9. Given an ellipse $C$ centered at the origin with foci on the $x$-axis, the eccentricity of the ellipse is $\frac{1}{2}$, and it passes through the point $M\left(1, \frac{3}{2}\right)$. A line $l$ passing through the point $P(2,1)$ intersects the ellipse $C$ at two distinct points $A, B$.
(1) Find the equation of the ellipse $C$;
(2) Does there exist a line $l$ such that $\overrightarrow{P A} \cdot \overrightarrow{P B}=\overrightarrow{P M}^{2}$? If it exists, find the equation of the line $l$; if not, explain the reason.
(1) According to the problem, $\left\{\begin{array}{l}e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{2}, \\ \frac{1}{a^{2}}+\frac{9}{4 b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=4, \\ b^{2}=3\end{array} \Rightarrow\right.\right.$ the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let the line $l:\left\{\begin{array}{l}x=2+t \cos \alpha, \\ y=1+t \sin \alpha,\end{array}\right.$ (with $t$ as the parameter), substituting into the ellipse equation gives
$$
\begin{array}{l}
3(2+t \cos \alpha)^{2}+4(1+t \sin \alpha)^{2}=12 \\
\Rightarrow\left(3 \cos ^{2} \alpha+4 \sin ^{2} \alpha\right) t^{2}+(12 \cos \alpha+8 \sin \alpha) t+4=0 .
\end{array}
$$

Since $|P M|^{2}=(2-1)^{2}+\left(1-\frac{3}{2}\right)^{2}=\frac{5}{4} \Rightarrow t_{1} t_{2}=\frac{4}{3 \cos ^{2} \alpha+4 \sin ^{2} \alpha}=\frac{5}{4}$
$\Rightarrow \sin ^{2} \alpha=\frac{1}{5} \Rightarrow \sin \alpha=\frac{\sqrt{5}}{5}, \cos \alpha= \pm \frac{2 \sqrt{5}}{5}$.
Also, $\Delta=(12 \cos \alpha+8 \sin \alpha)^{2}-16\left(3 \cos ^{2} \alpha+4 \sin ^{2} \alpha\right)$
$=16 \cdot 6 \cos \alpha(\cos \alpha+2 \sin \alpha)>0$ indicates that $\cos \alpha=-\frac{2 \sqrt{5}}{5}$ should be discarded.
Thus, $k_{l}=\frac{1}{2}$, and the equation of the line $l$ is $x-2 y=0$.