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JEE Main Previous Year Questions (2025): Hyperbola

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JEE Main Previous Year Questions 
(2025): Hyperbola 
Q1. Let one focus of the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be at ( v ???? , ?? ) and the 
corresponding directrix be ?? =
?? v ????
. If e and ?? respectively are the eccentricity and the 
length of the latus rectum of H , then ?? ( ?? ?? + ?? ) is equal to: 
A. 14 
B. 15 
C. 16 
D. 12 
Ans: C 
Solution: 
ae = v 10 and 
a
e
=
9
10
 ? a
2
= 9 and e =
v 10
3
 
 ( ae)
2
= a
2
+ b
2
10 = 9 + b
2
? b
2
= 1
 
 Now ?
2 b
2
a
=
2( 1)
3
 = 9( e
2
+ l)
 = 10 + 6
 = 16
 
 
Q2. If the equation of the hyperbola with foci ( ?? , ?? ) and ( ?? , ?? ) is ?? ?? ?? - ?? ?? - ???? +
???? + ?? = ?? , then ?? + ?? + ?? is equal to ____ . 
 
 
Ans: 141 
Page 2


JEE Main Previous Year Questions 
(2025): Hyperbola 
Q1. Let one focus of the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be at ( v ???? , ?? ) and the 
corresponding directrix be ?? =
?? v ????
. If e and ?? respectively are the eccentricity and the 
length of the latus rectum of H , then ?? ( ?? ?? + ?? ) is equal to: 
A. 14 
B. 15 
C. 16 
D. 12 
Ans: C 
Solution: 
ae = v 10 and 
a
e
=
9
10
 ? a
2
= 9 and e =
v 10
3
 
 ( ae)
2
= a
2
+ b
2
10 = 9 + b
2
? b
2
= 1
 
 Now ?
2 b
2
a
=
2( 1)
3
 = 9( e
2
+ l)
 = 10 + 6
 = 16
 
 
Q2. If the equation of the hyperbola with foci ( ?? , ?? ) and ( ?? , ?? ) is ?? ?? ?? - ?? ?? - ???? +
???? + ?? = ?? , then ?? + ?? + ?? is equal to ____ . 
 
 
Ans: 141 
Solution: 
Equation of hyperbola is 
( ?? - 6)
2
?? 2
-
( ?? - 2)
2
4 - ?? 2
= 1
 ? ( 4 - ?? 2
) ( ?? - 6)
2
- ?? 2
( ?? - 2)
2
= ?? 2
( 4 - ?? 2
)
 
comparing with 3?? 2
- ?? 2
- ???? + ???? + ?? = 0, we get a
2
= 1 and ?? = 36, ?? = 4 and ?? = 101 
? a + ß + ? = 141 
 
Q3. Let ?? ?? and ?? ?? be the eccentricities of the ellipse 
?? ?? ?? ?? +
?? ?? ????
= ?? and the hyperbola 
?? ?? ????
-
?? ?? ?? ?? = ?? , respectively. If ?? < ?? and ?? ?? ?? ?? = ?? , then the eccentricity of the ellipse 
having its axes along the coordinate axes and passing through all four foci (two of the 
ellipse and two of the hyperbola) is : 
A. 
4
5
 
B. 
3
5
 
C. 
v 7
4
 
D. 
v 3
2
 
Ans: B 
Solution: 
?? 1
2
= 1 -
?? 2
25
 ?? 2
2
= 1 -
?? 2
16
 ? ?? 1
2
?? 2
2
= 1
(1 -
?? 2
25
) (1 +
?? 2
16
) = 1
 ? 2 +
?? 2
16
-
?? 2
25
-
?? 2
400
= 1
 ?
9?? 2
400
=
?? 4
400
?? 2
= 9
?? 2
9
+
?? 2
25
= 1 
?? 2
16
-
?? 2
9
= 0
?? 1
v
1 -
9
25
?? 1
=
4
5
 
Page 3


JEE Main Previous Year Questions 
(2025): Hyperbola 
Q1. Let one focus of the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be at ( v ???? , ?? ) and the 
corresponding directrix be ?? =
?? v ????
. If e and ?? respectively are the eccentricity and the 
length of the latus rectum of H , then ?? ( ?? ?? + ?? ) is equal to: 
A. 14 
B. 15 
C. 16 
D. 12 
Ans: C 
Solution: 
ae = v 10 and 
a
e
=
9
10
 ? a
2
= 9 and e =
v 10
3
 
 ( ae)
2
= a
2
+ b
2
10 = 9 + b
2
? b
2
= 1
 
 Now ?
2 b
2
a
=
2( 1)
3
 = 9( e
2
+ l)
 = 10 + 6
 = 16
 
 
Q2. If the equation of the hyperbola with foci ( ?? , ?? ) and ( ?? , ?? ) is ?? ?? ?? - ?? ?? - ???? +
???? + ?? = ?? , then ?? + ?? + ?? is equal to ____ . 
 
