JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{E}_1: \frac{x^2}{9}+\frac{y^2}{4}=1\) be an ellipse. Ellipses \(\mathrm{E}_1\) '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 \(\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)\), is equal to
- A 50
- B 52
- C 54
- D 56
Answer & Solution
Correct Answer
(C) 54
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{E}_1=\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{4} \Rightarrow \mathrm{e}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \\ & \mathrm{E}_2: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{4}=1 \\ &…
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