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Ellipse

Section: Mathematics

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Question : 29 of 117

Marks: +1, -0

Let

E_1: \;\frac{x^2}{9}+\;\frac{y^2}{4}=1

be an ellipse. Ellipses

E_i

's is constructed such that their centres and eccentricities are same as that of

E_1

, and length of minor axis of

E_i

is the length of major axis of

E_{i+1}(i \ge 1)

A_i

is the area of the ellipse

E_i

\;\frac{5}{\pi} \left( \sum\limits_{i=1}^{\infty} A_i \right)

, is equal to ______

[28 Jan 2025 Shift 1]

Your Answer:

Solution:

E_i: \;\frac{x^2}{9}+\;\frac{y^2}{4}=1

Let

b_i

be minor axis of

E_i

and

a_i

be major axis of

E_i

\Rightarrow \;\; e_i=1-\;\frac{b_i^2}{a_i^2}

Now,

b_{i+1}

be minor axis of

E_{i+1}

and

a_{i+1}

be major axis of

E_{i+1}

\Rightarrow e_{i+1}=1-\;\frac{b_{i+1}^2}{a_{i+1}^2}

Also

a_{i+1}=b_i

and

e_i=e_{i+1}

\Rightarrow \;\; \frac{b_i^2}{a_i^2}=\;\frac{b_{i+1}^2}{b_i^2}

\Rightarrow \;\; b_{i+1}=\;\frac{b_i^2}{a_i}

⇒ Area of

E_i=S_i=\pi a_i b_i

\Rightarrow S_{i+1}=\pi a_{i+1} b_{i+1}

\;=\pi (b_i) \left( \;\frac{b_i^2}{a_i} \right)

\;=\pi (b_i a_i) \left( \frac{b_i}{a_i} \right)^2

\;S_{i+1}=S_i (1-e_i^2)

\;S_{i+1}=S_i \left(1- \left(1-\frac{4}{9}\right) \right) =S_i \left( \frac{4}{9} \right)

\;\Rightarrow S_1=6\pi,\; S_2=6\pi \left( \frac{4}{9} \right),\; S_3=6\pi \left( \frac{4}{9} \right)^2

\;\sum\limits_{k=1}^{\infty} S_i = \left( \frac{6\pi}{1-\frac{4}{9}} \right) = \frac{54\pi}{5}

\;\Rightarrow \frac{5}{\pi} \sum\limits_{k=1}^{\infty} S_i = \frac{5}{\pi} \cdot \frac{54\pi}{5} = 54

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