Question 162, Given $A(-2,2)$, it is known that $B$ is a moving point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, $F$ is the left focus, when $|A B|+$ $\frac{5}{3}|\mathrm{BF}|$ takes the minimum value, find the coordinates of $\mathrm{B}$.
Question 162, Answer: Let the semi-major axis, semi-minor axis, semi-focal distance of the ellipse be $a$, $b$, $c$, and the eccentricity be $e$, then $a=5, b=4, c=3, e=\frac{3}{5}$, the left directrix is $l: x=-\frac{25}{3}$. Draw a perpendicular from $B$ to the left directrix $l$, with the foot of the perpendicular being $N$. According to the second definition of the ellipse, $\frac{5}{3}|B F|=|B N|$. Therefore, $|A B|+\frac{5}{3}|B F|=|A B|+|B N| \geq d(A, l)=-$ $2-\left(-\frac{25}{3}\right)=\frac{19}{3}$, equality holds if and only if $A B \perp l$, and $B$ is between $A$ and $l$, at this time, the coordinates of point $B$ are $\left(-\frac{5 \sqrt{3}}{2}, 2\right)$