(11) (20 points) In the Cartesian coordinate system $x O y$, with the origin $O$ as the center, two circles are drawn with radii $a$ and $b$ ($a > b > 0$). Point $Q$ is the intersection of the radius $O P$ of the larger circle with the smaller circle. A perpendicular line $A N \perp O x$ is drawn from point $P$ to the x-axis, with the foot of the perpendicular being $N$. A perpendicular line $Q M \perp P N$ is drawn from point $Q$ to $P N$, with the foot of the perpendicular being $M$. Let the trajectory of point $M$ as the radius $O P$ rotates around point $O$ be the curve $E$.
(1) Find the equation of the curve $E$;
(2) Let $A$, $B$, and $C$ be three points on the curve $E$, and satisfy $\overrightarrow{O A} + \overrightarrow{O B} + \overrightarrow{O C} = \overrightarrow{0}$. Find the area of $\triangle A B C$.
11 (1) Let the coordinates of point $M$ be $(x, y)$, and take $\angle x O P$ as the parameter $\varphi$. Then, we have $\left\{\begin{array}{l}x=a \cos \varphi, \\ y=b \sin \varphi,\end{array}\right.$ Eliminating the parameter $\varphi$, we get $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which is the equation of curve $E$.
(2) Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively. From the above, we can also set the coordinates of $A$, $B$, and $C$ as $(a \cos \alpha, b \sin \alpha)$, $(a \cos \beta, b \sin \beta)$, and $(a \cos \gamma, b \sin \gamma)$, respectively. Then, from the condition
$$
\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{0}
$$

we get
$$
\left\{\begin{array}{l}
a \cos \alpha+a \cos \beta+a \cos \gamma=0, \\
b \sin \alpha+b \sin \beta+b \sin \gamma=0,
\end{array}\right.
$$

which can be simplified to
$$
\left\{\begin{array}{l}
\cos \alpha+\cos \beta+\cos \gamma=0, \\
\sin \alpha+\sin \beta+\sin \gamma=0.
\end{array}\right.
$$

Eliminating $\gamma$, we get $\cos (\beta-\alpha)=-\frac{1}{2}$, so $\sin (\beta-\alpha)=-\frac{\sqrt{3}}{2}$ or $\frac{\sqrt{3}}{2}$,
$$
\begin{aligned}
S_{\triangle A O B} & =\frac{1}{2}\left|x_{1} y_{2}-x_{2} y_{1}\right| \\
& =\frac{1}{2}|a b \cos \alpha \sin \beta-a b \sin \alpha \cos \beta| \\
& =\frac{a b}{2}|\sin (\beta-\alpha)| \\
& =\frac{\sqrt{3} a b}{4} .
\end{aligned}
$$

Similarly, we get
$$
S_{\triangle B O C}=S_{\triangle C O A}=\frac{\sqrt{3} a b}{4},
$$

Therefore,
$$
S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle B O C}+S_{\triangle C O A}=\frac{3 \sqrt{3} a b}{4} .
$$

Thus, the area of triangle $A B C$ is $\frac{3 \sqrt{3} a b}{4}$.