Mathematics · Hyperbola

JEE Main 2026 — 28 January, Evening Shift — Question 10

Let the ellipse $\mathrm{E}: \frac{\mathrm{x}^{2}}{144}+\frac{\mathrm{y}^{2}}{169}=1$ and the hyperbola $\mathrm{H}: \frac{\mathrm{x}^{2}}{16}-\frac{\mathrm{y}^{2}}{\lambda^{2}}=-1$ 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 $24(\mathrm{e}+\mathrm{L})$ is :

Option A:
296
Correct
Option B:
126
Option C:
148
Option D:
67

Answer: A

Step-by-step solution

Equation of hyperbola: $\frac{y^{2}}{\lambda^{2}}-\frac{x^{2}}{16}=1$

Equation of ellipse : $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$

$\mathrm{e}^{\prime}=\sqrt{1-\frac{144}{169}}=\frac{5}{13}$

focus $\Rightarrow(0,5)$

$\Rightarrow \lambda \sqrt{1+\frac{16}{\lambda^{2}}}=5$

$\Rightarrow \lambda^{2}+16=25$ $\lambda=3$

Eccentricity of hyperbola $=\sqrt{1+\frac{16}{\lambda^{2}}}=\frac{5}{3}$

Length of latus rectum of hyperbola $=\frac{2(16)}{3}=\frac{32}{3}$

$24(\mathrm{e}+\ell)=24\left[\frac{5}{3}+\frac{32}{3}\right]=8 \times 37=296$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Concyclic Points on a Hyperbola