3. The center of the hyperbola is at the origin, and the foci are on the coordinate axes. The distance from point \( P(-2,0) \) to its asymptotes is \( \frac{2 \sqrt{6}}{3} \). If a line with a slope of \( \frac{\sqrt{2}}{2} \) passing through point \( P \) intersects the hyperbola at points \( A \) and \( B \), and intersects the \( y \)-axis at point \( M \), and \( PM \) is the geometric mean of \( PA \) and \( PB \), then the semi-focal distance of the hyperbola is
3. $\sqrt{3}$

Analysis: Let the equation of the hyperbola be: $x^{2}-\frac{y^{2}}{k^{2}}=r$, then the equations of the asymptotes are: $y= \pm k x$. Therefore, $\frac{2|k|}{\sqrt{k^{2}+1}}=\frac{2 \sqrt{6}}{3}$, solving this gives $k= \pm \sqrt{2}$, which means the equation of the hyperbola is: $x^{2}-\frac{y^{2}}{2}=r$. Suppose the equation of the line passing through $P(-2,0)$ with a slope of $\frac{\sqrt{2}}{2}$ is: $\left\{\begin{array}{c}x=-2+\frac{\sqrt{6}}{3} t \\ y=\frac{\sqrt{3}}{3} t\end{array}\right.$, where $t$ is a parameter. Substituting the line equation into the hyperbola equation, simplifying yields: $\frac{1}{2} t^{2}-\frac{4 \sqrt{6}}{3} t+4-r=0$. It is easy to know that $P M=\sqrt{6}$, hence $2(4-r)=P A \times P B=P M^{2}=6$, which gives $r=1$. Therefore, the equation of the hyperbola is: $x^{2}-\frac{y^{2}}{2}=1$, and the semi-focal distance is $\sqrt{3}$.