Let $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ be an ellipse. Ellipses $E_i$ 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
Solution:
For $E_1$: $a_1 = 3, b_1 = 2$.
Eccentricity $e$:
$$ b_1^2 = a_1^2(1 – e^2) \Rightarrow 4 = 9(1 – e^2) \Rightarrow 1 – e^2 = \frac{4}{9} $$
Relation between ellipses:
Length of minor axis of $E_i$ = Length of major axis of $E_{i+1}$.
$$ 2b_i = 2a_{i+1} \Rightarrow b_i = a_{i+1} $$
Since all ellipses have the same eccentricity $e$:
$$ b_i = a_i \sqrt{1-e^2} = a_i \left(\frac{2}{3}\right) $$
So, $a_{i+1} = \frac{2}{3} a_i$.
This means $a_i$ forms a Geometric Progression (GP) with common ratio $r_a = 2/3$.
Area of ellipse $E_i$:
$$ A_i = \pi a_i b_i = \pi a_i (a_i \cdot \frac{2}{3}) = \frac{2\pi}{3} a_i^2 $$
Since $a_i$ is a GP with $r = 2/3$, $a_i^2$ is a GP with ratio $R = (2/3)^2 = 4/9$.
Sum of Areas $S = \sum_{i=1}^{\infty} A_i$:
$$ S = \frac{2\pi}{3} \sum_{i=1}^{\infty} a_i^2 $$
$$ \therefore S = \sum a_i^2 = \frac{a_1^2}{1 – 4/9} = \frac{9}{5/9} = \frac{81}{5} $$
$$ \therefore S = \frac{2\pi}{3} \times \frac{81}{5} = \frac{54\pi}{5} $$
$$ \therefore\frac{5}{\pi} S = \frac{5}{\pi} \left( \frac{54\pi}{5} \right) = 54 $$
Ans. 54