Question

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 \( 9|N_1P| = 4|N_1R| \)

Hint:

Use dot product for angle geometry, and coordinate geometry for line equations. Convert all geometry constraints into algebraic equations to solve efficiently.

Answer 1

The Correct Option is A, C, D

Step 1: Use angle information to find coordinates Given \( \angle ROM = \frac{\pi}{6} \) and \( \angle SOM = \frac{\pi}{3} \), use triangle geometry to find coordinates of \( R \) and \( S \) on the circle \( x^2 + y^2 = 9 \). Let \( M = (3, 0) \), \( O = (0, 0) \) From triangle \( \angle ROM = \frac{\pi}{6} \), use Law of Sines: \[ \sin(\angle ROM) = \frac{RM}{RO} \cdot \sin(\angle RMO) \Rightarrow \sin\left(\frac{\pi}{6}\right) = \frac{RM}{RO} \cdot \sin(\angle RMO) \] But more directly, since \( R \) lies on vertical line \( x = x_1 \) and on the circle \( x^2 + y^2 = 9 \), we can use coordinate geometry. Let \( R = (x_1, y_R) \in C \Rightarrow x_1^2 + y_R^2 = 9 \Rightarrow y_R = \sqrt{9 - x_1^2} \) Similarly, since angle \( \angle ROM = \frac{\pi}{6} \), use dot product formula: \[ \cos(\angle ROM) = \frac{\vec{OR} \cdot \vec{OM}}{|OR||OM|} = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] Let \( R = (x_1, y_R) \), \( M = (3, 0) \), \( O = (0, 0) \) So, \[ \vec{OR} \cdot \vec{OM} = x_1 \cdot 3 + y_R \cdot 0 = 3x_1 \\ |OR| = \sqrt{x_1^2 + y_R^2} = 3,\quad |OM| = 3 \] \[ \Rightarrow \frac{3x_1}{3 \cdot 3} = \frac{\sqrt{3}}{2} \Rightarrow x_1 = \frac{3\sqrt{3}}{2} \] Now use circle equation: \[ x_1^2 + y_R^2 = 9 \Rightarrow \left(\frac{9 \cdot 3}{4}\right) + y_R^2 = 9 \Rightarrow y_R^2 = 9 - \frac{27}{4} = \frac{9}{4} \Rightarrow y_R = \frac{3}{2} \] So \( R = \left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right) \) Similarly for \( S \), \( x_2 = \frac{3\sqrt{3}}{2} \), angle \( \angle SOM = \frac{\pi}{3} \), we get: \[ \cos\left( \frac{\pi}{3} \right) = \frac{3x_2}{9} = \frac{1}{2} \Rightarrow x_2 = \frac{3}{2} \Rightarrow y_S = \sqrt{9 - x_2^2} = \sqrt{9 - \frac{9}{4}} = \sqrt{\frac{27}{4}} = \frac{3\sqrt{3}}{2} \] So \( S = \left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right) \)
Step 2: Use ellipse equation to get points P and Q Given ellipse: \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) Let \( P = (x_1, y_1) \), \( x_1 = \frac{3\sqrt{3}}{2} \), solve for \( y_1 \): \[ \frac{27}{4 \cdot 9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{3}{4} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{y_1^2}{4} = \frac{1}{4} \Rightarrow y_1 = 1 \Rightarrow P = \left( \frac{3\sqrt{3}}{2}, 1 \right) \] Similarly for \( Q = (x_2, y_2) \), \( x_2 = \frac{3}{2} \Rightarrow \frac{1}{4} + \frac{y_2^2}{4} = 1 \Rightarrow y_2 = \sqrt{3} \) So \( Q = \left( \frac{3}{2}, \sqrt{3} \right) \)
Step 3: Equation of line PQ Two points: \[ P = \left( \frac{3\sqrt{3}}{2}, 1 \right),\quad Q = \left( \frac{3}{2}, \sqrt{3} \right) \] Use two-point form: Slope: \[ m = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3\sqrt{3}}{2}} = \frac{\sqrt{3} - 1}{\frac{3}{2}(1 - \sqrt{3})} = -\frac{2}{3} \] So line: \( y - 1 = -\frac{2}{3}(x - \frac{3\sqrt{3}}{2}) \) Multiply through: \[ 3y - 3 = -2x + 3\sqrt{3} \Rightarrow 2x + 3y = 3(1 + \sqrt{3}) \Rightarrow \text{(A) is correct} \]
Step 4: Check Option (C) Let \( N_2 = (x_2, 0) = \left( \frac{3}{2}, 0 \right),\ Q = \left( \frac{3}{2}, \sqrt{3} \right),\ S = \left( \frac{3}{2}, \frac{3\sqrt{3}}{2} \right) \) \[ |N_2Q| = \sqrt{(\sqrt{3})^2} = \sqrt{3} \\ |N_2S| = \frac{3\sqrt{3}}{2} \Rightarrow 2|N_2S| = 3\sqrt{3} \Rightarrow 3|N_2Q| = 3\sqrt{3} = 2|N_2S| \Rightarrow \text{(C) is correct} \] Step 5: Check Option (D) Similarly, \( N_1 = (x_1, 0) = \left( \frac{3\sqrt{3}}{2}, 0 \right),\ P = \left( \frac{3\sqrt{3}}{2}, 1 \right),\ R = \left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right) \) Then: \[ |N_1P| = 1,\quad |N_1R| = \frac{3}{2} \Rightarrow 9|N_1P| = 9,\quad 4|N_1R| = 6 \Rightarrow \text{Not equal ⇒ FALSE} \] Wait! Must check: \[ 9|N_1P| = 9 \quad ; \quad 4|N_1R| = 6 ⇒ \text{Not equal ⇒ (D) is False} \] Correction: \( |N_1P| = 1 \), \( |N_1R| = \frac{3}{2} \Rightarrow 9 \cdot 1 = 9,\ 4 \cdot \frac{3}{2} = 6 \) ⇒ False But option says \( 9|N_1P| = 4|N_1R| \Rightarrow 9 = 6 \)? ⇒ False % Final verdict: - (A) is TRUE - (B) is FALSE - (C) is TRUE - (D) is FALSE