Ellipse E1 With Minor Axis Equal to Next Ellipse’s Major Axis, Sum of Areas

JEE Main 2025 — 28 January, Shift 1. Previous Year Question.

Problem

Let $E_1: \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ be an ellipse. Ellipses $E_i$ are constructed such that their centres and eccentricities are the same as that of $E_1$, and the length of the minor axis of $E_i$ is the length of the major axis of $E_{i+1}$ ($i \geq 1$). If $A_i$ is the area of the ellipse $E_i$, then $\dfrac{5}{\pi}\left(\displaystyle\sum_{i=1}^{\infty} A_i\right)$ is equal to?

Key insight. Every $E_i$ shares the same eccentricity, which fixes a constant ratio between each ellipse’s semi-minor and semi-major axis. Combined with “minor axis of $E_i$ = major axis of $E_{i+1}$”, that ratio alone tells you exactly how fast the ellipses shrink — turning the whole construction into a single geometric series.

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Approach

First pin down the eccentricity of $E_1$ and use it to relate each ellipse’s semi-minor axis $b_i$ to its own semi-major axis $a_i$. Then use the given rule linking $b_i$ to $a_{i+1}$ to show that $a_i$ itself forms a geometric progression — which makes the areas $A_i = \pi a_ib_i$ a geometric series too, summable in closed form.

Solution

Step 1 — Find the eccentricity of $E_1$

For $E_1$, $a_1 = 3$, $b_1 = 2$, so:

$$e^2 = 1 – \frac{b_1^2}{a_1^2} = 1 – \frac{4}{9} = \frac{5}{9}$$

Since every $E_i$ shares this eccentricity, for every $i$:

$$\frac{b_i}{a_i} = \sqrt{1 – e^2} = \sqrt{\frac{4}{9}} = \frac{2}{3} \implies b_i = \frac{2}{3}a_i$$

Step 2 — Turn the axis rule into a recurrence for $a_i$

“Minor axis of $E_i$ = major axis of $E_{i+1}$” means $2b_i = 2a_{i+1}$, i.e. $b_i = a_{i+1}$. Combining with Step 1:

$$a_{i+1} = \frac{2}{3}a_i$$

So $a_1, a_2, a_3, \ldots$ is a G.P. with first term $a_1 = 3$ and common ratio $\dfrac{2}{3}$:

$$a_i = 3\left(\frac{2}{3}\right)^{i-1}$$

Step 3 — Write the area of each ellipse

$$A_i = \pi a_i b_i = \pi a_i \cdot \frac{2}{3}a_i = \frac{2\pi}{3}a_i^2$$

Since $a_i^2 = 9\left(\dfrac{4}{9}\right)^{i-1}$, the areas also form a geometric series with common ratio $\dfrac{4}{9}$.

Step 4 — Sum the infinite geometric series

$$\sum_{i=1}^{\infty} a_i^2 = 9 \cdot \frac{1}{1 – \frac{4}{9}} = 9 \cdot \frac{9}{5} = \frac{81}{5}$$

$$\sum_{i=1}^{\infty} A_i = \frac{2\pi}{3} \cdot \frac{81}{5} = \frac{54\pi}{5}$$

Step 5 — Apply the final $\frac{5}{\pi}$ factor

$$\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right) = \frac{5}{\pi} \cdot \frac{54\pi}{5} = 54$$

Answer

Common mistakes

• Assuming the ellipses shrink by a fixed area ratio directly, without first establishing that it’s the semi-major axis $a_i$ that forms a clean G.P. Working with $a_i$ first, then squaring, keeps the algebra simple.
• Mixing up axis length with semi-axis length. The problem gives minor axis = major axis of the next ellipse (full lengths, $2b_i = 2a_{i+1}$), which conveniently simplifies to the same relation between the semi-axes, but it’s worth checking that factor of 2 cancels rather than assuming it does.

Practise next

• Redo this problem starting from $E_1: \dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, and find $\dfrac{7}{\pi}\sum A_i$ using the same shrinking-ellipse construction.