11 (20 points) As shown in the figure, two perpendicular rays are drawn through the center $O$ of the hyperbola $x^{2}-$ $\frac{y^{2}}{4}=1$, intersecting the hyperbola at points $A$ and $B$. Try to find:
（1）the equation of the locus of the midpoint $P$ of the chord $A B$;
（2）the distance from the center $O$ of the hyperbola to the line $A B$.
11 (1) Let $P(x, y), A(x-m, y-n), B(x+m, y+n)$. From $A$, $B$ being on the hyperbola, we have
$$
\begin{array}{l}
4(x-m)^{2}-(y-n)^{2}=4, \\
4(x+m)^{2}-(y+n)^{2}=4 .
\end{array}
$$

Also, $O A \perp O B$, so
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=0 \Rightarrow(x-m)(x+m)+(y-n)(y+n)=0,
$$

which gives
$$
x^{2}+y^{2}=m^{2}+n^{2} .
$$

From (1), (2), and (3), eliminating $m, n$ yields
$$
3\left(4 x^{2}-y^{2}\right)^{2}=4\left(16 x^{2}+y^{2}\right) .
$$

This is the equation of the trajectory of point $P$.
(2) Draw $O C \perp A B$ at $C$, then $O C \cdot A B=O A \cdot O B$. From (3), we have
$$
A B^{2}=4\left(m^{2}+n^{2}\right)=4\left(x^{2}+y^{2}\right) \text {. }
$$

Thus,
$$
\begin{aligned}
O A^{2} \cdot O B^{2} & =\left[(x-m)^{2}+(y-n)^{2}\right]\left[(x+m)^{2}+(y+n)^{2}\right] \\
& =\left(x^{2}+y^{2}+m^{2}+n^{2}\right)^{2}-(2 x m+2 y n)^{2} \\
& =\left[2\left(x^{2}+y^{2}\right)^{2}\right]-4(m x+n y)^{2} .
\end{aligned}
$$

From (1), (2), and (3), we get $4 m x=n y, m^{2}=\frac{\left(x^{2}+y^{2}\right) y^{2}}{16 x^{2}+y^{2}}$, so
$$
\begin{aligned}
O A^{2} \cdot O B^{2} & =4\left(x^{2}+y^{2}\right)^{2}-100 x^{2} \cdot \frac{\left(x^{2}+y^{2}\right) y^{2}}{16 x^{2}+y^{2}} \\
& =\frac{4\left(x^{2}+y^{2}\right)\left(4 x^{2}-y^{2}\right)^{2}}{16 x^{2}+y^{2}} .
\end{aligned}
$$

Therefore,
$$
O C^{2}=\frac{O A^{2} \cdot O B^{2}}{A B^{2}}=\frac{\left(4 x^{2}-y^{2}\right)^{2}}{16 x^{2}+y^{2}} .
$$

Also,
$$
3\left(4 x^{2}-y^{2}\right)^{2}=4\left(16 x^{2}+y^{2}\right),
$$

so
$O C^{2}=\frac{4}{3} \Rightarrow O C=\frac{2 \sqrt{3}}{3}$. That is, the distance from $O$ to $A B$ is $\frac{2 \sqrt{3}}{3}$.