 
Ans: 141 
Solution: 
Equation of hyperbola is 
( ?? - 6)
2
?? 2
-
( ?? - 2)
2
4 - ?? 2
= 1
 ? ( 4 - ?? 2
) ( ?? - 6)
2
- ?? 2
( ?? - 2)
2
= ?? 2
( 4 - ?? 2
)
 
comparing with 3?? 2
- ?? 2
- ???? + ???? + ?? = 0, we get a
2
= 1 and ?? = 36, ?? = 4 and ?? = 101 
? a + ß + ? = 141 
 
Q3. Let ?? ?? and ?? ?? be the eccentricities of the ellipse 
?? ?? ?? ?? +
?? ?? ????
= ?? and the hyperbola 
?? ?? ????
-
?? ?? ?? ?? = ?? , respectively. If ?? < ?? and ?? ?? ?? ?? = ?? , then the eccentricity of the ellipse 
having its axes along the coordinate axes and passing through all four foci (two of the 
ellipse and two of the hyperbola) is : 
A. 
4
5
 
B. 
3
5
 
C. 
v 7
4
 
D. 
v 3
2
 
Ans: B 
Solution: 
?? 1
2
= 1 -
?? 2
25
 ?? 2
2
= 1 -
?? 2
16
 ? ?? 1
2
?? 2
2
= 1
(1 -
?? 2
25
) (1 +
?? 2
16
) = 1
 ? 2 +
?? 2
16
-
?? 2
25
-
?? 2
400
= 1
 ?
9?? 2
400
=
?? 4
400
?? 2
= 9
?? 2
9
+
?? 2
25
= 1 
?? 2
16
-
?? 2
9
= 0
?? 1
v
1 -
9
25
?? 1
=
4
5
 
Focii : - ( 0, ±4) ( ±5,0) 
ellipse passing through all four foci 
?? 2
25
+
?? 2
16
= 1
?? =
v
1 -
16
25
=
3
5
 
 
Q4. Let the product of the focal distances of the point ?? ( ?? , ?? v ?? ) on the hyperbola 
?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be 32 . 
Let the length of the conjugate axis of ?? be ?? and the length of its latus rectum be q . 
Then ?? ?? + ?? ?? is equal to 
Ans: 120  
Solution: 
?? 2
?? 2
-
?? 2
?? 2
= 1 … ( 1)
P( 4,2v 3)
PS
1
· PS
2
= 32
|PS
1
- PS
2
| = 2a
P( 4,2v 3) lies on H
 ?
16
a
2
-
12
 b
2
= 1
16?? 2
- 12?? 2
= ?? 2
?? 2
 … ( 2)
|PS
1
- PS
2
|
2
= 4a
2
PS
1
2
+ PS
2
 
2
- 2PS
1
· PS
2
= 4a
2
 ( ae - 4)
2
+ 12 + ( ae + 4)
2
+ 12 - 64 = 4a
2
2a
2
e
2
- 8 = 4a
2
a
2
+ b
2
- 4 = 2a
2
 b
2
- a
2
= 4
 ( 2) &( 3) ? 16( ?? 2
+ 4)- 12?? 2
= ?? 2
( ?? 2
+ 4)
 ? 16?? 2
+ 64 - 12?? 2
= ?? 4
+ 4?? 2
 ? ?? 4
= 64
 ? ?? 2
= 8
 ? b
2
= 12
p
2
+ q
2
= 4 b
2
+
4 b
4
a
2
 
= 120 
Page 4


JEE Main Previous Year Questions 
(2025): Hyperbola 
Q1. Let one focus of the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be at ( v ???? , ?? ) and the 
corresponding directrix be ?? =
?? v ????
. If e and ?? respectively are the eccentricity and the 
length of the latus rectum of H , then ?? ( ?? ?? + ?? ) is equal to: 
A. 14 
B. 15 
C. 16 
D. 12 
Ans: C 
Solution: 
ae = v 10 and 
a
e
=
9
10
 ? a
2
= 9 and e =
v 10
3
 
 ( ae)
2
= a
2
+ b
2
10 = 9 + b
2
? b
2
= 1
 
 Now ?
2 b
2
a
=
2( 1)
3
 = 9( e
2
+ l)
 = 10 + 6
 = 16
 
 
Q2. If the equation of the hyperbola with foci ( ?? , ?? ) and ( ?? , ?? ) is ?? ?? ?? - ?? ?? - ???? +
???? + ?? = ?? , then ?? + ?? + ?? is equal to ____ . 
 
