1. SOLVED SUBJECTIVE EXAMPLES
Example 1 :
A tangent to the hyperbola 1
b
y
a
x
2
2
2
2

 cuts the ellipse 1
b
y
a
x
2
2
2
2

 in points P and Q. Find the
locus ofthemidpoint ofPQ.
Solution :
Let M(x1
, y1
) be the midpoint ofthe chord PQ ofthe ellipse 1
b
y
a
x
2
2
2
2

 .
Equation ofPQ is
2 2 2 2 2
2
1 1 1 1 1 1 1
2 2 2 2 2 2 2
1 1
xx yy x y b xx x y
b
y
a b a b a y y a b
 
       
 
 
This is tangent to the hyperbola 1
b
y
a
x
2
2
2
2


if 2
2
1
4
2
1
4
2
2
2
2
1
2
2
1
2
1
4
b
y
a
x
b
a
b
y
a
x
y
b










  2
2
1
2
2
1
2
2
2
1
2
2
1
b
y
a
x
b
y
a
x











Hence locus of(x1
, y1
) is
2
2
2
2
2
2
2
2
2
b
y
a
x
b
y
a
x











Example 2 :
A straight line is drawn parallelto the conjugate axis ofthe hyperbola 1
b
y
a
x
2
2
2
2

 to meet it and
the conjugate hyperbola respectivelyinthepoint P and Q. Show that the normalsat P and Qto the
curves meet on the x-axis.
Solution :
Let P(a sec, b tan) be a point on the hyperbola, and Q(a tan , b sec ) be a point on the
conjugate hyperbola.  a sec = atan  sec = tan
Equationofthe normalto the hyperbola
1
b
y
a
x
2
2
2
2

 at P is y – b tan =



sec
b
tan
a
(x – b sec)
Equation ofthe normal to the conjugate hyperbola at Q is y– b sec = – )
tan
a
x
(
tan
b
sec
a




Eliminate x and use sec = tan
We get y (sec – tan ) = 0  y = 0
Hence the normals meet on the x-axis.

2. Example 3 :
Froma point G onthe transverse axis ofthe hyperbola 1
b
y
a
x
2
2
2
2

 , GLis drawnperpendicular to
one of its asymptotes. Also Gp is a normal to the curve at P. Prove that LP is parallel to the
conjugate axis.
Solution :
Let P(a sec, btan) be anypoint on the hyperbola
Equation of the normalat P is ax cos + bycot  = a2
+ b2
.
It meets the x-axis (transverse axis) at y= 0
 x = 

sec
a
b
a 2
2

2 2
a b
G sec ,0
a
 

 
 
 
The equation of line perpendicular to the asymptote bx – ay = 0 and passing through G, i.e.,
equation ofGL is y= – 









 sec
a
b
a
x
b
a 2
2
 ax + by = (a2
+ b2
) sec
Its intersection with the asymptote bx – ay = 0 gives x = a sec. So the x coordinate of
L is a sec, which is equalto the x-coordinate of the point P
 LP is parallelto the y-axis  LP is parallelto the conjugate axis.
Example 4 :
Avariablestraight line ofslope4 intersects the hyperbolaxy=1 at two points. Find thelocus ofthe
point which divides the line segment betweenthese points in the ratio 1 : 2. [IIT-1997]
Solution :
Let the line be y = 4x + c. It meets the curve xy= 1 at
x (4x + c) = 1  4x2
+ cx –1  x1
+ x2
= –c/4
Also y (y – c) = 4  y2
– cy – 4 = 0  y1
+ y2
= c
Let the point which divides the line segment in the ratio 1 : 2 be (h, k)
 h
3
x
2
x 2
1


 x2
= 3h + c/4  x1
= –c/2 – 3h
Also k
3
y
2
y 2
1


 y2
= 3k – c  y1
= –3k + 2c
Now (h, k) lies on the line y = 4x + c  k = 4h + c  c = k – 4h
 x1
= –k/2 + 2h – 3h = –h – k/2 and y1
= –3k + 2k – 8h = –k – 8h
 16h2
+ k2
+ 10hk = 2. Hence locus of (h, k) is 16x2
+ y2
+ 10 xy = 2
Example 5 :
Prove that if normal to the hyperbola xy = c2
at point t meets the curve again at a point t1
then
t3
t1
+ 1 = 0.
Solution :
Equationofnormalat point t i.e., (ct, c/t) is

3. y – xt2
=
t
c
(1 – t4
) ... (1)
It meets the curve again at t1
then (ct1
, c/t1
) must satisfy(1)

