Example 16 (1999 National High School Competition Question) Given point $A(-2,2)$, it is known that point $P$ is a moving point on $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $F$ is the left focus. Find the minimum value of $|P A|+\frac{5}{3}|P F|$ and the coordinates of point $P$.
The target difference in this problem is the value $|P A|+\frac{5}{3}|P F|$, with coefficients 1 and $\frac{5}{3}$. Therefore, we should seek the relationship between these two numbers or if they have any special significance. Recalling the known conditions, we know that the eccentricity of this ellipse $e=\frac{3}{5}$, so $\frac{5}{3}|P F|=\frac{|P F|}{e}=|P P_{1}|$. As shown in Figure 1-7, when $A, P, P_{1}$ are collinear, i.e., when $P$ is at the point $\left(\frac{-5 \sqrt{3}}{2}, 2\right)$, $|P A|+\frac{5}{3}|P F|=|P A|+|P P_{1}|=|P_{1} A|=\frac{19}{3}$.