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Hyperbola: JEE Mains Previous Year Questions (2021-2024) | Mathematics (Maths) for JEE Main & Advanced PDF Download

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JEE Mains Previous Year Questions 
(2021-2024): Hyperbola 
2024 
Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times 
eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : 
A. 
?? ?? 
B. 
?? ?? ????
 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: If the foci of a hyperbola are same as that of the ellipse 
?? ?? ?? +
?? ?? ????
= ?? and the eccentricity of the 
hyperbola is 
????
?? times the eccentricity of the ellipse, then the smaller focal distance of the point 
( v ?? ,
????
?? v
?? ?? ) on the hyperbola, is equal to 
A. ???? v
?? ?? -
?? ?? 
B. ?? v
?? ?? +
?? ?? 
C. ?? v
?? ?? -
?? ?? 
D. ???? v
?? ?? -
????
??             [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (c) 
x
2
9
+
y
2
25
= 1
a = 3 , b = 5
e =
v
1 -
9
25
=
4
5
? foci = ( 0 , ± be ) = ( 0 , ± 4 )
? ? e
H
=
4
5
×
15
8
=
3
2
 
Let equation hyperbola 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Hyperbola 
2024 
Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times 
eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : 
A. 
?? ?? 
B. 
?? ?? ????
 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: If the foci of a hyperbola are same as that of the ellipse 
?? ?? ?? +
?? ?? ????
= ?? and the eccentricity of the 
hyperbola is 
????
?? times the eccentricity of the ellipse, then the smaller focal distance of the point 
( v ?? ,
????
?? v
?? ?? ) on the hyperbola, is equal to 
A. ???? v
?? ?? -
?? ?? 
B. ?? v
?? ?? +
?? ?? 
C. ?? v
?? ?? -
?? ?? 
D. ???? v
?? ?? -
????
??             [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (c) 
x
2
9
+
y
2
25
= 1
a = 3 , b = 5
e =
v
1 -
9
25
=
4
5
? foci = ( 0 , ± be ) = ( 0 , ± 4 )
? ? e
H
=
4
5
×
15
8
=
3
2
 
Let equation hyperbola 
x
2
 A
2
-
y
2
 B
2
= - 1
? ? B · e
H
= 4 ? ? B =
8
3
? ? A
2
= B
2
( e
H
2
- 1 ) =
64
9
(
9
4
- 1 ) ? A
2
=
80
9
? ?
x
2
80
9
-
y
2
64
9
= - 1
 Directrix : y = ±
B
e
H
= ±
16
9
PS = e · PM =
3
2
|
14
3
·
v
2
5
-
16
9
|
? = 7
v
2
5
-
8
3
 
Q3: Let ?? be a point on the hyperbola ?? :
?? ?? ?? -
?? ?? ?? = ?? , in the first quadrant such that the area of 
triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the 
origin is 
A. 26 
B. 22 
C. 20 
D. 18         [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 
 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Hyperbola 
2024 
Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times 
eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : 
A. 
?? ?? 
B. 
?? ?? ????
 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: If the foci of a hyperbola are same as that of the ellipse 
?? ?? ?? +
?? ?? ????
= ?? and the eccentricity of the 
hyperbola is 
????
?? times the eccentricity of the ellipse, then the smaller focal distance of the point 
( v ?? ,
????
?? v
?? ?? ) on the hyperbola, is equal to 
A. ???? v
?? ?? -
?? ?? 
B. ?? v
?? ?? +
?? ?? 
C. ?? v
?? ?? -
?? ?? 
D. ???? v
?? ?? -
????
??             [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (c) 
x
2
9
+
y
2
25
= 1
a = 3 , b = 5
e =
v
1 -
9
25
=
4
5
? foci = ( 0 , ± be ) = ( 0 , ± 4 )
? ? e
H
=
4
5
×
15
8
=
3
2
 
Let equation hyperbola 
x
2
 A
2
-
y
2
 B
2
= - 1
? ? B · e
H
= 4 ? ? B =
8
3
? ? A
2
= B
2
( e
H
2
- 1 ) =
64
9
(
9
4
- 1 ) ? A
2
=
80
9
? ?
x
2
80
9
-
y
2
64
9
= - 1
 Directrix : y = ±
B
e
H
= ±
16
9
PS = e · PM =
3
2
|
14
3
·
v
2
5
-
16
9
|
? = 7
v
2
5
-
8
3
 
