The document provides a comprehensive overview of hyperbolas, including their definitions, properties, equations, and various examples. It explains key concepts such as vertices, axes, eccentricity, and latus rectum, alongside calculations for specific cases. Additionally, it discusses related topics like conjugate hyperbolas and the position of points in relation to the hyperbola.

1. Total no.of questions in Hyperbola are-
Inchapter examples..................................................10
Solved Examples............................................... ......14
Total no. of questions.............................................. 24
HYPERBOLA

2. (i) Definition hyperbola : A Hyperbola is the
locus of a point in a plane which moves in
the plane in such a way that the ratio of its
distance from a fixed point (called focus) in
the same plane to its distance from a fixed
line (called directrix) is always constant which
is always greater than unity.
(ii) Vertices : The point A and A where the
curve meets the line joining the foci S and S
are called vertices of hyperbola.
(iii) Transverse and Conjugate axes : The
straight line joining the vertices A and A is
called transverse axes of the hyperbola.
Straight line perpendicular to the transverse
axes and passes through its centre called
conjugate axes.
(iv) Latus Rectum : The chord of the hyperbola
which passes through the focus and is
perpendicular to its transverse axes is called
latus rectum. Length of latus rectum =
a
b
2 2
.
(v) Eccentricity : For the hyperbola 2
2
a
x
– 2
2
b
y
= 1,
b2 = a2 (e2 – 1)
e =
2
a
2
b
2
1 





 =
2
axes
Transverse
axes
Conjugate
1 








(vi) Focal distance : The distance of any point
on the hyperbola from the focus is called the
focal distance of the point.
Note : The difference of the focal distance of a point on
the hyperbola is constant and is equal to the length
of the transverse axes. |SP – SP| = 2a (const.)
z'
y z
P(x, y)
S
(ae, 0)
x
A
L'
L
M
M'
K' K
A'(–a, 0)
T '
–x
T
(a, 0)
(0, 0)
O
(–ae, 0)
S'
2. CONJUGATE HYPERBOLA
The hyperbola whose transverse and conjugate axes
are respectively the conjugate and transverse axes of
a given hyperbola is called conjugate hyperbola.
Note :
(i) If e1 and e2 are the eccentricities of the
hyperbola and its conjugate then 2
1
e
1
+
2
2
e
1
= 1
(ii) The focus of hyperbola and its conjugate are
concyclic.
Examples
based on Standard Equation and Difinitions
Ex.1 Find the equation of the hyperbola whose
directrix is 2x + y = 1, focus (1,2) and
eccentricity 3 .
Sol. Let P (x,y) be any point on the hyperbola.
Draw PM perpendicular from P on the
directrix.
Then by definition
SP = e PM
 (SP)2 = e2(PM)2
(x–1)2 + (y–2)2 = 3
2
1
4
1
y
x
2









5(x2 + y2 – 2x – 4y+ 5} =
3(4x2 + y2 + 1+ 4xy – 2y – 4x)
 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0
which is the required hyperbola.
Ex.2 Find the lengths of transverse axis and
conjugate axis, eccentricity and the co-
ordinates of foci and vertices; lengths of the
latus rectum, equations of the directrices of
the hyperbola 16x2 – 9y2 = –144
1. STANDARD EQUATION AND DEFINITIONS
Standard Equation of hyperbola is 2
2
a
x
– 2
2
b
y
= 1

3. Sol. The equation 16x2 – 9y2 = – 144 can be
written as
9
x2
–
16
y2
= – 1. This is of the form
2
2
a
x
– 2
2
b
y
= – 1
 a2 = 9, b2 = 16  a = 3, b = 4
Length of transverse axis :
The length of transverse axis = 2b = 8
Length of conjugate axis :
The length of conjugate axis = 2a = 6
Eccentricity : e = 2
2
b
a
1 =
16
9
1 =
4
5
Foci : the co- ordinates of the foci are (0,± be), i.e.,
(0, ± 4)
Length of Latus rectum :
The length of latus rectum =
b
a
2 2
=
4
)
3
(
2 2
=
2
9
Equation of directrices :
The equation of directrices are y = ±
e
b
y = ±
 
