Example 6 Given $A(-2,2)$, and $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, $F$ is the left focus. When $\mid A B \mid +\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of $B$.
(1999, National High School Mathematics Competition)
Let the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse be denoted as $a$, $b$, and $c$, respectively, and the eccentricity as $e$. Then, $a=5$, $b=4$, $c=\sqrt{a^{2}-b^{2}}=3$, $e=\frac{c}{a}=\frac{3}{5}$, and the left directrix is $x=-\frac{25}{3}$.

As shown in Figure 4, draw a perpendicular from point $B$ to the left directrix, with the foot of the perpendicular being $N$. Draw a perpendicular from $A$ to this directrix, with the foot of the perpendicular being $M$. By the definition of the ellipse, we have $|B N|=\frac{|B F|}{e}=\frac{5}{3}|B F|$. Therefore,
$$
\begin{array}{l}
|A B|+\frac{5}{3}|B F|=|A B|+|B N| \\
\geqslant|A N| \geqslant|A M| \text { (a constant). }
\end{array}
$$

Equality holds if and only if $B$ is the intersection point of $A M$ and the ellipse, at which point $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$.