Find Coordinates of External Point for Tangents to an Ellipse

To find the coordinates $S(p,q)$, we analyze the given conditions step-by-step:

1. Equation of the Ellipse and Key Features: The given ellipse is: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ Here, $a = 3$ and $b = 2$.

   • The center is $O(0,0)$.
   • The vertex with positive $x$-coordinate is $R(3,0)$.
   • The endpoints of the minor axis are $(0, 2)$ and $(0, -2)$.

2. First Tangent from $S(p,q)$: The point $S(p,q)$ lies in the first quadrant ($p > 0, q > 0$). One of the tangents drawn from $S$ touches the ellipse at an endpoint of the minor axis.

   • The tangent at $(0,2)$ is the horizontal line $y = 2$.
   • The tangent at $(0,-2)$ is the horizontal line $y = -2$.

   Since $S(p,q)$ is in the first quadrant ($q > 0$), it must lie on the line $y = 2$. Hence: $q = 2$

3. Finding the Coordinates of Point $T$: The point $T$ lies on the ellipse in the fourth quadrant, so its parametric coordinates can be written as: $T = (3\cos\theta, 2\sin\theta), \quad \text{where } \theta \in \left(-\frac{\pi}{2}, 0\right)$

   The area of triangle $\Delta ORT$ formed by $O(0,0)$, $R(3,0)$, and $T(3\cos\theta, 2\sin\theta)$ is: $\text{Area}(\Delta ORT) = \frac{1}{2} \big| x_R y_T - y_R x_T \big| = \frac{1}{2} \big| 3(2\sin\theta) - 0 \big| = 3|\sin\theta|$

   Given that the area is $\frac{3}{2}$: $3|\sin\theta| = \frac{3}{2} \implies |\sin\theta| = \frac{1}{2}$

   Since $T$ is in the fourth quadrant, $\sin\theta < 0$, which gives: $\sin\theta = -\frac{1}{2} \implies \cos\theta = \frac{\sqrt{3}}{2}$

   Therefore, the coordinates of $T$ are: $T = \left(\frac{3\sqrt{3}}{2}, -1\right)$

4. Equation of the Tangent at Point $T$: The equation of the tangent to the ellipse at $T(x_1, y_1)$ is given by: $\frac{x x_1}{9} + \frac{y y_1}{4} = 1$ Substituting $x_1 = \frac{3\sqrt{3}}{2}$ and $y_1 = -1$: $\frac{x\left(\frac{3\sqrt{3}}{2}\right)}{9} + \frac{y(-1)}{4} = 1 \implies \frac{\sqrt{3}x}{6} - \frac{y}{4} = 1$ Multiplying through by $12$: $2\sqrt{3}x - 3y = 12$

5. Finding $p$: Since $S(p,q) = (p, 2)$ lies on this tangent line, we substitute $x = p$ and $y = 2$: $2\sqrt{3}p - 3(2) = 12$ $2\sqrt{3}p - 6 = 12$ $2\sqrt{3}p = 18 \implies p = \frac{18}{2\sqrt{3}} = 3\sqrt{3}$

Thus, the values are $q = 2$ and $p = 3\sqrt{3}$, which corresponds to option (A).