4
/
5
4
= ±
5
16
Ex.3 Find the coordinates of foci, the eccentricity
and latus- rectum, equations of directrices
for the hyperbola 9x2–16y2–72x+96y–144=0
Sol. Equation can be rewritten as
2
2
4
)
4
x
( 
– 2
2
3
)
3
y
( 
= 1 so a = 4, b = 3
b2 = a2 (e2 – 1) gives e = 5/4
Foci : X = ± ae, Y = 0 gives the foci as
(9,3), (– 1,3)
Centre : X = 0, Y = 0 i.e. (4,3)
Directrices : X = ± a/e i.e. x – 4 = ± 16/5
 directrices are 5x – 36 = 0; 5x – 4 = 0
Latus- rectum = 2b2/ a = 2 x 9/4= 9/2
Ex.4 Two straight lines pass through the fixed
points (± a,0) and have gradients whose
product is k. Show that the locus of the
points of inter-section of the lines is a
hyperbola.
Sol. y= m1(x –a),y= m2(x+a) wherem1m2 =k, given
In order to find the locus of their point of
intersection we have to eliminate the unknown
m1 and m2. Multiplying, we get
y2 = m1m2 (x2 – a2) or y2 = k(x2–a2)
or
1
x2
–
k
y2
= a2 which represents a hyperbola.
Ex.5 The eccentricity of the conic represented by
x2 – y2 – 4x + 4y + 16 = 0 is
(A) 1 (B) 2 (C) 2 (D)1/2
Sol.[B] We have x2 – y2 – 4x +4y+16 = 0
or, (x2 – 4x)– (y2 – 4y) = 16
or, (x2 – 4x + 4) – (y2 – 4y+ 4) = – 16
or, (x–2)2 – (y – 2)2 = – 16
or, 2
2
4
)
2
x
( 
– 2
2
4
)
2
y
( 
= 1
Shifting the origin at (2,2) , we obtain
2
2
4
X
– 2
2
4
Y
= – 1, where x = X + 2, y = Y + 2
This is a rectangular hyperbola, whose
eccentricity is always 2 .
Ex.6 The equation 16x2 = 3y2 – 32x + 12y – 44= 0
represents a hyperbola
(A) the length of whose transverse axis is 3
4
(B) the length of whose conjugate axis is 4
(C) whose centre is (– 1,2)
(D) whose eccentricity is
3
19
Sol.[D] We have, 16(x2 – 2x) – 3 (y2 – 4y) = 44
 16 (x –1)2 – 3 (y– 2)2 = 48

3
)
1
x
( 2

–
16
)
2
y
( 2

= 1
This equation represents a hyperbola with
eccentricity given
e =
2
axis
Transverse
axis
Conjugate
1 







 =
2
3
4
1 







 =
3
19
3. Parametric equation of the Hyperbola
Let the equation of ellipse in standard form will be
given by 2
2
a
x
– 2
2
b
y
= 1

