Consider the ellipse frac{x^2}{9}+frac{y^2}{4 | vedclass

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(A) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. The vertex $R$ is $(3, 0)$ and the center $O$ is $(0, 0)$.Let point $T$ be $(3 \cos 	heta

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

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(A) The ellipse is $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. The vertex $R$ is $(3, 0)$ and the center $O$ is $(0, 0)$.Let point $T$ be $(3 \cos 	heta, 2 \sin 	heta)$. Since $T$ is in the fourth quadrant,we take $T = (3 \cos 	heta, -2 \sin 	heta)$ where $	heta$ is acute.The area of $	riangle ORT$ is given by $\frac{1}{2} |x_O(y_R - y_T) + x_R(y_T - y_O) + x_T(y_O - y_R)| = \frac{3}{2}$.$\frac{1}{2} |0(0 - (-2 \sin 	heta)) + 3(-2 \sin 	heta - 0) + 3 \cos 	heta(0 - 0)| = \frac{3}{2}$.$\frac{1}{2} |-6 \sin 	heta| = \frac{3}{2}$ $\Rightarrow 3 \sin 	heta = \frac{3}{2}$ $\Rightarrow \sin 	heta = \frac{1}{2}$.Thus,$	heta = 30^\circ = \frac{\pi}{6}$.So,$T = (3 \cos 30^\circ, -2 \sin 30^\circ) = (3 \cdot \frac{\sqrt{3}}{2}, -2 \cdot \frac{1}{2}) = (\frac{3 \sqrt{3}}{2}, -1)$.The tangent at $(0, 2)$ is $y = 2$.The tangent at $T(\frac{3 \sqrt{3}}{2}, -1)$ is $\frac{x(\frac{3 \sqrt{3}}{2})}{9} + \frac{y(-1)}{4} = 1 \Rightarrow \frac{x \sqrt{3}}{6} - \frac{y}{4} = 1$.Substituting $y = 2$ into the tangent equation: $\frac{x \sqrt{3}}{6} - \frac{2}{4} = 1 \Rightarrow \frac{x \sqrt{3}}{6} = 1 + \frac{1}{2} = \frac{3}{2}$.$x \sqrt{3} = 9 \Rightarrow x = \frac{9}{\sqrt{3}} = 3 \sqrt{3}$.Thus,$S(p, q) = (3 \sqrt{3}, 2)$.Therefore,$p = 3 \sqrt{3}$ and $q = 2$.

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