 
Ans: 141 
Solution: 
Equation of hyperbola is 
( ?? - 6)
2
?? 2
-
( ?? - 2)
2
4 - ?? 2
= 1
 ? ( 4 - ?? 2
) ( ?? - 6)
2
- ?? 2
( ?? - 2)
2
= ?? 2
( 4 - ?? 2
)
 
comparing with 3?? 2
- ?? 2
- ???? + ???? + ?? = 0, we get a
2
= 1 and ?? = 36, ?? = 4 and ?? = 101 
? a + ß + ? = 141 
 
Q3. Let ?? ?? and ?? ?? be the eccentricities of the ellipse 
?? ?? ?? ?? +
?? ?? ????
= ?? and the hyperbola 
?? ?? ????
-
?? ?? ?? ?? = ?? , respectively. If ?? < ?? and ?? ?? ?? ?? = ?? , then the eccentricity of the ellipse 
having its axes along the coordinate axes and passing through all four foci (two of the 
ellipse and two of the hyperbola) is : 
A. 
4
5
 
B. 
3
5
 
C. 
v 7
4
 
D. 
v 3
2
 
Ans: B 
Solution: 
?? 1
2
= 1 -
?? 2
25
 ?? 2
2
= 1 -
?? 2
16
 ? ?? 1
2
?? 2
2
= 1
(1 -
?? 2
25
) (1 +
?? 2
16
) = 1
 ? 2 +
?? 2
16
-
?? 2
25
-
?? 2
400
= 1
 ?
9?? 2
400
=
?? 4
400
?? 2
= 9
?? 2
9
+
?? 2
25
= 1 
?? 2
16
-
?? 2
9
= 0
?? 1
v
1 -
9
25
?? 1
=
4
5
 
Focii : - ( 0, ±4) ( ±5,0) 
ellipse passing through all four foci 
?? 2
25
+
?? 2
16
= 1
?? =
v
1 -
16
25
=
3
5
 
 
Q4. Let the product of the focal distances of the point ?? ( ?? , ?? v ?? ) on the hyperbola 
?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be 32 . 
Let the length of the conjugate axis of ?? be ?? and the length of its latus rectum be q . 
Then ?? ?? + ?? ?? is equal to 
Ans: 120  
Solution: 
?? 2
?? 2
-
?? 2
?? 2
= 1 … ( 1)
P( 4,2v 3)
PS
1
· PS
2
= 32
|PS
1
- PS
2
| = 2a
P( 4,2v 3) lies on H
 ?
16
a
2
-
12
 b
2
= 1
16?? 2
- 12?? 2
= ?? 2
?? 2
 … ( 2)
|PS
1
- PS
2
|
2
= 4a
2
PS
1
2
+ PS
2
 
2
- 2PS
1
· PS
2
= 4a
2
 ( ae - 4)
2
+ 12 + ( ae + 4)
2
+ 12 - 64 = 4a
2
2a
2
e
2
- 8 = 4a
2
a
2
+ b
2
- 4 = 2a
2
 b
2
- a
2
= 4
 ( 2) &( 3) ? 16( ?? 2
+ 4)- 12?? 2
= ?? 2
( ?? 2
+ 4)
 ? 16?? 2
+ 64 - 12?? 2
= ?? 4
+ 4?? 2
 ? ?? 4
= 64
 ? ?? 2
= 8
 ? b
2
= 12
p
2
+ q
2
= 4 b
2
+
4 b
4
a
2
 
