14. Let the lines $l_{1}: y=\sqrt{3} x, l_{2}: y=-\sqrt{3} x$. Points $A$ and $B$ move on lines $l_{1}$ and $l_{2}$ respectively, and $\overrightarrow{O A} \cdot \overrightarrow{O B}=-2$.
(1) Find the locus of the midpoint $M$ of $A B$;
(2) Let point $P(-2,0)$ have a symmetric point $Q$ with respect to the line $A B$. Prove that the line $M Q$ passes through a fixed point.
(1) Let $A\left(x_{1}, \sqrt{3} x_{1}\right), B\left(x_{2},-\sqrt{3} x_{2}\right), M\left(x_{0}, y_{0}\right)$,

Since $\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}-3 x_{1} x_{2}=-2 x_{1} x_{2}=-2 \Rightarrow x_{1} x_{2}=1$,
Combining $\left\{\begin{array}{l}x_{0}=\frac{x_{1}+x_{2}}{2}, \\ y_{0}=\frac{\sqrt{3}\left(x_{1}-x_{2}\right)}{2}\end{array} \Rightarrow x_{0}^{2}-\frac{y_{0}^{2}}{3}=x_{1} x_{2}=1\right.$.
Therefore, the trajectory of the midpoint $M$ of $A B$ is the hyperbola $x^{2}-\frac{y^{2}}{3}=1$.
(2) Let the slopes of $P M, Q M$ be $k_{1}, k_{2}$, and the slope of line $A B$ be $k$,

Then $k=\frac{\sqrt{3}\left(x_{1}+x_{2}\right)}{x_{1}-x_{2}}=\frac{3 x_{0}}{y_{0}}, k_{1}=\frac{y_{0}}{x_{0}+2}$, by the angle formula we get
$\frac{k_{2}-k}{1+k_{2} k}=\frac{k-k_{1}}{1+k k_{1}}=\frac{\frac{3 x_{0}}{y_{0}}-\frac{y_{0}}{x_{0}+2}}{1+\frac{3 x_{0}}{y_{0}} \cdot \frac{y_{0}}{x_{0}+2}}=\frac{3 x_{0}^{2}+6 x_{0}-y_{0}^{2}}{4 x_{0} y_{0}+2 y_{0}}=\frac{3}{2 y_{0}}$, thus
$$
k_{2}=\frac{2 y_{0} k+3}{2 y_{0}-3 k}=\frac{2 y_{0} \cdot \frac{3 x_{0}}{y_{0}}+3}{2 y_{0}-3 \cdot \frac{3 x_{0}}{y_{0}}}=\frac{6 x_{0}+3}{2 y_{0}-\frac{9 x_{0}}{y_{0}}}=\frac{y_{0}}{x_{0}-2} \Leftrightarrow 6 x_{0}^{2}-2 y_{0}^{2}=6 \text {. }
$$

Therefore, points $Q, M, R(2,0)$ are collinear, so line $M Q$ passes through the fixed point $R(2,0)$.