Let E_1: (x^2)/(9)+(y^2)/(4)=1 be an ellipse. Ellipses E_1 's are...

Let

\mathrm{E}_1: \frac{x^2}{9}+\frac{y^2}{4}=1

be an ellipse. Ellipses

\mathrm{E}_1

's are constructed such that their centres and eccentricities are same as that of

E_1

, and the length of minor axis of

E_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} A_i\right)

, is equal to Let E_1: (x^2)/(9)+(y^2)/(4)=1 be an ellipse. Ellipses E_1 's are constructed such that their centres and

Correct answer

\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}

\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}

\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}