JEE Main 2026 — 24 January, Morning Shift — Question 7

$2 \mathrm{ae}=8 \Rightarrow \mathrm{a}=5$

$\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)$

$\mathrm{b}^{2}=\mathrm{a}^{2} \times \frac{9}{25} \quad \mathrm{~b}^{2}=9$

$\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$

$\ell_{1}: \frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{2 \times 9}{5}=\frac{18}{5}$

$\mathrm{A}^{2}=\mathrm{B}^{2}\left(1-\mathrm{e}^{2}\right) \Rightarrow \mathrm{A}^{2}=\frac{9}{25} \mathrm{~B}^{2} \Rightarrow \mathrm{~A}=\frac{3}{5} \mathrm{~B}$

$2 \ell^{2}=9 \ell_{2} \Rightarrow 2\left(\frac{18}{5}\right)^{2}=9 \ell_{2} \Rightarrow \ell_{2}=\frac{4 \times 18}{25}$

$\frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{72}{25} \Rightarrow \mathrm{~A}^{2}= \frac{36}{25} \mathrm{~B}$

$\frac{9}{25} \mathrm{~B}^{2}=\frac{36 \mathrm{~B}}{25} \Rightarrow \mathrm{~B}=4$,

Distance between focii $2 \mathrm{Be}=2 \times \frac{4}{5} \times 4=\frac{32}{5}$