Question 4

The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$, so $a=3, b=2$. $O$ is $(0,0)$ and $R$ is $(3,0)$. One tangent from $S(p,q)$ (where $p, q > 0$) meets the ellipse at an endpoint of the minor axis. Since $S$ is in the first quadrant, the endpoint must be $(0,2)$. The equation of the tangent at $(0,2)$ is $y=2$. Thus, $q=2$. The other tangent meets the ellipse at $T(3\cos\theta, 2\sin\theta)$ in the fourth quadrant, so $\sin\theta < 0$ and $\cos\theta > 0$. The area of $\Delta ORT$ is $\frac{1}{2} |x_O(y_R - y_T) + x_R(y_T - y_O) + x_T(y_O - y_R)| = \frac{1}{2} |0 + 3(2\sin\theta - 0) + 0| = |3\sin\theta|$. Given area $= 3/2 \implies 3|\sin\theta| = 3/2 \implies |\sin\theta| = 1/2$. Since $T$ is in the fourth quadrant, $\sin\theta = -1/2$. Then $\cos\theta = \frac{\sqrt{3}}{2}$. The tangent at $T(3\cos\theta, 2\sin\theta)$ is $\frac{x\cos\theta}{3} + \frac{y\sin\theta}{2} = 1$. This passes through $S(p, 2)$: $\frac{p(\sqrt{3}/2)}{3} + \frac{2(-1/2)}{2} = 1 \implies \frac{p\sqrt{3}}{6} - \frac{1}{2} = 1 \implies \frac{p\sqrt{3}}{6} = \frac{3}{2} \implies p = \frac{18}{2\sqrt{3}} = 3\sqrt{3}$. So $q=2$ and $p=3\sqrt{3}$.