Problem 849

2. $\sqrt{3}$ or $\sqrt{21}$.

Let the equation of the asymptote be $y=k x$.
From the given condition, we have $\frac{|-2 k|}{\sqrt{1+k^{2}}}=\frac{2 \sqrt{6}}{3} \Rightarrow k= \pm \sqrt{2}$.
Thus, the equations of the asymptotes of the hyperbola are $y= \pm \sqrt{2} x$. Therefore, we can assume the equation of the hyperbola to be
$$2 x^{2}-y^{2}=\lambda(\lambda \neq 0) \text {. }$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$l_{A B}: y=\frac{\sqrt{2}}{2}(x+2) \text {. }$$

Substituting into the equation of the hyperbola and eliminating $y$, we get
$$3 x^{2}-4 x-2 \lambda-4=0 \text {. }$$

When $\Delta=16+12(2 \lambda+4)>0$, i.e., $\lambda>-\frac{8}{3}$, the above equation has exactly two real roots, and
$$\begin{array}{l} x_{1}+x_{2}=\frac{4}{3}, x_{1} x_{2}=-\frac{2}{3}(\lambda+2) . \\ \text { By } P M^{2}=P A \cdot P B \\ \Rightarrow\left|\left(x_{1}+2\right)\left(x_{2}+2\right)\right|=4 \\ \Rightarrow\left|-\frac{2}{3}(\lambda+2)+2 \times \frac{4}{3}+4\right|=4 . \end{array}$$

Solving, we get $\lambda=2$ or 14.
Therefore, the semi-focal distance of the hyperbola is
$$\sqrt{\frac{3 \lambda}{2}}=\sqrt{3} \text { or } \sqrt{21} \text {. }$$