= 120 
 
Q5. Let the sum of the focal distances of the point ?? ( ?? , ?? ) on the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be ?? v
?? ?? . If for ?? , the length of the latus rectum is ?? and the product of the focal 
distances of the point P is m , then ?? ?? ?? + ?? ?? is equal to 
A. 184 
B. 186 
C. 185 
D. 187 
Ans: C 
Solution: 
???? + ?? + ???? - ?? = 8
v
5
3
2???? = 8
v
5
3
2?? × 4 = 8
v
5
3
?? =
v
5
3
?? 2
= ?? 2
((
v 5
3
)
2
- 1)
?? 2
=
2
3
?? 2
16
?? 2
-
9
?? 2
= 1
 and ?? 2
=
2
3
?? 2
 ? ?? 2
=
5
2
 ?? 2
=
5
3
 
Now, 
Page 5


JEE Main Previous Year Questions 
(2025): Hyperbola 
Q1. Let one focus of the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be at ( v ???? , ?? ) and the 
corresponding directrix be ?? =
?? v ????
. If e and ?? respectively are the eccentricity and the 
length of the latus rectum of H , then ?? ( ?? ?? + ?? ) is equal to: 
A. 14 
B. 15 
C. 16 
D. 12 
Ans: C 
Solution: 
ae = v 10 and 
a
e
=
9
10
 ? a
2
= 9 and e =
v 10
3
 
 ( ae)
2
= a
2
+ b
2
10 = 9 + b
2
? b
2
= 1
 
 Now ?
2 b
2
a
=
2( 1)
3
 = 9( e
2
+ l)
 = 10 + 6
 = 16
 
 
Q2. If the equation of the hyperbola with foci ( ?? , ?? ) and ( ?? , ?? ) is ?? ?? ?? - ?? ?? - ???? +
???? + ?? = ?? , then ?? + ?? + ?? is equal to ____ . 
 
 
Ans: 141 
Solution: 
Equation of hyperbola is 
( ?? - 6)
2
?? 2
-
( ?? - 2)
2
4 - ?? 2
= 1
 ? ( 4 - ?? 2
) ( ?? - 6)
2
- ?? 2
( ?? - 2)
2
= ?? 2
( 4 - ?? 2
)
 
comparing with 3?? 2
- ?? 2
- ???? + ???? + ?? = 0, we get a
2
= 1 and ?? = 36, ?? = 4 and ?? = 101 
? a + ß + ? = 141 
 
Q3. Let ?? ?? and ?? ?? be the eccentricities of the ellipse 
?? ?? ?? ?? +
?? ?? ????
= ?? and the hyperbola 
?? ?? ????
-
?? ?? ?? ?? = ?? , respectively. If ?? < ?? and ?? ?? ?? ?? = ?? , then the eccentricity of the ellipse 
having its axes along the coordinate axes and passing through all four foci (two of the 
ellipse and two of the hyperbola) is : 
A. 
4
5
 
B. 
3
5
 
C. 
v 7
4
 
D. 
v 3
2
 
Ans: B 
Solution: 
?? 1
2
= 1 -
?? 2
25
 ?? 2
2
= 1 -
?? 2
16
 ? ?? 1
2
?? 2
2
= 1
(1 -
?? 2
25
) (1 +
?? 2
16
) = 1
 ? 2 +
?? 2
16
-
?? 2
25
-
?? 2
400
= 1
 ?
9?? 2
400
=
?? 4
400
?? 2
= 9
?? 2
9
+
?? 2
25
= 1 
?? 2
16
-
?? 2
9
= 0
?? 1
v
1 -
9
25
?? 1
=
4
5
 
Focii : - ( 0, ±4) ( ±5,0) 
ellipse passing through all four foci 
?? 2
25
+
?? 2
16
= 1
?? =
v
1 -
16
25
=
3
5
 
 
Q4. Let the product of the focal distances of the point ?? ( ?? , ?? v ?? ) on the hyperbola 
?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be 32 . 
Let the length of the conjugate axis of ?? be ?? and the length of its latus rectum be q . 
Then ?? ?? + ?? ?? is equal to 
Ans: 120  
Solution: 
?? 2
?? 2
-
?? 2
?? 2
= 1 … ( 1)
P( 4,2v 3)
PS
1
· PS
2
= 32
|PS
1
- PS
2
| = 2a
P( 4,2v 3) lies on H
 ?
16
a
2
-
12
 b
2
= 1
16?? 2
- 12?? 2
= ?? 2
?? 2
 … ( 2)
|PS
1
- PS
2
|
2
= 4a
2
PS
1
2
+ PS
2
 
