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 $\mathrm{b}^{2}$ is equal to $\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$, where $l$ and $\mathrm{m}$ are co-prime numbers, then $l^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}$ is equal to
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Answer (182)
Solution
LR subtends $60^{\circ}$ at centre
$\Rightarrow \tan 30^{\circ}=\frac{b^{2} / a}{a e}=\frac{b^{2}}{a^{2} e}=\frac{1}{\sqrt{3}}$
$\Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^{2}}{9}$
Also, $\mathrm{e}^{2}=1+\frac{\mathrm{b}^{2}}{9} \Rightarrow 1+\frac{\mathrm{b}^{2}}{9}=\frac{3 \mathrm{~b}^{4}}{81}$
$\Rightarrow \mathrm{b}^{4}=3 \mathrm{~b}^{2}+27$
$\Rightarrow \mathrm{b}^{4}-3 \mathrm{~b}^{2}-27=0$
$\Rightarrow \mathrm{b}^{2}=\frac{3}{2}(1+\sqrt{13})$
$\Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13$
$\Rightarrow \ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}=182$
