Let E_1: frac{x^2}{9}+frac{y^2}{4}=1 be an el | vedclass

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(A) For $E_1: \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,the semi-major axis $a_1 = 3$ and semi-minor axis $b_1 = 2$. The eccentricity $e = \sqrt{1 - \fra

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$54$

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(A) For $E_1: \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,the semi-major axis $a_1 = 3$ and semi-minor axis $b_1 = 2$. The eccentricity $e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.Since all ellipses $E_i$ have the same eccentricity $e = \frac{\sqrt{5}}{3}$,we have $e^2 = 1 - \frac{b_i^2}{a_i^2} = \frac{5}{9}$,which implies $\frac{b_i^2}{a_i^2} = \frac{4}{9}$,or $b_i = \frac{2}{3}a_i$.The problem states that the length of the minor axis of $E_i$ $(2b_i)$ is the length of the major axis of $E_{i+1}$ $(2a_{i+1})$,so $2b_i = 2a_{i+1}$,which means $a_{i+1} = b_i$.Substituting $b_i = \frac{2}{3}a_i$,we get $a_{i+1} = \frac{2}{3}a_i$. This is a geometric progression for the semi-major axes with common ratio $r = \frac{2}{3}$.The area of an ellipse is $A_i = \pi a_i b_i = \pi a_i (\frac{2}{3}a_i) = \frac{2\pi}{3} a_i^2$.Since $a_i$ forms a geometric progression with ratio $\frac{2}{3}$,$a_i^2$ forms a geometric progression with ratio $(\frac{2}{3})^2 = \frac{4}{9}$.Thus,$A_i$ forms a geometric progression with first term $A_1 = \pi(3)(2) = 6\pi$ and common ratio $R = \frac{4}{9}$.The sum of the infinite series is $\sum_{i=1}^{\infty} A_i = \frac{A_1}{1 - R} = \frac{6\pi}{1 - 4/9} = \frac{6\pi}{5/9} = \frac{54\pi}{5}$.Therefore,$\frac{5}{\pi} \sum_{i=1}^{\infty} A_i = \frac{5}{\pi} 	imes \frac{54\pi}{5} = 54$.

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