4. The center of the hyperbola is at the origin $O$, with foci on the $x$-axis. A line passing through the right focus of the hyperbola and with a slope of $\sqrt{\frac{3}{5}}$ intersects the hyperbola at points $P$ and $Q$. If $P O \perp O Q,|P Q|$ $=4$, find the equation of the hyperbola.
4. Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, and the equation of the line $P Q$ be $y=\sqrt{\frac{3}{5}}(x-c)$, where $c^{2}=a^{2}+b^{2}$. Then $\frac{3}{5} b^{4} x^{2}-\frac{3}{5} a^{2}\left(a^{2}+\frac{8}{3} b^{2}\right) y^{2}+2 \sqrt{\frac{3}{5}} a^{2} b^{2} x y=0$.

Given that $O P \perp O Q$, the above equation can be transformed into $\left(a^{2}+3 b^{2}\right)\left(3 a^{2}-b^{2}\right)=0$. Thus, $b=\sqrt{3} a$, which leads to $c=2 a, e=2$.
Therefore, the line with a slope of $\sqrt{\frac{3}{5}}$ intersects the hyperbola at points on the left and right branches.
Given that the line $P Q$ passes through the focus and $|P Q|=4$, we establish a polar coordinate system with $F$ as the pole and $F Z$ as the polar axis, then the equation of the hyperbola is $\rho=\frac{e p}{1-e \cos \theta}(\rho \in \mathbf{R})$. Hence, $|P Q|=|F P|-|F Q|=$ $\frac{2 e \rho}{e^{2} \cos ^{2} \theta-1}=4$, and by $e p=\frac{b^{2}}{a}, \cos ^{2} \theta=\frac{1}{1+\operatorname{tg}^{2} \theta}=\frac{5}{8}$, we get $\quad \frac{4 b^{2}}{3 a^{2}}=4$.
Thus, we deduce $a^{2}=1, b^{2}=3$. Therefore, the equation of the hyperbola is $x^{2}-\frac{y^{2}}{3}=1$.