2
- 2PS
1
· PS
2
= 4a
2
 ( ae - 4)
2
+ 12 + ( ae + 4)
2
+ 12 - 64 = 4a
2
2a
2
e
2
- 8 = 4a
2
a
2
+ b
2
- 4 = 2a
2
 b
2
- a
2
= 4
 ( 2) &( 3) ? 16( ?? 2
+ 4)- 12?? 2
= ?? 2
( ?? 2
+ 4)
 ? 16?? 2
+ 64 - 12?? 2
= ?? 4
+ 4?? 2
 ? ?? 4
= 64
 ? ?? 2
= 8
 ? b
2
= 12
p
2
+ q
2
= 4 b
2
+
4 b
4
a
2
 
= 120 
 
Q5. Let the sum of the focal distances of the point ?? ( ?? , ?? ) on the hyperbola ?? :
?? ?? ?? ?? -
?? ?? ?? ?? = ?? be ?? v
?? ?? . If for ?? , the length of the latus rectum is ?? and the product of the focal 
distances of the point P is m , then ?? ?? ?? + ?? ?? is equal to 
A. 184 
B. 186 
C. 185 
D. 187 
Ans: C 
Solution: 
???? + ?? + ???? - ?? = 8
v
5
3
2???? = 8
v
5
3
2?? × 4 = 8
v
5
3
?? =
v
5
3
?? 2
= ?? 2
((
v 5
3
)
2
- 1)
?? 2
=
2
3
?? 2
16
?? 2
-
9
?? 2
= 1
 and ?? 2
=
2
3
?? 2
 ? ?? 2
=
5
2
 ?? 2
=
5
3
 
Now, 
l =
2?? 2
?? l
2
=
4?? 4
?? 2
9l
2
= 36 ×
25
9 × 5
× 2
9l
2
= 40
?? = ( ???? + ?? ) ( ???? - ?? )
?? = ?? 2
?? 2
- ?? 2
 =
5
3
× 16 -
5
2
=
145
6
 = 6?? = 145
9l
2
+ 6?? 40 + 145 = 185
 
option (3) 
 
Q6. Consider the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? = ?? having one of its focus at ?? ( -?? , ?? ) . If the latus 
ractum through its other focus subtends a right angle at P and ?? ?? ?? ?? = ?? v ?? -
?? , ?? , ?? ? N. 
Ans: 1944 
Solution: 
f 1 = (-ae,0) = P(-3,0) 
? ae = 3
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FAQs on JEE Main Previous Year Questions (2025): Hyperbola

1. What is the standard form of a hyperbola?
Ans. The standard form of a hyperbola is given by the equations (x²/a²) - (y²/b²) = 1 for a hyperbola that opens horizontally, and (y²/a²) - (x²/b²) = 1 for a hyperbola that opens vertically. Here, 'a' and 'b' are constants that determine the shape and size of the hyperbola.
2. How do you find the foci of a hyperbola?
Ans. The foci of a hyperbola are found using the relationship c² = a² + b², where 'c' is the distance from the center to each focus, 'a' is the distance from the center to a vertex, and 'b' is related to the distance to the co-vertices. For a hyperbola centered at the origin with a horizontal transverse axis, the foci are located at (±c, 0), and for a vertical transverse axis, they are at (0, ±c).
3. What is the eccentricity of a hyperbola and how is it calculated?
Ans. The eccentricity (e) of a hyperbola is a measure of how "stretched" the hyperbola is, and it is calculated using the formula e = c/a, where 'c' is the distance from the center to a focus, and 'a' is the distance from the center to a vertex. The eccentricity of a hyperbola is always greater than 1.
4. Can you explain the asymptotes of a hyperbola?
Ans. The asymptotes of a hyperbola are straight lines that the hyperbola approaches but never touches. For a hyperbola in standard form (x²/a²) - (y²/b²) = 1, the equations of the asymptotes are y = (b/a)x and y = -(b/a)x. For (y²/a²) - (x²/b²) = 1, the asymptotes are y = (a/b)x and y = -(a/b)x.
5. How do transformations affect the graph of a hyperbola?
Ans. Transformations such as translations, reflections, and dilations can affect the graph of a hyperbola. A translation shifts the hyperbola's center to a new point (h, k), changing the standard form to (x-h)²/a² - (y-k)²/b² = 1 for horizontal hyperbolas and (y-k)²/a² - (x-h)²/b² = 1 for vertical ones. Reflections can flip the hyperbola across an axis, while dilations change the distance of the vertices and foci from the center, altering its overall shape.
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