JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
Let the latus ractum 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 $\frac{l}{m}\left(1+\sqrt{n}\right)$, where $l$ and $m$ are co-prime numbers, then $l^{2}+m^{2}+n^{2}$ is equal to $\_\_\_\_\_\_$ .
Hyperbola
Answer$182$
Solution
Here $a=3$. The latus rectum has end points $\left(ae,\pm\dfrac{b^2}{3}\right)$; it subtends $\tfrac\pi3$ at the centre, so $\tan\dfrac\pi6=\dfrac{b^2/3}{ae}$, with $ae=\sqrt{9+b^2}$. $b^2=\sqrt3\sqrt{9+b^2}\Rightarrow b^4-3b^2-27=0\Rightarrow b^2=\dfrac{3+3\sqrt{13}}{2}=\dfrac32\left(1+\sqrt{13}\right)$. So $l=3$, $m=2$, $n=13$ and $l^2+m^2+n^2=9+4+169=182$.