Hyperbola Solved Examples | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


Solved Examples on Hyperbola 
JEE Mains 
Q1.  The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? 
and with eccentricity v ?? is 
(a) ?? ?? - ?? ?? = ?? 
(b) ???? = ?? 
(c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? 
(d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? 
Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 
From definition, ???? = ?????? 
v( ?? - 1)
2
+ ( ?? + 1)
2
=
v 2 · ( ?? - ?? + 1)
v 1
2
+ 1
2
 
? ( ?? - 1)
2
+ ( ?? + 1)
2
= ( ?? - ?? + 1)
2
? 2???? - 4?? + 4?? + 1 = 0, which is the required 
equation of conic (Rectangular hyperbola) 
Q2.  
The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are 
(a) ( ±?? , ?? ) 
(b) ( ?? , ±?? ) 
(c) ( ±?? , ?? ) 
(d) ( ?? , ±?? ) 
Ans: (c) The equation of hyperbola is 
?? 2
16
-
?? 2
9
= 1. 
Now, ?? 2
= ?? 2
( ?? 2
- 1) ? 9 = 16( ?? 2
- 1) ? ?? =
5
4
. Hence foci are ( ±???? , 0) = (±4.
5
4
, 0) i.e., 
( ±5,0) 
Q3. If the foci of the ellipse 
?? ?? ????
+
?? ?? ?? ?? = ?? and the hyperbola 
?? ?? ??????
-
?? ?? ????
=
?? ????
 coincide, 
then the value of ?? ?? is 
(a) 1 
(b) 5 
(c) 7 
(d) 9 
Ans: (c) For hyperbola, 
?? 2
144
-
?? 2
81
=
1
25
 
?? =
v
144
25
, ?? =
v
81
25
, ?? 1
=
v
1 +
?? 2
?? 2
=
v
1 +
81
144
=
v
225
144
=
5
4
 
Therefore foci = ( ±?? ?? 1
, 0) = (±
12
5
·
5
4
, 0) = ( ±3,0) . Therefore foci of ellipse i.e., ( ±4?? , 0) =
( ±3,0) 
(For ellipse ?? = 4 ) 
? ?? =
3
4
, Hence ?? 2
= 16(1 -
9
16
) = 7. 
Page 2


Solved Examples on Hyperbola 
JEE Mains 
Q1.  The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? 
and with eccentricity v ?? is 
(a) ?? ?? - ?? ?? = ?? 
(b) ???? = ?? 
(c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? 
(d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? 
Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 
From definition, ???? = ?????? 
v( ?? - 1)
2
+ ( ?? + 1)
2
=
v 2 · ( ?? - ?? + 1)
v 1
2
+ 1
2
 
? ( ?? - 1)
2
+ ( ?? + 1)
2
= ( ?? - ?? + 1)
2
? 2???? - 4?? + 4?? + 1 = 0, which is the required 
equation of conic (Rectangular hyperbola) 
Q2.  
The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are 
(a) ( ±?? , ?? ) 
(b) ( ?? , ±?? ) 
(c) ( ±?? , ?? ) 
(d) ( ?? , ±?? ) 
Ans: (c) The equation of hyperbola is 
?? 2
16
-
?? 2
9
= 1. 
Now, ?? 2
= ?? 2
( ?? 2
- 1) ? 9 = 16( ?? 2
- 1) ? ?? =
5
4
. Hence foci are ( ±???? , 0) = (±4.
5
4
, 0) i.e., 
( ±5,0) 
Q3. If the foci of the ellipse 
?? ?? ????
+
?? ?? ?? ?? = ?? and the hyperbola 
?? ?? ??????
-
?? ?? ????
=
?? ????
 coincide, 
then the value of ?? ?? is 
(a) 1 
(b) 5 
(c) 7 
(d) 9 
Ans: (c) For hyperbola, 
?? 2
144
-
?? 2
81
=
1
25
 
?? =
v
144
25
, ?? =
v
81
25
, ?? 1
=
v
1 +
?? 2
?? 2
=
v
1 +
81
144
=
v
225
144
=
5
4
 
