13.4.1 * $P$ is a moving point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ in the first quadrant. Tangents $P A$ and $P B$ are drawn from point $P$ to the circle $x^{2}+y^{2}=$ 9, with points of tangency at $A$ and $B$. The line $A B$ intersects the $x$-axis and $y$-axis at points $M$ and $N$, respectively. Find the minimum value of $S_{\triangle M O N}$.
Let $P(4 \cos \theta, 3 \sin \theta), \theta \in\left(0, \frac{\pi}{2}\right)$, then the equation of the line $AB$ is $4 x \cos \theta + 3 y \sin \theta = 9$. Therefore, $|OM| = \frac{9}{4 \cos \theta}, |ON| = \frac{3}{\sin \theta}, S_{\triangle M O N} = \frac{1}{2} |OM| \cdot |ON| = \frac{27}{4 \sin 2 \theta} \geqslant \frac{27}{4}$. When $\theta = \frac{\pi}{4}$, i.e., point $P$ is $\left(2 \sqrt{2}, \frac{3 \sqrt{2}}{2}\right)$, the equality holds. Hence, the minimum value of $S_{\triangle M O N}$ is $\frac{27}{4}$.
* Given point $A(-2,2)$, and point $B$ is a moving point on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$, $F$ is the left focus. When $|AB| + \frac{5}{3}|BF|$ takes the minimum value, find the coordinates of point $B$.

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 $a = 5$, $b = 4$, $c = \sqrt{a^{2} - b^{2}} = \sqrt{5^{2} - 4^{2}} = 3$, and $e = \frac{c}{a} = \frac{3}{5}$. The left directrix is $x = -\frac{25}{3}$.

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 the same directrix, with the foot of the perpendicular being $M$. By the definition of the ellipse, $|BN| = \frac{|BF|}{e} = \frac{5}{3}|BF|$. Therefore, $|AB| + \frac{5}{3}|BF| = |AB| + |BN| \geqslant |AN| \geqslant |AM|$ (a constant), with equality holding if and only if $B$ is the intersection of $AM$ with the ellipse, at which point the coordinates of $B$ are $\left(-\frac{5}{2} \sqrt{3}, 2\right)$. Thus, when $|AB| + \frac{5}{3}|BF|$ takes the minimum value, the coordinates of $B$ are $\left(-\frac{5}{2} \sqrt{3}, 2\right)$.