Let E₁: x²/9 + y²/4 = 1 be an ellipse. Ellipses E_i's are constructed such that their centres and ec

Q: Why is the eccentricity constant? Does that mean the ellipses are similar?

Yes! Constant eccentricity means the ellipses are geometrically similar. They have the same 'shape' or 'flatness', just scaled differently. This is why the ratio $b_i/a_i$ remains constant for all $i$.

Q: I'm confused about the axes. Is the minor axis of $E_i$ always the major axis of $E_{i+1}$?

Exactly. The problem defines this as the bridge between successive ellipses. It's like a relay race where the finish line of one runner is the starting line of the next. Mathematically, $2b_i = 2a_{i+1}$, which simplifies to $a_{i+1} = b_i$.

Q: Why does the infinite sum converge?

Because the common ratio $r$ of the areas is $4/9$, which is strictly between $-1$ and $1$. Any geometric series with a common ratio $|r| < 1$ will converge to a finite sum. You are on the right track!

Question

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 $E_{i+1}(i\ge1).$ If $A_{i}$ is the area of the ellipse $E_{i},$ then $\frac{5}{\pi}(\sum_{i=1}^{\infty}A_{i})$, is equal to

Sigma Percentile · Accepted Answer

### The Infinite Dance of Ellipses

Imagine standing before a vast, empty canvas. We are about to construct a sequence of ellipses, each one smaller than the last, shrinking into the horizon of infinity. This isn't just a math problem; it's a study of geometric harmony.

### Phase 1: The Geometry of the First Ellipse

We begin with our foundational shape, $E_1$, defined by the equation:

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

By comparing this to the standard form $\frac{x^2}{a_1^2} + \frac{y^2}{b_1^2} = 1$, we immediately identify our parameters: $a_1^2 = 9$ and $b_1^2 = 4$.

This gives us a semi-major axis $a_1 = 3$ and a semi-minor axis $b_1 = 2$. This ellipse is our starting point, our anchor in the coordinate plane.

### Phase 2: The Constant Eccentricity

Now, here is the secret that binds this entire sequence together. The problem tells us that all ellipses $E_i$ share the same eccentricity $e$.

Recall that $e = \sqrt{1 - \frac{b_i^2}{a_i^2}}$. If the eccentricity is constant, the ratio $\frac{b_i}{a_i}$ must also be constant.

For our first ellipse, this ratio is $\frac{b_1}{a_1} = \frac{2}{3}$. Therefore, for any ellipse in our infinite sequence, the semi-minor axis will always be exactly two-thirds of the semi-major axis:

$$
b_i = \frac{2}{3} a_i
$$

This is the 'DNA' of our ellipses—no matter how small they get, their shape remains perfectly preserved.

### Phase 3: The Bridge Between Worlds

How do we move from one ellipse to the next? The problem provides the key: the minor axis of $E_i$ is the major axis of $E_{i+1}$.

Mathematically, this is $2b_i = 2a_{i+1}$, which simplifies beautifully to $a_{i+1} = b_i$. Now, we combine this with our DNA ratio.

Since $a_{i+1} = b_i$ and $b_i = \frac{2}{3} a_i$, we find that:

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

This reveals that the semi-major axes form a geometric progression with a common ratio $r = \frac{2}{3}$.

### Phase 4: The Infinite Summation

We are interested in the sum of the areas $A_i$. The area of an ellipse is $A = \pi ab$.

For $E_i$, this is:

$$
A_i = \pi a_i b_i = \pi a_i \left( \frac{2}{3} a_i ight) = \frac{2\pi}{3} a_i^2
$$

Since $a_i$ forms a geometric progression with ratio $\frac{2}{3}$, the area $A_i$ (which is proportional to $a_i^2$) forms a geometric progression with a common ratio $R = (\frac{2}{3})^2 = \frac{4}{9}$.

Our first area is $A_1 = \pi (3)(2) = 6\pi$. The sum of an infinite geometric series is $S = \frac{a}{1 - R}$.

Substituting our values, we get:

$$
\sum_{i=1}^{\infty} A_i = \frac{6\pi}{1 - \frac{4}{9}} = \frac{6\pi}{\frac{5}{9}} = \frac{54\pi}{5}
$$

### The Final Elegance

We have reached the summit. The question asks for the value of $\frac{5}{\pi} \left( \sum_{i=1}^{\infty} A_i ight)$.

Substituting our sum, we calculate:

$$
\frac{5}{\pi} 	imes \frac{54\pi}{5}
$$

The $\pi$ terms cancel, the $5$ terms cancel, and we are left with the clean, satisfying integer: **54**. You have successfully navigated the infinite, turning a complex sequence into a simple, elegant result.

Enter Numerical Value:

Visualized Solution

Analyzing the first ellipse $E_{1}$

Constant Eccentricity Condition

Linking Successive Ellipses

Forming the Geometric Progression

Area of the First Ellipse $A_{1}$

Ratio of Successive Areas

Sum of Infinite Areas

Evaluating the Infinite Sum

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Infinite Dance of Ellipses

Phase 1

Phase 2

Phase 3

Phase 4

The Final Elegance

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