1. Given ellipse and key points

The ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$ so its semi-major and semi-minor axes are $a=3,\quad b=2.$

Hence:

• Center: $O=(0,0)$
• Vertex with positive $x$-coordinate: $R=(3,0)$
• End points of minor axis: $(0,2)$ and $(0,-2)$

Since $S(p,q)$ is in the first quadrant and one tangent from $S$ touches the ellipse at an end point of the minor axis, that tangent must be the tangent at $(0,2)$ (not at $(0,-2)$).

2. Tangent at the end point of minor axis

For the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1,$ the tangent at $(0,2)$ is $\frac{0\cdot x}{9}+\frac{2y}{4}=1 \implies y=2.$

Since $S(p,q)$ lies on this tangent, we must have $q=2.$

So only options A and B remain.

3. Coordinates of point $T$ using area condition

Let $T=(x_T,y_T)$ be the point of tangency in the fourth quadrant. So $y_T<0, \quad \frac{x_T^2}{9}+\frac{y_T^2}{4}=1.$

Now area of triangle $\triangle ORT$ is given as $\frac{3}{2}$.

Points are: $O=(0,0),\quad R=(3,0),\quad T=(x_T,y_T).$

Using base $OR=3$, the height from $T$ to the $x$-axis is $|y_T|$. So $\text{Area} = \frac{1}{2}\cdot 3 \cdot |y_T| = \frac{3}{2}.$ Thus, $|y_T|=1.$

Since $T$ is in the fourth quadrant, $y_T=-1.$

Now substitute into the ellipse: $\frac{x_T^2}{9}+\frac{(-1)^2}{4}=1$ $\frac{x_T^2}{9}+\frac14=1$ $\frac{x_T^2}{9}=\frac34$ $x_T^2=\frac{27}{4}$ $x_T=\frac{3\sqrt3}{2}$ (since fourth quadrant implies $x_T>0$).

Therefore, $T=\left(\frac{3\sqrt3}{2},-1\right).$

4. Equation of tangent at $T$

For ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1,$ the tangent at $(x_1,y_1)$ is $\frac{xx_1}{9}+\frac{yy_1}{4}=1.$

At $T=\left(\frac{3\sqrt3}{2},-1\right),$ the tangent is $\frac{x\left(\frac{3\sqrt3}{2}\right)}{9}+\frac{y(-1)}{4}=1.$ Simplify: $\frac{\sqrt3}{6}x-\frac{y}{4}=1.$

Since $S(p,q)$ lies on this tangent and $q=2$, $\frac{\sqrt3}{6}p-\frac{2}{4}=1$ $\frac{\sqrt3}{6}p-\frac12=1$ $\frac{\sqrt3}{6}p=\frac32$ $p=\frac32\cdot \frac{6}{\sqrt3}=\frac{9}{\sqrt3}=3\sqrt3.$

Hence, $p=3\sqrt3,\quad q=2.$

5. Check external point condition

We verify that $S$ lies outside the ellipse: $\frac{p^2}{9}+\frac{q^2}{4}=\frac{(3\sqrt3)^2}{9}+\frac{2^2}{4}=\frac{27}{9}+1=3+1=4>1.$ So the condition is satisfied.

6. Evaluate options

• A: $q=2,\ p=3\sqrt3$ ✅
• B: $q=2,\ p=4\sqrt3$ ❌
• C: $q=1,\ p=5\sqrt3$ ❌
• D: $q=1,\ p=6\sqrt3$ ❌

Therefore the correct option is $\boxed{\text{A}}.$