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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______________.
$177$
$56$
$182$
$728$
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}^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$
