9. A line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M, N$. Point $E(m, 0)$ is on the $x$-axis. The extensions of $M E, N E$ intersect the parabola at points $P, Q$. If the slopes $k_{1}, k_{2}$ of $M N, P Q$ satisfy $k_{1}=3 k_{2}$, then the value of the real number $m$ is $\qquad$
Solve:
Given $M N: x=n y+1$, combining $\left\{\begin{array}{l}x=n y+1, \\ y^{2}=4 x\end{array}\right.$ $\Rightarrow y^{2}-4 n y-4=0$.
Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), P\left(x_{3}, y_{3}\right), Q\left(x_{4}, y_{4}\right)$,
then $k_{P M}=\frac{y_{1}}{x_{1}-m} \Rightarrow P M: x=\frac{x_{1}-m}{y_{1}} y+m$,
combining $\left\{\begin{array}{l}x=\frac{x_{1}-m}{y_{1}} y+m, \\ y^{2}=4 x\end{array}\right.$
$\Rightarrow y^{2}-\frac{4 x_{1}-4 m}{y_{1}} y-4 m=0 \Rightarrow y_{3}=-\frac{4 m}{y_{1}}$, similarly, we get $y_{4}=-\frac{4 m}{y_{2}}$.
Therefore, $y_{3}+y_{4}=-\frac{4 m\left(y_{1}+y_{2}\right)}{y_{1} y_{2}}=4 m n \Rightarrow k_{P Q}=\frac{4}{y_{3}+y_{4}}=\frac{1}{m n}=\frac{1}{3 n} \Rightarrow m=3$.