8. Draw a line with an inclination angle of $\frac{\pi}{4}$ through the point $P(11,0)$ intersecting the parabola $y^{2}=4 x$ at points $R, Q$. Then draw a line parallel to $R Q$ intersecting the parabolic arc $\overparen{R O Q}$ at points $M, N$ (as shown in the figure). Find the maximum value of the area of $\triangle P M N$.
8. Substituting the equation of line $M N$, $y=x+m$, into $y^{2}=4 x$, we get $|M N|=4 \sqrt{2} \cdot \sqrt{1-m}$. The distance from point $P$ to line $M N$ is $d=\frac{|m+11|}{\sqrt{2}}$, and $S=2 \sqrt{1-m}|m+11|$. Using the properties of inequalities, we find the maximum value to be 22.