* Given $P$ is a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the vertex of the minor axis $A(0$, $b)$, find the maximum value of $|P A|$.
Let point $P(x, y)$, then $y \in[-b, b]$, so
$$
\begin{aligned}
|P A|^{2} & =x^{2}+(y-b)^{2}=a^{2}-\frac{a^{2}}{b^{2}} y^{2}+y^{2}-2 b y+b^{2} \\
& =\frac{b^{2}-a^{2}}{b^{2}} y^{2}-2 b y+a^{2}+b^{2}=\frac{b^{2}-a^{2}}{b^{2}}\left(y+\frac{b^{3}}{a^{2}-b^{2}}\right)^{2}+\frac{a^{4}}{a^{2}-b^{2}},
\end{aligned}
$$

When $\frac{b^{3}}{a^{2}-b^{2}} \leqslant b$, i.e., $a \geqslant \sqrt{2} b$, $|P A|_{\max }=\frac{a^{2}}{\sqrt{a^{2}-b^{2}}}$, at this time $y=-\frac{b^{3}}{a^{2}-b^{2}}$; When $\frac{b^{3}}{a^{2}-b^{2}}$ $>b$, i.e., $b<a<\sqrt{2} b$, $|P A|_{\max }=2 b$, at this time $y=-b$. In summary,
$$
|P A|_{\max }=\left\{\begin{array}{ll}
\frac{a^{2}}{\sqrt{a^{2}-b^{2}}}, \quad a \geqslant \sqrt{2} b, \\
2 b, \quad b<a<\sqrt{2} b .
\end{array}\right.
$$