9. (16 points) $P$ is a moving point on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. A tangent line is drawn through point $P$ to the ellipse, intersecting the circle $\odot O$ : $x^{2}+y^{2}=12$ at points $M$ and $N$. The tangents to $\odot O$ at points $M$ and $N$ intersect at point $Q$.
(1) Find the equation of the locus of point $Q$;
(2) If $P$ is a point in the first quadrant, find the maximum value of the area of $\triangle O P Q$.
9. (1) Let points $P\left(x_{0}, y_{0}\right), Q\left(x_{1}, y_{1}\right)$. Since point $P$ is on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, we have
$$
\frac{x_{0}^{2}}{4}+\frac{y_{0}^{2}}{3}=1 .
$$

On one hand, it is easy to find that the equation of the tangent line to the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ at point $P$ is
$$
\frac{x_{0} x}{4}+\frac{y_{0} y}{3}=1 .
$$

On the other hand, $Q M, Q N$ are the two tangents drawn from point $Q$ to the circle $\odot O: x^{2}+y^{2}=12$, and the equation of the chord of contact $M N$ is
$$
x_{1} x+y_{1} y=12 \text {. }
$$

Since equations (2) and (3) represent the same line, we have
$$
\frac{\frac{x_{0}}{4}}{x_{1}}=\frac{\frac{y_{0}}{3}}{y_{1}}=\frac{1}{12}
$$

Thus, $x_{0}=\frac{x_{1}}{3}, y_{0}=\frac{y_{1}}{4}$. Substituting these into equation (1) and simplifying, we get $\frac{x_{1}^{2}}{36}+\frac{y_{1}^{2}}{48}=1$.
Therefore, the equation of the locus of point $Q$ is $\frac{x^{2}}{36}+\frac{y^{2}}{48}=1$.
(2) As shown in Figure 2, draw $P A \perp x$-axis and $Q B \perp x$-axis, with the feet of the perpendiculars being $A, B$, respectively. Then $A\left(x_{0}, 0\right)$, $Q\left(3 x_{0}, 4 y_{0}\right), B\left(3 x_{0}, 0\right)$, so
$$
\begin{aligned}
S_{\triangle O P Q} & =S_{\triangle O B Q}-S_{\triangle O P A}-S_{\text {trapezoid } P A B Q} \\
& =\frac{1}{2} \cdot 3 x_{0} \cdot 4 y_{0}-\frac{1}{2} \cdot x_{0} \cdot y_{0}-\frac{1}{2} \cdot\left(y_{0}+4 y_{0}\right) \cdot\left(3 x_{0}-x_{0}\right) \\
& =\frac{1}{2} \cdot x_{0} \cdot y_{0} .
\end{aligned}
$$

Also, since
$$
1=\frac{x_{0}^{2}}{4}+\frac{y_{0}^{2}}{3} \geqslant 2 \sqrt{\frac{x_{0}^{2} \cdot y_{0}^{2}}{12}}=\frac{x_{0} \cdot y_{0}}{\sqrt{3}},
$$

we have $x_{0} \cdot y_{0} \leqslant \sqrt{3}$, and thus $S_{\triangle O P Q} \leqslant \frac{\sqrt{3}}{2}$. Therefore, the maximum area of $\triangle O P Q$ is $\frac{\sqrt{3}}{2}$.