11. Given $\angle A O B=\theta\left(\theta\right.$ is a constant and $\left.0<\theta<\frac{\pi}{2}\right)$, moving points $P, Q$ are on rays $O A, O B$ respectively such that the area of $\triangle P O Q$ is always 36. Let the centroid of $\triangle P O Q$ be $G$, and point $M$ is on ray $O G$ such that $|O M|=\frac{3}{2}|O G|$.
(1) Find the minimum value of $|O G|$;
(2) Find the equation of the locus of the moving point $M$.
11. Solution: (1) Taking $O$ as the origin and the bisector of $\angle A O B$ as the $x$-axis to establish a rectangular coordinate system, we can set $P\left(a \cos \frac{\theta}{2}, a \sin \frac{\theta}{2}\right), Q\left(b \cos \frac{\theta}{2},-b \sin \frac{\theta}{2}\right)$. Thus, the coordinates of the centroid $G\left(x_{G}, y_{G}\right)$ of $\triangle O P Q$ are
$$
\begin{array}{l}
x_{G}=\frac{1}{3}\left(a \cos \frac{\theta}{2}+b \cos \frac{\theta}{2}+0\right)=\frac{1}{3}(a+b) \cos \frac{\theta}{2}, \\
y_{G}=\frac{1}{3}\left(a \sin \frac{\theta}{2}-b \sin \frac{\theta}{2}+0\right)=\frac{1}{3}(a-b) \sin \frac{\theta}{2} . \\
|O G|^{2}=x_{G}^{2}+y_{G}^{2}=\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b\left(\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}\right)=\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b \cos \theta \\
\geqslant \frac{1}{9} \cdot 2 a b+\frac{2}{9} a b \cos \theta=\frac{4}{9} a b \cos ^{2} \frac{\theta}{2} .
\end{array}
$$

Given that $S_{\triangle O P Q}=\frac{1}{2} a b \sin \theta=36$, we get $a b=\frac{72}{\sin \theta}$, thus $|O G| \geqslant \sqrt{\frac{4}{9} \cdot \frac{72}{\sin \theta} \cdot \cos ^{2} \frac{\theta}{2}}$ $=\sqrt{16 \cot \frac{\theta}{2}}=4 \sqrt{\cot \frac{\theta}{2}}$, and the equality holds when $a=b=\sqrt{\frac{72}{\sin \theta}}$, hence $|O G|_{\min }=4 \sqrt{\cot \frac{\theta}{2}}$.
(2) Let $M(x, y)$, then from $|O M|=\frac{3}{2}|O G|$ we get, $x=\frac{3}{2} x_{G}=\frac{1}{2}(a+b) \cos \frac{\theta}{2}>0$,
$$
y=\frac{3}{2} y_{G}=\frac{1}{2}(a-b) \sin \frac{\theta}{2},
$$

we get $b=\frac{x}{\cos \frac{\theta}{2}}-\frac{y}{\sin \frac{\theta}{2}}$, substituting into $a b=\frac{72}{\sin \theta}$ and simplifying, we get
$\frac{x^{2}}{36 \cot \frac{\theta}{2}}-\frac{y^{2}}{36 \tan \frac{\theta}{2}}=1(x>0)$, which is the equation of the trajectory of the moving point $M$.