11. (This question is worth 20 points) 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 $|P Q| \cdot|P R|$.
Solution: Let $P\left(t^{2}, 2 t\right)$, then the equation of the line $l$ is $y=x+2 t-t^{2}$, substituting into the equation of the curve $C_{2}$, we get,
$$
(x-4)^{2}+\left(x+2 t-t^{2}\right)^{2}=8 \text {, }
$$

Simplifying, we obtain
$$
2 x^{2}-2\left(t^{2}-2 t+4\right) x+\left(t^{2}-2 t\right)^{2}+8=0 \text {. }
$$

Since $l$ intersects $C_{2}$ at two distinct points, the discriminant $\Delta$ of the quadratic equation (1) in $x$ is positive. Calculating, we get,
$$
\begin{array}{c}
\frac{\Delta}{4}=\left(t^{2}-2 t+4\right)^{2}-2\left(\left(t^{2}-2 t\right)^{2}+8\right)=\left(t^{2}-2 t\right)^{2}-8\left(t^{2}-2 t\right)+16-2\left(t^{2}-2 t\right)^{2}-16 \\
=-\left(t^{2}-2 t\right)^{2}+8\left(t^{2}-2 t\right)=-\left(t^{2}-2 t\right)\left(t^{2}-2 t-8\right)=-t(t-2)(t+2)(t-4), \\
t \in(-2,0) \cup(2,4) .
\end{array}
$$

Therefore, we have
10 points
Let the x-coordinates of $Q, R$ be $x_{1}, x_{2}$, respectively. From (1), we know,
$$
x_{1}+x_{2}=t^{2}-2 t+4, x_{1} x_{2}=\frac{1}{2}\left(\left(t^{2}-2 t\right)^{2}+8\right),
$$

Thus, combining the fact that the slope angle of $l$ is $45^{\circ}$, we have,
$$
\begin{aligned}
|P Q| \cdot|P R|= & \sqrt{2}\left(x_{1}-t^{2}\right) \cdot \sqrt{2}\left(x_{2}-t^{2}\right)=2 x_{1} x_{2}-2 t^{2}\left(x_{1}+x_{2}\right)+2 t^{4} \\
& =\left(t^{2}-2 t\right)^{2}+8-2 t^{2}\left(t^{2}-2 t+4\right)+2 t^{4} \\
& =t^{4}-4 t^{3}+4 t^{2}+8-2 t^{4}+4 t^{3}-8 t^{2}+2 t^{4} \\
& =t^{4}-4 t^{2}+8=\left(t^{2}-2\right)^{2}+4 .
\end{aligned}
$$
$$
\text { .................. } 15 \text { points }
$$

From (2), we know that $t^{2}-2 \in(-2,2) \cup(2,14)$, hence $\left(t^{2}-2\right)^{2} \in[0,4) \cup(4,196)$, thus from (3) we get,
$$
|P Q| \cdot|P R|=\left(t^{2}-2\right)^{2}+4 \in[4,8) \cup(8,200) \text {. }
$$

Note 1: Using the fact that the distance from the center of $C_{2}$ to $l$ is less than the radius of $C_{2}$, we can set up the inequality $\left|\frac{4+2 t-t^{2}}{\sqrt{2}}\right|<2 \sqrt{2}$, which similarly yields the range of $t$ in (2).

Note 2: A simpler way to calculate $|P Q| \cdot|P R|$ is to use the power of a point theorem. In fact, the center of $C_{2}$ is $M(4,0)$, with radius $r=2 \sqrt{2}$, so
$$
|P Q| \cdot|P R|=|P M|^{2}-r^{2}=\left(t^{2}-4\right)^{2}+(2 t)^{2}-(2 \sqrt{2})^{2}=t^{4}-4 t^{2}+8 .
$$