Therefore foci = ( ±?? ?? 1
, 0) = (±
12
5
·
5
4
, 0) = ( ±3,0) . Therefore foci of ellipse i.e., ( ±4?? , 0) =
( ±3,0) 
(For ellipse ?? = 4 ) 
? ?? =
3
4
, Hence ?? 2
= 16(1 -
9
16
) = 7. 
Q4. 
The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are 
(a) ?? = ±?? 
(b) ?? = ±?? 
(c) ?? = ±v ?? 
(d) ?? = ±v ?? 
Ans: (c)  ( ?? + 1)
2
- ?? 2
- 1 + 5 = 0 ? -
( ?? +1)
2
4
+
?? 2
4
= 1 
Equation of directrices of 
?? 2
?? 2
-
?? 2
?? 2
= 1 are ?? = ±
?? ?? 
Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ±
2
v 2
? ?? = ±v 2. 
Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? 
is 
(a) ???? v ?? 
(b) v ?? 
(c) ?? v ?? 
(d) ?? , v ?? 
Ans: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ?
?? 8
= sec ?? ,
?? 8
= tan ?? 
? sec
2
 ?? - tan
2
 ?? = 1 ?
?? 2
8
2
-
?? 2
8
2
= 1 
Here ?? = 8, ?? = 8. Now ?? = v1 +
?? 2
?? 2
= v1 +
8
2
8
2
= v 2 
? Distance between directrices =
2?? ?? =
2×8
v 2
= 8v 2. 
Q6. 
The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is 
(a) ( ?? , ?? ) 
(b) ( ?? , ?? ) 
(c) ( ?? , -?? ) 
(d) None of these 
Ans: (a) 
The equation of line and hyperbola are ?? = ?? - 1 
....(i) and 3?? 2
- 4?? 2
= 12 
.....(ii) 
From (i) and (ii), we get 3?? 2
- 4( ?? - 1)
2
= 12 
? 3?? 2
- 4( ?? 2
- 2?? + 1) = 12 or ?? 2
- 8?? + 16 = 0 ? ?? = 4 
From (i), ?? = 3 so points of contact is ( 4,3) 
Trick: Points of contact are (±
?? 2
?? v?? 2
?? 2
-?? 2
, ±
?? 2
v?? 2
?? 2
-?? 2
). 
Here ?? 2
= 4, ?? 2
= 3 and ?? = 1. So the required points of contact is ( 4,3) . 
Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola 
?? ?? ?? -
?? ?? ?? = ?? is 
parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is 
(a) ????
°
 
(b) ????
°
 
Page 3


Solved Examples on Hyperbola 
JEE Mains 
Q1.  The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? 
and with eccentricity v ?? is 
(a) ?? ?? - ?? ?? = ?? 
(b) ???? = ?? 
(c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? 
(d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? 
Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 
From definition, ???? = ?????? 
v( ?? - 1)
2
+ ( ?? + 1)
2
=
v 2 · ( ?? - ?? + 1)
v 1
2
+ 1
2
 
? ( ?? - 1)
2
+ ( ?? + 1)
2
= ( ?? - ?? + 1)
2
? 2???? - 4?? + 4?? + 1 = 0, which is the required 
equation of conic (Rectangular hyperbola) 
Q2.  
The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are 
(a) ( ±?? , ?? ) 
(b) ( ?? , ±?? ) 
(c) ( ±?? , ?? ) 
(d) ( ?? , ±?? ) 
Ans: (c) The equation of hyperbola is 
?? 2
16
-
?? 2
9
= 1. 
Now, ?? 2
= ?? 2
( ?? 2
- 1) ? 9 = 16( ?? 2
- 1) ? ?? =
5
4
. Hence foci are ( ±???? , 0) = (±4.
5
4
, 0) i.e., 
( ±5,0) 
Q3. If the foci of the ellipse 
?? ?? ????
+
?? ?? ?? ?? = ?? and the hyperbola 
?? ?? ??????
-
?? ?? ????
=
?? ????
 coincide, 
then the value of ?? ?? is 
(a) 1 
(b) 5 
(c) 7 
(d) 9 
Ans: (c) For hyperbola, 
?? 2
144
-
?? 2
81
=
1
25
 
?? =
v
144
25
, ?? =
v
81
25
, ?? 1
=
v
1 +
?? 2
?? 2
=
v
1 +
81
144
=
v
225
144
=
5
4
 
Therefore foci = ( ±?? ?? 1
, 0) = (±
12
5
·
5
4
, 0) = ( ±3,0) . Therefore foci of ellipse i.e., ( ±4?? , 0) =
( ±3,0) 
(For ellipse ?? = 4 ) 
? ?? =
3
4
, Hence ?? 2
= 16(1 -
9
16
) = 7. 
Q4. 
The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are 
(a) ?? = ±?? 
(b) ?? = ±?? 
(c) ?? = ±v ?? 
(d) ?? = ±v ?? 
Ans: (c)  ( ?? + 1)
2
- ?? 2
- 1 + 5 = 0 ? -
( ?? +1)
2
4
+
?? 2
4
= 1 
Equation of directrices of 
?? 2
?? 2
-
?? 2
?? 2
= 1 are ?? = ±
?? ?? 
Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ±
2
v 2
? ?? = ±v 2. 
Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? 
is 
(a) ???? v ?? 
(b) v ?? 
(c) ?? v ?? 
(d) ?? , v ?? 
Ans: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ?
?? 8
= sec ?? ,
?? 8
= tan ?? 
? sec
2
 ?? - tan
2
 ?? = 1 ?
?? 2
8
2
-
?? 2
8
2
= 1 
Here ?? = 8, ?? = 8. Now ?? = v1 +
?? 2
?? 2
= v1 +
8
2
8
2
= v 2 
? Distance between directrices =
2?? ?? =
2×8
v 2
= 8v 2. 
Q6. 
The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is 
(a) ( ?? , ?? ) 
(b) ( ?? , ?? ) 
(c) ( ?? , -?? ) 
(d) None of these 
Ans: (a) 
The equation of line and hyperbola are ?? = ?? - 1 
....(i) and 3?? 2
- 4?? 2
= 12 
.....(ii) 
From (i) and (ii), we get 3?? 2
- 4( ?? - 1)
2
= 12 
? 3?? 2
- 4( ?? 2
- 2?? + 1) = 12 or ?? 2
- 8?? + 16 = 0 ? ?? = 4 
From (i), ?? = 3 so points of contact is ( 4,3) 
Trick: Points of contact are (±
?? 2
?? v?? 2
?? 2
-?? 2
, ±
?? 2
v?? 2
?? 2
-?? 2
). 
Here ?? 2
= 4, ?? 2
= 3 and ?? = 1. So the required points of contact is ( 4,3) . 
Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola 
?? ?? ?? -
?? ?? ?? = ?? is 
parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is 
(a) ????
°
 