Q3: Let ?? be a point on the hyperbola ?? :
?? ?? ?? -
?? ?? ?? = ?? , in the first quadrant such that the area of 
triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the 
origin is 
A. 26 
B. 22 
C. 20 
D. 18         [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 
 
?? 2
9
-
?? 2
4
= 1
?? 2
= 9 , ?? 2
= 4
?? 2
= ?? 2
( ?? 2
- 1 ) ? ?? 2
= 1 +
?? 2
?? 2
?? 2
= 1 +
4
9
=
13
9
?? =
v 13
3
? ?? 1
?? 2
= 2ae = 2 × 3 ×
v
13
3
= 2 v 13
 
Area of ? PS
1
 S
2
=
1
2
× ?? × S
1
 S
2
= 2 v 13 
? ?
1
2
× ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2
?? 2
9
-
?? 2
4
= 1 ?
?? 2
9
- 1 = 1 ? ?? 2
= 18 ? ?? = 3 v 2
 
Distance of P from origin = v ?? 2
+ ?? 2
 
= v 18 + 4 = v 22 
Q4: 4Let ?? ?? be the eccentricity of the hyperbola 
?? ?? ????
-
?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the 
chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : 
A. 
?? v ?? ?? 
B. ?? v ?? 
C. ?? v ?? 
D. 
???? v ?? ??    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (d) 
Explanation: Currently no explanation available 
Q5: Let the foci and length of the latus rectum of an ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , 
respectively. Then, the square of the eccentricity of the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? ?? ?? = ?? equals      [JEE Main 
2024 (Online) 31st January Morning Shift] 
Ans:  51 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Hyperbola 
2024 
Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times 
eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : 
A. 
?? ?? 
B. 
?? ?? ????
 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: If the foci of a hyperbola are same as that of the ellipse 
?? ?? ?? +
?? ?? ????
= ?? and the eccentricity of the 
hyperbola is 
????
?? times the eccentricity of the ellipse, then the smaller focal distance of the point 
( v ?? ,
????
?? v
?? ?? ) on the hyperbola, is equal to 
A. ???? v
?? ?? -
?? ?? 
B. ?? v
?? ?? +
?? ?? 
C. ?? v
?? ?? -
?? ?? 
D. ???? v
?? ?? -
????
??             [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (c) 
x
2
9
+
y
2
25
= 1
a = 3 , b = 5
e =
v
1 -
9
25
=
4
5
? foci = ( 0 , ± be ) = ( 0 , ± 4 )
? ? e
H
=
4
5
×
15
8
=
3
2
 
Let equation hyperbola 
x
2
 A
2
-
y
2
 B
2
= - 1
? ? B · e
H
= 4 ? ? B =
8
3
? ? A
2
= B
2
( e
H
2
- 1 ) =
64
9
(
9
4
- 1 ) ? A
2
=
80
9
? ?
x
2
80
9
-
y
2
64
9
= - 1
 Directrix : y = ±
B
e
H
= ±
16
9
PS = e · PM =
3
2
|
14
3
·
v
2
5
-
16
9
|
? = 7
v
2
5
-
8
3
 
Q3: Let ?? be a point on the hyperbola ?? :
?? ?? ?? -
?? ?? ?? = ?? , in the first quadrant such that the area of 
triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the 
origin is 
A. 26 
B. 22 
C. 20 
D. 18         [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 
 
?? 2
9
-
?? 2
4
= 1
?? 2
= 9 , ?? 2
= 4
?? 2
= ?? 2
( ?? 2
- 1 ) ? ?? 2
= 1 +
?? 2
?? 2
?? 2
= 1 +
4
9
=
13
9
?? =
v 13
3
? ?? 1
?? 2
= 2ae = 2 × 3 ×
v
13
3
= 2 v 13
 
Area of ? PS
1
 S
2
=
1
2
× ?? × S
1
 S
2
= 2 v 13 
? ?
1
2
× ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2
?? 2
9
-
?? 2
4
= 1 ?
?? 2
9
- 1 = 1 ? ?? 2
= 18 ? ?? = 3 v 2
 
