14. Given $A(-2,2)$, and it is known that $B$ 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 $B$.
Given the left directrix of the ellipse $l: x=-\frac{25}{3}$, draw perpendiculars from $A$ and $B$ to the left directrix, with the feet of the perpendiculars being $M$ and $N$. Thus, $\frac{|B F|}{|B M|}=e=\frac{3}{5} \Rightarrow|B M|=\frac{5}{3}|B F|$ $\Rightarrow|A B|+\frac{5}{3}|B F|=|A B|+|B M| \geqslant|A N|$. At this point, $B$ coincides with $C$. That is, the y-coordinates of points $A$ and $B$ are equal. It is easy to calculate that $B\left(-\frac{5 \sqrt{3}}{2}, 2\right)$.