Let H:x 2/a 2 -y 2/b 2 = 1 be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is 8/3 . If the line x = α intersects the hyperbola H at the points A and B such that the area of triangle AOB is 4√(15) where O is the origin, then α 2 equals

Mathematics · Hyperbola

JEE Main 2026 — 4 April, Evening Shift — Question 35

Let $H:\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$ be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is $\frac{8}{3}$ . If the line $x = \alpha$ intersects the hyperbola H at the points A and B such that the area of triangle AOB is $4\sqrt{15}$ where O is the origin, then $\alpha^2$ equals

Option A:
$12$
Option B:
$16$
Correct
Option C:
$24$
Option D:
$25$

Answer: B

Step-by-step solution

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Given: $2 \mathrm{ae}=6 \Rightarrow \mathrm{ae}=3$ $\frac{2 \mathrm{a}}{\mathrm{e}}=\frac{8}{3} \Rightarrow \frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{3}$ $\mathrm{a}^{2}=4 \mathrm{e}=\frac{3}{2} \mathrm{~b}^{2}=5$ $\frac{x^{2}}{4}-\frac{y^{2}}{5}=1 \quad$ solve with $x=\alpha$ $y= \pm \sqrt{\frac{5\left(\alpha^{2}-4\right)}{4}}$

A\left(\alpha, \sqrt{\frac{5\left(\alpha^{2}-4\right)}{4}}\right) B\left(\alpha,-\sqrt{\frac{5\left(\alpha^{2}-4\right)}{4}}\right)

Ar. of $\Delta \mathrm{OAB}=\frac{1}{2}|\alpha| \sqrt{5\left(\alpha^{2}-4\right)}=4 \sqrt{15}$ $\Rightarrow 16 \times 15=\frac{1}{4} \alpha^{2}(5)\left(\alpha^{2}-4\right)$ $\Rightarrow \alpha^{2}=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola