11. (20 points) Given that $O$ is the origin, points $A$ and $B$ lie on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, and $|A B|=\frac{3}{2}$. The tangents to the ellipse at points $A$ and $B$ intersect at point $C$. Find the range of the area of quadrilateral $O A C B$.
11. When $A B \perp x$-axis, let's assume $A\left(\frac{\sqrt{13}}{2}, \frac{3}{4}\right), B\left(\frac{\sqrt{13}}{2},-\frac{3}{4}\right)$.

Thus, $C\left(\frac{8}{\sqrt{13}}, 0\right)$.
Let the area of quadrilateral $O A C B$ be $S$.
Then $S=\frac{1}{2} \times \frac{3}{2} \times \frac{8}{\sqrt{13}}=\frac{6 \sqrt{13}}{13}$.
When $A B$ is not perpendicular to the $x$-axis, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
$l_{A B}: y=k x+b$.
Substituting into the ellipse equation gives
$$
\begin{array}{l}
\left(3+4 k^{2}\right) x^{2}+8 k b x+4 b^{2}-12=0 \\
\Rightarrow x_{1}+x_{2}=\frac{-8 k b}{3+4 k^{2}}, x_{1} x_{2}=\frac{4 b^{2}-12}{3+4 k^{2}} .
\end{array}
$$

Thus, $|A B|=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\sqrt{1+k^{2}} \frac{4 \sqrt{12 k^{2}-3 b^{2}+9}}{3+4 k^{2}}=\frac{3}{2} \\
\Rightarrow b^{2}=3+4 k^{2}-\frac{3\left(3+4 k^{2}\right)^{2}}{64\left(1+k^{2}\right)} .
\end{array}
$$

Let $\frac{4 k^{2}+3}{1+k^{2}}=t \in[3,4)$. Then
$$
\frac{b^{2}}{1+k^{2}}=t-\frac{3}{64} t^{2} \text {. }
$$

Let $C\left(x_{0}, y_{0}\right)$. Then
the line $l_{A B}: y=-\frac{3 x_{0}}{4 y_{0}} x+\frac{3}{y_{0}}$.
Combining equations (1) and (2) gives $y_{0}=\frac{3}{b}, x_{0}=\frac{-4 k}{b}$.
Thus, the distance from point $C$ to $A B$ is
$$
d_{1}=\frac{\left|k x_{0}-y_{0}+b\right|}{\sqrt{1+k^{2}}}=\frac{\left|\frac{4 k^{2}+3-b^{2}}{b}\right|}{\sqrt{1+k^{2}}},
$$

and the distance from point $O$ to $A B$ is $d_{2}=\frac{|b|}{\sqrt{1+k^{2}}}$.
$$
\begin{aligned}
\text { Then } S= & \frac{3}{4}\left(d_{1}+d_{2}\right) \\
= & \frac{3}{4} \times \frac{1}{\sqrt{1+k^{2}}}\left(|b|+\left|\frac{4 k^{2}+3-b^{2}}{b}\right|\right) \\
\Rightarrow & S^{2}=\frac{9}{16} \times \frac{1}{1+k^{2}}\left(b^{2}+2\left(4 k^{2}+3-b^{2}\right)+\right. \\
& \left.=\frac{\left(4 k^{2}+3-b^{2}\right)^{2}}{b^{2}}\right) \\
= & \left.\frac{9}{16} \times \frac{64 t}{64} t^{2}+\frac{9}{64^{2}} t^{4} \cdot \frac{1}{t-\frac{3}{64} t^{2}}\right) \\
\Rightarrow & S \in\left[\frac{6 \sqrt{165}}{55}, \frac{6 \sqrt{13}}{13}\right) .
\end{aligned}
$$

In summary, the range of the area of quadrilateral $O A C B$ is
$$
\left[\frac{6 \sqrt{165}}{55}, \frac{6 \sqrt{13}}{13}\right] .
$$