QJEEM24Ma799c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma799c Solution: Let the latus rectum of the hyperbola subtend an angle of at…

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

$\begin{aligned}& \Rightarrow \tan 30^{\circ}=\frac{\mathrm{b}^2 / \mathrm{a}}{\mathrm{ae}}=\frac{\mathrm{b}^2}{\mathrm{a}^2 \mathrm{e}}=\frac{1}{\sqrt{3}} \\& \Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^2}{9}\end{aligned}$

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

$\begin{aligned}& \Rightarrow \mathrm{b}^4=3 \mathrm{~b}^2+27 \\& \Rightarrow \mathrm{b}^4-3 \mathrm{~b}^2-27=0 \\& \Rightarrow \mathrm{b}^2=\frac{3}{2}(1+\sqrt{13}) \\& \Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13 \\& \Rightarrow \ell^2+\mathrm{m}^2+\mathrm{n}^2=182\end{aligned}$