Let $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ be an ellipse. Ellipses $\mathrm{E}_{\mathrm{i}}$'s are constructed such that their centres and eccentricities are same as that of $\mathrm{E}_{1}$, and the length of minor axis of $\mathrm{E}_{\mathrm{i}}$ is the length of major axis of $\mathrm{E}_{\mathrm{i}+1}(\mathrm{i} \geq 1)$. If $\mathrm{A}_{\mathrm{i}}$ is the area of the ellipse $\mathrm{E}_{\mathrm{i}}$, then $\frac{5}{\pi}\left(\sum_{\mathrm{i}=1}^{\infty} \mathrm{A}_{\mathrm{i}}\right)$, is equal to

Ellipse — JEE Main · 2025 · 28 Jan 2025 · Shift 1 Question with Solution

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

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

Answer: 54

For

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

, semi-axes are

3, 2

, so eccentricity

e = 1 - \frac{4}{9} = \frac{5}{3}

.

The minor axis of

E_{1}

has length

4

, which becomes the major axis of

E_{2}

2 a_{2} = 4 \Rightarrow a_{2} = 2

.

Keeping the same eccentricity:

\frac{5}{3} = 1 - \frac{b _{2}^{2}}{4} \Rightarrow b_{2}^{2} = \frac{16}{9}

, so minor axis of

E_{2} = 2 b_{2} = \frac{8}{3}

.

This becomes the major axis of

E_{3}

a_{3} = \frac{4}{3}

, and similarly

b_{3}^{2} = \frac{64}{81}

.

Areas:

A_{1} = π (3) (2) = 6 π

A_{2} = π (2) (\frac{4}{3}) = \frac{8 π}{3}

A_{3} = π (\frac{4}{3}) (\frac{8}{9}) = \frac{32 π}{27}

Each ratio

\frac{A _{i + 1}}{A _{i}} = \frac{4}{9}

, a geometric series with first term

6 π

and common ratio

\frac{4}{9}

i = 1 \sum \infty A_{i} = \frac{6 π}{1 - \frac{4}{9}} = \frac{6 π}{5/9} = \frac{54 π}{5}

\frac{5}{π} i = 1 \sum \infty A_{i} = \frac{5}{π} \cdot \frac{54 π}{5} = 54

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Integer answer