5. 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, the coordinates of $B$ are $\qquad$.
5. $\left(-\frac{5}{2} \sqrt{3}, 2\right)$. It is easy to see that the eccentricity $e=\frac{3}{5}$, and the left directrix is $x=-\frac{25}{3}$. Draw a perpendicular from $B$ to the left directrix, with the foot of the perpendicular being $N$, and draw a perpendicular from $A$ to the left directrix, with the foot of the perpendicular being $M$. By the second definition of an ellipse, we know that $|B N|=\frac{5}{3}|B F|$. Therefore, $|A B|+\frac{5}{3}|B F|=$ $|A B|+|B N| \geqslant|A N| \geqslant|A M|=$ a constant. Equality holds when $B\left(-\frac{5}{2} \sqrt{3}, 2\right)$.