11. (15 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 $y=k(x-2)+3$.
When combined with the ellipse, we get
$$
\left(1+4 k^{2}\right) x^{2}+8 k(3-2 k) x+4\left(4 k^{2}-12 k+8\right)=0 \text {. }
$$

Therefore, $\Delta=64(3 k-2)>0 \Rightarrow k>\frac{2}{3}$.
Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$. Then the tangent equations of the ellipse at $P$ and $Q$ 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 by $y_{1}=k\left(x_{1}-2\right)+3$ and $y_{2}=k\left(x_{2}-2\right)+3$, we get
$$
y=\frac{1}{3-2 k}\left(k \neq \frac{3}{2}\right) \text {. }
$$

Similarly, $x=\frac{-4 k}{3-2 k}\left(k \neq \frac{3}{2}\right)$.
Thus, the trajectory equation of point $M$ is $x+6 y-2=0$ (the part outside the ellipse).
(2) Note that
$$
|P Q|=\frac{\sqrt{64(3 k-2)\left(1+k^{2}\right)}}{1+4 k^{2}} \text {, }
$$
the distance from $O$ to $P Q$ is $d_{1}=\frac{|3-2 k|}{\sqrt{1+k^{2}}}$,
the distance from $M$ to $P Q$ is $d_{2}=\frac{4|3 k-2|}{|3-2 k| \sqrt{1+k^{2}}}$.
Then $d_{1}+d_{2}=\frac{4 k^{2}+1}{|3-2 k| \sqrt{1+k^{2}}}$.
Therefore, the area of quadrilateral $P O Q M$ is
$$
S=\frac{1}{2}\left(d_{1}+d_{2}\right)|P Q|=\frac{4 \sqrt{3 k-2}}{|3-2 k|} \text {. }
$$

When $S=4$, solving gives $k=1$ or $k=\frac{11}{4}$.
Thus, the line $l$ is
$x-y+1=0$ or $11 x-4 y-10=0$.