Let be an ellipse. Ellipse are constructed such that their centres and eccentricities are same as that of , and the length of minor axis of is the length of major axis of . If is the area of the ellipse , then , is equal to __________. [2025]

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Solution

Q.
Let $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ be an ellipse. Ellipse $E_{i}' s$ 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 $\frac{5}{π} (\sum_{i = 1}^{\infty} A_{i})$ , is equal to __________. [2025]

Ans.
(54)

Given, $E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{\sqrt{5}}{3}$

As length of minor axis of $E_{i}$ is the length of major axis of $E_{i + 1}$ .

$E_{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1 \Rightarrow e = \sqrt{1 - \frac{a^{2}}{4^{2}}} = \frac{\sqrt{5}}{3}$ [ $∵$ Eccentricities are same]

$\Rightarrow a = \frac{4}{3}$ $∴ E_{2} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{4} = 1$

Now, $E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{b^{2}}{\frac{16}{9}}} = \frac{\sqrt{5}}{3}$

$\Rightarrow b^{2} = \frac{64}{81}$ $∴ E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{\frac{64}{81}} = 1$

Now, $A_{1} = π a b = π \times 3 \times 2 = 6 π$ ,

$A_{2} = π \times \frac{4}{3} \times 2 = \frac{8}{3} π$

and $A_{3} = π \times \frac{4}{3} \times \frac{8}{9} = \frac{32}{27} π$

$∴ \sum_{i = 1}^{\infty} A_{i} = 6 π + \frac{8}{3} π + \frac{32}{27} π + . . . \infty$

$= \frac{6 π}{1 - \frac{4}{9}} = \frac{54 π}{5}$ $[∵ S_{\infty} = \frac{a}{1 - r}]$

So, $\frac{5}{π} \sum_{i = 1}^{\infty} A_{i} = \frac{5}{π} \times \frac{54 π}{5} = 54$ .

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