JEE Main 2025 — Mathematics Coordinate Geometry

$$\begin{aligned} & e x+a+e x-a=8 \sqrt{\frac{5}{3}} \\ & 2 e x=8 \sqrt{\frac{5}{3}} \\ & 2 e \times 4=8 \sqrt{\frac{5}{3}} \\ & e=\sqrt{\frac{5}{3}} \\ & b^2=a^2\left(\left(\frac{\sqrt{5}}{3}\right)^2-1\right) \\ & b^2=\frac{2}{3} a^2\end{aligned}$$​ex+a+ex−a=835​​2ex=835​​2e×4=835​​e=35​​b2=a2​(35​​)2−1​b2=32​a2​

$$\frac{16}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1$$a216​− b29​=1

and $b^2=\frac{2}{3} a^2$b2=32​a2

$$\Rightarrow \mathrm{a}^2=\frac{5}{2} \quad \mathrm{~b}^2=\frac{5}{3}$$⇒a2=25​ b2=35​

Now,

$\begin{aligned}
& \ell=\frac{2 \mathrm{b}^2}{\mathrm{a}} \
& \ell^2=\frac{4 \mathrm{b}^4}{\mathrm{a}^2} \
& 9 \ell^2=36 \times \frac{25}{9 \times 5} \times 2 \
& 9 \ell^2=40 \
& \mathrm{~m}=(\mathrm{ex}+\mathrm{a})(\mathrm{ex}-\mathrm{a}) \
& \mathrm{m}=\mathrm{e}^2 \mathrm{x}^2-\mathrm{a}^2 \
& =\frac{5}{3} \times 16-\frac{5}{2}=\frac{145}{6}
\end{aligned}$

$\begin{aligned}
& =6 \mathrm{m}=145 \
& 9 \ell^2+6 \mathrm{m} \
& 40+145=185
\end{aligned}$

option (3)