11. (20 points) Let point $A(-2,0)$ and $\odot O: x^{2}+y^{2}=4$, $AB$ is the diameter of $\odot O$, from left to right $M$, $O$, $N$ are the four equal division points of $AB$, $P$ (different from $A$, $B$) is a moving point on $\odot O$, $PD \perp AB$ at point $D$, $\overrightarrow{PE}=\lambda \overrightarrow{ED}$, line $PA$ intersects $BE$ at point $C$, $|CM|+|CN|$ is a constant.
(1) Find the value of $\lambda$ and the equation of the trajectory curve $E$ of point $C$;
(2) If points $Q$, $R$ are different points on curve $E$, and $PQ$, $PR$ are tangent to curve $E$, find the minimum value of the area of $\triangle OQR$.
11. (1) It is easy to get the points $B(2,0), M(-1,0), N(1,0)$.

Let the points $P\left(x_{0}, y_{0}\right), C(x, y)$. Then $E\left(x_{0}, \frac{y_{0}}{1+\lambda}\right)$.
Since the line $P A$ intersects $B E$ at point $C$, we have $x_{0} \neq \pm 2$, $\frac{y}{x+2}=\frac{y_{0}}{x_{0}+2}$,

and $\frac{y}{x-2}=\frac{\frac{y_{0}}{1+\lambda}}{x_{0}-2}$.
(1) $\times$ (2) gives $\frac{y^{2}}{x^{2}-4}=\frac{\frac{y_{0}^{2}}{1+\lambda}}{x_{0}^{2}-4}$.
Since point $P$ (different from $A, B$) is a moving point on $\odot O$, then $\frac{y^{2}}{x^{2}-4}=-\frac{1}{1+\lambda} \Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{\frac{4}{1+\lambda}}=1$.
To make $|C M|+|C N|$ a constant, then
$4-\frac{4}{1+\lambda}=1 \Rightarrow \lambda=\frac{1}{3}$,

At this time, the equation of the trajectory curve $E$ of point $C$ is
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=1(x \neq \pm 2) \text {. }
$$
(2) Let the points $Q\left(x_{1}, y_{1}\right), R\left(x_{2}, y_{2}\right)$.

By the chord of contact formula, we have
$$
l_{Q R}: \frac{x_{0} x}{4}+\frac{y_{0} y}{3}=1 \text {. }
$$

Substituting into equation (3) gives
$$
\left(4 y_{0}^{2}+3 x_{0}^{2}\right) x^{2}-24 x_{0} x+48-16 y_{0}^{2}=0 \text {. }
$$

By Vieta's formulas, we get
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{24 x_{0}}{4 y_{0}^{2}+3 x_{0}^{2}}=\frac{24 x_{0}}{y_{0}^{2}+12}, \\
x_{1} x_{2}=\frac{48-16 y_{0}^{2}}{4 y_{0}^{2}+3 x_{0}^{2}}=\frac{48-16 y_{0}^{2}}{y_{0}^{2}+12} .
\end{array}
$$

Thus, the area of $\triangle O Q R$ is
$$
\begin{aligned}
S & =\frac{1}{2}|Q R| d_{O-Q R}=\frac{3}{2\left|y_{0}\right|}\left|x_{1}-x_{2}\right| \\
& =3 \sqrt{\frac{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}{4 y_{0}^{2}}},
\end{aligned}
$$

where $d_{O-Q R}$ represents the distance from point $O$ to the line $Q R$.
Substituting equations (4) and (5) and simplifying, we get $S=\frac{12\left|y_{0}\right|}{y_{0}^{2}+12}$.
Let $t=\left|y_{0}\right|$. Then $t \in(0,2]$.
It is easy to know that the function $f(t)=\frac{12 t}{t^{2}+12}=\frac{12}{t+\frac{12}{t}}$ is increasing on $(0,2]$, thus,
$$
f(t)_{\max }=f(2)=\frac{3}{2} \text {. }
$$

Therefore, when $t=\left|y_{0}\right|=2$, i.e., $P(0, \pm 2)$, the maximum area of $\triangle O Q R$ is $\frac{3}{2}$.