˜ k ËÁ a2 b2 ¯˜
fi y = x - - fi h2 = - 2h
k2 a

which is a tangent to the circle x2 + y2 = c2. So fi h = -2
c2 = a2(1 + m2) k2 a

fi Ê b2 Ê h2 - k2 ˆˆ 2 = c 2 Ê + b4h2 ˆ fi k 2 = - ah
ËÁ k ËÁ a2 b2 ¯˜ ¯˜ ÁË1 a4k 2 ¯˜ 2

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

fi b4 Ê h2 - k2 ˆ 2 = c2 Ê b4h2 ˆ y2 = - ax
k2 ÁË a2 b2 ¯˜ ËÁ1 + a4k 2 ¯˜ 2

which is a parabola.

Hyperbola 6.47

16. The equation of the chord of contact of the circle Let the co-ordinates of the point P be (a, b).
Since PM and PN are perpendicular to the axes, so the
x2 + y2 = a2 is hx + ky = a2 …(i) co-ordinates of P are

and the equation of the normal chord of the hyperbola Ê Ê a2 + b2 ˆ Ê a2 + b2 ˆ ˆ
is ÁË ÁË a ˜¯ ËÁ b ¯˜ j ˜¯

ax cos j + by cot j = a2 + b2 …(ii) secj, tan

Solving, we get

a cosj = b cotj = a2 + b2 Therefore,
h k a2
a = Ê a2 + b2 ˆ secj and b = Ê a2 + b2 ˆ tan j
ËÁ a ˜¯ ËÁ b ˜¯
fi cosj = (a2 + b2 )h , cotj = (a2 + b2 )k
a3 a2b
fi a Ê a ˆ = secj and b Ê b ˆ = tan j
(a2 + b2 )h bh ÁË a2 + b2 ˜¯ ÁË a2 + b2 ¯˜
fi cos j = a3 , sin j = ak

As we know that, sec2 j – tan2 j = 1

We know that sin2 j + cos2 j = 1 Ê a ˆ 2 Ê b ˆ 2
ÁË + ˜¯ ËÁ + ˜¯
2 Ê (a2 + b2)hˆ 2 a 2 - b2 = 1
ÁË a3 ˜¯
fi Ê bh ˆ + = 1 a2 b2 a2 b2
ËÁ ak ¯˜
fi a2a2 – b2b2 = (a2 + b2)2

h2 Ê b2 Ê a2 + b2 ˆ2ˆ Hence, the locus of (a, b) is
a2 Á k2 ËÁ a2
fi ËÁ + ¯˜ ˜ =1 a2x2 – b2y2 = (a2 + b2)2
¯˜
18. Let the equation of the hyperbola be

fi Ê b2 + Ê a2 + b2 ˆ2ˆ = a2 x2 - y2 =1 and the point of intersection be P(h, k).
Á k2 ËÁ a2 ˜ h2 a2 b2
ËÁ ¯˜ ¯˜
The equation of any tangent to the hyperbola is

Ê a2 b2 ˆ Ê a2 + b2 ˆ2 y = mx + a2m2 - b2
ÁË h2 k2 ˜¯ ÁË a2 ˜¯
fi - = which is passing through P(h, k).

Hence, the locus of M(h, k) is k = mh + a2m2 - b2

Ê a2 b2 ˆ Ê a2 + b2 ˆ 2 fi (k – mh)2 = (a2m2 – b2)
ÁË x2 y2 ˜¯ ËÁ a2 ¯˜
- = fi (h2 – a2)m2 – 2(kh)m + (k2 + b2) = 0

It has two roots, say m1 and m2

17. The equation of any normal to the hyperbola Thus, m1 + m2 = 2hk
(h2 - a2 )
x2 - y2 =1 is
a2 b2 k2 b2
and m1m2 = h2 + a2
ax cos j + by cot j = (a2 + b2) …(i) -

Y Clearly, b = q1 – q2
NP

Q fi tan b = tan (q1 – q2)

X¢ MX = tanq1 - tan q2
O 1+ tanq1tanq2

= m1 - m2
1+ m1m2

= (m1 + m2 )2 - 4 m1m2
1+ m1m2

Y¢ 4h2k 2 Ê k2 + b2 ˆ
(h2 - a2 )2 - 4 ËÁ h2 - a2 ¯˜
Since the normal (i) meets the x-axis at M and y-axis at

N respectively. Then, =

M = Ê a2 + b2 , ˆ and N = Ê Ê a2 + b2 ˆ ˆ 1 + Ê k2 + b2 ˆ
ËÁ a cosj 0¯˜ ÁË 0, ÁË b ¯˜ tan j ˜¯ ÁË h2 - a2 ˜¯

6.48 Coordinate Geometry Booster

Ê h2 + k2 + b2 - a2 ˆ 2 Ê k , b k 2 - a2 ˆ
ÁË h2 - a2 ¯˜ ËÁ a ˜¯
fi tan 2 b Let P be

4h2k 2 - 4(k 2 + b2 )(h2 - a2 ) and Q be Ê k, b k 2 + a2 ˆ
(h2 - a2 )2 ÁË a ¯˜
=

The equation of the tangent at P to the hyperbola is

fi (h2 + k2 + b2 – a2) tan2 b xk - y k2 - a2 = 1 …(i)
a2 ab
= 4(h2k2 – (k2 + b2)(h2 – a2))

fi (h2 + k2 + b2 – a2)2 tan2 b = 4(a2k2 + h2b2 + a2b2) and the equation of the tangent at Q to the conjugate

fi (h2 + k2 + b2 – a2)2 = 4 cot b(a2k2 – h2b2 + a2b2) hyperbola is

Hence, the locus of P(h, k) is xk - y k 2 + a2 = -1 …(ii)
(x2 + y2 + b2 – a2)2 = 4 cot2 b(a2y2 – b2x2 + a2b2) a2 ab

x2 y2 Squaring and adding Eqs (i) and (ii), we get
a2 b2
19. Let the hyperbola be - =1 …(i) k2 = 1

