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Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the tirst quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. I wo tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac{3}{2}$, then which of the following options is correct?
$q=2, p=3 \sqrt{3}$
$q=2, p=4 \sqrt{3}$
$q=1, p=5 \sqrt{3}$
$q=1, p=6 \sqrt{3}$
Solution
$\operatorname{Ar}(\triangle ORT )=\frac{3}{2}$
$\left|\frac{1}{2} \times 3 \times 2 \sin \theta\right|=\frac{3}{2}$
$\sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{11 \pi}{6}$
$T \left(\frac{3 \sqrt{3}}{2},-1\right)$
Tanget at $(0,2) \frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2$ $. . . . .(1)$
Tangent at $\left(\frac{3 \sqrt{3}}{2},-1\right) \frac{x\left(\frac{3 \sqrt{3}}{2}\right)}{9}+\frac{y(-1)}{4}=1$ $. . . . .(2)$
$\therefore$ By solving $(1)$ & $(2)$ $\Rightarrow p=3 \sqrt{3}, q=2$
$\Rightarrow$ Option $(A)$ is Correct.
