8. Let $x_{0}>0, x_{0} \neq \sqrt{3}, Q\left(x_{0}, 0\right), P(0,4)$, the line $P Q$ intersects the hyperbola $x^{2}-\frac{y^{2}}{3}=1$ at points $A, B$. If $\overrightarrow{P Q}=t \overrightarrow{Q A}=(2-t) \overrightarrow{Q B}$, then $x_{0}=$ $\qquad$ .
Given $P Q: x=-\frac{x_{0}}{4}(y-4), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, combining $\left\{\begin{array}{l}x=-\frac{x_{0}}{4}(y-4), \\ 3 x^{2}-y^{2}=3\end{array}\right.$ $\Rightarrow \frac{3 x_{0}^{2}-16}{16} y^{2}-\frac{3 x_{0}^{2}}{2} y+3 x_{0}^{2}-3=0$. According to the problem, $\overrightarrow{P Q}=\left(x_{0},-4\right)$,
$$
\overrightarrow{Q A}=\left(x_{1}-x_{0}, y_{1}\right), \overrightarrow{Q B}=\left(x_{2}-x_{0}, y_{2}\right) \Rightarrow\left\{\begin{array}{l}
-4=t y_{1}, \\
-4=(2-t) y_{2}
\end{array} \quad \Rightarrow \frac{1}{y_{1}}+\frac{1}{y_{2}}=-\frac{1}{2}\right. \text {. }
$$

Thus $\frac{\frac{3 x_{0}^{2}}{2}}{3 x_{0}^{2}-3}=-\frac{1}{2} \Rightarrow x_{0}^{2}=\frac{1}{2} \Rightarrow x_{0}=\frac{\sqrt{2}}{2}$.
Note: This problem is incorrect, as the line $P Q$ does not intersect the hyperbola.