Let one focus of the hyperbola H : frac x 2 a 2 -frac y 2 b 2 =1 be at (√(10), 0) and the corresponding directrix be x=frac 9 √(10) . If e and I respectively are the eccentricity and the length of the latus rectum of H , then 9 (e 2 +l ) is equal to

Mathematics · Hyperbola

JEE Main 2025 — 2 April, Morning Shift — Question 40

Let one focus of the hyperbola $\mathrm{H}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be

$x=\frac{9}{\sqrt{10}}$ . If $e$ and $I$ respectively are the eccentricity and the length of the latus rectum of H , then $9\left(e^{2}+l\right)$ is equal to

Option A:
$15$
Option B:
$14$
Option C:
$12$
Option D:
$16$
Correct

Answer: D

Step-by-step solution

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Directrix : $x=\frac{9}{\sqrt{10}}=\frac{a}{e} …(i)$

Focus: $(\sqrt{10}, 0) \equiv(a e, 0)$

\begin{gathered} a e=\sqrt{10} …(ii)\end{gathered}

(i) $\times$ (ii)

$\Rightarrow a^{2}=9 \Rightarrow a=3$

Substitute in (ii)

$e=\frac{\sqrt{10}}{3}$

Now $e^{2}=1+\frac{b^{2}}{a^{2}}$

$\frac{10}{9}=1+\frac{b^{2}}{a}$

$\Rightarrow \quad b=1$

$I=\frac{2 b^{2}}{a}=\frac{2 \times 1}{3}=\frac{2}{3}$

$a\left[e^{2}+I\right]=9\left[\frac{10}{9}+\frac{2}{3}\right]=10+6$

$=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola