JEE Main 2025 — Hyperbola Question with Solution

$\begin{array}{rl}& ex+a+ex-a=8\sqrt{\frac{5}{3}}\\ & 2ex=8\sqrt{\frac{5}{3}}\\ & 2e\times 4=8\sqrt{\frac{5}{3}}\\ & e=\sqrt{\frac{5}{3}}\\ & b^2=a^2\left({\left(\frac{\sqrt{5}}{3}\right)}^2-1\right)\\ & b^2=\frac{2}{3}{a}^2\end{array}$
$\frac{16}{{\mathrm{a}}^2}-\frac{9}{{\mathrm{b}}^2}=1$
and $b^2=\frac{2}{3}{a}^2$
$\Rightarrow {\mathrm{a}}^2=\frac{5}{2}{\mathrm{b}}^2=\frac{5}{3}$
Now,
$\begin{aligned}
& \ell=\frac{2 \mathrm{~b}^2}{\mathrm{a}} \\ & \ell^2=\frac{4 \mathrm{~b}^4}{\mathrm{a}^2} \\ & 9 \ell^2=36 \times \frac{25}{9 \times 5} \times 2 \\ & 9 \ell^2=40 \\ & \mathrm{~m}=(\mathrm{ex}+\mathrm{a})(\mathrm{ex}-\mathrm{a}) \\ & \mathrm{m}=\mathrm{e}^2 \mathrm{x}^2-\mathrm{a}^2 \\ & =\frac{5}{3} \times 16-\frac{5}{2}=\frac{145}{6}
\end{aligned}$<br/>$\begin{aligned}
& =6 \mathrm{~m}=145 \\ & 9 \ell^2+6 \mathrm{~m} \\ & 40+145=185
\end{aligned}$
option (3)