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Let the ellipse E: x^2 144+ y^2 169=1 and the hyperbola H: x^2 16- y^2 λ^2=-1 have the same foci. If e and L respectively denote the eccentricity and…

JEE Main 2026 — Mathematics Coordinate Geometry

2026mcqmedium

Let the ellipse E:x2144+y2169=1\mathrm{E}: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1E:144x2​+169y2​=1 and the hyperbola H:x216−y2λ2=−1\mathrm{H}: \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1H:16x2​−λ2y2​=−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)24(\mathrm{e}+\mathrm{L})24(e+L) is :

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

  1. A

    148

  2. B

    126

  3. C

    67

  4. D

    296

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