4. Then the equation of ellipse in the parametric form
will be given by x= a sec , y = b tan  where  is
the eccentric angle whose value vary from 0   <
2. Therefore coordinate of any point P on the ellipse
will be given by (a sec, b tan ).
4. POSITION OFAPOINT P(x1, y1) WITH RESPECT
TO HYPERBOLA
The quantity 2
2
1
a
x
– 2
2
1
b
y
= 1 is positive, zero or
negative according as the point (x1, y1) lies inside on
or outside the hyperbola 2
2
a
x
– 2
2
b
y
= 1.
5. LINE & HYPERBOLA
‘’The straight line y = mx + c is a sacant, a tangent or
passes outside the hyperbola 2
2
a
x
– 2
2
b
y
= 1 according
as c2 > = < a2m2 – b2
Examples
based on Line and Hyperbola
Ex.7 Find the position of the point (5, – 4) relative
to the hyperbola 9x2 – y2 = 1.
Sol. Since 9(5)2 –(–4)2 –1=225–16 –1 =208 > 0
so the point ( 5,–4) lies inside the hyperbola
9x2– y2 = 1
S.No. Particulars Hyperbola Conjugate Hyperbola
2
2
a
x
– 2
2
b
y
= 1 – 2
2
a
x
+ 2
2
b
y
= 1
1. Co-ordinate of the centre (0, 0) (0, 0)
2. Co-ordinate of the vertices (a, 0) & (–a, 0) (0, b) & (0, –b)
3. Co-ordinate of foci (+ ae, 0) (0, + be)
4. Length of the transverse axes 2a 2b
5. Length of the conjugate axes 2b 2a
6. Equation of directrix x = + a/e y = + b/e
7. Eccentricity e = 2
2
a
b
1 e = 2
2
b
a
1
8. Length of latus rectum
a
b
2 2
b
a
2 2
9. Equation of transverse axes y = 0 x = 0
10. Equation of conjugate axes x = 0 y = 0
Ex.8 Show that the line x cos  + y sin  = p
touches the hyperbola 2
2
a
x
– 2
2
b
y
= 1 if
a2 cos2  – b2 sin2  = p2.
Sol. The given line is
x cos  + y sin  = p
 y sin  = – x cos  + p
y = – x cot  + p cosec 
Comparing this line with y = mx + c
m = – cot , c = p cosec 
Since the given line touches the hyperbola
2
2
a
x
– 2
2
b
y
= 1 then
c2 = a2m2 – b2
p2 cosec2  = a2 cot2 – b2
or p2 = a2 cos2  – b2 sin2 
Ex.9 The line 5x + 12y = 9 touches the hyperbola
x2 – 9y2 = 9 at the point
(A) (– 5,4/3) (B) (5, – 4/3)
(C) (3, – 1/2) (D) None of these
Sol.[B] We have : m = Slope of the tangent = –
12
5
If a line of slope m is tangent to the hyperbola
2
2
a
x
– 2
2
b
y
= 1, then the coordinates of the point
of contact are

5. l
2
x
x
a
+
l
2
y
y
b
= a2 + b2 = a2 e2.
(ii) The equation of normal at (a sec , b tan )
to the ellipse 2
2
a
x
– 2
2
b
y
= 1 is
ax cos  + by cot  = a2 + b2.
(iii) The equation of the normal to the hyperbola
2
2
a
x
– 2
2
b
y
= 1 in terms of the slope m of the
normal is given by y = mx – 2
2
2
2
2
m
b
a
)
b
a
(
m


Note : In general four normals can be drawn from a
point (x1, y1) to the hyperbola 2
2
a
x
– 2
2
b
y
= 1
8. EQUATION OF PAIR OF TANGENTS
If P (x1, y1) be any point out side the hyperbola
2
2
a
x
– 2
2
b
y
= 1 and a pair of tangents PA, PB can be
drawn to it from P.
then the equation of pair of tangents of PA and PB
is SS1 = T2
where S  2
2
a
x
– 2
2
b
y
– 1 = 0
S1 = 2
2
1
a
x
– 2
2
1
b
y
– 1 = 0
T = 2
1
a
x
x
– 2
1
b
y
y
– 1 = 0












2
2
2
2
2
2
2
2
b
m
a
b
,
b
m
a
m
a
Here, a2 = 9, b2 = 1 and m = – 5/12
So, points of contact are 







3
4
,
5
i.e. 






