Consider the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$. Let $S(p, q)$ be a point in the first quadrant such that $\frac{p^{2}}{9} + \frac{q^{2}}{4} > 1$. Two 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 $\Delta ORT$ is $\frac{3}{2}$, then which of the following options is correct?

Consider the ellipse x²/9 + y²/4 = 1. Let S(p, q) be a point in the first quadrant such that p²/9 + q²/4 > 1. Two 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 Δ ORT is 3/2, then which of the follow

$E:\ \dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1,\quad S(p,q),\ p,q>0$

$\dfrac{p^{2}}{9}+\dfrac{q^{2}}{4}>1 \ \Rightarrow\ S \text{ is outside } E$

R=(3,0), O=(0,0), [triangle ORT]=3/2

$\text{Tangent } 1: \text{ at a minor-axis end point}$

$\text{Tangent } 2: \text{ at } T \in \text{Q4}$

$\text{To find: option giving } (p,q)$

$\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1 \Rightarrow a=3,\ b=2$

$R=(3,0),\quad \text{minor-axis ends } (0,\pm 2)$

$1.\ \text{tangent at a minor-axis end} \Rightarrow q$

$2.\ T=(3\cos\theta,2\sin\theta) \Rightarrow \text{tangent at } T$

3. [triangle ORT] ⇒ θ ⇒ p

$\text{Tangent at } (0,2):\ \dfrac{y\cdot 2}{4}=1 \Rightarrow y=2$

$S(p,q)\in y=2 \Rightarrow q=2$

$T=(3\cos\theta,\ 2\sin\theta)$

$\text{Tangent at } T:\ \dfrac{x\cos\theta}{3}+\dfrac{y\sin\theta}{2}=1$

$S(p,2)\in \text{tangent}_T:\ \dfrac{p\cos\theta}{3}+\sin\theta=1$

[triangle ORT]=1/2|x_R y_T-x_T y_R|=1/2|3· 2sinθ|=3|sinθ|

$3|\sin\theta|=\dfrac{3}{2} \Rightarrow |\sin\theta|=\dfrac{1}{2}$

$T\in \text{Q4}:\ \sin\theta=-\dfrac{1}{2},\quad \cos\theta=\dfrac{\sqrt{3}}{2}$

$\dfrac{p(\sqrt{3}/2)}{3}-\dfrac{1}{2}=1 \Rightarrow \dfrac{p\sqrt{3}}{6}=\dfrac{3}{2}$

$p\sqrt{3}=9 \Rightarrow p=3\sqrt{3}$

$q=2,\ p=3\sqrt{3}$

$\text{Option } (1)$