ધારો કે E_1: frac{x^2}{9}+frac{y^2}{4}=1 એક ઉ | vedclass

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Question

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$E_1=\frac{x^2}{9}+\frac{y^2}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$
$E_2: \frac{x^2}{a^2}+\frac{y^2}{4}=1$
$e=\frac{\sqrt

Vedclass · Accepted Answer

$54$

Vedclass · Answer

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$E_1=\frac{x^2}{9}+\frac{y^2}{4} \Rightarrow e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$
$E_2: \frac{x^2}{a^2}+\frac{y^2}{4}=1$
$e=\frac{\sqrt{5}}{3}=\sqrt{1-\frac{a^2}{4}} \Rightarrow \frac{5}{9}=1-\frac{a^2}{4}$
$ a ^2=\frac{16}{9}$
$E _2: \frac{ x ^2}{\frac{16}{9}}+\frac{ y ^2}{4}=1$
$E _3: \frac{ x ^2}{\frac{16}{9}}+\frac{ y ^2}{b^2}=1$
$e =\frac{\sqrt{5}}{3}=\sqrt{1-\frac{ b ^2}{\frac{16}{9}}} \Rightarrow b^2=\frac{64}{81}$
$ E _3=\frac{ x ^2}{\frac{16}{9}}+\frac{ y ^2}{\frac{64}{81}}=1$
$A_1=\pi 	imes 3 	imes 2 \Rightarrow 6 \pi$
$A_2=\pi 	imes \frac{4}{3} 	imes 2=\frac{8 \pi}{3}$
$A_3=\pi 	imes \frac{4}{3} 	imes \frac{8}{9}=\frac{32 \pi}{81}$
$\sum_{ i =1}^{\infty} A _{ i }=6 \pi+\frac{8 \pi}{3}+\frac{32 \pi}{81}+\ldots \infty \Rightarrow \frac{6 \pi}{1-\frac{4}{9}} \Rightarrow \frac{54 \pi}{5}$
$	herefore \frac{5}{\pi} \sum_{ i =1}^{\infty} A _{ i } \Rightarrow \frac{5}{\pi} 	imes \frac{54 \pi}{5}=54$

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