(b) ????
°
 
(c) ????
°
 
(d) ????
°
 
Ans: (c) Here ?? = 2sec ?? and ?? = 3tan ?? 
Differentiating w.r.t. ?? 
????
????
= 2sec ?? tan ?? and 
????
????
= 3sec
2
 ?? 
? Gradient of tangent 
????
????
=
???? /????
???? /????
=
3sec
2
 ?? 2sec ?? tan ?? ; ? 
????
????
=
3
2
cosec ?? 
But tangent is parallel to 3?? - ?? + 4 = 0; ? Gradient ?? = 3 
From (i) and (ii), 
3
2
cosec ?? = 3 ? cosec ?? = 2, ? ?? = 30
°
 
Q8.  If the normal at ' ?? ' on the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? meets transverse axis at ?? , 
then ???? · ?? '
?? = (Where ?? and ?? '
 are the vertices of the hyperbola) 
(a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) 
(b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) 
(c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) 
(d) None of these 
Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? +
???? cot ?? = ( ?? 2
+ ?? 2
) 
This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are ((
?? 2
+?? 2
?? )sec ?? , 0) 
and the co-ordinates of the vertices ?? and ?? '
 are ?? ( ?? , 0) and ?? '
( -?? , 0) respectively. 
? ???? · ?? '
?? = (-?? + (
?? 2
+ ?? 2
?? ) sec ?? ) (?? + (
?? 2
+ ?? 2
?? ) sec ?? ) = (
?? 2
+ ?? 2
?? )
2
sec
2
 ?? - ?? 2
= ( ?? ?? 2
)
2
sec
2
 ?? - ?? 2
= ?? 2
( ?? 4
sec
2
 ?? - 1) 
Q9.  What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose 
mid point is ( ?? , ?? ) 
(a) ?????? ?? - ?????? ?? = ???? 
(b) ?????? ?? - ???? ?? = ?????? 
(c) ?????? ?? + ???? ?? = ?????? 
(d) ???? ?? + ?????? ?? = ?????? 
Ans: (b) According to question, ?? = 25?? 2
- 16?? 2
- 400 = 0 
Equation of required chord is ?? 1
= ?? 
Here ?? 1
= 25( 5)
2
- 16( 3)
2
- 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1
- 16?? ?? 1
- 400, 
where ?? 1
= 5, ?? 1
= 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 
So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. 
 
Q10.  If the polar of a point w.r.t. 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? touches the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? , 
then the locus of the point is 
(a) Given hyperbola 
(b) Ellipse 
(c) Circle 
(d) None of these 
Page 4


Solved Examples on Hyperbola 
JEE Mains 
Q1.  The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? 
and with eccentricity v ?? is 
(a) ?? ?? - ?? ?? = ?? 
(b) ???? = ?? 
(c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? 
(d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? 
Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 
From definition, ???? = ?????? 
v( ?? - 1)
2
+ ( ?? + 1)
2
=
v 2 · ( ?? - ?? + 1)
v 1
2
+ 1
2
 
? ( ?? - 1)
2
+ ( ?? + 1)
2
= ( ?? - ?? + 1)
2
? 2???? - 4?? + 4?? + 1 = 0, which is the required 
equation of conic (Rectangular hyperbola) 
Q2.  
The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are 
(a) ( ±?? , ?? ) 
(b) ( ?? , ±?? ) 
(c) ( ±?? , ?? ) 
(d) ( ?? , ±?? ) 
Ans: (c) The equation of hyperbola is 
?? 2
16
-
?? 2
9
= 1. 
Now, ?? 2
= ?? 2
( ?? 2
- 1) ? 9 = 16( ?? 2
- 1) ? ?? =
5
4
. Hence foci are ( ±???? , 0) = (±4.
5
4
, 0) i.e., 
( ±5,0) 
Q3. If the foci of the ellipse 
?? ?? ????
+
?? ?? ?? ?? = ?? and the hyperbola 
?? ?? ??????
-
?? ?? ????
=
?? ????
 coincide, 
then the value of ?? ?? is 
(a) 1 
(b) 5 
(c) 7 
(d) 9 
Ans: (c) For hyperbola, 
?? 2
144
-
?? 2
81
=
1
25
 
?? =
v
144
25
, ?? =
v
81
25
, ?? 1
=
v
1 +
?? 2
?? 2
=
v
1 +
81
144
=
v
225
144
=
5
4
 
