18 Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\left(a, b \in \mathbf{R}^{+}\right)$ with semi-focal distance $c$, and $b^{2}=a c . P$ and $Q$ are any two points on the hyperbola, $M$ is the midpoint of $P Q$, when the slopes $k_{P Q}$ and $k_{O M}$ of $P Q$ and $O M$ both exist, find the value of $k_{P Q} \cdot k_{O M}$.
Given that $M$ is the midpoint of $P Q$, we can set the coordinates of point $M$ as $\left(x_{0}, y_{0}\right)$, the coordinates of point $P$ as $\left(x_{0}+\alpha, y_{0}+\beta\right)$, and the coordinates of point $Q$ as $\left(x_{0}-\alpha, y_{0}-\beta\right)$. Thus, $k_{O M}=\frac{y_{0}}{x_{0}}, k_{P Q}=\frac{\beta}{\alpha}$. Since $P$ and $Q$ are both on the hyperbola, we have
$$
\left\{\begin{array}{l}
b^{2}\left(x_{0}+\alpha\right)^{2}-a^{2}\left(y_{0}+\beta\right)^{2} \cdot \dot{a}^{2} b^{2}, \\
b^{2}\left(x_{0}-\alpha\right)^{2}-a^{2}\left(y_{0}-\beta\right)^{2}=a^{2} b^{2},
\end{array}\right.
$$

Subtracting the two equations yields $4 b^{2} \alpha x_{0}-4 a^{2} \beta y_{0}=0$, hence $k_{O M} \cdot k_{P Q}=\frac{\beta y_{0}}{a x_{0}}=\frac{b^{2}}{a^{2}}=\frac{a c}{a^{2}}=\frac{c}{a}$. From $\left\{\begin{array}{l}c^{2}=a^{2}+b^{2}, \\ b^{2}=a c,\end{array}\right.$, eliminating $b^{2}$ gives $c^{2}=a^{2}+a c$, solving for $\frac{c}{a}=\frac{1+\sqrt{5}}{2}$ (discarding the negative root). Therefore, $k_{O M} \cdot k_{P Q}=\frac{1+\sqrt{5}}{2}$.