Question
Let the ellipse E : x^ 2 144 + y^ 2 169 =1 and the hyperbola H : x^ 2 16 - y^ 2 ^ 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 :
JEE Main
Mathematics
Ellipse
2026
JEE Main 2026 (28 January Shift 2)
JEE Main Mathematics Question (2026) — Solution
Options
- A. 148
- B. 126
- C. 67
- D. 296
Answer
D. 296
Step-by-step solution
Ellipse E: x^2 144 + y^2 169 = 1 has a^2 = 169, b^2 = 144, so c_E = 25 = 5 with foci at (0, 5). Hyperbola H: y^2 x^2 - x^2 16 = 1 has the same foci, so c_H^2 = x^2 + 16 = 25, giving x^2 = 9. Thus H is y^2 9 - x^2 16 = 1 with a = 3, b = 4, c = 5. Eccentricity: e = 5 3 . Latus rectum: L = 2b^2 a = 32 3 . Therefore 24(e + L) = 24 ( 5 3 + 32 3 ) = 24 37 3 = 296
Related: Mathematics — Ellipse · All PYQ Banks