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Hyperbola: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (2021-2026): 
Hyperbola  
 
(January 2026) 
Q1: Let the ellipse E : x²/144 + y²/169 = 1 and the hyperbola H : x²/16 - y²/?² = -1 
have the same foci. If e and L respectively denote the eccentricity and the length 
of the latus rectum of H, then the value of 24(e + L) is: 
A: 296 
B: 126 
C: 67 
D: 148 
Answer: A 
Explanation: 
 
The equation of the ellipse is: 
E : x²/144 + y²/169 = 1 
Comparing this with the standard form x²/a² + y²/b² = 1, we have: 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Hyperbola  
 
(January 2026) 
Q1: Let the ellipse E : x²/144 + y²/169 = 1 and the hyperbola H : x²/16 - y²/?² = -1 
have the same foci. If e and L respectively denote the eccentricity and the length 
of the latus rectum of H, then the value of 24(e + L) is: 
A: 296 
B: 126 
C: 67 
D: 148 
Answer: A 
Explanation: 
 
The equation of the ellipse is: 
E : x²/144 + y²/169 = 1 
Comparing this with the standard form x²/a² + y²/b² = 1, we have: 
a² = 144 ? a = 12 
b² = 169 ? b = 13 
Since b > a, the ellipse is vertically oriented. The distance from the center to the foci (c) 
is given by: 
c² = b² - a² 
c² = 169 - 144 = 25 ? c = 5 
The foci of the ellipse are at (0, ±5). 
The equation of the hyperbola is: 
H : x²/16 - y²/?² = -1, which is equivalent to y²/?² - x²/16 = 1 
This is a vertical hyperbola. 
Let B² = ?² and A² = 16. 
Since it shares the same foci as the ellipse, its focal distance c is also 5. For a 
hyperbola: 
c² = A² + B² 
25 = 16 + ?² ? ?² = 9 ? ? = 3 
Calculate Eccentricity (e) and Latus Rectum (L) of H: 
Eccentricity (e): 
e = c/B = 5/3 
Length of Latus Rectum (L): For a vertical hyperbola y²/B² - x²/A² = 1, the length of the 
latus rectum is: 
L = 2A²/B = (2 × 16)/3 = 32/3 
We need to find 24(e + L): 
e + L = 5/3 + 32/3 = 37/3 
? 24(e + L) = 24 × 37/3 = 8 × 37 = 296 
 
Q2: Let PQ be a chord of the hyperbola x²/4 - y²/b² = 1, perpendicular to the x-axis 
such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If 
the eccentricity of the hyperbola is v3, then the area of the triangle OPQ is 
A: 2v3 
B: 11/5 
C: 8v3/5 
D: 9/5 
Answer: C 
Explanation: 
 
 
 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Hyperbola  
 
(January 2026) 
Q1: Let the ellipse E : x²/144 + y²/169 = 1 and the hyperbola H : x²/16 - y²/?² = -1 
have the same foci. If e and L respectively denote the eccentricity and the length 
of the latus rectum of H, then the value of 24(e + L) is: 
A: 296 
B: 126 
C: 67 
D: 148 
Answer: A 
Explanation: 
 
The equation of the ellipse is: 
E : x²/144 + y²/169 = 1 
Comparing this with the standard form x²/a² + y²/b² = 1, we have: 
a² = 144 ? a = 12 
b² = 169 ? b = 13 
Since b > a, the ellipse is vertically oriented. The distance from the center to the foci (c) 
is given by: 
c² = b² - a² 
c² = 169 - 144 = 25 ? c = 5 
The foci of the ellipse are at (0, ±5). 
The equation of the hyperbola is: 
H : x²/16 - y²/?² = -1, which is equivalent to y²/?² - x²/16 = 1 
This is a vertical hyperbola. 
Let B² = ?² and A² = 16. 
Since it shares the same foci as the ellipse, its focal distance c is also 5. For a 
hyperbola: 
c² = A² + B² 
25 = 16 + ?² ? ?² = 9 ? ? = 3 
Calculate Eccentricity (e) and Latus Rectum (L) of H: 
Eccentricity (e): 
e = c/B = 5/3 
Length of Latus Rectum (L): For a vertical hyperbola y²/B² - x²/A² = 1, the length of the 
latus rectum is: 
L = 2A²/B = (2 × 16)/3 = 32/3 
We need to find 24(e + L): 
e + L = 5/3 + 32/3 = 37/3 
? 24(e + L) = 24 × 37/3 = 8 × 37 = 296 
 
