Question 7

MCQHARD

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two distinct points on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ such that $y_1 > 0$ and $y_2 > 0$ . Let $C$ denote the circle $x^2 + y^2 = 9$ , and $M$ be the point $(3, 0)$ . Suppose the line $x = x_1$ intersects $C$ at $R$ , and the line $x = x_2$ intersects $C$ at $S$ , such that the $y$ -coordinates of $R$ and $S$ are positive. Let $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$ , where $O$ denotes the origin $(0, 0)$ . Let $|XY|$ denote the length of the line segment $XY$ . Then which of the following statements is (are) TRUE?

(A)

The equation of the line joining $P$ and $Q$ is $2x + 3y = 3(1 + \sqrt{3})$

(B)

The equation of the line joining $P$ and $Q$ is $2x + y = 3(1 + \sqrt{3})$

(C)

If $N_2 = (x_2, 0)$ , then $3|N_2Q| = 2|N_2S|$

(D)

If $N_1 = (x_1, 0)$ , then $9|N_1P| = 4|N_1R|$

Detailed Solution

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).

Question 7

Detailed Solution

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