Let P(x_1, y_1) and Q(x_2, y_2) be two distin | vedclass

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(A, C) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Any point on the ellipse can be represented as $(3 \cos 	heta, 2 \sin 	heta)$.Given $

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$A, D$

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(A, C) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Any point on the ellipse can be represented as $(3 \cos 	heta, 2 \sin 	heta)$.Given $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$,where $M=(3,0)$ and $O=(0,0)$,the points $R$ and $S$ lie on the circle $x^2+y^2=9$. Thus,$R = (3 \cos \frac{\pi}{6}, 3 \sin \frac{\pi}{6}) = (3 \frac{\sqrt{3}}{2}, 3 \frac{1}{2}) = (\frac{3\sqrt{3}}{2}, \frac{3}{2})$ and $S = (3 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}) = (3 \frac{1}{2}, 3 \frac{\sqrt{3}}{2}) = (\frac{3}{2}, \frac{3\sqrt{3}}{2})$.Since $x_1$ is the $x$-coordinate of $R$,$x_1 = \frac{3\sqrt{3}}{2}$. Since $P$ is on the ellipse,$P = (x_1, y_1) = (\frac{3\sqrt{3}}{2}, 2 \sin 	heta_1)$. Since $\frac{x_1^2}{9} + \frac{y_1^2}{4} = 1$,we have $\frac{27/4}{9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{3}{4} + \frac{y_1^2}{4} = 1 \Rightarrow y_1^2 = 1 \Rightarrow y_1 = 1$ (as $y_1 > 0$). So $P = (\frac{3\sqrt{3}}{2}, 1)$.Similarly,$x_2 = \frac{3}{2}$. For $Q = (x_2, y_2)$,$\frac{9/4}{9} + \frac{y_2^2}{4} = 1 \Rightarrow \frac{1}{4} + \frac{y_2^2}{4} = 1 \Rightarrow y_2^2 = 3 \Rightarrow y_2 = \sqrt{3}$. So $Q = (\frac{3}{2}, \sqrt{3})$.The line joining $P$ and $Q$ has slope $m = \frac{\sqrt{3}-1}{3/2 - 3\sqrt{3}/2} = \frac{\sqrt{3}-1}{-\frac{3}{2}(\sqrt{3}-1)} = -\frac{2}{3}$.The equation is $y - \sqrt{3} = -\frac{2}{3}(x - \frac{3}{2}) \Rightarrow 3y - 3\sqrt{3} = -2x + 3 \Rightarrow 2x + 3y = 3(1+\sqrt{3})$. Thus,$(A)$ is true.For $(C)$,$N_2 = (x_2, 0) = (\frac{3}{2}, 0)$. $|N_2Q| = y_2 = \sqrt{3}$ and $|N_2S| = \frac{3\sqrt{3}}{2}$. $3|N_2Q| = 3\sqrt{3}$ and $2|N_2S| = 2(\frac{3\sqrt{3}}{2}) = 3\sqrt{3}$. So $3|N_2Q| = 2|N_2S|$,$(C)$ is true.

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