Let P(x_1, y_1) and Q(x_2, y_2) be two distin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A, C) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Any point on the ellipse can be represented as $(3 \cos 	heta, 2 \sin 	heta)$.Given $

Vedclass · Accepted Answer

$A, D$

Vedclass · Answer

(A, C) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Any point on the ellipse can be represented as $(3 \cos 	heta, 2 \sin 	heta)$.Given $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$,where $M=(3,0)$ and $O=(0,0)$,the points $R$ and $S$ lie on the circle $x^2+y^2=9$. Thus,$R = (3 \cos \frac{\pi}{6}, 3 \sin \frac{\pi}{6}) = (3 \frac{\sqrt{3}}{2}, 3 \frac{1}{2}) = (\frac{3\sqrt{3}}{2}, \frac{3}{2})$ and $S = (3 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}) = (3 \frac{1}{2}, 3 \frac{\sqrt{3}}{2}) = (\frac{3}{2}, \frac{3\sqrt{3}}{2})$.Since $x_1$ is the $x$-coordinate of $R$,$x_1 = \frac{3\sqrt{3}}{2}$. Since $P$ is on the ellipse,$P = (x_1, y_1) = (\frac{3\sqrt{3}}{2}, 2 \sin 	heta_1)$. Since $\frac{x_1^2}{9} + \frac{y_1^2}{4} = 1$,we have $\frac{27/4}{9} + \frac{y_1^2}{4} = 1 \Rightarrow \frac{3}{4} + \frac{y_1^2}{4} = 1 \Rightarrow y_1^2 = 1 \Rightarrow y_1 = 1$ (as $y_1 > 0$). So $P = (\frac{3\sqrt{3}}{2}, 1)$.Similarly,$x_2 = \frac{3}{2}$. For $Q = (x_2, y_2)$,$\frac{9/4}{9} + \frac{y_2^2}{4} = 1 \Rightarrow \frac{1}{4} + \frac{y_2^2}{4} = 1 \Rightarrow y_2^2 = 3 \Rightarrow y_2 = \sqrt{3}$. So $Q = (\frac{3}{2}, \sqrt{3})$.The line joining $P$ and $Q$ has slope $m = \frac{\sqrt{3}-1}{3/2 - 3\sqrt{3}/2} = \frac{\sqrt{3}-1}{-\frac{3}{2}(\sqrt{3}-1)} = -\frac{2}{3}$.The equation is $y - \sqrt{3} = -\frac{2}{3}(x - \frac{3}{2}) \Rightarrow 3y - 3\sqrt{3} = -2x + 3 \Rightarrow 2x + 3y = 3(1+\sqrt{3})$. Thus,$(A)$ is true.For $(C)$,$N_2 = (x_2, 0) = (\frac{3}{2}, 0)$. $|N_2Q| = y_2 = \sqrt{3}$ and $|N_2S| = \frac{3\sqrt{3}}{2}$. $3|N_2Q| = 3\sqrt{3}$ and $2|N_2S| = 2(\frac{3\sqrt{3}}{2}) = 3\sqrt{3}$. So $3|N_2Q| = 2|N_2S|$,$(C)$ is true.

Similar Questions

Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle $60^{\circ}$ at the foci,then the square of the sum of the lengths of its minor and major axes is equal to $...........$.

The eccentricity of the ellipse $25x^2 + 16y^2 - 150x - 175 = 0$ is

The product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

If an ellipse with foci at $(3,3)$ and $(-4,4)$ is passing through the origin,then the eccentricity of that ellipse is

The sum of the focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $9x^2+4y^2=36$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module