and its conjugate be x2 - y2 = -1 …(ii) x2 + y2
a2 b2 a4 a2b2

The equation of any tangent, say AB at (p, q) is fi k= 1

px - qy = -1 …(iii) x2 + y2
a2 b2 a4 a2b2

p2 q2 Solving, we get
a2 b2
where - +1= 0 1 Ê ay ˆ 2
2k ÁË b ¯˜
x =

i.e. b2p2 – a2q2 + a2b2 = 0

Eliminating y between Eqs (i) and (iii), we get 1 Ê ay ˆ 4
4k2 ËÁ b ¯˜
xp ˆ 2 Ê b2 ˆ 2 fi x2 =
a2 ˜¯ ËÁ q ¯˜
x2 - 1 ÊÁË1 + =1 1Ê ay ˆ 4 Ê x2 y2 ˆ
a2 b2 4 ËÁ b ¯˜ ËÁ a4 a2b2 ¯˜
fi x2 = ¥ +
p2b2 2pb2 b2
fi Ê 1 - a4q2 ˆ x2 - Ê a2q2 ˆ x - Ê p2 ˆ = 0 4x2 y4 Ê x2 y2 ˆ
ËÁ a2 ˜¯ ÁË ˜¯ ÁË + 1¯˜ a2 b4 ¥ ËÁ a2 b2 ˜¯
fi = +

fi Ê a2b2 ˆ x2 - Ê 2pb2 ˆ x - Ê b2 ˆ = 0 21. Let the mid-point be M(h, k).
ÁË a4q2 ¯˜ ÁË a2q2 ˜¯ ËÁ p2 + 1˜¯ The equation of the chord is
T = S1
Let its roots are x1 and x2. Then fi hx + ky = h2 + k2
fi ky = –hx + (h2 + k2)
x1 + x2 = 2p ◊ b2 ∏ a2b2 = 2p
q2 a2 a4q2

fi x1 + x2 = p Ê h ˆ Ê h2 + k 2 ˆ
2 ËÁ k ¯˜ ÁË k ¯˜
fi y = - x + …(i)
…(ii)
Similarly y1 + y2 = q which is a tangent to the hyperbola
2
9x2 – 16y2 = 144

Hence, the point of contact is the mid-point of the fi x2 - y2 = 1
16 9
chord AB.

20. Let the equation of the hyperbola be So, c2 = a2m2 – b2

x2 - y2 =1 Ê h2 + k2 ˆ 2 Ê hˆ 2
a2 b2 ËÁ k ˜¯ ÁË k ˜¯
fi = 16 - - 9

and its conjugate be x2 - y2 = -1 . fi (h2 + k2)2 = 16h2 – 9k2
a2 b2 Hence, the locus of M(h, k) is

The equation of any line parallel to the conjugate axis be (x2 + y2)2 = 16x2 – 9y2

x = k.

Hyperbola 6.49

22. Given hyperbola is Also, OC = OR

2x2 + 5xy + 2y2 + 4x + 5y = 0 …(i) Ê acosj ˆ 2 Ê bcosj ˆ 2
ËÁ 1+ sinj ¯˜ ËÁ 1+ sinj ¯˜
The equation of the asymptote of the above hyperbola fi h2 + k2 = h - + k +

is Ê cos j ˆ
ËÁ 1+ sin j ˜¯
2x2 + 5xy + 2y2 + 4x + 5y + k = 0 …(ii) fi 2(ah - bk ) = (a2 + b2 ) …(v)

If (ii) is an asymptote of (i), then

abc + 2fgh – af2 – bg2 – ch2 = 0 Multiplying Eqs (iv) and (v), we get

fi 2◊2◊k + 2◊ 5 ◊ 4 ◊ 5 - 2◊ 25 - 2◊4 - k ◊ 25 = 0 4(a2h2 – b2k2) = (a2 + b2)2
222 4 4 Hence, the locus of (h, k) is

fi 16 ◊ k + 100 – 50 – 32 – 25k = 0 4(a2x2 – b2y2) = (a2 + b2)2

fi 9k = 18 25. The area of the DPQR

fi k=2 x1 c2 1
x1
Putting k = 2 in Eq. (ii), we get

2x2 + 5xy + 2y2 + 4x + 5y + 2 = 0 1 c2
2 x2
fi (2x + y + 2)(x + 2y + 1) = 0 = x2 1

fi (2x + y + 2) = 0, (x + 2y + 1) = 0

Hence, the equation of the hyperbola is x3 c2 1
x3
(2x + y + 2)(x + 2y + 1) = c

23. Let the equation of the hyperbola is x12
x22
(x + 2y + 3)(3x + 4y + 5) = c c2 x32 1 x1
2 1 x2
which is passing through (1, –1). So = ¥ 1 1 x3
x1x2 x3
(1 – 2 + 3)(3 – 4 + 5) = c

fi c = 2.4 = 8

Hence, the equation of the hyperbola is = c2 ¥ (x1 - x2 )(x2 - x3)(x3 - x1)
2 x1x2 x3
(x + 2y + 3)(3x + 4y + 5) = 8

fi 3x2 + 10xy + 8y2 + 14x + 22y + 7 = 0

24. Let the point P be (a sec j, b tan j). LEVEL IV

The equation of the tangent to the hyperbola at P is

x secj - y tanj = 1 …(i) 1. The equation of any tangent to the parabola
ab x2 = 4ay is x = my + a
m
and the equation of the asymptotes to the hyperbola
fi mx – m2y – a = 0
x2 - y2 =1 is Let the mid-point be M(h, k). …(i)
a2 b2 The equation of the chord of the hyperbola …(ii)

x= y …(ii) xy = c2 is
ab …(iii) xk + yh – 2c2 = 0
and x = - y Since the lines (i) and (ii) are the same line, so
ab
m = - m2 = a
Solving Eqs (i) and (ii), we get k h 2c2

