10. From $\left\{\begin{array}{l}x=t y+n, \\ y^{2}=4 x,\end{array}\right.$ eliminating $x$ yields
$$
y^{2}-4 t y-4 n=0 \text {. }
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$, $D\left(x_{4}, y_{4}\right)$, and the slopes of lines $l_{A B}$ and $l_{C D}$ be $k_{1}$ and $k_{2}$, respectively.
In equation (1), let $t=\frac{1}{k_{1}}, n=1$, then
$$
y_{1} y_{2}=-4 \Rightarrow x_{1} x_{2}=1 \text {. }
$$

In equation (1), let $n=4$, then
$$
\begin{array}{l}
y_{1} y_{3}=-16, y_{2} y_{4}=-16 \\
\Rightarrow y_{3}=-\frac{16}{y_{1}}, y_{4}=-\frac{16}{y_{2}} .
\end{array}
$$

Thus, $k_{1}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{y_{1}-y_{2}}{\frac{y_{1}^{2}}{4}-\frac{y_{2}^{2}}{4}}=\frac{4}{y_{1}+y_{2}}$,
$$
k_{2}=\frac{4}{y_{3}+y_{4}}=\frac{-y_{1} y_{2}}{4\left(y_{1}+y_{2}\right)} \text {. }
$$

Therefore, $\frac{k_{1}}{k_{2}}=\frac{16}{-y_{1} y_{2}}=4$.
Thus, $l_{C D}: x=4 t y+16$.
Let $N(16,0)$. Hence
$$
\begin{array}{l}
S_{\triangle C D M}=\frac{1}{2}|M N|\left|y_{3}-y_{4}\right|=96 \sqrt{t^{2}+1}, \\
S_{\triangle A B M}=\frac{1}{2}|M F|\left|y_{1}-y_{2}\right|=6 \sqrt{t^{2}+1} .
\end{array}
$$

Therefore, $\frac{S_{\triangle C D M}}{S_{\triangle A B M}}=16$.
10. (20 points) Through the focus $F$ of the parabola $y^{2}=4 x$, draw a line $l$ intersecting the parabola at points $A$ and $B$. Let $M(4,0)$. Extend $A M$ and $B M$ to intersect the parabola at points $C$ and $D$, respectively. Find the value of $\frac{S_{\triangle C D M}}{S_{\triangle A B M}}$.