11. In the plane rectangular coordinate system $x O y$, the curve $C_{1}: y^{2}=4 x$, the curve $C_{2}:(x-4)^{2}+y^{2}=8$. Through a point $P$ on $C_{1}$, a line $l$ with an inclination angle of $45^{\circ}$ is drawn, intersecting $C_{2}$ at two different points $Q, R$. Find the range of $|P Q| \cdot|P R|$.
Given the point $P$ to draw a tangent line $P M$ of $\odot C_{2}$, with the tangent point at $M$, connect $P M, P C_{2}, M C_{2}$. Let $P\left(t^{2}, 2 t\right), C_{2}(4,0)$, and note that the radius of $\odot C_{2}$ is $r=2 \sqrt{2}$. By the secant-tangent theorem, we have $|P Q| \cdot|P R|=|P M|^{2}=\left|P C_{2}\right|^{2}-\left|M C_{2}\right|^{2}$ $=\left(t^{2}-4\right)^{2}+(2 t)^{2}-8=t^{4}-4 t^{2}+8=\left(t^{2}-2\right)^{2}+4$.
Let the line $l: y-2 t=x-t^{2} \Rightarrow x-y-t^{2}+2 t=0$, and since the line $l$ intersects $\odot C_{2}$, we get
$$
\begin{array}{l}
\frac{\left|4-0-t^{2}+2 t\right|}{\sqrt{2}}-4, \\
t^{2}-2 t-4 0 , } \\
{ t ^ { 2 } - 2 t - 8  0 , } \\
{ ( t + 2 ) ( t - 4 ) 2, \\
-2<t<4
\end{array}\right.\right.\right. \\
\Rightarrow t \in(-2,0) \cup(2,4) \Rightarrow t^{2} \in(0,4) \cup(4,16) .
\end{array}
$$

Thus, $t^{2}-2 \in(-2,2) \bigcup(2,14) \Rightarrow\left(t^{2}-2\right)^{2} \in[0,4) \bigcup(4,196)$,
Therefore, $|P Q| \cdot|P R|=\left(t^{2}-2\right)^{2}+4 \in[4,8) \cup(8,200)$.