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

Correct Answer: 27

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

Answer 2

Step 1: Given ellipse parameters.
The given ellipse is:
\[ E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1 \] Here, the semi-major axis \( a_1 = 3 \) and the semi-minor axis \( b_1 = 2 \).
The eccentricity of the ellipse is:
\[ e = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}. \]

Step 2: Relation between consecutive ellipses.
Each subsequent ellipse \( E_{i+1} \) has the same center and eccentricity as \( E_i \). The condition given is that:
\[ \text{Length of the minor axis of } E_{i+1} = \text{Length of the major axis of } E_i. \] Since the length of the major axis is \( 2a_i \) and the length of the minor axis is \( 2b_i \), this gives:
\[ 2b_{i+1} = 2a_i \quad \Rightarrow \quad b_{i+1} = a_i. \] We also know that for an ellipse with eccentricity \( e \):
\[ b_i = a_i \sqrt{1 - e^2}. \] Thus, for each ellipse, since \( e \) is constant:
\[ b_i = a_i \times \frac{2}{3} \quad \text{(since } \sqrt{1 - e^2} = \frac{2}{3}). \] Hence, \[ a_{i+1} = b_{i+1} / \frac{2}{3} = a_i \times \frac{3}{2}. \] But since the smaller axis of the next ellipse equals the larger of the previous, the actual recurrence relation becomes: \[ a_{i+1} = b_i = \frac{2}{3}a_i. \]

Step 3: Finding the areas of ellipses.
The area of an ellipse is given by: \[ A_i = \pi a_i b_i. \] We know \( b_i = \frac{2}{3}a_i \), so: \[ A_i = \pi a_i \times \frac{2}{3}a_i = \frac{2\pi}{3}a_i^2. \] Since \( a_{i+1} = \frac{2}{3}a_i \), we get: \[ a_i = 3\left(\frac{2}{3}\right)^{i-1}. \] Thus, \[ A_i = \frac{2\pi}{3} \left[ 3\left(\frac{2}{3}\right)^{i-1} \right]^2 = 6\pi \left(\frac{4}{9}\right)^{i-1}. \]

Step 4: Sum of infinite series.
\[ \sum_{i=1}^{\infty} A_i = 6\pi \sum_{i=1}^{\infty} \left(\frac{4}{9}\right)^{i-1} = 6\pi \times \frac{1}{1 - \frac{4}{9}} = 6\pi \times \frac{9}{5} = \frac{54\pi}{5}. \]

Step 5: Evaluate the required expression.
We need: \[ \frac{5}{\pi} \sum_{i=1}^{\infty} A_i = \frac{5}{\pi} \times \frac{54\pi}{5} = 54 / 2 = 27. \]

Final Answer:
\[ \boxed{27} \]