Q = Ê a cosj , b cosj ˆ fi m= ak and m2 = - ah
ÁË 1- sinj 1- sinj ¯˜ 2c2 2c2

and R = Ê a cosj , -b cosj ˆ fi Ê ak ˆ2 = - ah
ËÁ 1+ sinj 1+ sinj ¯˜ ËÁ 2c2 ¯˜ 2c2

Let O be the centre of the circle passing through C, Q fi a2k 2 = - 2ah
and R having its co-ordinate as (h, k). c4 c2
Thus, OC = OQ = OR
Now, OC = OQ fi k2 = Ê 2c2 ˆ h
- ËÁ a ¯˜
2 2
Ê acosj ˆ Ê bcosj ˆ
fi h2 + k2 = ÁË h - 1- sinj ¯˜ + ÁË k - 1 - sinj ˜¯ Hence, the locus of M is

(a2 b2 Ê cos j ˆ y2 = Ê 2c2 ˆ x
ÁË - sin j ¯˜ - ÁË a ˜¯
fi 2(ah + bk ) = + ) …(iv)
1
which is a parabola

6.50 Coordinate Geometry Booster

2. Given hyperbola is 25x2 – 16y2 = 400 Solving, we get
x2 - y2 = 1
16 25 Ê h2 k2 ˆ 2
ÁË a2 b2 ¯˜
The equation of the chord of the hyperbola bisected at +
(6, 2) is (k
(a2m2 - b2 ) = /b2 )2
T = S1
fi 6x - 2y = 36 - 4 Ê k ˆ 2 Ê Ê hb2 ˆ 2 ˆ Ê h2 k2 ˆ 2
ËÁ b2 ˜¯ Á a2 ÁË ka2 ˜¯ - b2˜ ÁË a2 b2 ˜¯
16 25 16 25 fi ËÁ = +
fi 150x – 32y = 900 – 64 ˜¯
fi 150x – 32y = 836
fi 75x – 16y = 418 Ê h2 k2 ˆ Ê h2 k2 ˆ 2
ÁË a2 b2 ¯˜ ËÁ a2 b2 ¯˜
3. Let the mid-point be M(h, k). fi - = +
The equation of the chord bisected at M to the given
circle is Hence, the locus of M is
hx + ky = h2 + k2
fi ky = –hx + (h2 + k2) Ê x2 - y2 ˆ = Ê x2 + y2 ˆ 2
ËÁ a2 b2 ˜¯ ËÁ a2 b2 ¯˜

5. Let the parameters of the vertices A, B and C of the

points on the hyperbola xy = c2 be t1, t2 and t3 respec-
tively.

Ê h2 2 ˆ Now the equation of the side BC is
ËÁ ¯˜
fi y = - Ê hˆ x + + k x + yt2t3 – c(t2 + t3) = 0
ÁË k ¯˜ k Any line through A perpendicular to BC is

which is a tangent to the given hyperbola. So y - c = t2t3 ( x - ct1)
t1
c2 = a2m2 – b2
c
Ê h2 + k 2 ˆ 2 Ê hˆ 2 fi y - x t2t3 = t1 - ct1t2t3 …(i)
ÁË k ˜¯ ÁË k ˜¯
fi = 16 - 9 Similarly, any line through B perpendicular to AC is

fi (h2 + k2)2 = 16h2 – 9k2 y - xt1t3 = c - ct1t2t3 …(ii)
Hence, the locus of (h, k) is t2

(x2 + y2)2 = 16x2 – 9y2 Solving Eqs (i) and (ii), we get, the orthocentre as
4. The equation of any tangent to the given hyperbola is
Ê - c ,- ˆ .
y = mx + a2m2 - b2 ÁË t1t2t3 c t1t2t3¯˜

Clearly, it satisfies the hyperbola

fi mx - y + a2m2 - b2 = 0 …(i) xy = c2
…(ii)
Let the mid-point be M(h, k). Hence, the result.

The equation of the chord of the ellipse is 6. The equation of the normal to the rectangular hyper-

T = S1 bola is

h2 k2 xt3 – yt = c(t4 – 1) …(i)
a2 b2
fi hx + ky = + Let the pole be M(h, k).
a2 b2
The equation of the polar at M is

Since the lines (i) and (ii) are the same line. So xk + yh = 2c2 …(ii)

a2m2 - b2 Since the lines (i) and (ii) are the same line, so

m = -1 = Ê h2 k2 ˆ t3 = -t = c(t4 -1)
(h/a2 ) (k /b2 ) ËÁ a2 b2 ¯˜ k h 2c2
+

fi m = - (h/a2 ) fi t3 = -t = (t4 -1)
(k /b2 ) k h 2c

Solving, we get

Ê h2 + k2 ˆ t2 = - k, t2 = (t4 -1)2
ÁË a2 b2 ¯˜ h h2 4c2

and a2m2 - b2 = - (k /b2 ) fi - k = (t4 -1)2
h3 4c2

Hyperbola 6.51

Ê k2 ˆ2 Y
ËÁ h2 - 1¯˜
fi - k = A
h3 4c2
E
C

fi –4c2kh = (k2 – h2)2 X¢ X

fi (h2 – k2)2 + 4c2hk = 0 OD

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

(x2 – y2)2 + 4c2xy = 0

7. Let the equation of the circle be Y¢

x2 + y2 + 2gx + 2fy + k = 0 …(i) Slope of AD = t2t3
Now the equation of the altitude AD is
and the equation of the rectangular hyperbola is
c
xy = c2 …(ii) y - t1 = t2t3 ( x - ct1) …(i)
…(ii)
Putting x = ct and yc/t, then Similarly, equation of the altitude BE is

c2t 2 + c2 + 2g (ct ) + 2f Ê cˆ + k = 0 y - c = t1t3 ( x - ct2 )
t2 ÁË t ˜¯ t2

fi c2t4 + c2 + 2g(ct3) + 2f(ct) + kt2 = 0 Solving, Eqs (i) and (ii), we get the co-ordinates of the

orthocentre as

fi c2t4 + 2gct3 + kt2 + 2fct + c2 = 0 Ê - c , - ct1t2t3 ˆ
ËÁ t1t2t3 ˜¯
which is a bi-quadratic equation of t. So, it has four
roots t1, t2, t3 and t4. Then which lies on xy = c2.

