Mindyard | Let  P ( x 1 , y 1 ) and Q ( x 2 , y 2 ) be two distinct points on the ellipse x 2 9 + 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  ∠ R O M = π 6 and  ∠ S O M = π 3 , where  O  denotes the origin ( 0 , 0 ) . Let | X Y | denote the length of the line segment X Y . Then which of the following statements is (are) TRUE? The equation of the line joining P and Q  is  2x + 3y = 3 ( 1 + 3 ) , The equation of the line joining  P  and  Q  is 2x + y = 3 ( 1 + 3 ) , If  N 2 = ( x = 0 ) , then  3 | N 2 Q | = 2 | N 2 S | , If  N 1 = ( x = 0 ) , then 9 | N 1 P | = 4

$Let P (x_{1}, y_{1}) and Q (x_{2}, y_{2}) be two distinct points on the ellipse \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 such that y_{1} > 0 and y_{2} > 0 .$
$Let C denote the circle x^{2} + y^{2} = 9, and M be the point (3, 0) . Suppose the line x = x_{1} intersects C at R, and$

$the line x = x_{2} intersects C at S, such that the y-coordinates of R and S are positive.$ $Let ∠ R O M = \frac{π}{6} and ∠ S O M = \frac{π}{3}, where O denotes the origin (0, 0) . Let | X Y | denote the length of the line segment X Y .$
$Then which of the following statements is (are) TRUE?$

Options

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

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

$If N_{2} = (x = 0), then 3 | N_{2} Q | = 2 | N_{2} S |$

$If N_{1} = (x = 0), then 9 | N_{1} P | = 4 | N_{1} R |$

Correct Answer :

Option C

$If N_{2} = (x = 0), then 3 | N_{2} Q | = 2 | N_{2} S |$

Option A

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

Solution :

We are given:
The equation of the ellipse is:

$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

Here, the semi-major axis is a = 3 and the semi-minor axis is b = 2.
The auxiliary circle of this ellipse is:

$C : x^{2} + y^{2} = 9$

Let the points R and S lie on the auxiliary circle C such that their y-coordinates are positive. We are given the angles:

$∠ R O M = \frac{π}{6} and ∠ S O M = \frac{π}{3}$

where O is the origin (0, 0) and M is the point (3, 0) on the positive x-axis.
Therefore, the coordinates of R and S are:

$R = (3 \cos \frac{π}{6}, 3 \sin \frac{π}{6}) = (\frac{3 \sqrt{3}}{2}, \frac{3}{2})$

$S = (3 \cos \frac{π}{3}, 3 \sin \frac{π}{3}) = (\frac{3}{2}, \frac{3 \sqrt{3}}{2})$

The vertical line $x = x_{1}$ passes through R, which gives:

$x_{1} = \frac{3 \sqrt{3}}{2}$

Similarly, the vertical line

x = x_{2}

passes through S, which gives:

$x_{2} = \frac{3}{2}$

Since P and Q lie on the ellipse with positive y-coordinates ( $y_{1} > 0, y_{2} > 0$ ), their coordinates can be written in terms of the eccentric angles of the auxiliary points:

$P = (3 \cos \frac{π}{6}, 2 \sin \frac{π}{6}) = (\frac{3 \sqrt{3}}{2}, 1)$

$Q = (3 \cos \frac{π}{3}, 2 \sin \frac{π}{3}) = (\frac{3}{2}, \sqrt{3})$

Step 1: Check the equation of the line joining P and Q
The slope of the line joining P and Q is:

$m = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3 \sqrt{3}}{2}} = \frac{2 (\sqrt{3} - 1)}{3 (1 - \sqrt{3})} = - \frac{2}{3}$

Using the point-slope form with P:

$y - 1 = - \frac{2}{3} (x - \frac{3 \sqrt{3}}{2})$

$3 (y - 1) = - 2 x + 3 \sqrt{3}$

$2 x + 3 y = 3 + 3 \sqrt{3} = 3 (1 + \sqrt{3})$

Thus, the equation of the line joining P and Q is indeed

2 x + 3 y = 3 (1 + \sqrt{3})

. This statement is TRUE.

Step 2: Check the statement involving N₂
Here, $N_{2}$ denotes the line $x = 0$ (the y-axis). Let $N_{2} Q$ and $N_{2} S$ denote the distances from the y-axis to the points Q and S along their respective y-coordinates (which corresponds to their y-coordinates):