JEE Main 2025 — Mathematics Coordinate Geometry

$\begin{aligned} & \mathrm{E}_1=\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{4} \Rightarrow \mathrm{e}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\ & \mathrm{E}_2: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{4}=1 \\ & \mathrm{e}=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{a}^2}{4}} \Rightarrow \frac{5}{9}=1-\frac{\mathrm{a}^2}{4} \\ & \mathrm{a}^2=\frac{16}{9}\end{aligned}$​E1​=9x2​+4y2​⇒e=1−94​​=35​​E2​:a2x2​+4y2​=1e=35​​=1−4a2​​⇒95​=1−4a2​a2=916​​ $\begin{aligned} & \mathrm{E}_2: \frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{4}=1 \\ & \mathrm{E}_3: \frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\ & \mathrm{e}=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{b}^2}{\frac{16}{9}}} \Rightarrow b^2=\frac{64}{81}\end{aligned}$​E2​:916​x2​+4y2​=1E3​:916​x2​+ b2y2​=1e=35​​=1−916​b2​​⇒b2=8164​​ $\displaystyle \begin{aligned} & \mathrm{E}_3=\frac{\mathrm{x}^2}{\frac{16}{9}}+\frac{\mathrm{y}^2}{\frac{64}{81}}=1 \\ & \mathrm{~A}_1=\pi \times 3 \times 2 \Rightarrow 6 \pi \\ & \mathrm{~A}_2=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3} \\ & \mathrm{~A}_3=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{81} \\ & \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots . \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5} \\ & \therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54\end{aligned}$​E3​=916​x2​+8164​y2​=1 A1​=π×3×2⇒6π A2​=π×34​×2=38π​ A3​=π×34​×98​=8132π​i=1∑∞​Ai​=6π+38π​+8132π​+….∞⇒1−94​6π​⇒554π​∴π5​i=1∑∞​Ai​⇒π5​×554π​=54​