2g 9. Let the equation of the circle be
c
S t1 = - x2 + y2 + 2gx + 2fy + k = 0 …(i)

k and the equation of the rectangular hyperbola is
c2
S t1t2 = xy = c2 …(ii)

Putting x = ct and yc/t, then

S t1t2t3 = - 2f c2t 2 + c2 + 2g (ct ) + 2f Ê cˆ + k = 0
c t2 ÁË t ˜¯

and S t1t2t3t4 = 1 fi c2t4 + c2 + 2g(ct3) + 2f(ct) + kt2 = 0

 ÂAlso, 1 = t1t2t3 = - 2f fi c2t4 + 2gct3 + kt2 + 2fct + c2 = 0

which is a bi-quadratic equation of t. So, it has four

Ât1 t1t2t3t4 c roots, say t1, t2, t3 and t4.

Now, (–g, –f) Then  t1 = - 2g
c

= Ê c Ê - 2g ˆ , c Ê - 2f ˆ ˆ Â t1t2 = k
ÁË 2 ËÁ c ˜¯ 2 ËÁ c ˜¯ ¯˜ c2

Ê c cÊ1 1 1 1 ˆˆ  t1t2t3 = - 2f
ÁË 2 2 ÁË t1 t2 t3 t4 ¯˜ ¯˜ c
= (t1 + t2 + t3 + t4 ), + + +

Âand t1t2t3t4 = 1

Ê c Ê 1 ˆ cÊ 1 1 1 ˆˆ  ÂÂAlso,
= ÁË 2 ËÁ t1 + t2 + t3 + t1t2t3 ¯˜ , 2 ËÁ t1 + t2 + t3 + t1t2t3˜¯ ¯˜ 1 = t1t2t3 = - 2f
t1 t1t2t3t4 c

8. Let t1, t2 and t3 are the vertices of DABC described on The centre of the mean position of the four points is
the rectangular hyperbola xy = c2.
Ê c (t1 + t2 + t3 + t4 ), c Ê 1 + 1 + 1 + 1 ˆˆ
So the co-ordinates of A, B and C are ÁË 4 4 ÁË t1 t2 t3 t4 ˜¯ ˜¯

Ê ct1, c ˆ , Ê ct2 , c ˆ and Ê ct3 , c ˆ respectively = Ê c  t1, c  1 ˆ
ÁË t1 ¯˜ ÁË t2 ¯˜ ËÁ t3 ˜¯ ËÁ 4 4 t1 ¯˜

Now, slope of BC = - 1 = Ê - g ,- fˆ
t2t3 ËÁ 2 2 ˜¯

6.52 Coordinate Geometry Booster

Thus, the centres of the circle and the rectangular 1 1 = 1+1
e12 e22
hyperbola are (–g, –f) and (0, 0). 3. Clearly, + Ê b2 ˆ Ê a2 ˆ
ËÁ1 + a2 ¯˜ ËÁ1 + b2 ˜¯
and the mid-points of the centres of the circle and the

hyperbola is Ê - g , - fˆ a2 b2
ËÁ 2 2 ¯˜ a2 + b2 a2 + b2
= +
Hence, the result.

10. Let the equation of the circle be …(i) = a2 + b2 =1
x2 + y2 – 6ax – 6by + k = 0 a2 + b2

and the equation of the rectangular hyperbola is

x2 – y2 = 9a2 …(ii) Ê 1 1 ˆ
ÁË e12 e22 3˜¯
Eliminating y between Eqs (i) and (ii), we get Hence, the value of + + is 4.

(x2 + x2 – 9a2 – 6ax + k)2 = 36b2(x2 – 9a)2

fi 4x4 – 24ax3 + (…)x2 + (…) x +.. = 0 4. Clearly, the point (4, 3) lies on the hyperbola. So, the

which is a bi-quadratic equation of x. number of tangents is 1.
5. The director circle of the given circle x2 + y2 = a2 is
Let the abscissae of four points P, Q, R, S be x1, x2, x3
and x4, respectively. x2 + y2 = 2a2
So the radius of the circle is a 2 , whereas the length
Thus, x1 + x2 + x3 + x4 = 6a
of the transverse axis is a 3 .
Similarly, y1 + y2 + y3 + y4 = 6b So, the director circle and the hyperbola will never in-
Let (h, k) be the centroid of DPQR. So tersect.
So, the number of points is zero.
h = x1 + x2 + x3 , k = y1 + y2 + y3 6. Given hyperbola is
33
9x2 – 16y2 – 18x – 32y – 151 = 0
fi h = 6a - x4 , k = 6b - y4 fi 9(x2 – 2x) – (y2 + 2y) = 151
33 fi 9(x – 1)2 – 16(y + 1)2 = 144

fi x4 = 6a – 3h, y4 = 6b – 3k fi (x - 1)2 - ( y + 1)2 = 1
Since (x4, y4) lies on the curve, so 16 9

x42 - y42 = 9a2

fi (6a – 3h)2 – (6b – 3k)2 = 9a2 Latus rectum = m = 2b2 = 2.9 = 9.
fi (2a – h)2 – (2b – k)2 = a2 a 4 2

Hence, the locus of (h, k) is Hence, the value of 2m – 3 = 9 – 3 = 6.
(2a – x)2 – (2b – y)2 = a2 7. No real tangent can be drawn.
(x –2a)2 – (y – 2b)2 = a2
So, the value of m is zero.
Integer Type Questions Hence, the value of (m + 4) is 4.
8. Given hyperbola is
1. e= 1 + b2 = 1+ 12 = 4=2
a2 4 16x2 – 9y2 = 144

2. We have, fi x2 - y2 = 1
9 16
b2 1+ 81 = 225 = 15 = 5
a2 144 144 12 4 The equation of any tangent to the given hyperbola is

e= 1+ = y = mx + a2m2 - b2

Also, a2 = 144 fi y = mx + 9 m2 -16
25 It is given that,

Thus, the foci are (9 m2 -16) = 2 5

(±ae, 0) = Ê ± 12 ¥ 5 , 0˜¯ˆ = (± 3, 0) fi (9m2 – 16) = 20
ËÁ 5 4 fi 9m2 = 36
fi m2 = 4
Now, for the ellipse, fi m = ±2
So, a + b = 2 – 2 = 0
ae = 3 Hence, the value of (a + b + 3) is 3.
9. Given curves are xy = c, (c > 0)
fi a2c2 = 9 and x2 + y2 = 1

Thus, b2 = a2(1 – e2) = a2 – a2e2

= 16 – 9 = 7

Hence, the value of (b2 + 1) is 8.

