Example 3 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 $l_{P}$ connecting the two points of tangency is perpendicular to $P O$. Let the intersection points of the line $l_{P}$ with $P O$ and the $x$-axis be $Q$ and $R$, respectively.
(1) Prove: $R$ is a fixed point;
(2) Find the minimum value of $\frac{|P Q|}{|Q R|}$.
(2019, National High School Mathematics Joint Competition (A Volume))
(1) Let $P(a, b)(b \neq 0)$.
By Proposition 4, we know the line $l_{P}: b y=2(x+a)$, and the slopes of lines $P O$ and $l_{P}$ are $\frac{b}{a}$ and $\frac{2}{b}$, respectively.
Since $P O \perp l_{P}$, we have
$\frac{b}{a} \cdot \frac{2}{b}=-1 \Rightarrow a=-2$.
Thus, the intersection point $R$ of $l_{P}$ with the $x$-axis is the fixed point $(2,0)$.
(2) The slopes of lines $P O$ and $P R$ are
$k_{1}=-\frac{b}{2}, k_{2}=-\frac{b}{4}$.
Let $\angle O P R=\alpha$. Then
$$
\begin{array}{l}
\frac{|P Q|}{|Q R|}=\frac{1}{\tan \alpha}=\left|\frac{1+k_{1} k_{2}}{k_{1}-k_{2}}\right| \\
=\left|\frac{1+\left(-\frac{b}{2}\right)\left(-\frac{b}{4}\right)}{-\frac{b}{2}+\frac{b}{4}}\right|=\frac{8+b^{2}}{2|b|} \\
\geqslant \frac{2 \sqrt{8 b^{2}}}{2|b|}=2 \sqrt{2} .
\end{array}
$$

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