JEE Main 2026 — 6 April, Evening Shift — Question 30

$\mathrm{e}=\frac{\sqrt{3}}{2}$ $\frac{\mathrm{a}}{\mathrm{e}}=\frac{4 \sqrt{6}}{3}$ $\mathrm{a}=2 \sqrt{2}$ $(\mathrm{ae})^{2}=\mathrm{a}^{2}-\mathrm{b}^{2}$ $\mathrm{b}^{2}=2$ so $E: \frac{x^{2}}{8}+\frac{y^{2}}{2}=1$ Now $\mathrm{e}_{\mathrm{H}}=2 \sqrt{2} \Rightarrow \mathrm{~b}^{2}=7 \mathrm{a}^{2}$ $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2 \sqrt{2} \Rightarrow \mathrm{a}=\frac{\sqrt{2}}{7}$ $\left(\mathrm{ae}_{\mathrm{H}}\right)^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}=8 \mathrm{a}^{2}=\frac{16}{49}$ $\mathrm{ae}_{\mathrm{H}}=\frac{4}{7}$ Distance between focii $=2 \mathrm{ae}_{\mathrm{H}}=\frac{8}{7}$