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JEE Main 28-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 14 Marks
Question
Let $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ be an ellipse. Ellipses $\mathrm{E}_{\mathrm{i}}^{\prime}$ 's are constructed such that their centres and eccentricities are same as that of $E_{1}$, and the length of minor axis of $\mathrm{E}_{\mathrm{i}}$ is the length of major axis of $E_{i+1}(i \geq 1)$. If $A_{i}$ is the area of the ellipse $E_{i}$, then $\frac{5}{\pi}\left(\sum_{i=1}^{\infty} \mathrm{A}_{\mathrm{i}}\right)$, is equal to $\qquad$
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Answer

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$E_{1}=\frac{x^{2}}{9}+\frac{y^{2}}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$
$\mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{4}=1$
$\mathrm{e}=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{a}^{2}}{4}} \Rightarrow \frac{5}{9}=1-\frac{\mathrm{a}^{2}}{4}$
$a^{2}=\frac{16}{9}$
$E_{2}: \frac{x^{2}}{\frac{16}{9}}+\frac{y^{2}}{4}=1$
$\mathrm{E}_{3}: \frac{\mathrm{x}^{2}}{\frac{16}{9}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$
$\mathrm{e}=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{\mathrm{b}^{2}}{\frac{16}{9}}} \Rightarrow \mathrm{~b}^{2}=\frac{64}{81}$
$E_{3}=\frac{x^{2}}{\frac{16}{9}}+\frac{y^{2}}{\frac{64}{81}}=1$
$\mathrm{A}_{1}=\pi \times 3 \times 2 \Rightarrow 6 \pi$
$\mathrm{A}_{2}=\pi \times \frac{4}{3} \times 2=\frac{8 \pi}{3}$
$\mathrm{A}_{3}=\pi \times \frac{4}{3} \times \frac{8}{9}=\frac{32 \pi}{81}$
$\sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots . \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5}$
$\therefore \frac{5}{\pi} \sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}} \Rightarrow \frac{5}{\pi} \times \frac{54 \pi}{5}=54$

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