Therefore foci = ( ±?? ?? 1
, 0) = (±
12
5
·
5
4
, 0) = ( ±3,0) . Therefore foci of ellipse i.e., ( ±4?? , 0) =
( ±3,0) 
(For ellipse ?? = 4 ) 
? ?? =
3
4
, Hence ?? 2
= 16(1 -
9
16
) = 7. 
Q4. 
The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are 
(a) ?? = ±?? 
(b) ?? = ±?? 
(c) ?? = ±v ?? 
(d) ?? = ±v ?? 
Ans: (c)  ( ?? + 1)
2
- ?? 2
- 1 + 5 = 0 ? -
( ?? +1)
2
4
+
?? 2
4
= 1 
Equation of directrices of 
?? 2
?? 2
-
?? 2
?? 2
= 1 are ?? = ±
?? ?? 
Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ±
2
v 2
? ?? = ±v 2. 
Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? 
is 
(a) ???? v ?? 
(b) v ?? 
(c) ?? v ?? 
(d) ?? , v ?? 
Ans: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ?
?? 8
= sec ?? ,
?? 8
= tan ?? 
? sec
2
 ?? - tan
2
 ?? = 1 ?
?? 2
8
2
-
?? 2
8
2
= 1 
Here ?? = 8, ?? = 8. Now ?? = v1 +
?? 2
?? 2
= v1 +
8
2
8
2
= v 2 
? Distance between directrices =
2?? ?? =
2×8
v 2
= 8v 2. 
Q6. 
The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is 
(a) ( ?? , ?? ) 
(b) ( ?? , ?? ) 
(c) ( ?? , -?? ) 
(d) None of these 
Ans: (a) 
The equation of line and hyperbola are ?? = ?? - 1 
....(i) and 3?? 2
- 4?? 2
= 12 
.....(ii) 
From (i) and (ii), we get 3?? 2
- 4( ?? - 1)
2
= 12 
? 3?? 2
- 4( ?? 2
- 2?? + 1) = 12 or ?? 2
- 8?? + 16 = 0 ? ?? = 4 
From (i), ?? = 3 so points of contact is ( 4,3) 
Trick: Points of contact are (±
?? 2
?? v?? 2
?? 2
-?? 2
, ±
?? 2
v?? 2
?? 2
-?? 2
). 
Here ?? 2
= 4, ?? 2
= 3 and ?? = 1. So the required points of contact is ( 4,3) . 
Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola 
?? ?? ?? -
?? ?? ?? = ?? is 
parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is 
(a) ????
°
 
(b) ????
°
 
(c) ????
°
 
(d) ????
°
 
Ans: (c) Here ?? = 2sec ?? and ?? = 3tan ?? 
Differentiating w.r.t. ?? 
????
????
= 2sec ?? tan ?? and 
????
????
= 3sec
2
 ?? 
? Gradient of tangent 
????
????
=
???? /????
???? /????
=
3sec
2
 ?? 2sec ?? tan ?? ; ? 
????
????
=
3
2
cosec ?? 
But tangent is parallel to 3?? - ?? + 4 = 0; ? Gradient ?? = 3 
From (i) and (ii), 
3
2
cosec ?? = 3 ? cosec ?? = 2, ? ?? = 30
°
 
Q8.  If the normal at ' ?? ' on the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? meets transverse axis at ?? , 
then ???? · ?? '
?? = (Where ?? and ?? '
 are the vertices of the hyperbola) 
(a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) 
(b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) 
(c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) 
(d) None of these 
Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? +
???? cot ?? = ( ?? 2
+ ?? 2
) 
This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are ((
?? 2
+?? 2
?? )sec ?? , 0) 
and the co-ordinates of the vertices ?? and ?? '
 are ?? ( ?? , 0) and ?? '
( -?? , 0) respectively. 
? ???? · ?? '
?? = (-?? + (
?? 2
+ ?? 2
?? ) sec ?? ) (?? + (
?? 2
+ ?? 2
?? ) sec ?? ) = (
?? 2
+ ?? 2
?? )
2
sec
2
 ?? - ?? 2
= ( ?? ?? 2
)
2
sec
2
 ?? - ?? 2
= ?? 2
( ?? 4
sec
2
 ?? - 1) 
Q9.  What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose 
mid point is ( ?? , ?? ) 
(a) ?????? ?? - ?????? ?? = ???? 
(b) ?????? ?? - ???? ?? = ?????? 
(c) ?????? ?? + ???? ?? = ?????? 
(d) ???? ?? + ?????? ?? = ?????? 
Ans: (b) According to question, ?? = 25?? 2
- 16?? 2
- 400 = 0 
Equation of required chord is ?? 1
= ?? 
Here ?? 1
= 25( 5)
2
- 16( 3)
2
- 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1
- 16?? ?? 1
- 400, 
where ?? 1
= 5, ?? 1
= 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 
So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. 
 
