8 . Given that $\mathbf{A} 、 \mathbf{B}$ are points on $C_{1}: x^{2}-y+1=0$ and $C_{2}: y^{2}-x+1=0$ respectively, then the minimum value of $|A B|$ is

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【Answer】 $\frac{3 \sqrt{2}}{4}$
【Solution】【Detailed】For the symmetry points $A^{\prime} 、 B^{\prime}$ of $y=x$, they also lie on the other parabola, and $|A B|=\left|A^{\prime} B^{\prime}\right|$. Let $\mathrm{AB}$ intersect $A^{\prime} 、 B^{\prime}$ at point $\mathrm{M}$, then $|A B|+\left|A^{\prime} B^{\prime}\right|=|M A|+\left|M A^{\prime}\right|+\left|M B^{\prime}\right|>\left|A A^{\prime}\right|+\left|B B^{\prime}\right|$, so one of $\left|A A^{\prime}\right|$ and $\left|B B^{\\prime}\right|$ must be less than $|A B|$, which is a contradiction. Therefore, we only need to find twice the minimum distance from point $\mathrm{A}$ to the line $y=x$, then $\mathrm{A}$ is the tangent point of the line parallel to $y=x$ and $y=x^{2}+1$, solving this gives $A\left(\frac{15}{2}, \frac{1}{4}\right)$, hence the minimum value of $|A B|$ is $\frac{3 \sqrt{2}}{4}$. Therefore, the answer is: