JEE Main 2025 — 28 January, Morning Shift — Question 25

$E_{1}=\frac{x^{2}}{9}+\frac{y^{2}}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$

$\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{4}=1$

$e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{a}^{2}}{4}} \Rightarrow \frac{5}{9}=1-\frac{\mathrm{a}^{2}}{4}$

$\mathrm{a}^{2}=\frac{16}{9}$

$E_{2}: \frac{x^{2}}{\frac{16}{9}}+\frac{y^{2}}{4}=1$

$E_{3}: \frac{x^{2}}{\frac{16}{9}}+\frac{y^{2}}{b^{2}}=1$

$\mathrm{e}=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{b}^{2}}{\frac{16}{9}}} \Rightarrow \mathrm{~b}^{2}=\frac{64}{81}$

$E_{3}=\frac{x^{2}}{\frac{16}{9}}+\frac{y^{2}}{\frac{64}{81}}=1$

$\mathrm{A}_{1}=\pi \times 3 \times 2 \Rightarrow 6 \pi$

$\mathrm{A}_{2}=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3}$

$\mathrm{A}_{3}=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{27}$

$\sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{27}+\ldots \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5}$

$\therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54$