Distance of P from origin = v ?? 2
+ ?? 2
 
= v 18 + 4 = v 22 
Q4: 4Let ?? ?? be the eccentricity of the hyperbola 
?? ?? ????
-
?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the 
chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : 
A. 
?? v ?? ?? 
B. ?? v ?? 
C. ?? v ?? 
D. 
???? v ?? ??    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (d) 
Explanation: Currently no explanation available 
Q5: Let the foci and length of the latus rectum of an ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , 
respectively. Then, the square of the eccentricity of the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? ?? ?? = ?? equals      [JEE Main 
2024 (Online) 31st January Morning Shift] 
Ans:  51 
 focii = ( ± 5 , 0 ) ;
2 ?? 2
?? = v 50
?? = 5 ? ?? 2
=
5 v 2 ?? 2
?? 2
= ?? 2
( 1 - ?? 2
) =
5 v 2 ?? 2
? ? a ( 1 - e
2
) =
5 v 2
2
? ?
5
e
( 1 - e
2
) =
5 v 2
2
? ? v 2 - v 2 e
2
= e
? ? v 2 e
2
+ e - v 2 = 0
? ? v 2 e
2
+ 2e - e - v 2 = 0
? ? v 2 e ( e + v 2 ) - 1 ( 1 + v 2 ) = 0
? ? ( e + v 2 ) ( v 2 e - 1 ) = 0
? ? e ? - v 2 ; e =
1
v 2
x
2
 b
2
-
y
2
a
2
 b
2
= 1 ? a = 5 v 2
?? = 5
?? 2
?? 2
= ?? 2
( ?? 1
2
- 1 ) ? ?? 1
2
= 51
 
Q6: Let the latus rectum of the hyperbola 
?? ?? ?? -
?? ?? ?? ?? = ?? subtend an angle of 
?? ?? at the centre of the 
hyperbola. If ?? ?? is equal to 
?? ?? ( ?? + v ?? ) , where ?? and ?? are coprime numbers, then ?? ?? + ?? ?? + ?? ?? is 
equal to       [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: 182 
LR subtends 60
°
 at centre 
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Hyperbola 
2024 
Q1: For ?? < ?? < ?? / ?? , if the eccentricity of the hyperbola ?? ?? - ?? ?? ?? ???? ???? ?? ? ?? = ?? is v ?? times 
eccentricity of the ellipse ?? ?? ?? ???? ???? ?? ? ?? + ?? ?? = ?? , then the value of ?? is : 
A. 
?? ?? 
B. 
?? ?? ????
 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (c) 
Explanation: Currently no explanation available 
Q2: If the foci of a hyperbola are same as that of the ellipse 
?? ?? ?? +
?? ?? ????
= ?? and the eccentricity of the 
hyperbola is 
????
?? times the eccentricity of the ellipse, then the smaller focal distance of the point 
( v ?? ,
????
?? v
?? ?? ) on the hyperbola, is equal to 
A. ???? v
?? ?? -
?? ?? 
B. ?? v
?? ?? +
?? ?? 
C. ?? v
?? ?? -
?? ?? 
D. ???? v
?? ?? -
????
??             [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (c) 
x
2
9
+
y
2
25
= 1
a = 3 , b = 5
e =
v
1 -
9
25
=
4
5
? foci = ( 0 , ± be ) = ( 0 , ± 4 )
? ? e
H
=
4
5
×
15
8
=
3
2
 
Let equation hyperbola 
x
2
 A
2
-
y
2
 B
2
= - 1
? ? B · e
H
= 4 ? ? B =
8
3
? ? A
2
= B
2
( e
H
2
- 1 ) =
64
9
(
9
4
- 1 ) ? A
2
=
80
9
? ?
x
2
80
9
-
y
2
64
9
= - 1
 Directrix : y = ±
B
e
H
= ±
16
9
PS = e · PM =
3
2
|
14
3
·
v
2
5
-
16
9
|
? = 7
v
2
5
-
8
3
 
Q3: Let ?? be a point on the hyperbola ?? :
?? ?? ?? -
?? ?? ?? = ?? , in the first quadrant such that the area of 
triangle formed by ?? and the two foci of ?? is ?? v ???? . Then, the square of the distance of ?? from the 
origin is 
A. 26 
B. 22 
C. 20 
D. 18         [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 
 
