11. (20 points) Through the point $(2,3)$, draw a moving line $l$ intersecting the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at two distinct points $P$ and $Q$. Draw the tangents to the ellipse at $P$ and $Q$, and let the intersection of these tangents be $M$.
(1) Find the equation of the locus of point $M$;
(2) Let $O$ be the origin. When the area of quadrilateral $P O Q M$ is 4, find the equation of line $l$.
11. (1) According to the problem, let the equation of line $l$ be
$$
y=k(x-2)+3 \text {, }
$$

Combining with the ellipse equation, we get
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k(3-2 k) x+4\left(4 k^{2}-12 k+8\right) \\
=0 .
\end{array}
$$

From $\Delta=64(3 k-2)>0$, we get $k>\frac{2}{3}$.
Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$. Then the tangent line equations at points $P$ and $Q$ on the ellipse are
$$
\begin{array}{l}
\frac{x_{1} x}{4}+y_{1} y=1, \\
\frac{x_{2} x}{4}+y_{2} y=1 .
\end{array}
$$
(1) $\times x_{2}$ - (2) $\times x_{1}$, and using $y_{1}=k\left(x_{1}-2\right)+3$ and $y_{2}=k\left(x_{2}-2\right)+3$, we get
$$
y=\frac{1}{3-2 k}\left(k \neq \frac{3}{2}\right) .
$$

Similarly, $x=\frac{-4 k}{3-2 k}\left(k \neq \frac{3}{2}\right)$.
Thus, the trajectory equation of point $M$ is $x+6 y-2=0$
$$
\left(x \in\left(-\infty,-\frac{8}{5}\right) \cup(2,+\infty)\right) \text {. }
$$
(2) Note that
$$
|P Q|=\frac{\sqrt{64(3 k-2)\left(1+k^{2}\right)}}{1+4 k^{2}},
$$

The distance from point $O$ to $P Q$ is
$$
d_{1}=\frac{|3-2 k|}{\sqrt{1+k^{2}}},
$$

The distance from point $M$ to $P Q$ is
$$
d_{2}=\frac{4|3 k-2|}{|3-2 k| \sqrt{1+k^{2}}} .
$$

Thus, $d_{1}+d_{2}=\frac{4 k^{2}+1}{|3-2 k| \sqrt{1+k^{2}}}$.
Therefore, $S_{\text {quadrilateral } \operatorname{EN} P Q Q M}=\frac{1}{2}\left(d_{1}+d_{2}\right)|P Q|$
$$
=\frac{4 \sqrt{3 k-2}}{|3-2 k|}=4 \text {. }
$$

Solving, we get $k=1$ or $\frac{11}{4}$.
Thus, the equation of line $l$ is
$$
x-y+1=0 \text { or } 11 x-4 y-10=0 \text {. }
$$