9. (16 points) In the plane rectangular coordinate system $x O y$, $P$ is a moving point not on the $x$-axis, and it satisfies that two tangent lines to the parabola $y^{2}=4 x$ can be drawn through point $P$. The line connecting the two points of tangency $l_{P}$ is perpendicular to $P O$, and the intersection points of line $l_{P}$ with $P O$ and the $x$-axis are $Q$ and $R$, respectively.
(1) Prove: $R$ is a fixed point;
(2) Find the minimum value of $\frac{|P Q|}{|Q R|}$.


(1) Let point $P(a, b)(b \neq 0)$. It is easy to know that $a \neq 0$.
As shown in Figure 5, let the points of tangency be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $l_{P A}: y y_{1}=2\left(x+x_{1}\right)$,
$l_{P B}: y y_{2}=2\left(x+x_{2}\right)$.
Since point $P(a, b)$ satisfies equations (1) and (2), the coordinates of points $A$ and $B$ both satisfy the equation
$$
b y=2(x+a) \text {. }
$$

Thus, equation (3) is the equation of line $A B$.
Note that, $k_{P O}=\frac{b}{a}, k_{A B}=\frac{2}{b}$.
Since $P O \perp A B$, we have
$$
\frac{b}{a} \cdot \frac{2}{b}=-1 \Rightarrow a=-2 \text {. }
$$

Therefore, equation (3) is $y=\frac{2}{b}(x-2)$.
Thus, the intersection point $R$ of $A B$ with the $x$-axis is the fixed point $(2,0)$.
(2) Since $a=-2$, the slope of line $P Q$ 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
$$
\begin{array}{l}
\frac{|P Q|}{|Q R|}=\frac{1}{\tan \alpha}=\left|\frac{1+k_{1} k_{2}}{k_{1}-k_{2}}\right| \\
=\left|\frac{1+\left(-\frac{b}{2}\right)\left(-\frac{b}{4}\right)}{-\frac{b}{2}+\frac{b}{4}}\right|=\frac{8+b^{2}}{2|b|} \\
\geqslant \frac{2 \sqrt{8 b^{2}}}{2|b|}=2 \sqrt{2} .
\end{array}
$$

When $b= \pm 2 \sqrt{2}$, the minimum value of $\frac{|P Q|}{|Q R|}$ is $2 \sqrt{2}$.