3
2
1
1
4
2
1
1
t
t
1
t
t
t
1
)
t
1
(
t
c
t
ct
t
c







 0
)
t
t
(
t
t
1
t
1
1
2
1



 
1
1
tt
)
t
t
( 
(1 + t3
t1
) = 0
Clearly 1
t
t   t3
t1
+ 1 = 0.
Example 6 :
The angle betweena pair oftangents drawnfroma point P to the parabola y2
= 4ax is45°. Show
that the locusofthe point P is a hyperbola. [IIT-1998]
Solution :
Let P ( 
, ) be anypoint on the locus. Equation ofpair oftangents fromP( 
, ) to the parabola
y2
= 4ax is
[ )
ax
4
y
(
)
a
4
(
]
)
x
(
a
2
y 2
2
2









[T2
= SS1
] ... (i)
A = coefficient ofx2
= 4a2
2H = coefficient ofxy= –4
and B = coefficient of y2
=  2
– ( 2
– 4a ) = 4a .
Since the angle between the two lines of(1) is 45°, we have
1 = tan45° =
2
2 H AB
A B


 (A + B)2
= 4 (H2
–AB)  (4a2
+ 4 
a )2
= 4[a2 2
 – (4a)2
(4a )]
 0
a
a
6 2
2
2






 or 2
2
2
a
8
)
a
3
( 




The equation of required locus is (x + 3a)2
– y2
= 8a2
which is a hyperbola.
Alternate Solution
Equation ofanytangent to hyperbola y2
= 4ax is
y = mx + a/m
which passes through ( ,  )if
 = 
m + a/m or m2
 – m + a = 0 ... (1)
If m1
and m2
are roots of (1).
m1
+ m2
= 
/ and m1
m2
= 
/
a we have 1 = tan45° =
2
1
2
1
m
m
1
m
m


 (1 + m1
m2
)2
= (m1
– m2
)2
 (1 + m1
m2
)2
= (m1
+ m2
)2
– 4m1
m2
 (1 + a/ )2
= 2
)
/
( 
 – 4a/
 ( + a)2
=  2
– 4a or ( + 3a)2
– 2
 = 8a2
The required locus is (x + 3a)2
–y2
= 8a2
which is a hyperbola.

4. Example 7 :
Find the centre, eccentricity, foci, directories and the lengths ofthe transverseand conjugate axes
of the hyperbola, whose equation is (x – 1)2
–2 (y – 2)2
+ 6 = 0
Solution :
The equation ofthe hyperbola can be written as (x – 1)2
– 2(y–2)2
+ 6 = 0
or
   
1
3
)
2
y
(
6
)
1
x
(
– 2
2
2
2




or
   
2 2
2 2
Y x
1
3 6
 
Where Y= (y–2) and x = (x–1) ... (1)
 centre: X = 0, Y = 0 i.e., (x – 1) = 0, x = 1 & (y – 2) = 0, y = 2.
So a = 3 and b = 6 so transverse axis = 3
2 , and conjugate axis = 6
2 .
Also b2
= a2
(e2
– 1)
 6 = 3 (e2
– 1) i.e., e = 3
In (X, Y) coordinates, fociare (0,  ae)
i.e., (0,  3).  foci are (1, 2  3) i.e., (1, 5) and (1, –1)
Equations ofdirectories,Y=  a/e.
 directrices y – 2 =  1
3
/
3 
 or y = 3, y = 1
Example 8 :
Find the equationand angle between the asymptotes ofthe hyperbola
x2
+ 2xy – 3y2
+ x + 7y + 9 = 0
Solution :
Let the combined equation of asymptotes x2
+ 2xy – 3y2
+ x + 7y +  = 0
Ifit representspair ofstraight lines
abc + 2fgh – af 2
– bg2
– ch2
= 0
 = –23/16
 Asymptotes x2
+ 2xy – 3y2
+ 7y – 23/16 = 0
Required angle =  tan–1
2.
Example 9 :
Prove that thelocus ofa point whose chord ofcontact touches the circle inscribed on the straight
line joining the foci of the hyperbola x2
/a2
– y2
/ b2
= 1 as diameter is x2
/a4
+ y2
/ b4
= 1/(a2
+ b2
).
Solution :
Circle on the join of foci (ae, 0) and (–ae, 0) diameter is (x – ae) (x + ae) + (y – 0) (y – 0) = 0
i.e., x2
+ y2
= a2
e2
= a2
+ b2
... (i) [ a2
e2
= a2
+ b2
]
Let chord ofcontact ofP (x1
, y1
) touchthe cirlce (i)
Equation ofchord ofcontact ofP is [T = 0]
xx1
/a2
– yy1
/b2
= 1 i.e., b2
x1
x–a2
y1
y – a2
b2
= 0 ... (ii)

)
b
a
(
)
y
a
x
b
(
b
a 2
2
2
1
4
2
1
4
2
2




Hence locus of P (x1
, y1
) is (b4
x2
+ a4
y2
) (a2
+ b2
) = a4
b4
.

