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 $P$, with the foot of the perpendicular being $N$. A perpendicular line $Q M \perp P N$ is drawn from point $Q$, 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 $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\mathbf{0}$. Find the area of $\triangle A B C$.
(1) Let point $M(x, y)$, take $\angle x O P$ as parameter $\varphi$. Then
$$
\left\{\begin{array}{l}
x=a \cos \varphi, \\
y=b \sin \varphi .
\end{array}\right.
$$

Eliminating parameter $\varphi$ yields $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which is the equation of curve $E$.
(2) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$.
From the above, we can also set
$A(a \cos \alpha, b \sin \alpha)$,
$B(a \cos \beta, b \sin \beta)$,
$C(a \cos \gamma, b \sin \gamma)$.
By the condition $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\mathbf{0}$, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
a \cos \alpha+a \cos \beta+a \cos \gamma=0, \\
b \sin \alpha+b \sin \beta+b \sin \gamma=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\cos \alpha+\cos \beta+\cos \gamma=0, \\
\sin \alpha+\sin \beta+\sin \gamma=0 .
\end{array}\right.
\end{array}
$$

Eliminating $\gamma$ yields $\cos (\beta-\alpha)=-\frac{1}{2}$.
Thus, $\sin (\beta-\alpha)= \pm \frac{\sqrt{3}}{2}$.
Then $S_{\triangle A O B}=\frac{1}{2}\left|x_{1} y_{2}-x_{2} y_{1}\right|$
$=\frac{1}{2}|a b \cos \alpha \cdot \sin \beta-a b \sin \alpha \cdot \cos \beta|$
$=\frac{a b}{2}|\sin (\beta-\alpha)|=\frac{\sqrt{3} a b}{4}$.
Similarly, $S_{\triangle B O C}=S_{\triangle C O A}=\frac{\sqrt{3} a b}{4}$.
Therefore, $S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle B O C}+S_{\triangle C O A}=\frac{3 \sqrt{3} a b}{4}$.