Ellipse and Auxiliary Circle Intersections and Line Equations

To determine which statements are true, we start by finding the coordinates of the points on the circle $C$ and the ellipse.

1. Coordinates of Points $R$ and $S$ on Circle $C$

The circle equation is given by: $C: x^2 + y^2 = 9$ This circle has radius $r = 3$ and centered at the origin $O(0, 0)$. The point $M$ is $(3, 0)$, lying on the positive $x$-axis.

Since $R$ lies on the circle $C$ with $y_R > 0$ and $\angle ROM = \frac{\pi}{6}$: $R = \left(3 \cos \frac{\pi}{6}, 3 \sin \frac{\pi}{6}\right) = \left(3 \cdot \frac{\sqrt{3}}{2}, 3 \cdot \frac{1}{2}\right) = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$

Since $S$ lies on the circle $C$ with $y_S > 0$ and $\angle SOM = \frac{\pi}{3}$: $S = \left(3 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}\right) = \left(3 \cdot \frac{1}{2}, 3 \cdot \frac{\sqrt{3}}{2}\right) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$

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2. Coordinates of Points $P$ and $Q$ on the Ellipse

The ellipse equation is: $\frac{x^2}{9} + \frac{y^2}{4} = 1$

The vertical line $x = x_1 = \frac{3\sqrt{3}}{2}$ intersects the ellipse at $P(x_1, y_1)$ with $y_1 > 0$: $\frac{\left(\frac{3\sqrt{3}}{2}\right)^2}{9} + \frac{y_1^2}{4} = 1 \implies \frac{27/4}{9} + \frac{y_1^2}{4} = 1 \implies \frac{3}{4} + \frac{y_1^2}{4} = 1 \implies y_1 = 1$ Thus, $P = \left(\frac{3\sqrt{3}}{2}, 1\right)$.

The vertical line $x = x_2 = \frac{3}{2}$ intersects the ellipse at $Q(x_2, y_2)$ with $y_2 > 0$: $\frac{\left(\frac{3}{2}\right)^2}{9} + \frac{y_2^2}{4} = 1 \implies \frac{9/4}{9} + \frac{y_2^2}{4} = 1 \implies \frac{1}{4} + \frac{y_2^2}{4} = 1 \implies y_2 = \sqrt{3}$ Thus, $Q = \left(\frac{3}{2}, \sqrt{3}\right)$.

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3. Equation of the Line Joining $P$ and $Q$

The slope $m$ of the line segment $PQ$ is: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3\sqrt{3}}{2}} = \frac{2(\sqrt{3} - 1)}{3(1 - \sqrt{3})} = -\frac{2}{3}$

Using point-slope form with point $P\left(\frac{3\sqrt{3}}{2}, 1\right)$: $y - 1 = -\frac{2}{3} \left(x - \frac{3\sqrt{3}}{2}\right)$ $3(y - 1) = -2x + 3\sqrt{3}$ $2x + 3y = 3 + 3\sqrt{3} = 3(1 + \sqrt{3})$

This matches Option A (and refutes Option B).

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4. Checking Options C and D

For point $N_2 = (x_2, 0) = \left(\frac{3}{2}, 0\right)$:

• $|N_2 Q| = y_2 = \sqrt{3}$
• $|N_2 S| = y_S = \frac{3\sqrt{3}}{2}$

Evaluating the ratio for Option C: $3 |N_2 Q| = 3\sqrt{3}$ $2 |N_2 S| = 2 \left(\frac{3\sqrt{3}}{2}\right) = 3\sqrt{3}$ Since $3 |N_2 Q| = 2 |N_2 S|$, Option C is TRUE.

For point $N_1 = (x_1, 0) = \left(\frac{3\sqrt{3}}{2}, 0\right)$:

• $|N_1 P| = y_1 = 1$
• $|N_1 R| = y_R = \frac{3}{2}$

Evaluating the ratio for Option D: $9 |N_1 P| = 9(1) = 9$ $4 |N_1 R| = 4\left(\frac{3}{2}\right) = 6 \neq 9$ Thus, Option D is FALSE.

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Conclusion

The correct statements are A and C.