Question

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

Hint:

Understanding the recursive properties of geometric series, especially in the context of areas of similar shapes, can simplify complex infinite sum calculations.

Answer 1

Step 1: Calculate the area of \(E_1\).

The area of the ellipse is given by:

\[ A_1 = \pi \times \text{semi-major axis} \times \text{semi-minor axis} = \pi \times \frac{3}{2} \times 2 = 3\pi \]

Step 2: Recursive relationship for areas.

Each successive ellipse \(E_{i+1}\) switches the axes, making the area:

\[ A_{i+1} = \pi \times \left(\frac{\text{semi-minor axis of } E_i}{2}\right)^2 \times \text{semi-major axis of } E_i \]

Since the minor axis becomes the major axis, the area relation forms a geometric series where each term is \(\left(\frac{2}{3}\right)^2\) of the previous term.

Step 3: Sum the infinite series.

\[ \sum_{i=1}^{\infty} A_i = A_1 + \left(\frac{4}{9}\right)A_1 + \left(\frac{4}{9}\right)^2A_1 + \ldots = 3\pi \left(\frac{1}{1-\frac{4}{9}}\right) = \frac{27\pi}{5} \]

Step 4: Compute the final result.

\[ \frac{5}{\pi} \sum_{i=1}^{\infty} A_i = \frac{5}{\pi} \times \frac{27\pi}{5} = 27 \]