17 Let the line $L: x+m y+n=0$ intersect the parabola $y^{2}=4 a x$ at points $P, Q$. $F$ is the focus of the parabola, and $P F, Q F$ intersect the parabola at points $R, S$. Find the equation of the line $R S$.
17 Let $P\left(a t_{1}^{2}, 2 a t_{1}\right), Q\left(a t_{2}^{2}, 2 a t_{2}\right), R\left(a t_{1}^{\prime 2}, 2 a t_{1}^{\prime}\right), S\left(a t_{2}^{\prime 2}, 2 a t_{2}^{\prime}\right)$, then the equation of $P R$ is
$$
2 x-\left(t_{1}+t_{1}^{\prime}\right) y+2 a t_{1} t_{1}^{\prime}=0 .
$$

Since $P R$ passes through $F(a, 0)$, we have $2 a t_{1} t_{1}^{\prime}+2 a=0$, i.e.,
$$
t_{1}^{\prime}=-\frac{1}{t} .
$$

Similarly, we get,
$$
t_{2}^{\prime}=-\frac{1}{t_{2}}
$$

Thus, the coordinates of $R, S$ are $\left(\frac{a}{t_{1}^{2}},-\frac{2 a}{t_{1}}\right),\left(\frac{a}{t_{2}^{2}},-\frac{2 a}{t_{2}}\right)$.
Since points $P, Q$ lie on the line $l x+m y+n=0$, we have
$$
\begin{array}{l}
l a t_{1}^{2}+2 a m t_{1}+n=0, \\
l a t_{2}^{2}+2 a m t_{2}+n=0 .
\end{array}
$$

Thus, $t_{1}+t_{2}=-\frac{2 m}{l}, t_{1} t_{2}=\frac{n}{a l}$.

If the equation of the line $R S$ is $A x+B y+C=0$, then
$$
\frac{A a}{t_{1}^{2}}-\frac{2 a B}{t_{1}}+C=0, \frac{A a}{t_{2}^{2}}-\frac{2 a B}{t_{2}}+C=0 .
$$

This indicates that $t_{1}, t_{2}$ are the two real roots of the equation $C^{2}-2 a B t+A a=0$. Therefore,
$$
t_{1}+t_{2}=\frac{2 a B}{C}, t_{1} t_{2}=\frac{A a}{C} .
$$

Thus, $-\frac{2 m}{l}=\frac{2 a B}{C}, \frac{n}{a l}=\frac{A a}{C}$. Therefore,
$$
\frac{B}{C}=-\frac{m}{l a}, \frac{A}{C}=\frac{n}{l a^{2}} \text {. }
$$

Hence, the equation of the required line is $\frac{n x}{l a^{2}}-\frac{m}{l a} y+1=0$, i.e.,
$$
n x-m a y+l a^{2}=0 .
$$