8. If point $P$ is on the curve $y=-x^{2}-1$, and point $Q$ is on the curve $x=1+y^{2}$, then the minimum value of $|P Q|$ is ( ).
(A) $3 \sqrt{2}$
(B) $\frac{3 \sqrt{2}}{2}$
(C) $\frac{3 \sqrt{2}}{4}$
(D) $\frac{3 \sqrt{2}}{8}$
8. C.

Given that the two parabolas are symmetric about the line $y=-x$, the minimum value of $|P Q|$ is twice the minimum distance from point $P\left(x,-x^{2}-1\right)$ to the line $y=-x$. Therefore,
$$
\begin{array}{l}
d=\frac{\left|x-x^{2}-1\right|}{\sqrt{2}}=\frac{x^{2}-x+1}{\sqrt{2}}, \\
d_{\min }=\frac{3 \sqrt{2}}{8} .
\end{array}
$$

Thus, $|P Q|_{\text {min }}=\frac{3 \sqrt{2}}{4}$.