Example 3 (1999 National High School Competition Question) Given $A(-2,2)$, it is known that $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $F$ is the left focus. When $|A B|+\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of point $B$.
Solve $a=5, b=4, c=\sqrt{a^{2}-b^{2}}=3, e=\frac{3}{5}$, the equation of the left directrix is $x=-\frac{25}{3}$. As shown in Figure 6-1, draw perpendiculars from $A$ and $B$ to the left directrix, with the feet of the perpendiculars being $M$ and $N$, respectively. By the definition of an ellipse, we have $|B N|=\frac{|B F|}{e}=\frac{5}{3}|B F|$, thus $|A B|+\frac{5}{3}|B F|=$ $|A B|+|B N| \geqslant|A N| \geqslant|A M|$ (a constant), equality holds if and only if $B$ is the intersection point of $A M$ and the ellipse, at this time $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$, therefore, when $|A B|+\frac{5}{3}|B F|$ takes the minimum value, the coordinates of point $B$ are $\left(-\frac{5}{2} \sqrt{3}, 2\right)$.