QJEEM25Ma2f5f: JEE Main 2025 Mathematics - Class 11 JEEM25Ma2f5f Solution: Let be an ellipse. Ellipses 's are constructed such that their…

$\begin{aligned}& E_1=\frac{x^2}{9}+\frac{y^2}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\& E_2: \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\& e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{a^2}{4}} \Rightarrow \frac{5}{9}=1-\frac{a^2}{4} \\& a^2=\frac{16}{9} \\& E_2: \frac{x^2}{16}+\frac{y^2}{4}=1 \\& E_3: \frac{x^2}{\frac{16}{9}}+\frac{y^2}{b^2}=1 \\& e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{b^2}{16}} \Rightarrow b^2=\frac{64}{81}\end{aligned}$

$\begin{aligned}& \mathrm{E}_3=\frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{\frac{64}{81}}=1 \\& \mathrm{~A}_1=\pi \times 3 \times 2 \Rightarrow 6 \pi \\& \mathrm{~A}_2=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3} \\& \mathrm{~A}_3=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{81} \\& \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5} \\& \therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54\end{aligned}$