3
4
,
5 and 






3
4
,
5 .
Out of these two points 






3
4
,
5 lies on the
line 5x + 12y = 9. Hence, 






3
4
,
5 is the
required point.
6. EQUATION OFTANGENT
(i) The equation of tangents of slope m to the
hyperbola 2
2
a
x
– 2
2
b
y
= 1 are
y = mx + 2
2
2
b
m
a  and the co-ordinates
of the point of contacts are












2
2
2
2
2
2
2
2
b
m
a
b
,
b
m
a
m
a
(ii) Equation of tangent to the hyperbola
2
2
a
x
– 2
2
b
y
= 1 at the point (x1, y1) is
2
1
a
xx
– 2
1
b
yy
= 1
(iii) Equation of tangent to the hyperbola
2
2
a
x
– 2
2
b
y
= 1 at the point
(a sec, b tan) is
a
x
sec –
b
y
tan = 1
Note : In general two tangents can be drawn from
an external point (x1, y1) to the hyperbola
and they are y – y1 = m1 (x – x1) and
y – y1 = m2 (x – x1), where m1 and m2 are
roots of (x1
2 – a2) m2 – 2x1y1 + y1
2 + b2 = 0
7. EQUATION OF NORMAL
(i) The equation of normal to the hyperbola
2
2
a
x
– 2
2
b
y
= 1 at (x1, y1) is

6. 9. CHORD OF CONTACT
If PA and PB be the tangents through point
P (x1, y1) to the hyperbola 2
2
a
x
– 2
2
b
y
= 1, then
the equation of the chord of contact AB is
2
1
a
x
x
– 2
1
b
y
y
= 1 or T = 0 (at x1, y1)
10. EQUATION OFACHORDWHOSE MIDDLE POINT
IS GIVEN
Equation of the chord of the hyperbola
2
2
a
x
– 2
2
b
y
= 1, bisected at the given point (x1, y1) is
T = S1.
2
1
a
xx
– 2
1
b
yy
– 1 = 2
2
1
a
x
– 2
2
1
b
y
– 1
11. DIRECTOR CIRCLE
The locus of the intersection of tangents which are
at right angles is known as director circle of the
hyperbola. The equation of the director circle is x2 +
y2 = a2 – b2.
Note: The circle described on transverse axis of the
hyperbola as diameter is called auxiliary circle and so
its equation is x2 + y2 = a2
Examples
based on Director Circle
Ex.10 Find the locus of the mid points of the chords
of the circle x2 + y2 = 16, which are tangent
to the hyperbola 9x2 – 16y2 = 144.
Sol. Any tangent to hyperbola
16
x2
–
9
y2
= 1 is
y = mx + (16m2 – 9) ...(1)
Let (x1,y1) be the mid- point of the chord of
the circle x2 +y2 = 16, then equation of the
chord is (T= S1)
xx1 + yy1 – (x1
2 + y1
2) = 0 ...(2)
Since (1) and (2) are same, comparing,
we get
1
x
m
= –
1
y
1
=
)
y
x
(
9
m
16
2
1
2
1
2



 m = –
1
1
y
x
, ( 2
1
x + 2
1
y )2 = 2
1
y (15m2 – 9)
Eliminating m and generalizing (x1, y1),
required locus is (x2 + y2)2 = 16x2 – 9y2

7. Ex.1 If e and e be the eccentricities of a hyperbola
and its conjugate then the value of 2
e
1
+ 2
'
e
1
=
(A) 0 (B) 1 (C) 2 (D) 4
Sol.[B] Hyperbola 2
2
a
x
– 2
2
b
y
= 1
b2
= a2
(e2
– 1)
or e2
= 1 + 2
2
a
b
= 2
2
2
a
b
a 
Conjugate hyperbola 2
2
a
x
– 2
2
b
y
= –1
i.e.Transverse axis is along y-axis and conjugate
along x-axis.
a2
= b2
(e'2
– 1)
e'2
= 2
2
2
b
b
a 
 2
e
1
+ 2
'
e
1
= 2
2
2
2
b
a
b
a


