9. In the Cartesian 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 $\frac{|P Q|}{|Q R|}$.
(1) Let $P(a, b), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then $P A: y y_{1}=2\left(x+x_{1}\right)$,
$P B: y y_{2}=2\left(x+x_{2}\right)$. Thus $\left\{\begin{array}{l}b y_{1}=2\left(a+x_{1}\right), \\ b y_{2}=2\left(a+x_{2}\right)\end{array} \Rightarrow A B: b y=2(a+x)\right.$ $\Rightarrow A B: 2 x-b y+2 a=0$. Also, $P O: b x-a y=0$, by $A B \perp P O \Rightarrow 2 b+a b=0 \Rightarrow a=-2$. Therefore, in the equation of line $A B$, setting $y=0$ gives $x=2 \Rightarrow R(2,0)$, which means $R$ is a fixed point.
(2) By symmetry, let $P$ be in the second quadrant, i.e., $b>0$.
Let the inclination angle of line $A B$ be $\theta \Rightarrow \sin \theta=\frac{2}{\sqrt{b^{2}+4}}, \cos \theta=\frac{b}{\sqrt{b^{2}+4}}$,
then $|Q R|=|O R| \cos \theta=\frac{2 b}{\sqrt{b^{2}+4}},|P Q|=|P O|+|O Q|=\sqrt{b^{2}+4}+\frac{4}{\sqrt{b^{2}+4}}$.
Thus $\frac{|P Q|}{|Q R|}=\frac{\sqrt{b^{2}+4}+\frac{4}{\sqrt{b^{2}+4}}}{\frac{2 b}{\sqrt{b^{2}+4}}}=\frac{b^{2}+8}{2 b}=\frac{1}{2}\left(b+\frac{8}{b}\right) \geqslant 2 \sqrt{2}$,
equality holds when $b=2 \sqrt{2}$. Therefore, the minimum value of $\frac{|P Q|}{|Q R|}$ is $2 \sqrt{2}$.