Q2: Let PQ be a chord of the hyperbola x²/4 - y²/b² = 1, perpendicular to the x-axis 
such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If 
the eccentricity of the hyperbola is v3, then the area of the triangle OPQ is 
A: 2v3 
B: 11/5 
C: 8v3/5 
D: 9/5 
Answer: C 
Explanation: 
 
 
 
 
PQ is a chord perpendicular to x-axis of hyperbola x²/4 - y²/b² = 1 
 
equation become x²/4 - y²/8 = 1 ...(1) 
let coordinate of point P(x 1, y1) 
PQ is perpendicular to x-axis so Q(x1, -y 1) 
It is given that OPQ is equilateral triangle where O is origin 
so, ?POQ = 60° ? ?POX = 30° 
tan 30° = slope of OP = y1/x 1 
y1/x 1 = 1/v3 ? y1 = x1/v3 
Point P(x 1, y1) satisfy hyperbola. 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Hyperbola  
 
(January 2026) 
Q1: Let the ellipse E : x²/144 + y²/169 = 1 and the hyperbola H : x²/16 - y²/?² = -1 
have the same foci. If e and L respectively denote the eccentricity and the length 
of the latus rectum of H, then the value of 24(e + L) is: 
A: 296 
B: 126 
C: 67 
D: 148 
Answer: A 
Explanation: 
 
The equation of the ellipse is: 
E : x²/144 + y²/169 = 1 
Comparing this with the standard form x²/a² + y²/b² = 1, we have: 
a² = 144 ? a = 12 
b² = 169 ? b = 13 
Since b > a, the ellipse is vertically oriented. The distance from the center to the foci (c) 
is given by: 
c² = b² - a² 
c² = 169 - 144 = 25 ? c = 5 
The foci of the ellipse are at (0, ±5). 
The equation of the hyperbola is: 
H : x²/16 - y²/?² = -1, which is equivalent to y²/?² - x²/16 = 1 
This is a vertical hyperbola. 
Let B² = ?² and A² = 16. 
Since it shares the same foci as the ellipse, its focal distance c is also 5. For a 
hyperbola: 
c² = A² + B² 
25 = 16 + ?² ? ?² = 9 ? ? = 3 
Calculate Eccentricity (e) and Latus Rectum (L) of H: 
Eccentricity (e): 
e = c/B = 5/3 
Length of Latus Rectum (L): For a vertical hyperbola y²/B² - x²/A² = 1, the length of the 
latus rectum is: 
L = 2A²/B = (2 × 16)/3 = 32/3 
We need to find 24(e + L): 
e + L = 5/3 + 32/3 = 37/3 
? 24(e + L) = 24 × 37/3 = 8 × 37 = 296 
 
Q2: Let PQ be a chord of the hyperbola x²/4 - y²/b² = 1, perpendicular to the x-axis 
such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If 
the eccentricity of the hyperbola is v3, then the area of the triangle OPQ is 
A: 2v3 
B: 11/5 
C: 8v3/5 
D: 9/5 
Answer: C 
Explanation: 
 
 
 
 
PQ is a chord perpendicular to x-axis of hyperbola x²/4 - y²/b² = 1 
 
equation become x²/4 - y²/8 = 1 ...(1) 
let coordinate of point P(x 1, y1) 
PQ is perpendicular to x-axis so Q(x1, -y 1) 
It is given that OPQ is equilateral triangle where O is origin 
so, ?POQ = 60° ? ?POX = 30° 
tan 30° = slope of OP = y1/x 1 
y1/x 1 = 1/v3 ? y1 = x1/v3 
Point P(x 1, y1) satisfy hyperbola. 
 