Hyperbola 6.53

Y Ê c 0ˆ¯˜ Ê c ct 3 ˆ
ÁË t3 ËÁ t ˜¯
Thus, N = ct - , and N ¢ = 0, - .

Now, D = ar(DPNT)

X¢ O X ct c1
t
= 1 ct - c 01
2 t3 01

2ct

Y¢

Hence, the distance between the points of contact = = c2 Ê t + 1 ˆ
2t ËÁ t3 ˜¯
diameter of a circle = 2

10. The equation of the asymptotes of x2 - y2 =1 are = c2 ËÁÊ1 + 1 ˆ
a2 b2 2 t4 ˜¯

x + y = 0, x - y = 0 and D = ar(DPN¢T¢)
ab ab

i.e. bx + ay = 0, bx – ay = 0 ct c 1
Let any point on the hyperbola be =1 0 1
t 1
P(a sec j, b tan j). 2 c - ct3
It is given that, p1p2 = 6 0 t

fi absecj + ab tanj absecj - ab tanj = 6 2c

t

a2 + b2 a2 + b2 = ct Ê c - ct 3 - 2c ˆ
2 ËÁ t t ˜¯
a2b2 (sec2j - tan2j)
fi (a2 + b2) = 6 ct Ê cˆ
2 ËÁ t ˜¯
= ct 3 +

fi a2b2 ) = 6 = c2 (t4 +1)
(a2 + b2 2

fi a2 ◊2a2 = 6 Now, 1 + 1 = 2 Ê t4 ˆ + 2 Ê 1ˆ
(a2 + 2a2) D D¢ c2 ÁË t4 + 1¯˜ c2 ÁË t4 + 1˜¯

fi 2a2 = 18 2 Ê t4 + 1ˆ
c2 ÁË t4 + 1¯˜
fi a2 = 9 =

fi 2a = 6 2
c2
\ Length of the transverse axis = 6 =

11. The equation of the tangent to the hyperbola xy = c2 at

Ê ct, c ˆ is Hence, the value of Ê c2 + c2 + ˆ is 6.
ÁË t ¯˜ ËÁ D D¢ 4¯˜

x + yt2 = 2ct …(i) 12. The equation of the given hyperbola is x2 - y2 = 1 .
94
Thus, T = (2ct, 0) and T ¢ = Ê 0, 2c ˆ .
ËÁ t ˜¯ Thus, the equations of the asymptotes are

The equation of the normal to the hyperbola xy = c2 at Ê x + yˆ Ê x - yˆ = 0
ËÁ 3 2 ˜¯ ÁË 3 2 ˜¯
Ê ct, c ˆ is
ÁË t ˜¯
Ê x yˆ Ê x yˆ
xt3 – yt – ct4 + c = 0 …(ii) fi ÁË 3 + 2 ¯˜ =0 and ÁË 3 - 2 ¯˜ =0

6.54 Coordinate Geometry Booster
Y
fi 2x + 3y = 0 and 2x – 3y = 0 P(1/2. 1)
The equation of any tangent to the hyperbola
x2 - y2 = 1 is
94

x secj - y tanj = 1 X¢ O X
32
Let the points of intersection of 2x + 3y = 0, 2x – 3y = x = –1 x=1
0 and
x secj - y tanj = 1 are Y¢
32
O, P and Q respectively. Thus, x = 1 is nearer to P(1/2, 1).
Therefore, the directrix of the ellipse is x = 1.
Therefore, O = (0, 0), Let Q(x, y) be any point on the ellipse.
Now, the length of the perpendicular from Q to the di-
P = Ê secj 3 tanj , - secj 2 tanj ˆ rectrix x – 1 = 0 is
ËÁ + + ¯˜
|x – 1|.
and Q = Ê secj 3 tanj , secj 2 tanj ˆ By the definition of the ellipse, we have
ÁË - - ¯˜
QP = e|x – 1|

Hence, the area of DOPQ fi Ê x - 1ˆ 2 + ( y -1)2 = e| x -1|
ÁË 2¯˜
00

3 -2 fi Ê x - 1ˆ 2 + ( y -1)2 = e2 x -12
= 1 secj + tanj secj + tanj ÁË 2˜¯
2
23 secj - tanj Ê 1ˆ 2 1
secj - tanj ËÁ 2¯˜ 4
fi x - + (y - 1)2 = (x2 - 2x + 1)

00

= 1 (6 + 6) = 6 sq. u. fi x2 - x + 1 + y2 - 2y + 1 = x2 - 1 x + 1
2 4 42 4

Previous Years’ JEE-Advanced Examinations fi x2 - x + y2 - 2y + 1 = x2 - x
42
1. Given curve is
x2 - y2 = 1, r > 1 fi 4x2 – 4x + 4y2 – 8y + 4 = x2 – 2x
1- r 1+ r fi 3x2 + 4y2 – 2x – 8y + 4 = 0
fi (3x2 – 2x) + (4y2 – 8y) + 4 = 0
fi (1 + r) x2 – (1 – r)y2 = (1 – r)(1 + r)
fi (1 + r) x2 – (1 – r)y2 + (r2 – 1) = 0 fi 3ÊËÁ x2 - 2 xˆ˜¯ + 4( y2 - 2y) + 4 = 0
Now, h2 – ab = 0 – (1 + r)(r – 1) 3