= 1
Ex. 2. If the foci of the ellipse
16
x2
+ 2
2
b
y
= 1 and hyperbola
144
x 2
–
81
y2
=
25
1
coincide, then the value of b2
is-
(A) 1 (B) 5 (C) 7 (D) 9
Sol.[C] For hyperbola
144
x 2
–
81
y2
=
25
1
e2
= 1 + 2
2
a
b
= 1 +
144
81
=
144
225
e =
12
15
=
4
5
i.e., e > 1
Hence the foci are
(±ae, 0) 





 0
,
4
5
.
5
12
= (±3, 0)
Now the foci coincide therefore for ellipse
ae = 3 or a2
e2
= 9
or a2








 2
2
a
b
1 = 9
a2
– b2
= 9
or 16 – b2
= 9
 b2
= 7
Ex. 3. The equation of the common tangents to the
parabola y2
= 8x and the hyperbola 3x2
– y2
= 3 is-
(A) 2x± y+1 = 0 (B) x ± y + 1 = 0
(C) x ± 2y+1 = 0 (D) x ± y + 2 = 0
Sol.[A] Parabola y2
= 8x
 4a = 8  a = 2
Any tangent to the parabola is
y = mx +
m
2
...(i)
If it is also tangent to the hyperbola
1
x2
–
3
y2
= 1 i.e. a2
= 1, b2
= 3 then
c2
= a2
m2
– b2

2
m
2






= 1.m2
– 3
or m4
– 3m2
– 4 = 0  (m2
–4) (m2
+ 1) = 0
 m = ±2 puting for m in (i), we get the tangents
as 2x ± y + 1 = 0
Ex. 4. The locus of the point of intersection of the
lines 3 x–y–4 3 k=0 and
3 kx +ky – 4 3 = 0 for different values of k
is-
(A) Ellipse (B) Parabola
(C) Circle (D) Hyperbola
Sol.[D] 3 x – y = 4 3 k ...(i)
K( 3 x + y) = 4 3 ...(ii)
To find the locus of their point of intersection
eliminate the variable K between the equations
from (i) K =
3
4
y
x
3 
and putting in (ii), we get
( 3 x – y) ( 3 x + y) = 3 (4)2
3x2
– y2
= 48
or
16
x2
–
48
y2
= 1
Hence the locus is hyperbola
Ex. 5. The area of a triangle formed by the lines
x – y = 0, x + y = 0 and any tangent to the
hyperbola x2
– y2
= a2
is-
(A) a2
(B) 2a2
(C) 3a2
(D) 4a2
Sol.[A] Any tangent to the hyperbola at
P(a sec , a tan ) is
x sec  – y tan  = a ...(i)
Also x – y = 0 ...(ii)
x + y = 0 ...(iii)
Solving the above three lines in pairs, we get the
point A, B, C as











 tan
sec
a
,
tan
sec
a
,
SOLVED EXAMPLES

8. 











 tan
sec
a
,
tan
sec
a
and (0, 0)
Since the one vertex is the origin therefore the
area of the triangle ABC is
2
1
(x1
y2
– x2
y1
)
=
2
a 2














2
2
2
2
tan
sec
1
tan
sec
1
=
2
a 2
(–2) = –a2
= a2
Ex.6. The locus of the mid point ofthe chords of the
circle x2
+ y2
= 16, which are tangent to the
hyperbola 9x2
– 16y2
= 144 is-
(A) x2
+ y2
= a2
– b2
(B) (x2
+ y2
)2
= a2
– b2
(C) (x2
+ y2
)2
= a2
x2
– b2
y2
(D) (x2
+ y2
)2
= a2
+ b2
Sol.[C] Let (h, k) be the mid point of the chord of the
circle x2
+ y2
= a2
, so that its equation by
T = S1
is
hx + ky = h2
+ k2
or y = –
k
h
x +
k
k
h 2
2