 
Q3: Let the domain of the function f(x) = log 3 log 5 log 7 (9x - x² - 13) be the interval 
(m, n). Let the hyperbola x²/a² - y²/b² = 1 have eccentricity n/3 and the length of 
the latus rectum 8m/3. Then b² - a² is equal to: 
A: 7 
B: 9 
C: 11 
D: 5 
Answer: A 
Explanation: 
log 5 (log 7 (9x - x² - 13)) > 0 
? 9x - x² - 13 > 7 
? x ? (4, 5) = (m, n) 
? eccentricity = 5/3 
L(LR) = 32/3 
25/9 = 1 + b²/a² ...(1) 
2b²/a = 32/3 ? b² = 16a/3 
Substitute in (1): 
16/9 = 16a/3a² ? a = 3 
b² = 16 
? b² - a² = 16 - 9 = 7 
 
Q4: Let P(10, 2v15) be a point on the hyperbola x²/a² - y²/b² = 1, whose foci are S 
and S'. If the length of its latus rectum is 8, then the square of the area of ?PSS' 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Hyperbola  
 
(January 2026) 
Q1: Let the ellipse E : x²/144 + y²/169 = 1 and the hyperbola H : x²/16 - y²/?² = -1 
have the same foci. If e and L respectively denote the eccentricity and the length 
of the latus rectum of H, then the value of 24(e + L) is: 
A: 296 
B: 126 
C: 67 
D: 148 
Answer: A 
Explanation: 
 
The equation of the ellipse is: 
E : x²/144 + y²/169 = 1 
Comparing this with the standard form x²/a² + y²/b² = 1, we have: 
a² = 144 ? a = 12 
b² = 169 ? b = 13 
Since b > a, the ellipse is vertically oriented. The distance from the center to the foci (c) 
is given by: 
c² = b² - a² 
c² = 169 - 144 = 25 ? c = 5 
The foci of the ellipse are at (0, ±5). 
The equation of the hyperbola is: 
H : x²/16 - y²/?² = -1, which is equivalent to y²/?² - x²/16 = 1 
This is a vertical hyperbola. 
Let B² = ?² and A² = 16. 
Since it shares the same foci as the ellipse, its focal distance c is also 5. For a 
hyperbola: 
c² = A² + B² 
25 = 16 + ?² ? ?² = 9 ? ? = 3 
Calculate Eccentricity (e) and Latus Rectum (L) of H: 
Eccentricity (e): 
e = c/B = 5/3 
Length of Latus Rectum (L): For a vertical hyperbola y²/B² - x²/A² = 1, the length of the 
latus rectum is: 
L = 2A²/B = (2 × 16)/3 = 32/3 
We need to find 24(e + L): 
e + L = 5/3 + 32/3 = 37/3 
? 24(e + L) = 24 × 37/3 = 8 × 37 = 296 
 
Q2: Let PQ be a chord of the hyperbola x²/4 - y²/b² = 1, perpendicular to the x-axis 
such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If 
the eccentricity of the hyperbola is v3, then the area of the triangle OPQ is 
A: 2v3 
B: 11/5 
C: 8v3/5 
D: 9/5 
Answer: C 
Explanation: 
 
 
 
 
PQ is a chord perpendicular to x-axis of hyperbola x²/4 - y²/b² = 1 
 
equation become x²/4 - y²/8 = 1 ...(1) 
let coordinate of point P(x 1, y1) 
PQ is perpendicular to x-axis so Q(x1, -y 1) 
It is given that OPQ is equilateral triangle where O is origin 
so, ?POQ = 60° ? ?POX = 30° 
tan 30° = slope of OP = y1/x 1 
y1/x 1 = 1/v3 ? y1 = x1/v3 
Point P(x 1, y1) satisfy hyperbola. 
 