Q10.  If the polar of a point w.r.t. 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? touches the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? , 
then the locus of the point is 
(a) Given hyperbola 
(b) Ellipse 
(c) Circle 
(d) None of these 
Ans: (a) Let ( ?? 1
, ?? 1
) be the given point. 
Its polar w.r.t. 
?? 2
?? 2
+
?? 2
?? 2
= 1 is 
?? ?? 1
?? 2
+
?? ?? 1
?? 2
= 1 i.e., ?? =
?? 2
?? 1
(1 -
?? ?? 1
?? 2
) = -
?? 2
?? 1
?? 2
?? 1
?? +
?? 2
?? 1
 
This touches 
?? 2
?? 2
-
?? 2
?? 2
= 1 if (
?? 2
?? 1
)
2
= ?? 2
· (
?? 2
?? 1
?? 2
?? 1
) - ?? 2
?
?? 4
?? 1
2
=
?? 2
?? 4
?? 1
2
?? 4
?? 1
2
- ?? 2
?
?? 2
?? 1
2
=
?? 2
?? 1
2
?? 2
?? 1
2
- 1 ?
?? 1
2
?? 2
-
?? 1
2
?? 2
= 1 
? Locus of ( ?? 1
, ?? 1
) is 
?? 2
?? 2
-
?? 2
?? 2
= 1. Which is the same hyperbola. 
Q11.  The combined equation of the asymptotes of the hyperbola ?? ?? ?? + ?? ???? +
?? ?? ?? + ?? ?? + ?? ?? = ?? 
(a) ?? ?? ?? + ?? ???? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ???? + ?? ?? ?? - ?? ?? + ?? ?? + ?? = ?? = ?? 
(c) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? - ?? = ?? 
(d) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? + ?? = ?? 
Ans: (d) Given, equation of hyperbola 2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? = 0 and equation of 
asymptotes 
2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? + ?? = 0 
(i) which is the equation of a pair of straight lines. We know that the standard equation of a 
pair of straight lines is ?? ?? 2
+ 2h???? + ?? ?? 2
+ 2???? + 2???? + ?? = 0 
Comparing equation (i) with standard equation, we get ?? = 2, ?? = 2, h =
5
2
, ?? = 2, ?? =
5
2
 and 
?? = ?? . 
We also know that the condition for a pair of straight lines is ?????? + 2???? h - ?? ?? 2
- ?? ?? 2
-
?? h
2
= 0. 
Therefore, 4?? + 25 -
25
2
- 8 -
25
4
?? = 0 or 
-9?? 4
+
9
2
= 0 or ?? = 2 
Substituting value of ?? in equation (i), we get 2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? + 2 = 0. 
Q12.  If the normal at (???? ,
?? ?? ) on the curve ???? = ?? ?? meets the curve again in ?? '
, then 
(a) ?? '
= -
?? ?? ?? 
(b) ?? '
= -
?? ?? 
(c) ?? '
=
?? ?? ?? 
(d) ?? '?? = -
?? ?? ?? 
Ans: (a) The equation of the tangent at (???? ,
?? ?? ) is ???? = ?? 3
?? - ?? ?? 4
+ ?? 
If it passes through (?? ?? '
,
?? ?? '
) then 
?
????
?? '
= ?? 3
?? ?? '
- ?? ?? 4
+ ?? ? ?? = ?? 3
?? '2
- ?? 4
?? '
+ ?? '
? ?? - ?? '
= ?? 3
?? '
( ?? '
- ?? ) ? ?? '
= -
1
?? 3
 
Q13.  
A variable straight line of slope 4 intersects the hyperbola ???? = ?? at two points. 
The locus of the point which divides the line segment between these two points 
in the ratio ?? : ?? is 
(a) ???? ?? ?? + ???? ???? + ?? ?? = ?? 
(b) ???? ?? ?? - ???? ???? + ?? ?? = ?? 
(c) ???? ?? ?? + ???? ???? + ?? ?? = ?? 
(d) None of these 
Page 5


Solved Examples on Hyperbola 
JEE Mains 
Q1.  The equation of the conic with focus at ( ?? , -?? ) , directrix along ?? - ?? + ?? = ?? 
and with eccentricity v ?? is 
(a) ?? ?? - ?? ?? = ?? 
(b) ???? = ?? 
(c) ?? ???? - ?? ?? + ?? ?? + ?? = ?? 
(d) ?? ???? + ?? ?? - ?? ?? - ?? = ?? 
Ans: (c) Here, focus ( ?? ) = ( 1, -1) , eccentricity ( ?? ) = v 2 
From definition, ???? = ?????? 
v( ?? - 1)
2
+ ( ?? + 1)
2
=
v 2 · ( ?? - ?? + 1)
v 1
2
+ 1
2
 
? ( ?? - 1)
2
+ ( ?? + 1)
2
= ( ?? - ?? + 1)
2
? 2???? - 4?? + 4?? + 1 = 0, which is the required 
equation of conic (Rectangular hyperbola) 
Q2.  
The foci of the hyperbola ?? ?? ?? - ???? ?? ?? = ?????? are 
(a) ( ±?? , ?? ) 
(b) ( ?? , ±?? ) 
(c) ( ±?? , ?? ) 
(d) ( ?? , ±?? ) 
Ans: (c) The equation of hyperbola is 
?? 2
16
-
?? 2
9
= 1. 
Now, ?? 2
= ?? 2
( ?? 2
- 1) ? 9 = 16( ?? 2
- 1) ? ?? =
5
4
. Hence foci are ( ±???? , 0) = (±4.
5
4
, 0) i.e., 
( ±5,0) 
Q3. If the foci of the ellipse 
?? ?? ????
+
?? ?? ?? ?? = ?? and the hyperbola 
?? ?? ??????
-
?? ?? ????
=
?? ????
 coincide, 
then the value of ?? ?? is 
(a) 1 
(b) 5 
(c) 7 
(d) 9 
Ans: (c) For hyperbola, 
?? 2
144
-
?? 2
81
=
1
25
 