i.e.
the form y = mx + c
It will touch the hyperbola if c2
= a2
m2
– b2

2
2
2
k
k
h







 
= a2
2
k
h






 – b2
(h2
+ k2
)2
= a2
h2
– b2
k2
Generalising, the locus of the mid-point
(h, k) is (x2
+ y2
)2
= a2
x2
– b2
y2
Ex.7 The eccentricity of the conic represented by
x2
– y2
– 4x + 4y + 16 = 0 is-
(A) 1 (B) 2 (C) 2 (D)
2
1
Sol.[B] We have x2
– y2
– 4x + 4y + 16 = 0
or (x2
– 4x) – (y2
– 4y) = – 16
or (x2
– 4x + 4) – (y2
– 4y + 4) = – 16
or (x – 2)2
– (y – 2)2
= – 16
or 2
2
4
)
2
x
( 
– 2
2
4
)
2
y
( 
= – 1
Shifting the origin at (2, 2), we obtain
2
2
4
x
– 2
2
4
y
= –1,
where x = X + 2, y = Y + 2
This is rectangular hyperbola, whose eccentricity
is always 2 .
i.e. e2
= 1+ 2
2
a
b
= 1 + 2
2
4
4
; e = 2
Ex.8 The equation 9x2
– 16y2
– 18x + 32y – 151 = 0
represent a hyperbola -
(A) The length of the transverse axes is 4
(B) Length of latus rectum is 9
(C) Equation of directrix is x=
5
21
and x = –
5
11
(D) None of these
Sol.[C] We have 9x2
– 16y2
– 18x + 32y – 151 = 0
9(x2
– 2x) – 16(y2
– 2y) = 151
9(x2
–2x+1) – 16(y2
– 2y + 1) = 144
9(x – 1)2
– 16(y – 1)2
= 144
16
)
1
x
( 2

–
9
)
1
y
( 2

= 1
Shifting the origin at (1, 1) without rotating the
axes
16
x2
–
9
y2
= 1
where x = X + 1 and y = Y + 1
This is of the form 2
2
a
x
– 2
2
b
y
= 1
where a2
= 16 and b2
= 9 so
The length of the transverse axes = 2a = 8
The length of the letus rectum =
a
b
2 2
=
2
a
The equaiton of the directrix x = ±
a
e
x – 1 = ±
5
16
 x = ±
5
16
+ 1
x =
5
21
; x = –
5
11
Ex.9 The equations to the common tangents to the
two hyperbola 2
2
a
x
– 2
2
b
y
= 1 & 2
2
a
y
– 2
2
b
x
= 1 are
(A) y = ± x ± 2
2
a
b 
(B) y = ± x ± 2
2
b
a 
(C) y = ± x ± (a2
– b2
)
(D) y = ± x ± 2
2
b
a 
Sol.[B] Any tangent to the hyperbola 2
2
a
x
– 2
2
b
y
= 1 is
y = mx ± 2
2
2
b
m
a 
or y = mx + c
where c = ± 2
2
2
b
m
a 
This will touch the hyperbola 2
2
a
y
– 2
2
b
x
= 1
if the equation 2
2
a
)
c
mx
( 
– 2
2
b
x
= 1 has equal
roots or x2
(b2
m2
– a2
) + 2b2
mcx + (c2
– a2
)b2
= 0 is an
quadratic equation have equal roots

