Five. (20 points) Through the center $O$ of the hyperbola $x^{2}-\frac{y^{2}}{4}=1$, draw two mutually perpendicular rays, intersecting the hyperbola at points $A$ and $B$. Try to find:
(1) The equation of the trajectory of the midpoint $P$ of chord $A B$;
(2) The distance from the center $O$ of the hyperbola to the line $A B$;
(3) The minimum length of chord $A B$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
(1) Let $P(x, y)$, $A(x-m, y-n)$, and $B(x+m, y+n)$, then we have
$$
\begin{array}{l}
4(x-m)^{2}-(y-n)^{2}=4, \\
4(x+m)^{2}-(y+n)^{2}=4 .
\end{array}
$$

From $O A \perp O B$ we get
$$
x^{2}+y^{2}=m^{2}+n^{2} \text {. }
$$
(2) - (1) gives
$4 m x=n y$,
(2) + (1) gives
$$
4 x^{2}-y^{2}+4 m^{2}-n^{2}=4 \text {. }
$$

From equations (3) and (4), we solve to get
$$
m^{2}=\frac{y^{2}\left(x^{2}+y^{2}\right)}{16 x^{2}+y^{2}}, n^{2}=\frac{16 x^{2}\left(x^{2}+y^{2}\right)}{16 x^{2}+y^{2}} \text {. }
$$

Substituting into equation (5) gives $3\left(4 x^{2}-y^{2}\right)^{2}=4\left(16 x^{2}+y^{2}\right)$,
where $x^{2}+y^{2} \neq 0$.
This is the required equation of the trajectory.
(2) As shown in Figure 5, draw $O C \perp$
$A B$ at $C$, then it is easy to know
$$
\begin{array}{l}
O C \cdot A B \\
=O A \cdot O B .
\end{array}
$$

From equations (3) and (4) we get
$$
\begin{array}{l}
A B^{2}=4\left(m^{2}+n^{2}\right) \\
=4\left(x^{2}+y^{2}\right), \\
O A^{2} \cdot O B^{2} \\
=\left[(x-m)^{2}+(y-n)^{2}\right]\left[(x+m)^{2}+(y+n)^{2}\right] \\
=4\left(x^{2}+y^{2}\right)^{2}-4(m x+n y)^{2} \\
=4\left(x^{2}+y^{2}\right)^{2}-4(5 m x)^{2} \\
=4\left(x^{2}+y^{2}\right)^{2}-\frac{100 x^{2} y^{2}\left(x^{2}+y^{2}\right)}{16 x^{2}+y^{2}} \\
=\frac{4\left(x^{2}+y^{2}\right)\left(4 x^{2}-y^{2}\right)^{2}}{16 x^{2}+y^{2}} .
\end{array}
$$

Then $O C^{2}=\frac{(O A \cdot O B)^{2}}{A B^{2}}=\frac{\left(4 x^{2}-y^{2}\right)^{2}}{16 x^{2}+y^{2}}=\frac{4}{3}$.
Thus, $O C=\frac{2 \sqrt{3}}{3}$.
(3) From equation (6) we get
$$
O C^{2}=\frac{(O A \cdot O B)^{2}}{A B^{2}}=\frac{(O A \cdot O B)^{2}}{O A^{2}+O B^{2}} \text {. }
$$

Then $\frac{1}{O C^{2}}=\frac{1}{O A^{2}}+\frac{1}{O B^{2}}=\frac{3}{4}$.
Also, $\left(O A^{2}+O B^{2}\right)\left(\frac{1}{O A^{2}}+\frac{1}{O B^{2}}\right) \geqslant 4$, thus,
$$
A B^{2}=O A^{2}+O B^{2} \geqslant \frac{16}{3}, A B \geqslant \frac{4 \sqrt{3}}{3} \text {. }
$$

And the equality holds only when $O A=O B=\frac{2 \sqrt{6}}{3}$.
Therefore, the minimum value of the chord $A B$ is $\frac{4 \sqrt{3}}{3}$.