8. Given the hyperbola $H: x^{2}-y^{2}=1$ and a point $M$ in the first quadrant on $H$, the line $l$ is tangent to $H$ at point $M$ and intersects the asymptotes of $H$ at points $P, Q$ (with $P$ in the first quadrant). If point $R$ is on the same asymptote as $Q$, then the minimum value of $\overrightarrow{R P} \cdot \overrightarrow{R Q}$ is $\qquad$
Given $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right), M\left(x_{0}, y_{0}\right)$,
$l: x_{0} x-y_{0} y=1$, solving the system $\left\{\begin{array}{l}x^{2}-y^{2}=0, \\ x_{0} x-y_{0} y=1\end{array}\right.$
$$
\Rightarrow x^{2}-2 x_{0} x+1=0 \text {. }
$$

Then $\frac{x_{1}+x_{2}}{2}=x_{0} \Rightarrow$ point $M$ is the midpoint of segment $P Q$, thus by the polarization identity we get
$$
\begin{array}{l}
\overrightarrow{R P} \cdot \overrightarrow{R Q}=R M^{2}-M Q^{2} \geqslant M N^{2}-M Q^{2} \\
=\frac{1}{4} P O^{2}-\frac{1}{4} P Q^{2}=-\frac{1}{4} O Q^{2} .
\end{array}
$$

Noting that $x_{1} x_{2}=1$, and $P$ is in the first quadrant, therefore $x_{1} \geqslant 1, x_{2} \leqslant 1 \Rightarrow O Q^{2}=2 x_{2}^{2} \leqslant 2$, equality holds when $M(1,0)$.
In summary, the minimum value of $\overrightarrow{R P} \cdot \overrightarrow{R Q}$ is $-\frac{1}{2}$.