$8 \mathrm{C}$ Hint: The two parabolas $y=-x^{2}-1$, $x=1+y^{2}$ are symmetric with respect to the line $y=-x$. The minimum value of $|P Q|$ is twice the minimum distance from a point on the parabola $y=-x^{2}-1$ to the line $y=-x$. Let $P\left(x,-x^{2}-1\right)$ be any point on $y=-x^{2}-1$, then
$$
\begin{aligned}
d & =\frac{\left|x-x^{2}-1\right|}{\sqrt{2}}=\frac{x^{2}-x+1}{\sqrt{2}}, \\
d_{\min } & =\frac{3 \sqrt{2}}{8},|P Q|_{\min }=\frac{3 \sqrt{2}}{4} .
\end{aligned}
$$
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) $\frac{3 \sqrt{2}}{8}$