Let $E_1 : \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ 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 $E_{i+1}$ $(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$, then $\dfrac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)$ is equal to _____.
Step-by-Step Solution
Key Concept: Each $E_{i+1}$ has semi-major axis $= $ semi-minor axis of $E_i$; same eccentricity $e=\tfrac{\sqrt{5}}{3}$ determines the new semi-minor axis; the areas form a geometric series with ratio $\tfrac{4}{9}$.
$E_1$: $a_1=3$, $b_1=2$, $e=\sqrt{1-4/9}=\tfrac{\sqrt{5}}{3}$. Minor axis of $E_1=2b_1=4$ becomes major axis of $E_2$, so $a_2=2$, $b_2=a_2\sqrt{1-e^2}=2\cdot\tfrac{2}{3}=\tfrac{4}{3}$. Generally $a_{i+1}=b_i$. Areas: $A_1=6\pi$, $A_2=\pi\cdot2\cdot\tfrac{4}{3}=\tfrac{8\pi}{3}$, $A_3=\pi\cdot\tfrac{4}{3}\cdot\tfrac{8}{9}=\tfrac{32\pi}{27}$, \ldots with common ratio $r=A_2/A_1=\tfrac{4}{9}$. $$\sum_{i=1}^\infty A_i=\frac{6\pi}{1-4/9}=\frac{54\pi}{5}.$$ Hence $\dfrac{5}{\pi}\cdot\dfrac{54\pi}{5}=54$.
Correct Answer: 54