?? 2
9
-
?? 2
4
= 1
?? 2
= 9 , ?? 2
= 4
?? 2
= ?? 2
( ?? 2
- 1 ) ? ?? 2
= 1 +
?? 2
?? 2
?? 2
= 1 +
4
9
=
13
9
?? =
v 13
3
? ?? 1
?? 2
= 2ae = 2 × 3 ×
v
13
3
= 2 v 13
 
Area of ? PS
1
 S
2
=
1
2
× ?? × S
1
 S
2
= 2 v 13 
? ?
1
2
× ?? × ( 2 v 13 ) = 2 v 13 ? ?? = 2
?? 2
9
-
?? 2
4
= 1 ?
?? 2
9
- 1 = 1 ? ?? 2
= 18 ? ?? = 3 v 2
 
Distance of P from origin = v ?? 2
+ ?? 2
 
= v 18 + 4 = v 22 
Q4: 4Let ?? ?? be the eccentricity of the hyperbola 
?? ?? ????
-
?? ?? ?? = ?? and ?? ?? be the eccentricity of the ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?? , which passes through the foci of the hyperbola. If ?? ?? ?? ?? = ?? , then the length of the 
chord of the ellipse parallel to the ?? -axis and passing through ( ?? , ?? ) is : 
A. 
?? v ?? ?? 
B. ?? v ?? 
C. ?? v ?? 
D. 
???? v ?? ??    [JEE Main 2024 (Online) 27th January Evening Shift] 
Ans: (d) 
Explanation: Currently no explanation available 
Q5: Let the foci and length of the latus rectum of an ellipse 
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?????? ( ± ?? , ?? ) and v ???? , 
respectively. Then, the square of the eccentricity of the hyperbola 
?? ?? ?? ?? -
?? ?? ?? ?? ?? ?? = ?? equals      [JEE Main 
2024 (Online) 31st January Morning Shift] 
Ans:  51 
 focii = ( ± 5 , 0 ) ;
2 ?? 2
?? = v 50
?? = 5 ? ?? 2
=
5 v 2 ?? 2
?? 2
= ?? 2
( 1 - ?? 2
) =
5 v 2 ?? 2
? ? a ( 1 - e
2
) =
5 v 2
2
? ?
5
e
( 1 - e
2
) =
5 v 2
2
? ? v 2 - v 2 e
2
= e
? ? v 2 e
2
+ e - v 2 = 0
? ? v 2 e
2
+ 2e - e - v 2 = 0
? ? v 2 e ( e + v 2 ) - 1 ( 1 + v 2 ) = 0
? ? ( e + v 2 ) ( v 2 e - 1 ) = 0
? ? e ? - v 2 ; e =
1
v 2
x
2
 b
2
-
y
2
a
2
 b
2
= 1 ? a = 5 v 2
?? = 5
?? 2
?? 2
= ?? 2
( ?? 1
2
- 1 ) ? ?? 1
2
= 51
 
Q6: Let the latus rectum of the hyperbola 
?? ?? ?? -
?? ?? ?? ?? = ?? subtend an angle of 
?? ?? at the centre of the 
hyperbola. If ?? ?? is equal to 
?? ?? ( ?? + v ?? ) , where ?? and ?? are coprime numbers, then ?? ?? + ?? ?? + ?? ?? is 
equal to       [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: 182 
LR subtends 60
°
 at centre 
 
? ? t a n ? 30
°
=
b
2
/ a
ae
=
b
2
a
2
e
=
1
v 3
? ? e =
v 3 b
2
9
 
Also, e
2
= 1 +
b
2
9
? 1 +
b
2
9
=
3 b
4
81
 
? ? ?? 4
= 3 ?? 2
+ 27
? ? ?? 4
- 3 ?? 2
- 27 = 0
? ? ?? 2
=
3
2
( 1 + v 13 )
? ? l = 3 , m = 2 , n = 13
? ? l
2
+ m
2
+ n
2
= 182
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Document Description: Hyperbola: JEE Mains Previous Year Questions (2021-2024) for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for Hyperbola: JEE Mains Previous Year Questions (2021-2024) have been prepared according to the JEE exam syllabus. Information about Hyperbola: JEE Mains Previous Year Questions (2021-2024) covers topics like and Hyperbola: JEE Mains Previous Year Questions (2021-2024) Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Hyperbola: JEE Mains Previous Year Questions (2021-2024).
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