?? =
v
144
25
, ?? =
v
81
25
, ?? 1
=
v
1 +
?? 2
?? 2
=
v
1 +
81
144
=
v
225
144
=
5
4
 
Therefore foci = ( ±?? ?? 1
, 0) = (±
12
5
·
5
4
, 0) = ( ±3,0) . Therefore foci of ellipse i.e., ( ±4?? , 0) =
( ±3,0) 
(For ellipse ?? = 4 ) 
? ?? =
3
4
, Hence ?? 2
= 16(1 -
9
16
) = 7. 
Q4. 
The equations of the directrices of the conic ?? ?? + ?? ?? - ?? ?? + ?? = ?? are 
(a) ?? = ±?? 
(b) ?? = ±?? 
(c) ?? = ±v ?? 
(d) ?? = ±v ?? 
Ans: (c)  ( ?? + 1)
2
- ?? 2
- 1 + 5 = 0 ? -
( ?? +1)
2
4
+
?? 2
4
= 1 
Equation of directrices of 
?? 2
?? 2
-
?? 2
?? 2
= 1 are ?? = ±
?? ?? 
Here ?? = 2, ?? = v 1 + 1 = v 2. Hence, ?? = ±
2
v 2
? ?? = ±v 2. 
Q5. The distance between the directrices of the hyperbola ?? = ???????? ?? , ?? = ???????? ?? 
is 
(a) ???? v ?? 
(b) v ?? 
(c) ?? v ?? 
(d) ?? , v ?? 
Ans: Equation of hyperbola is ?? = 8sec ?? , ?? = 8tan ?? ?
?? 8
= sec ?? ,
?? 8
= tan ?? 
? sec
2
 ?? - tan
2
 ?? = 1 ?
?? 2
8
2
-
?? 2
8
2
= 1 
Here ?? = 8, ?? = 8. Now ?? = v1 +
?? 2
?? 2
= v1 +
8
2
8
2
= v 2 
? Distance between directrices =
2?? ?? =
2×8
v 2
= 8v 2. 
Q6. 
The points of contact of the line ?? = ?? - ?? with ?? ?? ?? - ?? ?? ?? = ???? is 
(a) ( ?? , ?? ) 
(b) ( ?? , ?? ) 
(c) ( ?? , -?? ) 
(d) None of these 
Ans: (a) 
The equation of line and hyperbola are ?? = ?? - 1 
....(i) and 3?? 2
- 4?? 2
= 12 
.....(ii) 
From (i) and (ii), we get 3?? 2
- 4( ?? - 1)
2
= 12 
? 3?? 2
- 4( ?? 2
- 2?? + 1) = 12 or ?? 2
- 8?? + 16 = 0 ? ?? = 4 
From (i), ?? = 3 so points of contact is ( 4,3) 
Trick: Points of contact are (±
?? 2
?? v?? 2
?? 2
-?? 2
, ±
?? 2
v?? 2
?? 2
-?? 2
). 
Here ?? 2
= 4, ?? 2
= 3 and ?? = 1. So the required points of contact is ( 4,3) . 
Q7. If the tangent at the point ( ???????? ?? , ???????? ?? ) on the hyperbola 
?? ?? ?? -
?? ?? ?? = ?? is 
parallel to ?? ?? - ?? + ?? = ?? , then the value of ?? is 
(a) ????
°
 
(b) ????
°
 
(c) ????
°
 
(d) ????
°
 
Ans: (c) Here ?? = 2sec ?? and ?? = 3tan ?? 
Differentiating w.r.t. ?? 
????
????
= 2sec ?? tan ?? and 
????
????
= 3sec
2
 ?? 
? Gradient of tangent 
????
????
=
???? /????
???? /????
=
3sec
2
 ?? 2sec ?? tan ?? ; ? 
????
????
=
3
2
cosec ?? 
But tangent is parallel to 3?? - ?? + 4 = 0; ? Gradient ?? = 3 
From (i) and (ii), 
3
2
cosec ?? = 3 ? cosec ?? = 2, ? ?? = 30
°
 