 
Q3: Let the domain of the function f(x) = log 3 log 5 log 7 (9x - x² - 13) be the interval 
(m, n). Let the hyperbola x²/a² - y²/b² = 1 have eccentricity n/3 and the length of 
the latus rectum 8m/3. Then b² - a² is equal to: 
A: 7 
B: 9 
C: 11 
D: 5 
Answer: A 
Explanation: 
log 5 (log 7 (9x - x² - 13)) > 0 
? 9x - x² - 13 > 7 
? x ? (4, 5) = (m, n) 
? eccentricity = 5/3 
L(LR) = 32/3 
25/9 = 1 + b²/a² ...(1) 
2b²/a = 32/3 ? b² = 16a/3 
Substitute in (1): 
16/9 = 16a/3a² ? a = 3 
b² = 16 
? b² - a² = 16 - 9 = 7 
 
Q4: Let P(10, 2v15) be a point on the hyperbola x²/a² - y²/b² = 1, whose foci are S 
and S'. If the length of its latus rectum is 8, then the square of the area of ?PSS' 
is equal to: 
A: 4200 
B: 1462 
C: 900 
D: 2700 
Answer: D 
Explanation: 
 
From (1) & (2): 
100/a² - 60/4a = 1 
400 - 60a = 4a² 
4a² + 60a - 400 = 0 
a² + 15a - 100 = 0 
a = 5 & -20 (rejected) 
? b = v20 
? Hyperbola is x²/25 - y²/20 = 1 
 
 
Q5: If the line ax + 2y = 1, where a ? R, does not meet the hyperbola x² - 9y² = 9, 
then a possible value of a is: 
A: 0.6 
B: 0.7 
C: 0.8 
D: 0.5 
Answer: C 
Explanation: 
ax + 2y = 1, a ? R 
x² - 9y² = 9 
y = (1 - ax) / 2
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FAQs on Hyperbola: JEE Main Previous Year Questions (2021-2026)

1. What is a hyperbola?
Ans. A hyperbola is a type of conic section that is formed by the intersection of a plane and a double cone, where the angle of the plane is less than that of the cone's side. It consists of two separate curves called branches, which open away from each other. The standard equation for a hyperbola centred at the origin is (x²/a²) - (y²/b²) = 1, where 'a' and 'b' are real numbers that determine the shape and size of the hyperbola.
2. What are the key characteristics of a hyperbola?
Ans. Key characteristics of a hyperbola include its two branches that are mirror images of each other, the transverse axis which connects the vertices of the branches, and the conjugate axis which is perpendicular to the transverse axis at the centre. Additionally, hyperbolas have asymptotes, which are straight lines that the branches approach but never intersect. The foci of a hyperbola are located along the transverse axis and are used to define its eccentricity.
3. How do you find the foci of a hyperbola?
Ans. The foci of a hyperbola can be found using the relationship c² = a² + b², where 'c' is the distance from the centre to each focus, 'a' is the distance from the centre to each vertex along the transverse axis, and 'b' is the distance from the centre to the vertices of the conjugate axis. For a hyperbola of the form (x²/a²) - (y²/b²) = 1, the foci are located at (±c, 0), while for the form (y²/a²) - (x²/b²) = 1, the foci are at (0, ±c).
4. What role do asymptotes play in the graph of a hyperbola?
Ans. Asymptotes are crucial in the graph of a hyperbola as they provide a framework for the shape of the hyperbola. They are straight lines that the branches of the hyperbola approach as they extend towards infinity. The equations of the asymptotes for the hyperbola (x²/a²) - (y²/b²) = 1 are given by y = (±b/a)x, while for (y²/a²) - (x²/b²) = 1, they are given by y = (±a/b)x. Understanding the asymptotes helps in sketching the hyperbola accurately.
5. How is the equation of a hyperbola derived from the definition of distance?
Ans. The equation of a hyperbola can be derived from the definition that states the difference in distances from any point on the hyperbola to the two foci is constant. If the foci are located at (±c, 0), then for any point (x, y) on the hyperbola, the equation can be established as |√((x - c)² + y²) - √((x + c)² + y²)| = 2a, where 'a' is the distance to the vertices. Squaring both sides and simplifying leads to the standard form of the hyperbola equation.
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