Hyperbola Question 4
Question 4 - 2024 (30 Jan Shift 1)
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 $b^{2}$ 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
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Answer (182)
Solution
LR subtends $60^{\circ}$ at centre
$\Rightarrow \tan 30^{\circ}=\frac{b^{2} / a}{a e}=\frac{b^{2}}{a^{2} e}=\frac{1}{\sqrt{3}}$
$\Rightarrow e=\frac{\sqrt{3} b^{2}}{9}$
Also, $e^{2}=1+\frac{b^{2}}{9} \Rightarrow 1+\frac{b^{2}}{9}=\frac{3 b^{4}}{81}$
$\Rightarrow b^{4}=3 b^{2}+27$
$\Rightarrow b^{4}-3 b^{2}-27=0$
$\Rightarrow b^{2}=\frac{3}{2}(1+\sqrt{13})$
$\Rightarrow \ell=3, m=2, n=13$
$\Rightarrow \ell^{2}+m^{2}+n^{2}=182$