= (1 – r2) < 0, as r > 1 3 ÏÔÌÓÔÊÁË 1ˆ 2 1 Ô¸
So, it represents an ellipse. { }fi 3¯˜ 9 ˝˛Ô
2. Hints x - - + 4 ( y -1)2 -1 +4=0
3. Given curve is
3ÊËÁ 1ˆ 2 1
2x2 + 3y2 – 8x – 18y + 35 = k 3˜¯ 3
fi 2(x2 – 4x) + 3(y2 – 6y) = k = 35 fi x - + 4( y - 1)2 = -4 + + 4
fi 2{(x –2)2 – 4} + 3 {(y – 3)2 – 9} = k – 35
fi 2(x – 2)2 + 3(y – 3)2 = k – 35 + 8 + 27 3ÊÁË 1ˆ 2 1
fi 2(x – 2)2 + 3(y – 3)2 = k 3¯˜ 3
It represents a point if k = 0. fi x - + 4( y - 1)2 =

4. Clearly common tangents of the given curves are Ê x - 1ˆ 2 (y - 1)2
x = 1 and x = –1, respectively. ËÁ 3¯˜
fi + =1
1/9 1/12

Hyperbola 6.55

5. Let M(h, k) be the point. fi x4 + c4 = a2x2
The equation of any line through M(h, k) having slope
4 is fi x4 – a2x2 + c4 = 0
y – k = 4(x – h)
Suppose the line meets the curve xy = 1 at P(x1, y1) and Let its roots be x1, x2, x3 and x4.
Q(x2, y2). Thus, x1 + x2 + x3 + x4 = 0
and x1x2x3x4 = c4
Y Similarly, y1 + y2 + y3 + y4 = 0
and y1y2y3y4 = c4
7. Let the point P be (h, k)

P(x1, y1) The equation of any tangent to the parabola is
X¢ X y = mx + a which is passing through P(h, k).
m
O
M(h, k) \ k = mh + a
m
Q(x2, y2)
fi m2h – km + a = 0

Since it has two roots say m1 and m2. Thus,

m1 + m2 = k and m1m2 = a
m h

Y¢ Now, tan(45°) = m2 - m1
1+ m1m2
Now,
y – k = 4(x – h) Ê m2 - m1 ˆ 2
ÁË 1+ m1m2 ˜¯
fi 1 - k = 4(x - h) fi = 1
x
fi (m2 – m1)2 = (1 + m1m2)2
fi 4x2 – (4h – k) x – 1 = 0 fi (m2 + m1)2 – 4m1m2 = (1 + m1m2)2

Let its roots be x1, x2. Ê kˆ2 Ê aˆ ËÁÊ1 + aˆ2
ËÁ h ¯˜ ËÁ h ¯˜ h ˜¯
\ x1 + x2 = 4h - k …(i) fi - 4 =
4
k2 - 4ah (a + h)2
1 fi h2 = h2
and x1x2 = - 4 …(ii)

Also, h = 2x1 + x2 fi (a + h)2 = k2 – 4ah
3
fi h2 + 6ah + a2 = k2
2x1 + x2 = 3h
From Eqs (i) and (iii), we get …(iii) fi (h + 3a)2 = k2 – 8a2

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

(x + 3a)2 = y2 – 8a2

x1 = 3h - 4h - k = 8h + k 8. Let P(h, k) be the point whose chord of contact w.r.t.
4 4
the hyperbola x2 – y2 = 9 is

8h + k (2h + k) x=9 …(i)
2 2
fi x2 = 3h - = - Also, the equation of the chord contact of the tangents

Putting the values of x1 and x2 in Eq. (ii), we get from P(h, k) is
(8h + k) ¥ - (2h + k) = - 1
4 24 hx – ky = 9 …(ii)

Since the Eqs (i) and (ii) are identical, so

h = 1 and k = 0

fi (8h + k)(2h + k) = 2 Thus, the point P is (1, 0).

fi 16h2 + 10hk + k2 – 2 = 0 The equations of the pair of tangents is

Hence, the locus of M(h, k) is SS1 = T2
fi (x2 – y2 – 9) (1 – 0 – 9) = (x – 9)2
16x2 + 10xy + y2 – 2 = 0.
fi –8(x2 – y2 – 9) = x2 – 18x + 81

6. Given x2 + y2 = a2 and xy = c2 fi 9x2 – 8y2 – 18x + 9 = 0

We have, 9. The equation of the normal at P(a sec q, b tan q) is

Ê c2 ˆ 2 ax cos q + by cot q = a2 + b2.
ÁË x ˜¯
x2 + = a2 fi ax + by cosec q = (a2 + b2) sec q

and the equation of the normal at Q(a sec j, b tan j) is

6.56 Coordinate Geometry Booster

ax cos j + by cot j = a2 + b2 12. Given hyperbola is x2 - y2 = 1 .
fi ax + by cosec j = (a2 + b2) sec j 94
Solving, we get
The equation of any point on the above hyperbola is
b(cosec q – cosec j) y = (a2 + b2)(sec q – sec j)
P(3 sec q, 2 tan q).

fi y = Ê a2 + b2 ˆ Ê secq - secj ˆ and the equation of the chord of contact of the circle
ÁË b ˜¯ ËÁ cosecq - cosecj ¯˜
x2 + y2 = 9 relative to the point P is