9. 4b4
m2
c2
= 4(b2
m2
– a2
) (c2
– a2
)b2
c2
= a2
– b2
m2
a2
m2
– b2
= a2
– b2
m2
m2
(a2
+ b2
) = a2
+ b2
 m = ±1
Hence, the equations of the common tangents
are y = ±x ± 2
2
b
a 
Ex.10 For what value of  does the line y = 2x +  touches
the hyperbola 16x2
– 9y2
= 144?
Sol.  Equation of hyperbola is 16x2
– 9y2
= 144
or
9
x2
–
16
y2
= 1 comparing this with
2
2
a
x
– 2
2
b
y
= 1, we get a2
= 9, b2
= 16 and comparing
this line y = 2x +  with m = mx + c ;
m = 2 & c = 
If the line y = 2x +  touches the hyperbola
16x2
– 9y2
= 144
then c2
= a2
m2
– b2
  = 9(2)2
– 16
= 36 – 16 = 20;   = ±25
Ex.11 Find the equation of the tangent to the hyperbola
x2
– 4y2
= 36 which is perpendicular to the line
x – y + 4 = 0.
Sol. Let m be the slope of the tangent since the
tangent is perpendicular to the line
x – y + 4 = 0.
 m x 1 = –1  m = –1
since x2
– 4y2
= 36
or
36
x2
–
9
y2
= 1
Comparing this with 2
2
a
x
+ 2
2
b
y
= 1;
 a2
= 36 & b2
= 9 so the equation of tangents
are
y = (– 1) x ± 9
)
1
(
x
36 2


 y = –x ±27 or x + y ± 33 = 0
EX.12 Find the locus of the point of intersection of
tangents to the hyperbola 4x2
– 9y2
= 36 which
meet at a constant angle /4.
Sol. Let the point of intersection of tangents be
P (x1
,y1
). Then the equation of pair of tangents
from P(x1
,y1
) to the given hyperbola is
(4x2
– 9y2
– 36) (4x1
2
– 9y1
2
– 36)
= [4x1
x – 9y1
y – 36]2
. ...(1)
[SS1
= T2
]
or x2
(y1
2
+4) +2x1
y1
xy+y2
(x1
2
– 9)+..= 0
...(2)
Since angle between the tangent is /4.
tan 




 
4
=
  
 
9
x
4
y
9
x
4
y
y
x
2
2
1
2
1
2
1
2
1
2
1
2
1






Hence locus of P (x1
, y1
) is (x2
+ y2
– 5)2
= 4(9y2
– 4x2
+ 36)
Ex.13 Prove that the locus of the middle points of
chords of hyperbola 2
2
a
x
– 2
2
b
y
=1, passing
through the fixed point (h,k) is a hyperbola whose
centre is the point (
2
1
h,
2
1
k)
Sol. Using T= S1
, equation of chord having (x1
,y1
) as
mid point is 2
1
a
x
x
– 2
1
b
y
y
=
2
2
1
a
x
– 2
2
1
b
y
It passes through (h,k) so 2
1
a
h
x
– 2
1
b
k
y
=
2
2
1
a
x
– 2
2
1
b
y
 Locus of (x, y) is
2
2
a
h
2
1
x 






– 2
2
b
k
2
1
y 






=
4
1








 2
2
2
2
b
k
a
h
which is a hyperbola whose centre is 





k
2
1
,
h
2
1
.
Ex.14 From the points on the circle x2
+ y2
= a2
, tangents
are drawn to the hyperbola
x2
– y2
= a2
; prove that the locus of the middle
points of the chords of contact is the curve (x2
– y2
)2
= a2
(x2
+ y2
).
Sol. Since any point on the circle x2
+ y2
= a2
is
(a cos, a sin) chord of contact of this point
w.r.t hyperbola x2
– y2
= a2
is
x (a cos) – y(a sin) = a2
or x cos – y sin = a ...(1)
If its mid point be ( h,k) , then it is same as
T = S1
i.e. hx – ky– a2
= h2
–k2
– a2
or hx – ky = h2
– k2
...(2)
Comparing (1) & (2) , we get
h
cos 
=
k
sin 
=
)
k
h
(
a
2
2

or (h2
– k2
) cos = ah ...(3)
and (h2
– k2
) sin= ak ...(4)
Squaring and adding (3) and (4), we get
(h2
– k2
)2
= a2
h2
+ a2
k2
 (h2
– k2
)2
= a2
(h2
+ k2
)
Hence the required locus is
(x2
– y2
)2
= a2
(x2
+ y2
)