Let E1:x29+y24=1 be an ellipse. Ellipses Ei's are constructed such that their centres and eccentricities are same as that of E1, and the length of minor axis of Ei is the length of major axis of Ei+1 (i≥ 1). If Ai is the area of the ellipse Ei, then 5π∑i=1∞ Ai, , is equal to ……

E1=x29+y24⇒e=1−49=53E2:x2a2+y24=1e=53=1−a24⇒59=1−a24a2=169E2:x2169+y24=1E3:x2169+y2b2=1e=53=1−b216⇒b2=6481E3=x2169+y26481=1A1=π×3×2⇒6πA2=π×43×2=8π3A3=π×43×89=32π81∑i=1∞ Ai=6π+8π3+32π81+…∞⇒6π1−49⇒54π5∴5π∑i=1∞ Ai⇒5π×54π5=54

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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 E₁, and the length of minor axis of E_i is the length of major axis of E_i+1 (i $\geq$ 1). If Ai is the area of the ellipse Ei, then $\frac{5}{π} (\sum_{i = 1}^{\infty} A_{i}),$ , is equal to ……

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answer is 54.

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Detailed Solution

$\begin{array}{l} E_{1} = \frac{x^{2}}{9} + \frac{y^{2}}{4} \Rightarrow e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} \\ E_{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1 \\ e = \frac{\sqrt{5}}{3} = \sqrt{1 - \frac{a^{2}}{4}} \Rightarrow \frac{5}{9} = 1 - \frac{a^{2}}{4} \\ a^{2} = \frac{16}{9} \\ E_{2} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{4} = 1 \\ E_{3} : \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{b^{2}} = 1 \\ e = \frac{\sqrt{5}}{3} = \sqrt{1 - \frac{b^{2}}{16}} \Rightarrow b^{2} = \frac{64}{81} \\ E_{3} = \frac{x^{2}}{\frac{16}{9}} + \frac{y^{2}}{\frac{64}{81}} = 1 \\ A_{1} = π \times 3 \times 2 \Rightarrow 6 π \\ A_{2} = π \times \frac{4}{3} \times 2 = \frac{8 π}{3} \end{array}$

$\begin{array}{l} A_{3} = π \times \frac{4}{3} \times \frac{8}{9} = \frac{32 π}{81} \\ \sum_{i = 1}^{\infty} A_{i} = 6 π + \frac{8 π}{3} + \frac{32 π}{81} + \dots \infty \Rightarrow \frac{6 π}{1 - \frac{4}{9}} \Rightarrow \frac{54 π}{5} \\ ∴ \frac{5}{π} \sum_{i = 1}^{\infty} A_{i} \Rightarrow \frac{5}{π} \times \frac{54 π}{5} = 54 \end{array}$