3x sec q + 2y sin q = 9 …(i)

fi y = Ê a2 + b2 ˆ Ê cosecj - cosecq ˆ Let M(h, k) be the mid-point of (i).
ËÁ b ¯˜ ÁË cosecq - cosecj ¯˜
The equation of the chord of the circle bisected at M is

fi y = - Ê a2 + b2 ˆ T = S1 …(ii)
ÁË b ¯˜ fi hx + ky = h2 + k2

Y

Ê a2 + b2 ˆ AP
- ËÁ b ¯˜
Thus, k =

M(h, k)

x2 y2 X¢ O X
cos2a sin 2a
10. Given curve is - =1

We have B

b2 = a2 (e2 – 1)

fi a2e2 = a2 + b2 Y¢

fi a2e2 = cos2 a + sin2 a = 1 Clearly, Eqs (i) and (ii) are identical. So

fi (ae)2 = 1 3secq 2 tanq 9
h k + k2
fi (ae) = ±1 = = h2

Abscissae of foci are ±1 irrespective of the value of a. 3h 9k
h2 + k2 2(h2 +
11. Given hyperbola is x2 – 2y2 = 4 fi secq = , tanq = k2)

fi x2 - y2 = 1 …(i) As we know that,
42 sec2 q – tan2 q = 1

Given line is 2

Ê 3h ˆ 2 Ê 9k ˆ
ÁË h2 + k2 ¯˜ ËÁ 2(h2 + 2 )¯˜
2x + 6 y = 2 fi - = 1

k

fi 6 y = -2x + 2 fi 36h2 – 81k2 = (h2 +k2)2
Hence, the locus of M(h, k) is
fi y=- 2 x+ 2
66 36x2 – 81y2 = (x2 + y2)2
13. Let e be the eccentricity of the given ellipse x2 + y2 = 1
fi y=- 2 x+ 2
33 25 16

As we know that if the line y = mx + c be a tangent Thus, e = 1 - b2 = 1- 16 = 3
a2 25 5
x2 y2 Therefore, the eccentricity of the hyperbola is 5 .
to the hyperbola a2 - b2 =1, the co-ordinates of the 3

Ê a2m , b2 ˆ Also, ae = 5◊ 3 = 3
ÁË ± c c ¯˜ 5
point of contact is ± .
Thus, the hyperbola passes through the focus, i.e. (3, 0)

Ê Ê 2ˆ ˆ of the given ellipse.
4ÁË - 3 ¯˜
Á ˜ So, the semi-transverse axis is 3, i.e. a = 3
˜
= Á ± , ± 2 So, the semi conjugate axis is 4, i.e. b = 4
Á
2 2˜ Hence, the equation of the hyperbola is

Á 3 3 ˜ x2 - y2 =1
Ë ¯ a2 b2

= (± (-4), ± 6) fi x2 - y2 = 1
9 16
= (4, - 6) satisfies the given curve.

Hyperbola 6.57

14. Given, the transverse axis = 2 sin q fi x - 2 = 6, y + 2 =1
fi 2a = 2 sin q
fi a = sin q fi x = 2 + 6, y =1- 2
Given ellipse is
3x2 + 4y2 = 12 So, B = ( 2 + 6,1 - 2) .

fi x2 + y2 = 1 Now, ar(DABC) = 1 ¥( 6 - 2) ¥1
43 2

The eccentricity of the ellipse, Ê 3 ˆ
= ÁË 2 - 1¯˜
b2 1- 3 = 1
e= 1 - a2 = 42 16. Given hyperbola is
2x2 – 2y2 = 1
and the foci of the ellipse = (±ae, 0)
fi x2 - y2 = 1
= Ê ± 2◊ 1 , 0¯ˆ˜ = (±1, 0) 1/2 1
ËÁ 2

Let e1 be the eccentricity of the hyperbola. Thus, e = 1 + b2 = 1+ 2 = 2
Now, b2 = a2 (e12 -1) a2 2

= a2e12 - a2 The eccentricity of the ellipse is 1 .
= 1 – sin2 q = cos2 q 2

Hence, the equation of the hyperbola is a2 ÊÁË1 - 1ˆ a2
2˜¯ 2
x2 y2 Also, b2 = a2 (1 - e2 ) = =
a2 b2
- =1

Thus, the equation of the ellipse is

fi x2 - y2 =1 x2 + y2 =1
sin 2q cos2q a2 b2

fi x2 cosec2 q – y2 sec2 q = 1 x2 y2
a2 a2/2
15. Given curve is fi + = 1

x2 - 2y2 - 2 2x - 4 2 y - 6 = 0

fi (x2 - 2 2 x) - 2( y2 + 2 2 y) = 6 fi x2 + 2y2 = a2

{ } { }fi (x - 2)2 - 2 - 2 ( y + 2)2 - 2 = 6 Let P(h, k) be the point of intersection of the ellipse and

the hyperbola.

fi (x - 2)2 - 2( y + 2)2 = 6 + 2 - 4 Thus, h2 + 2k2 = a2 and 2h2 – 2k2 = 1 …(i)
fi (x - 2)2 - 2(y + 2)2 = 4
The Equations of tangents at P(h, k) to the ellipse and
fi (x - 2)2 - (y + 2)2 = 1
42 hyperbola are

Vertex: (a, 0) hx + 2ky = a2 and 2hx – 2ky = 1
fi x - 2 = 2, y + 2 = 0
Now, m(ET ) = - h and m(HT ) = h
2k k

fi x = 2 + 2, y = - 2 Since both the curves cut at right angles, so
m(ET) ¥ m(HT) = –1

fi - h ¥ h = -1
2k k

Thus, A( 2 + 2, - 2) fi h2 = 2k2 …(ii)
From Eqs (i) and (ii), we get,
Focus: x - 2 = ae, y + 2 = 0
2h2 – h2 = 1
fi x = 2 + 2◊ 3 , y = - 2 fi h2 = 1
2 and h2 + 2k2 = a2
fi h2 + h2 = a2
fi x = 2 + 6, y = - 2 fi a2 = 1 + 1 = 2
Thus, the equation of the ellipse is
Therefore, the focus is C: ( 2 + 6, - 2) fi x2 + 2y2 = 2
and its foci are (±ae, 0) = (±1, 0).
End-point of L.R. = Ê b2 ˆ
ÁË ae, a ˜¯

