Let E_1:x^29+y^24=1 be an ellipse.

Question

Let ${E}_{1}:\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1$ be an ellipse. Ellipses ${E}_{i}$ '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 ${E}_{i+1}(i\ge 1)$. If ${A}_{i}$ is the area of the ellipse ${E}_{i}$, then $\frac{5}{\pi }(\sum _{i=1}^{\infty}{A}_{i})$, is equal to \_\_\_\_\_\_\_.

Answer: 54

Accepted Answer

$\begin{gathered} {E}_{1}=\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}\Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \ {E}_{2}:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{4}=1 \ e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{{a}^{2}}{4}}\Rightarrow \frac{5}{9}=1-\frac{{a}^{2}}{4} \ {a}^{2}=\frac{16}{9} \ {E}_{2}:\frac{{x}^{2}}{16}+\frac{{y}^{2}}{4}=1 \ {E}_{3}:\frac{{x}^{2}}{\frac{16}{9}}+\frac{{y}^{2}}{{b}^{2}}=1 \ e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{{b}^{2}}{16}}\Rightarrow {b}^{2}=\frac{64}{81} \end{gathered}$

$\begin{gathered} {E}_{3}=\frac{{x}^{2}}{\frac{16}{9}}+\frac{{y}^{2}}{\frac{64}{81}}=1 \ {A}_{1}=\pi 	imes 3	imes 2\Rightarrow 6\pi \ {A}_{2}=\pi 	imes \frac{4}{3}	imes 2=\frac{8\pi }{3} \ {A}_{3}=\pi 	imes \frac{4}{3}	imes \frac{8}{9}=\frac{32\pi }{81} \ \sum _{i=1}^{\infty}{A}_{i}=6\pi +\frac{8\pi }{3}+\frac{32\pi }{81}+\ldots \infty\Rightarrow \frac{6\pi }{1-\frac{4}{9}}\Rightarrow \frac{54\pi }{5} \ 	herefore \frac{5}{\pi }\sum _{i=1}^{\infty}{A}_{i}\Rightarrow \frac{5}{\pi }	imes \frac{54\pi }{5}=54 \end{gathered}$

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