11. (20 points) In the Cartesian coordinate system $x O y$, the curve $C_{1}: y^{2}=4 x$,
curve $C_{2}:(x-4)^{2}+y^{2}=8$,
a line $l$ with an inclination angle of $45^{\circ}$ is drawn through a point $P$ on the curve $C_{1}$, intersecting the curve $C_{2}$ at two different points $Q$ and $R$. Find the range of $|P Q||P R|$.
11. Let $P\left(t^{2}, 2 t\right)$. Then the line $l: y=x+2 t-t^{2}$.

Substituting into the equation of curve $C_{2}$ and rearranging, we get
$$
2 x^{2}-2\left(t^{2}-2 t+4\right) x+\left(t^{2}-2 t\right)^{2}+8=0 \text {. }
$$

Since $l$ intersects curve $C_{2}$ at two distinct points, we have
$$
\begin{array}{l}
\frac{\Delta}{4}=\left(t^{2}-2 t+4\right)^{2}-2\left(\left(t^{2}-2 t\right)^{2}+8\right) \\
=-t(t-2)(t+2)(t-4)>0 \\
\Rightarrow t \in(-2,0) \cup(2,4) .
\end{array}
$$

Let the x-coordinates of points $Q$ and $R$ be $x_{1}$ and $x_{2}$, respectively. From equation (1), we know
$$
\begin{array}{l}
x_{1}+x_{2}=t^{2}-2 t+4, \\
x_{1} x_{2}=\frac{1}{2}\left(\left(t^{2}-2 t\right)^{2}+8\right) .
\end{array}
$$

Thus, combining with the fact that the inclination angle of $l$ is $45^{\circ}$, we have
$$
\begin{array}{l}
|P Q||P R|=\sqrt{2}\left(x_{1}-t^{2}\right) \sqrt{2}\left(x_{2}-t^{2}\right) \\
=2 x_{1} x_{2}-2 t^{2}\left(x_{1}+x_{2}\right)+2 t^{4} \\
=\left(t^{2}-2 t\right)^{2}+8-2 t^{2}\left(t^{2}-2 t+4\right)+2 t^{4} \\
=\left(t^{2}-2\right)^{2}+4 .
\end{array}
$$

From conclusion (2), we know
$$
\begin{array}{l}
|P Q||P R|=\left(t^{2}-2\right)^{2}+4 \\
\in[4,8) \cup(8,200)
\end{array}
$$