5. Example 10 :
An ellipse has eccentricity
2
1
and one focus at the point 





1
,
2
1
P . Its one directrix is common
tangent to the circle x2
+ y2
= 1 and the hyperbola x2
– y2
= 1, nearer to P. The equation of the
ellipse inthe standard form.
Solution :
The circle x2
+ y2
= 1 and the hyperbola x2
– y2
= 1 touch eachother at the points ( 1, 0)and the
common tangent at these point are x =  1. Since x = 1 is nearer to the focus 





1
,
2
1
P , this is the
directrix ofthe required ellipse.
Therefore, the major axis is parallel to the axis passing through the focus 





1
,
2
1
P . Hence the
equation ofthe major axis is y= 1.
Let a bethelengthofthesemimajor axis ofthe ellipse and let thecoordinates ofthe centreC ofthe
ellipse be )
1
,
( .
Then CP = 

2
1
= a.e = a ×
2
1
... (i) 






2
1
e

and the distance ofthe directrix fromthe centre =
e
a
.
 1 –  = a
2
e
a
 ... (ii)
x – y = 1
2 2
x = 1
x + y = 1
2 2
From(i) and (ii) we get a =
1 1
ande
3 2
 .
Ifb is the length of the semi minor axis ofthe ellipse, then b2
= a2
(1 – e2
)
 b2
=
12
1
4
1
1
9
1








Hence the required equation oftheellipse is 1
12
1
)
1
y
(
3
1
3
1
x 2
2
2
















or
2
3
1
x
9 





 + 12(y – 1)2
= 1

6. SOLVED OBJECTIVE EXAMPLES
Example 1 :
The equation ofa line passing throughthe centre ofarectangular hyperbola is x– y– 1 =0. Ifone
ofits asymptotes is 3x – 4y– 6 = 0, the equation of the other asymptote is
(A) 4x – 3y + 17 = 0 (B) –4x – 3y + 17 = 0
(C) –4x + 3y + 1 = 0 (D) 4x + 3y + 17 = 0
Solution :
Weknowthat asymptotesofrectangularhyperbolaaremutuallyperpendicular,thusotherasymptote
should be 4x+ 3y+  = 0. Intersection point ofasymptotes is also the centre ofthe hyperbola.
Hence intersection point of4x + 3y +  = 0 and 3x – 4y– 6 = 0 should lie on the line x–y–1 = 0,
using it  canbe easilyobtained.
Hence (D) is the correct answer.
Example 2 :
The locus ofthe middle points ofchords of hyperbola 3x2
– 2y2
+ 4x – 6y= 0 parallelto y= 2x is
(A) 3x – 4y = 4 (B) 3x – 4y + 4 = 0
(C) 4x – 4y = 3 (D) 3x – 4y = 2
Solution :
Let the mid point be (h, k). Equation ofa chord whose mid point is (h, k) would be T = S1
or 3x h – 2yk + 2(x + h) – 3(y+k) = 3h2
– 2k2
+ 4h – 6k
 x (3h + 2) –y (2k + 3) – (2h + 3k) – 3h2
+ 2k2
= 0
Its slope is
3
k
2
2
h
3


= 2 (given)
 3h = 4k + 4
 Required locus is 3x – 4y= 4
Hence (A) is the correct answer.
Example 3 :
The tangent at a point P on the hyperbola 2
2
2
2
b
y
a
x
 =1 meets one of the directrix in F. If PF
subtends an angle  at the corresponding focus, then  equals
(A) 4
/
 (B) 2
/

(C) 4
/
3 (D) 
Solution :
Let directrix be x= a/e and focus be S(ae, 0). Let P (a 
sec , b 
tan ) be any point onthe curve.
Equation of tangent at P is
b
tan
y
a
sec
x 


= 1. Let F be the intersection point of tangent of
directrix, then F = 








tan
e
)
e
(sec
b
,
e
/
a

7.  SF PS
2
b(sec e) b tan
m ,m
etan (a 1) a(sec e)
 
 
      mSF
.mPS
= –1
Hence (B) is the correct answer.
Example 4 :
The line lx + my+ n = 0 will be a normal to the hyperbola b2
x2
– a2
y2
= a2
b2
if
(A) 2
2
2
2
2
2
2
2
n
)
b
a
(
m
b
a 


l
(B)
2 2 2 2 2
2 2 2
a b (a b )
m n

 
l
(C)
n
)
b
a
(
m
b
a 2
2
2
2
2
2
2



l
(D) noneofthese
Solution :
Equation ofnormalat (a sec, btan) is
ax cos + by cot  = a2
+ b2
Comparing it with lx + my + n = 0 we get
l