Q8.  If the normal at ' ?? ' on the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? meets transverse axis at ?? , 
then ???? · ?? '
?? = (Where ?? and ?? '
 are the vertices of the hyperbola) 
(a) ?? ?? ( ?? ?? ?????? ?? ?? - ?? ) 
(b) ( ?? ?? ?? ?? ?????? ?? ?? - ?? ) 
(c) ?? ?? ( ?? - ?? ?? ?????? ?? ?? ) 
(d) None of these 
Ans: (a) The equation of normal at ( ?? sec ?? , ?? tan ?? ) to the given hyperbola is ???? cos ?? +
???? cot ?? = ( ?? 2
+ ?? 2
) 
This meets the transverse axis i.e., ?? -axis at ?? . So the co-ordinates of ?? are ((
?? 2
+?? 2
?? )sec ?? , 0) 
and the co-ordinates of the vertices ?? and ?? '
 are ?? ( ?? , 0) and ?? '
( -?? , 0) respectively. 
? ???? · ?? '
?? = (-?? + (
?? 2
+ ?? 2
?? ) sec ?? ) (?? + (
?? 2
+ ?? 2
?? ) sec ?? ) = (
?? 2
+ ?? 2
?? )
2
sec
2
 ?? - ?? 2
= ( ?? ?? 2
)
2
sec
2
 ?? - ?? 2
= ?? 2
( ?? 4
sec
2
 ?? - 1) 
Q9.  What will be equation of that chord of hyperbola ???? ?? ?? - ???? ?? ?? = ?????? , whose 
mid point is ( ?? , ?? ) 
(a) ?????? ?? - ?????? ?? = ???? 
(b) ?????? ?? - ???? ?? = ?????? 
(c) ?????? ?? + ???? ?? = ?????? 
(d) ???? ?? + ?????? ?? = ?????? 
Ans: (b) According to question, ?? = 25?? 2
- 16?? 2
- 400 = 0 
Equation of required chord is ?? 1
= ?? 
Here ?? 1
= 25( 5)
2
- 16( 3)
2
- 400 = 625 - 144- 400 = 81 and ?? = 25?? ?? 1
- 16?? ?? 1
- 400, 
where ?? 1
= 5, ?? 1
= 3 ? 25?? ( 5)- 16?? ( 3)- 400 = 125?? - 48?? - 400 
So, from (i) required chord is 125?? - 48?? - 400 = 81 ? 125?? - 48?? = 481. 
 
Q10.  If the polar of a point w.r.t. 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? touches the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? , 
then the locus of the point is 
(a) Given hyperbola 
(b) Ellipse 
(c) Circle 
(d) None of these 
Ans: (a) Let ( ?? 1
, ?? 1
) be the given point. 
Its polar w.r.t. 
?? 2
?? 2
+
?? 2
?? 2
= 1 is 
?? ?? 1
?? 2
+
?? ?? 1
?? 2
= 1 i.e., ?? =
?? 2
?? 1
(1 -
?? ?? 1
?? 2
) = -
?? 2
?? 1
?? 2
?? 1
?? +
?? 2
?? 1
 
This touches 
?? 2
?? 2
-
?? 2
?? 2
= 1 if (
?? 2
?? 1
)
2
= ?? 2
· (
?? 2
?? 1
?? 2
?? 1
) - ?? 2
?
?? 4
?? 1
2
=
?? 2
?? 4
?? 1
2
?? 4
?? 1
2
- ?? 2
?
?? 2
?? 1
2
=
?? 2
?? 1
2
?? 2
?? 1
2
- 1 ?
?? 1
2
?? 2
-
?? 1
2
?? 2
= 1 
? Locus of ( ?? 1
, ?? 1
) is 
?? 2
?? 2
-
?? 2
?? 2
= 1. Which is the same hyperbola. 
Q11.  The combined equation of the asymptotes of the hyperbola ?? ?? ?? + ?? ???? +
?? ?? ?? + ?? ?? + ?? ?? = ?? 
(a) ?? ?? ?? + ?? ???? + ?? ?? ?? = ?? 
(b) ?? ?? ?? + ?? ???? + ?? ?? ?? - ?? ?? + ?? ?? + ?? = ?? = ?? 
(c) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? - ?? = ?? 
(d) ?? ?? ?? + ?? ???? + ?? ?? ?? + ?? ?? + ?? ?? + ?? = ?? 
Ans: (d) Given, equation of hyperbola 2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? = 0 and equation of 
asymptotes 
2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? + ?? = 0 
(i) which is the equation of a pair of straight lines. We know that the standard equation of a 
pair of straight lines is ?? ?? 2
+ 2h???? + ?? ?? 2
+ 2???? + 2???? + ?? = 0 
Comparing equation (i) with standard equation, we get ?? = 2, ?? = 2, h =
5
2
, ?? = 2, ?? =
5
2
 and 
?? = ?? . 
We also know that the condition for a pair of straight lines is ?????? + 2???? h - ?? ?? 2
- ?? ?? 2
-
?? h
2
= 0. 
Therefore, 4?? + 25 -
25
2
- 8 -
25
4
?? = 0 or 
-9?? 4
+
9
2
= 0 or ?? = 2 
Substituting value of ?? in equation (i), we get 2?? 2
+ 5???? + 2?? 2
+ 4?? + 5?? + 2 = 0. 
Q12.  If the normal at (???? ,
?? ?? ) on the curve ???? = ?? ?? meets the curve again in ?? '
, then 
(a) ?? '
= -
?? ?? ?? 
(b) ?? '
= -
?? ?? 
(c) ?? '
=
?? ?? ?? 
(d) ?? '?? = -
?? ?? ?? 
Ans: (a) The equation of the tangent at (???? ,
?? ?? ) is ???? = ?? 3
?? - ?? ?? 4
+ ?? 
If it passes through (?? ?? '
,
?? ?? '
) then 
?
????
?? '
= ?? 3
?? ?? '
- ?? ?? 4
+ ?? ? ?? = ?? 3
?? '2
- ?? 4
?? '
+ ?? '
? ?? - ?? '
= ?? 3
?? '
( ?? '
- ?? ) ? ?? '
= -
1
?? 3
 