6.58 Coordinate Geometry Booster

17. (P) As hx + ky –1 = 0 touches the circle x2 + y2 = 4, so, Now, CM = 4

0+ 0-1 = 2 fi 4m - 0 + 9m2 - 4 = 4
h2 + k2 m2 +1

fi (h2 + k2) = 1 fi (4m + 9m2 - 4)2 = 16(m2 + 1)
4
fi 16m2 + 9m2 - 4 + 8m 9m2 - 4
Thus, (h, k) lies on (x2 + y2 ) = 1 = 16m2 + 16
4
fi (9m2 - 20)2 = (-8m 9m2 - 4)2
(Q) z lies on the hyperbola, since fi 81m4 + 400 – 360m = 64m(9m2 – 4)
|SP – S¢P| = 2a. fi 495m4 + 104m2 – 400 = 0
fi (99m2 + 100)(5m2 – 4) = 0
(R) Ê x ˆ2 + y2 = Ê1- t2 ˆ2 + Ê 2t ˆ2 fi (5m2 – 4) = 0
ËÁ 3 ¯˜ ËÁ 1+ t2 ˜¯ ÁË 1+ t2 ˜¯ fi m = 2 , since m > 0

= (1- t2 )2 + 4t2 5
(1+ t2 )2 Thus, the equation of the tangent is

= (1 + t2 )2 y= 2 x+ 4
(1 + t2 )2 55

=1 fi 2x - 5y + 4 = 0

which represents an ellipse. 19. Y
(S) For x > 1, the given conic is a hyperbola.
MA
For x = 1, the conic is a parabola.
(T) Let z = x + iy X¢ OC X

Given Re(z + 1)2 = |z|2 + 1 B
fi (x+ 1)2 = x2 + y2 + 1
fi x2 + 2x + 1 = x2 + y2 + 1 Y¢
fi y2 = 2x
which represents a parabola. Let the co-ordinates of A be
21. Given circle is (3 sec q, 2 tan q)

x2 + y2 – 8x = 0 As A lies on the circle, so
fi (x – 4)2 + y2 = 16 9 sec2 q = 4 tan2 q – 24 sec q = 0

Y fi 9 sec2 q + 4(sec2 q – 1) – 24 sec q = 0
fi 13 sec2 q – 24 sec q – 4 = 0
MA fi (13 sec q + 2) (sec q – 2) = 0
fi sec q = 2, since |sec q| ≥ 1
X¢ OC X Thus, the points A and B are

(6, 2 3) and (6, - 2 3)

B The equation of the circle with AB as diameter is
(x – 6)2 + y2 = 12
Y¢
fi x2 + y2 – 12x + 24 = 0.
The centre is (4, 0) and the radius = 4.
Given hyperbola is x2 - y2 = 1 20. Given line is 2x + y = 1 …(i)

94 which is passing through Ê a , 0ˆ˜¯
The equation of any tangent to the given hyperbola is ÁË e

y = mx + a2m2 - b2 fi 2◊ a = 1
e
fi y = mx + 9 m2 - 4
fi a= e.
2

Hyperbola 6.59

Since the equation (i) is a tangent to the given hy- The hyperbola passing through the focus of the

perbola, so ellipse, so

c2 = a2m2 – b2 3 - 0 = 1
a2
fi 1 = a2, 4 – b2

fi 4a2 – b2 = 1 fi a2 = 3
Now, b2 = a2 (e2 –1)
fi 4a2 – a2(e2 – 1) = 1

fi 5a2 – a2e2 = 1

fi a2(5 – e2) = 1 fi b2 = a 2 Ê 4 - 1˜¯ˆ = a2 =1
ÁË 3 3
fi e2 (5 - e2 ) = 1
4 Hence, the equation of the hyperbola is

fi 5e2 – e4 = 4 x2 - y2 = 1
31
fi e4 – 5e2 + 4 = 0

fi (e2 – 1)(e2 – 4) = 0 fi x2 – 3y2 = 3
\ Focus of a hyperbola = (2, 0)
fi (e2 – 4) = 0, as e π 1 23. Given line is 2x – y = 1. So,

fi e=2 m=2
The equation of any tangent to the given hyper-
21. The equation of the normal at P is bola is

a2x + b2 y = a2 + b2
x1 y1

fi a2x + b2 y = a2 + b2 y = mx ± a2m2 - b2
63

which is passing through (9, 0). So fi y = 2 x ± 9.4 - 4

a2 ◊9 + b2 ◊0 = a2 + b2 fi y = 2 x ± 32
63
fi y = 2x±4 2
fi 3a2 = a2 + b2
2 fi 2 x - y + 4 2 = 0, 2 x - y - 4 2 = 0

fi 3a2 = a2 + a2 (e2 -1) fi x - y = 1 and - x + y = 1
2 22 42 22 42

fi 3 = 1+ (e2 -1) Comparing it with xx1 - yy1 = 1 . we get the points
2 94
of contact as
fi (e2 -1) = 3 -1 = 1
22 Ê 9 1ˆ Ê 9 1ˆ
ËÁ 2 2 ˜¯ ËÁ 2 ˜¯
fi e2 = 1+ 1 = 3 2 , and - 2 2 , - .
22

fi e= 3 . 24. Tangent at P is xx1 – yy1 = 1 intersects the x-axis at
2
M Ê 1 , 0¯ˆ˜
Hence, the eccentricity is ÁË x1
3.
22. Given ellipse is 2 Slope of normal = - y1 = y1 - 0
x2 + 4y2 = 4 x1 x1 - x2

fi x2 = 2x1

x2 + y2 = 1 Thus, N = (2x1, 0)
fi 41
3x1 + 1 y1
x1
The eccentricity, For centroid, l = , m =
33
b2 1- 1 = 3
e= 1 - a2 = 4 2 fi dl =1- 1
dx1 3x12
2.
Thus, the eccentricity of the hyperbola is 3 and dm = 1 , dm = 1 dy1 = x1
dy1 3 dx1 3 dx1 3 x12 - 1
Foci of the ellipse = (±ae, 0) = (± 3 , 0) .