Four, (Full marks 20 points) 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 $|A B|+\frac{5}{3}|B F|$ takes the minimum value, find the coordinates of $B$.

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保留了原文的换行和格式。
$$
\left(-\frac{5}{2} \sqrt{3}, 2\right)
$$

Four, 【Analysis and Solution】Let the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse be $a$, $b$, and $c$ respectively, and the eccentricity be $e$. Then $a=5, b=4, c=\sqrt{a^{2}-b^{2}}=\sqrt{5^{2}-4^{2}}=3, e=\frac{c}{a}=\frac{3}{5}$, and the left directrix is $x=-\frac{25}{3}$.

As shown in the figure, draw a perpendicular from point $B$ to the left directrix $x=-\frac{25}{3}$, 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, $|B N|=\frac{|B F|}{e}=\frac{5}{3}|B F|$. Therefore, $|A B|+\frac{5}{3}|B F|=|A B|+|B N| \geqslant|A N| \geqslant|A M|$ (a constant), with equality holding if and only if $B$ is the intersection point of $A M$ and the ellipse, at which time $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$.