Sequences of Inscribed Ellipses and Rectangles of Maximum Area

To determine the correct options, we analyze the sequences of ellipses $\{E_n\}$ and rectangles $\{R_n\}$ step-by-step.

1. Finding the General Terms for $E_n$ and $R_n$

Given the first ellipse: $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ Here, the semi-major axis is $a_1 = 3$ and the semi-minor axis is $b_1 = 2$.

A rectangle inscribed in the ellipse $\frac{x^2}{a_1^2} + \frac{y^2}{b_1^2} = 1$ with sides parallel to the coordinate axes has vertices at $(\pm a_1 \cos\theta, \pm b_1 \sin\theta)$. Its area is given by: $\text{Area} = (2a_1 \cos\theta)(2b_1 \sin\theta) = 2a_1 b_1 \sin(2\theta)$ The area is maximized when $\sin(2\theta) = 1$, i.e., $\theta = \frac{\pi}{4}$. Thus, the vertices of the maximum-area rectangle $R_1$ are: $\left(\pm \frac{a_1}{\sqrt{2}}, \pm \frac{b_1}{\sqrt{2}}\right)$ The dimensions of $R_1$ are $2 \cdot \frac{a_1}{\sqrt{2}} \times 2 \cdot \frac{b_1}{\sqrt{2}}$, and its area is: $\text{Area}(R_1) = 2 a_1 b_1 = 2 \times 3 \times 2 = 12$

Next, $E_2$ is the ellipse of largest area inscribed in the rectangle $R_1$. The ellipse inscribed in a rectangle of sides $2A \times 2B$ centered at the origin has maximum area when its semi-axes equal the half-lengths of the sides of the rectangle: $a_2 = \frac{a_1}{\sqrt{2}}, \quad b_2 = \frac{b_1}{\sqrt{2}}$

Continuing this process for any integer $n \ge 1$: $a_n = \frac{a_1}{(\sqrt{2})^{n-1}} = \frac{3}{2^{(n-1)/2}}$ $b_n = \frac{b_1}{(\sqrt{2})^{n-1}} = \frac{2}{2^{(n-1)/2}}$

The area of rectangle $R_n$ inscribed in $E_n$ is: $\text{Area}(R_n) = 2 a_n b_n = \frac{2 a_1 b_1}{\left(\sqrt{2}\right)^{2(n-1)}} = \frac{12}{2^{n-1}}$

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2. Evaluation of the Options

• Option (A): The eccentricity of an ellipse $E_n$ is: $e_n = \sqrt{1 - \frac{b_n^2}{a_n^2}} = \sqrt{1 - \frac{b_1^2}{a_1^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ Since the ratio $\frac{b_n}{a_n} = \frac{2}{3}$ is constant for all $n$, the eccentricity is identical for all ellipses $E_n$. Thus, the eccentricities of $E_{18}$ and $E_{19}$ are equal. Therefore, Option (A) is incorrect.

• Option (B): The sum of the areas of the first $N$ rectangles is: $\sum_{n=1}^N \text{Area}(R_n) = \sum_{n=1}^N \frac{12}{2^{n-1}} = 12 \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{N-1}} \right) = 12 \left( \frac{1 - (1/2)^N}{1 - 1/2} \right) = 24 \left(1 - \frac{1}{2^N}\right)$ Since $\frac{1}{2^N} > 0$ for all positive integers $N$, it follows that: $\sum_{n=1}^N \text{Area}(R_n) < 24 \quad \text{for each positive integer } N$ Therefore, Option (B) is correct.

• Option (C): For $n = 9$: $a_9 = \frac{3}{2^{(9-1)/2}} = \frac{3}{2^4} = \frac{3}{16}$ $b_9 = \frac{2}{2^{(9-1)/2}} = \frac{2}{2^4} = \frac{2}{16} = \frac{1}{8}$ The length of the latus rectum of $E_9$ is given by: $\text{L.R.} = \frac{2 b_9^2}{a_9} = \frac{2 \left(\frac{1}{8}\right)^2}{\frac{3}{16}} = \frac{2 \times \frac{1}{64}}{\frac{3}{16}} = \frac{\frac{1}{32}}{\frac{3}{16}} = \frac{1}{6}$ Therefore, Option (C) is correct.

• Option (D): The distance of a focus from the centre in $E_9$ is given by $a_9 e_9$: $\text{Distance} = a_9 e_9 = \left(\frac{3}{16}\right) \left(\frac{\sqrt{5}}{3}\right) = \frac{\sqrt{5}}{16} \neq \frac{\sqrt{5}}{32}$ Therefore, Option (D) is incorrect.

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Conclusion

The correct options are B and C.