Define the collections {E_1, E_2, E_3, ldots} | vedclass

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(D) For $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$,let a point on the ellipse be $(3 \cos 	heta, 2 \sin 	heta)$. The area of the inscribed rectangle $

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$3, 4$

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(D) For $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$,let a point on the ellipse be $(3 \cos 	heta, 2 \sin 	heta)$. The area of the inscribed rectangle $R_1$ is $A_1 = (2 \cdot 3 \cos 	heta)(2 \cdot 2 \sin 	heta) = 24 \sin 	heta \cos 	heta = 12 \sin 2 	heta$. This is maximum when $\sin 2 	heta = 1$,i.e.,$	heta = \frac{\pi}{4}$.Thus,the vertices of $R_1$ are $(\pm \frac{3}{\sqrt{2}}, \pm \frac{2}{\sqrt{2}})$.For $E_n$ inscribed in $R_{n-1}$,the semi-axes $a_n, b_n$ satisfy $a_n = \frac{a_{n-1}}{\sqrt{2}}$ and $b_n = \frac{b_{n-1}}{\sqrt{2}}$.Since the ratio $\frac{b_n}{a_n} = \frac{b_1}{a_1} = \frac{2}{3}$ is constant for all $n$,the eccentricity $e = \sqrt{1 - \frac{b_n^2}{a_n^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ is constant for all $E_n$. Thus,$(1)$ is incorrect.For $E_9$,$a_9 = \frac{3}{(\sqrt{2})^8} = \frac{3}{16}$ and $b_9 = \frac{2}{(\sqrt{2})^8} = \frac{2}{16} = \frac{1}{8}$.The distance of a focus from the centre is $a_9 e = \frac{3}{16} \cdot \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{16}$. Thus,$(2)$ is incorrect.The length of the latus rectum of $E_9$ is $\frac{2 b_9^2}{a_9} = \frac{2 (1/8)^2}{3/16} = \frac{2/64}{3/16} = \frac{1}{32} \cdot \frac{16}{3} = \frac{1}{6}$. Thus,$(3)$ is correct.The area of $R_n$ is $A_n = 4 a_n b_n = 4 \cdot \frac{3}{(\sqrt{2})^{n-1}} \cdot \frac{2}{(\sqrt{2})^{n-1}} = \frac{24}{2^{n-1}}$.The sum $\sum_{n=1}^N A_n = 24 (1 + \frac{1}{2} + \dots + \frac{1}{2^{N-1}}) = 24 \cdot \frac{1 - (1/2)^N}{1 - 1/2} = 48 (1 - \frac{1}{2^N}) = 48 - \frac{48}{2^N} = 48 - \frac{3 \cdot 16}{2^N}$. This is always less than $48$,but the question asks for comparison with $24$. The sum is $24(2 - 2^{1-N}) < 48$. Actually,$\sum_{n=1}^N A_n < 48$. Checking the options,$(3)$ and $(4)$ are correct.

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