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 $$\frac{p^2}{9} + \frac{q^2}{4} \gt 1$$. 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 $$\triangle ORT$$ is $$\frac{3}{2}$$, then which of the following options is correct?

The ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$.
Hence $$a=3,\;b=2$$ and its centre is $$O(0,0)$$. The vertex with positive $$x$$-coordinate is $$R(3,0)$$.

Let the second point of contact in the fourth quadrant be $$T(x_T,y_T)$$.
Since $$O$$ and $$R$$ both lie on the $$x$$-axis, the area of $$\triangle ORT$$ equals

$$\text{Area}=\frac12\,(OR)\,\bigl|y_T\bigr| = \frac12\,(3)\,\bigl|y_T\bigr|.$$

Given $$\text{Area}= \frac32,$$ we get
$$\frac12\,(3)\,\bigl|y_T\bigr|=\frac32 \;\Longrightarrow\; |y_T|=1.$$ Because $$T$$ is in the fourth quadrant, $$y_T=-1$$.

Substituting $$y_T=-1$$ in the ellipse,

$$\frac{x_T^{2}}{9}+\frac{(-1)^{2}}{4}=1 \;\Longrightarrow\; \frac{x_T^{2}}{9}+\frac14=1 \;\Longrightarrow\; \frac{x_T^{2}}{9}=\frac34 \;\Longrightarrow\; x_T^{2}=\frac{27}{4} \;\Longrightarrow\; x_T=\frac{3\sqrt3}{2}\;(\text{positive in quadrant IV}).$$

Thus $$T\Bigl(\dfrac{3\sqrt3}{2},\,-1\Bigr).$$

Equation of the two tangents
For a point $$(x_1,y_1)$$ on the ellipse, the tangent is $$\frac{xx_1}{9}+\frac{yy_1}{4}=1.$$

• At the upper end of the minor axis $$(0,2):$$ $$\frac{x\cdot0}{9}+\frac{y\cdot2}{4}=1\;\Longrightarrow\;y=2.$$
• At $$T\Bigl(\dfrac{3\sqrt3}{2},-1\Bigr):$$ $$\frac{x\left(\dfrac{3\sqrt3}{2}\right)}{9}+\frac{y(-1)}{4}=1 \;\Longrightarrow\; \frac{x\sqrt3}{6}-\frac{y}{4}=1 \;\Longrightarrow\; y=\frac{2\sqrt3}{3}\,x-4.$$

Both tangents meet at the external point $$S(p,q)$$, so $$S$$ is the intersection of

$$y=2 \quad\text{and}\quad y=\frac{2\sqrt3}{3}\,x-4.$$

Putting $$y=2$$ in the second equation:

$$2=\frac{2\sqrt3}{3}\,x-4 \;\Longrightarrow\; \frac{2\sqrt3}{3}\,x=6 \;\Longrightarrow\; x=\frac{6\cdot3}{2\sqrt3}=3\sqrt3.$$

Hence $$S(3\sqrt3,\,2).$$

Verification that $$S$$ lies outside the ellipse

$$\frac{p^{2}}{9}+\frac{q^{2}}{4} =\frac{(3\sqrt3)^{2}}{9}+\frac{2^{2}}{4} =\frac{27}{9}+1=3+1=4\gt1,$$ so $$S$$ is indeed outside the ellipse.

The obtained values match Option A:

Option A which is: $$q = 2,\; p = 3\sqrt{3}$$