Q13.  
A variable straight line of slope 4 intersects the hyperbola ???? = ?? at two points. 
The locus of the point which divides the line segment between these two points 
in the ratio ?? : ?? is 
(a) ???? ?? ?? + ???? ???? + ?? ?? = ?? 
(b) ???? ?? ?? - ???? ???? + ?? ?? = ?? 
(c) ???? ?? ?? + ???? ???? + ?? ?? = ?? 
(d) None of these 
Ans: (a) 
Let ?? ( h, ?? ) be any point on the locus. Equation of the line through ?? and having slope 4 is 
?? - ?? = 4( ?? - h) 
Suppose this meets ???? = 1 
..(ii) in ?? ( ?? 1
, ?? 1
) and ?? ( ?? 2
, ?? 2
) 
Eliminating ?? between (i) and (ii), we get 
1
?? - ?? = 4( ?? - h) 
? 1 - ???? = 4?? 2
- 4h?? ? 4?? 2
- ( 4h - ?? ) ?? - 1 = 0 
This has two roots say ?? 1
, ?? 2
; ?? 1
+ ?? 2
=
4h-?? 4
 
(iv) and ?? 1
?? 2
= -
1
4
 
Also, 
2?? 1
+?? 2
3
= h 
[? ?? divides ???? in the ratio 1: 2] 
i.e., 2?? 1
+ ?? 2
= 3h 
(vi) - (iv) gives, ?? 1
= 3h -
4h-?? 4
=
8h+?? 4
 and ?? 2
= 3h - 2 ·
8h+?? 4
= -
2h+?? 2
 
Putting in (v), we get 
8h+?? 4
(-
2h+?? 2
) = -
1
4
 
? ( 8h + ?? ) ( 2h + ?? ) = 2 ? 16h
2
+ 10h?? + ?? 2
= 2 
?  Required locus of ?? ( h, ?? ) is 16?? 2
+ 10???? + ?? 2
= 2. 
Q14.  ???? and ???? are two perpendicular chords of the rectangular hyperbola ???? =
?? ?? . If ?? is the centre of the rectangular hyperbola, then the product of the slopes 
of ???? , ???? , ???? and ???? is equal to 
(a) -1 
(b) 1 
(c) 0 
(d) None of these 
Ans: (b) Let ?? 1
, ?? 2
, ?? 3
, ?? 4
 be the parameters of the points ?? , ?? , ?? and ?? respectively. Then, the 
coordinates of ?? , ?? , ?? and ?? are (?? ?? 1
,
?? ?? 1
), (?? ?? 2
,
?? ?? 2
), (?? ?? 3
,
?? ?? 3
) and (?? ?? 4
,
?? ?? 4
) respectively. 
Now, ???? ? ???? ?
?? ?? 2
-
?? ?? 1
?? ?? 2
-?? ?? 1
×
?? ?? 4
-
?? ?? 3
?? ?? 4
-?? ?? 3
= -1 ? -
1
?? 1
?? 2
× -
1
?? 3
?? 4
= -1 ? ?? 1
?? 2
?? 3
?? 4
= -1 
?  Product of the slopes of ???? , ???? , ???? and ???? 
1
?? 1
2
×
1
?? 2
2
×
1
?? 3
2
×
1
?? 4
2
=
1
?? 1
2
?? 2
2
?? 3
2
?? 4
2
= 1 
[Using (i)] 
Q15. If the circle ?? ?? + ?? ?? = ?? ?? intersects the hyperbola ???? = ?? ?? in four points 
?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) then 
(a) ?? ?? + ?? ?? + ?? ?? + ?? ?? = ?? 
(b) ?? ?? + ?? ?? + ?? ?? + ?? ?? = ?? 
(c) ?? ?? ?? ?? ?? ?? ?? ?? = ?? ?? 
(d) ?? ?? ?? ?? ?? ?? ?? ?? = ?? ?? 
Ans: (a,b,c,d) Given, circle is ?? 2
+ ?? 2
= ?? 2
 
(i) and hyperbola be ???? = ?? 2
 
from (ii) ?? =
?? 2
?? . Putting in (i), we get ?? 2
+
?? 4
?? 2
= ?? 2
? ?? 4
- ?? 2
?? 2
+ ?? 4
= 0 
? ?? 1
+ ?? 2
+ ?? 3
+ ?? 4
= 0, ?? 1
?? 2
?? 3
?? 4
= ?? 4
 
Since both the curves are symmetric in ?? and ?? , ? ?? 1
+ ?? 2
+ ?? 3
+ ?? 4
= 0; ?? 1
?? 2
?? 3
?? 4
= ?? 4
.