Let the latus rectum of the hyperbola frac x 2 9 -frac y 2 b 2 =1 subtend an angle of π/3 at the centre of the hyperbola. If b 2 is equal to frac l m (1+sqrt n ) , where l and m are co-prime numbers, then l 2 + m 2 + n 2 is equal to

Mathematics · Hyperbola

JEE Main 2024 — 30 January, Shift 1 — Question 23

Let the latus rectum of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{b^{2}}=1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $\mathrm{b}^{2}$ is equal to $\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$ , where $l$ and m are co-prime numbers, then $l^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}$ is equal to $\qquad$

Answer: 182

Numerical answer — enter this value.

Step-by-step solution

LR subtends $60^{\circ}$ at centre

$\Rightarrow \tan 30^{\circ}=\frac{\mathrm{b}^{2} / \mathrm{a}}{\mathrm{ae}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2} \mathrm{e}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^{2}}{9}$

Also, $\mathrm{e}^{2}=1+\frac{\mathrm{b}^{2}}{9} \Rightarrow 1+\frac{\mathrm{b}^{2}}{9}=\frac{3 \mathrm{~b}^{4}}{81}$

$\Rightarrow b^{4}=3 b^{2}+27$

$\Rightarrow b^{4}-3 b^{2}-27=0$

$\Rightarrow \mathrm{b}^{2}=\frac{3}{2}(1+\sqrt{13})$

$\Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13$

$\Rightarrow \ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=182$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola