Example 4 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 point $B$.
(1999, National High School Mathematics Competition)
Analysis: Since the eccentricity of the ellipse $e=\frac{3}{5}$, we have $|A B|+\frac{5}{3}|B F|=|A B|+\frac{1}{e}|B F|$.
And $\frac{|B F|}{e}$ is the distance from the moving point $B$ to the left directrix. Therefore, this problem is transformed into:

Finding a point $B$ on the ellipse such that the sum of its distances to point $A$ and the left directrix is minimized.

Let the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse be $a$, $b$, and $c$ respectively, then $a=5$, $b=4$, $c=3$, $e=\frac{3}{5}$, and the left directrix $l: x=-\frac{25}{3}$.

Draw $B N \perp l$ at point $N$, and $A M \perp l$ at point $M$. By the definition of the ellipse, we have $|B N|=\frac{|B F|}{e}=\frac{5}{3}|B F|$, thus,
$$
\begin{array}{l}
|A B|+\frac{5}{3}|B F|=|A B|+|B N| \\
\geqslant|A N| \geqslant|A M|
\end{array}
$$

is a constant, with equality holding if and only if $B$ is the intersection of $A M$ and the ellipse, i.e., $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$.

Therefore, when $|A B|+\frac{5}{3}|B F|$ is minimized, the coordinates of point $B$ are $\left(-\frac{5}{2} \sqrt{3}, 2\right)$.