13. 5 * ** Given that the center of the hyperbola is at the origin, the foci are on the $x$-axis, and a line passing through the right focus with a slope of $\sqrt{\frac{3}{5}}$ intersects the hyperbola at points $P$ and $Q$. If $O P \perp O Q$ and $|P Q|=4$, find the equation of the hyperbola.
According to the problem, we can set the equation of the hyperbola as \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), with the right focus at \((c, 0)\), where \(a, b, c > 0\). Then, the line \(l_{P Q}\) is given by \(y=\sqrt{\frac{3}{5}}(x-c)\). By solving the system of equations 
\[
\left\{\begin{array}{l}
y=\sqrt{\frac{3}{5}}(x-c), \\
b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2},
\end{array}\right.
\]
we obtain
\[
\left(5 b^{2}-3 a^{2}\right) x^{2}+6 a^{2} c x-a^{2}\left(3 c^{2}+5 b^{2}\right)=0.
\]

Let \(P\left(x_{1}, \sqrt{\frac{3}{5}}\left(x_{1}-c\right)\right)\) and \(Q\left(x_{2}, \sqrt{\frac{3}{5}}\left(x_{2}-c\right)\right)\). Since \(O P \perp O Q\), we have
\[
\begin{aligned}
0 & =\stackrel{\rightharpoonup}{O P} \cdot \overrightarrow{O Q}=x_{1} x_{2}+\frac{3}{5}\left(x_{1}-c\right)\left(x_{2}-c\right) \\
& =\frac{8}{5} x_{1} x_{2}-\frac{3}{5} c\left(x_{1}+x_{2}\right)+\frac{3}{5} c^{2}.
\end{aligned}
\]

Thus, 
\[
8 \frac{-a^{2}\left(3 c^{2}+5 b^{2}\right)}{5 b^{2}-3 a^{2}}-3 c \frac{-6 a^{2} c}{5 b^{2}-3 a^{2}}+3 c^{2}=0,
\]
which simplifies to \(3 a^{4}+8 a^{2} b^{2}-3 b^{4}=0\). Solving this, we get \(b^{2}-3 a^{2}\), and \(c=\sqrt{a^{2}+b^{2}}=2 a\). Substituting into the equation, we get \(4 x^{2}+4 a x-9 a^{2}=0\). Then, \(4=|P Q|=\sqrt{1+\frac{3}{5}}\left|x_{1}-x_{2}\right|=\sqrt{\frac{8}{5}} \sqrt{a^{2}+9 a^{2}}=4 a\), so \(a^{2}=1\) and \(b^{2}=3\). Therefore, the equation of the hyperbola is \(x^{2}-\frac{y^{2}}{3}=1\).