Four. (20 points) Given the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ with the left vertex $C$ and the right focus $F$. A line $l$ passing through $F$ intersects the ellipse at points $A$ and $B$. The lines $C A$ and $C B$ intersect the line $x=10$ at points $M$ and $N$, respectively. Try to find the expression for the length of segment $M N$ in terms of the inclination angle $\theta$ of line $l$, and the minimum length of segment $M N$.
Four, the problem can be transformed into finding the expression and minimum value of the length of the segment $M^{\prime} N^{\prime}$ intercepted by the lines $C A, C B$ on the directrix.

As shown in Figure 4, draw $A P$ perpendicular to the left directrix $l_{1}$ at $P$, $B Q \perp l_{1}$ at $Q$, $T$ is the intersection of $l_{1}$ and the $x$-axis, $\theta \in (0, \pi)$. Then
$$
\begin{array}{l}
|F A| \\
=\frac{e F T}{1+e \cos \theta} \\
=\frac{16}{5+3 \cos \theta}, \\
|F B|=\frac{16}{5-3 \cos \theta},|T C|=\frac{40}{3} .
\end{array}
$$

Since $\frac{|A P|}{|T C|}=\frac{\left|P M^{\prime}\right|}{\left|T M^{\prime}\right|}=1-\frac{|T P|}{\left|T M^{\prime}\right|}$,
$$
|A P|=\frac{|A F|}{e},|T P|=|A F| \sin \theta,
$$

Therefore, $\left|T M^{\prime}\right|=\frac{16 \sin \theta}{3(1+\cos \theta)}$.
Similarly, $\left|T N^{\prime}\right|=\frac{16 \sin \theta}{3(1-\cos \theta)}$.
Thus, $\left|M^{\prime} N^{\prime}\right|=\left|M^{\prime} T\right|+\left|T N^{\prime}\right|=\frac{32}{3 \sin \theta} \geqslant \frac{32}{3}$.
The minimum value is achieved when $A B \perp x$-axis.
Since $\frac{\left|M^{\prime} N^{\prime}\right|}{|M N|}=\frac{|T C|}{10+5}=\frac{8}{9}$,
thus $|M N|=\frac{12}{\sin \theta}$, and the minimum value of $|M N|$ is 12.