9. (This question is worth 16 points) In the plane rectangular coordinate system $x O y$, $P$ is a moving point not on the $x$-axis, satisfying the condition: through $P$, two tangent lines can be drawn to the parabola $y^{2}=4 x$, and the line connecting the two points of tangency $l_{P}$ is perpendicular to $P O$.
Let the intersection points of the line $l_{P}$ with the line $P O$ and the $x$-axis be $Q$ and $R$, respectively.
(1) Prove that $R$ is a fixed point;
(2) Find the minimum value of $\frac{|P Q|}{|Q R|}$.
Solution: (1) Let the coordinates of point $P$ be $(a, b) (b \neq 0)$, it is easy to know that $a \neq 0$. Let the coordinates of the two tangent points $A, B$ be $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right)$, respectively, then the equations of $P A, P B$ are
$$
\begin{array}{l}
y y_{1}=2\left(x+x_{1}\right), \\
y y_{2}=2\left(x+x_{2}\right),
\end{array}
$$

Since the coordinates of point $P$ $(a, b)$ satisfy (1) and (2), the coordinates of $A, B$ $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right)$ both satisfy the equation
$$
b y=2(x+a).
$$

Thus, (3) is the equation of line $A B$.
The slopes of lines $P O$ and $A B$ are $\frac{b}{a}$ and $\frac{2}{b}$, respectively. Since $P O \perp A B$, we have $\frac{b}{a} \cdot \frac{2}{b}=-1$, hence $a=-2$.
$\qquad$
Therefore, (3) is $y=\frac{2}{b}(x-2)$, so the intersection point $R$ of $A B$ with the $x$-axis is the fixed point $(2,0)$. $\qquad$
(2) Since $a=-2$, the slope of line $P O$ is $k_{1}=-\frac{b}{2}$, and the slope of line $P R$ is $k_{2}=-\frac{b}{4}$. Let $\angle O P R=\alpha$, then $\alpha$ is an acute angle, and
$$
\frac{|P Q|}{|Q R|}=\frac{1}{\tan \alpha}=\left|\frac{1+k_{1} k_{2}}{k_{1}-k_{2}}\right|=\left|\frac{1+\left(-\frac{b}{2}\right)\left(-\frac{b}{4}\right)}{-\frac{b}{2}+\frac{b}{4}}\right|=\frac{8+b^{2}}{2|b|} \geq \frac{2 \sqrt{8 \cdot b^{2}}}{2|b|}=2 \sqrt{2}.
$$

When $b= \pm 2 \sqrt{2}$, $\frac{|P Q|}{|Q R|}$ reaches its minimum value of $2 \sqrt{2}$.
16 points