cos
a
=
2 2
bcot (a b )
m n
 


 cos =
nb
)
b
a
(
m
cot
and
an
)
b
a
( 2
2
2
2






l
 sin =
am
bl
Thus 2
2
2
2
m
a
b l
+ 1
n
a
)
b
a
(
2
2
2
2
2
2


l
or, 2
2
2
2
2
2
2
2
n
)
b
a
(
m
b
a 


l
Hence (B) is the correct answer.
Example 5 :
If (a sec, btan) and (asec , btan ) be the coordinate of the ends of a focal chord of
2
2
2
2
b
y
a
x
 = 1, then tan
2
tan
2


equals to
(A)
1
e
1
e


(B)
e
1
e
1


(C)
c
1
e
1


(D)
1
e
1
e


Solution :
Equation ofchord connecting the points (asec, b tan) and (asec, b 
tan ) is

8. x y
cos sin cos
a 2 b 2 2
      
     
 
     
     
Ifit passes through (ae, 0);we have, ecos cos
2 2
    
   

   
   
 e =
cos 1 tan . tan
2 2 2
1 tan tan
cos
2 2
2
  
   

 
  
 
  
  
 
 
 tan
e
1
e
1
2
tan
.
2 




Hence (B) is the correct answer.
Example 6 :
The point ofintersectionofthe curves whose parametric equations arex= t3
+ 1, y= 2t andx= 2s,
y =
2
s
, is given by
(A) (1, –3) (B) (2, 2)
(C) (–2, 4) (D) (1, 2)
Solution :
x = t2
+ 1, y = 2t  x – 1 = y2
/2
x = 2s, y = 2/s  xy = 4
For the point ofintersection we have 0
16
y
4
y
4
y
1
y
4 3
2





  y= 2  x = 2
Hence (B) is the correct answer.
Example 7 :
The equation ofthe hyperbola whose fociare (6, 5), (–4, 5) and eccentricity
4
5
is
(A) 1
9
)
5
y
(
16
)
1
x
( 2
2




(B) 1
9
y
16
x 2
2


(C) 1
9
)
5
y
(
16
)
1
x
( 2
2





(D) None ofthese
Solution :
S1  (6, 5); S2 (–4, 5), e = 5/4
S1
S2
= 10  2ae = 10  a = 4 and b2
= a2
(e2
– 1) = 16 9
1
16
25








Centre of the hyperbola is (1, 5)
 Equation ofrequiredhyperbola is 1
9
)
5
y
(
16
)
1
x
( 2
2





9. Example 8 :
The equation (x –  )2
+ (y –  )2
= k(lx + my + n)2
represents
(A) a parabola for k < (l2
+ m2
)–1
(B) an ellipse for 0 < k < (l2
+ m2
)–1
(C) a hyperbola for k > (l2
+ m2
)–1
(D) a point circle for k = 0.
Solution :
(x –  )2
+ 2
)
( 

 = k (lx + my + n)2
= k (l2
+ m2
)
2
2
2
m
n
my
x











l
l

PS
k
PM
 (l2
+ m2
). If k(l2
+ m2
) = 1, P lies on parabola
If k(l2
+ m2
) < 1, P lies on ellipse
If k(l2
+ m2
) > 1, P lies on hyperbola
Ifk = 0, P lies on a point circle
Hence (B), (C), (D) are correct.
Example 9 :
The point on the hyperbola 1
18
y
24
x 2
2

 which is nearest to the line 3x + 2y+ 1 = 0 is
(A) (6, 3) (B) (–6, 3)
(C) (6, –3) (D) (–6, –3)
Solution :
Equation oftangent is )
tan
18
,
sec
24
( 
 is
1
18
tan
y
24
sec
x



 , then point is nearest to the line 3x + 2y + 1 = 0.
so its slope = –
2
3

2
3
tan
18
24
sec




 sin = –
3
1
Hence the point is (6, –3)
Hence (C) is the correct answer.
Example 10 :
The locus ofa point, fromwhere tangents to the rectangularhyperbola x2
– y2
= a2
containanangle
of45° is
(A) (x2
+ y2
) + a2
(x2
– y2
) = 4a4
(B) 2(x2
+ y2
) + 4a2
(x2
– y2
) = 4a2
(C) (x2
+ y2
) + 4a2
(x2
– y2
) = 4a4
(D) (x2
+ y2
) + a2
(x2
– y2
) = a4
Solution :
Let y= mx 2
2
2
a
a
m 
 be two tangent and passes through (h, k) then (k – mk)2
= m2
a2
– a2
 m2
(h2
– a2
) – 2khm + k2
+ a2
= 0
 m1
+ m2
= 2
2
a
h
kh
2

and m1
m2
=
2
2
2
2
a
h
a
k

 , using tan45°=
1 2
1 2
m m
1 m m


Hence (C) is the correct answer.