Question 7

The circle $C$ is $x^2 + y^2 = 9$ (radius 3). $R$ and $S$ are on $C$. $R = (3\cos(\pi/6), 3\sin(\pi/6)) = (\frac{3\sqrt{3}}{2}, \frac{3}{2})$, so $x_1 = \frac{3\sqrt{3}}{2}$. $S = (3\cos(\pi/3), 3\sin(\pi/3)) = (\frac{3}{2}, \frac{3\sqrt{3}}{2})$, so $x_2 = \frac{3}{2}$. $P$ is on the ellipse with $x = x_1$: $\frac{27/4}{9} + \frac{y_1^2}{4} = 1 \Rightarrow y_1 = 1 \Rightarrow P = (\frac{3\sqrt{3}}{2}, 1)$. $Q$ is on the ellipse with $x = x_2$: $\frac{9/4}{9} + \frac{y_2^2}{4} = 1 \Rightarrow y_2 = \sqrt{3} \Rightarrow Q = (\frac{3}{2}, \sqrt{3})$. The slope of $PQ$ is $m = \frac{\sqrt{3}-1}{3/2 - 3\sqrt{3}/2} = -\frac{2}{3}$. Line $PQ$: $y - 1 = -\frac{2}{3}(x - \frac{3\sqrt{3}}{2}) \Rightarrow 3y - 3 = -2x + 3\sqrt{3} \Rightarrow 2x + 3y = 3(1 + \sqrt{3})$. Statement (A) is true. $|N_2Q| = y_2 = \sqrt{3}$ and $|N_2S| = y_S = \frac{3\sqrt{3}}{2}$. Thus $3|N_2Q| = 3\sqrt{3}$ and $2|N_2S| = 3\sqrt{3}$, so (C) is true. For $N_1$, $|N_1P| = 1$ and $|N_1R| = 3/2$, so $3|N_1P| = 2|N_1R|$, which contradicts (D).