One, (25 points) Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $c$ is the semi-focal distance, and chord $AB$ passes through the right focus $F_{2}$. Find the maximum value of the area of $\triangle F_{1}AB$.
Given the points \(A\left(x_{1}, y_{1}\right)\), \(B\left(x_{2}, y_{2}\right)\), and the angle \(a\) between \(AB\) and the \(x\)-axis, and \(c^{2}=a^{2}-b^{2}\).

Since
\[
\begin{array}{l}
S_{\triangle F_{1} A B} \\
=S_{\triangle A F_{1} F_{2}}+S_{\triangle B F_{1} F_{2}} \\
=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{1}-y_{2}\right| \\
=c\left|y_{1}-y_{2}\right|,
\end{array}
\]

and \(c\) is a constant, the maximum value of \(S_{\triangle F_{1} A B}\) occurs when \(\left|y_{1}-y_{2}\right|\) is maximized. The line \(AB\) passing through \(F_{2}\) has the equation \(y = k(x - c)\) (if \(k\) exists), so \(x = -\frac{y}{k} + c\). Substituting and simplifying the equation, we get
\[
\begin{array}{l}
\left(\frac{1}{a^{2} k^{2}}+\frac{1}{b^{2}}\right) y^{2}+\frac{2 c}{a^{2} k} y+\frac{c^{2}}{a^{2}}-1=0 . \\
\left\lvert\, \begin{aligned}
y_{1}-\left.y_{2}\right|^{2} & =\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2} \\
& =\frac{4 a^{2} b^{4} k^{2}\left(1+k^{2}\right)}{\left[b^{2}+\left(b^{2}+c^{2}\right) k^{2}\right]} \\
& =\left(\frac{2 a b^{2}}{c^{2} \sin \alpha+\frac{b^{2}}{\sin \alpha}}\right)^{2} .
\end{aligned}\right.
\end{array}
\]

(1) When \(b \leqslant c\), and \(\sin \alpha > 0\), we have
\[
\begin{array}{l}
\left|y_{1}-y_{2}\right| \leqslant \frac{2 a b^{2}}{2 \sqrt{c^{2} \sin \alpha \cdot \frac{b^{2}}{\sin \alpha}}}=\frac{a b}{c}, \\
S_{\triangle F_{1} A B} \leqslant c \cdot \frac{a b}{c}=a b .
\end{array}
\]

Thus, the maximum value of \(S_{\triangle F_{1} A B}\) is \(a b\) when and only when \(c^{2} \sin \alpha = \frac{b^{2}}{\sin \alpha}\), i.e., \(\sin \alpha = \frac{b}{c} \leqslant 1\).

(2) When \(b > c\),
\[
\begin{aligned}
c^{2} \sin \alpha + \frac{b^{2}}{\sin \alpha} & = c^{2} \sin \alpha + \frac{c^{2}}{\sin \alpha} + \frac{b^{2} - c^{2}}{\sin \alpha} \\
& \geqslant 2 c^{2} + \frac{b^{2} - c^{2}}{\sin \alpha} \\
& \geqslant 2 c^{2} + b^{2} - c^{2} = a^{2} .
\end{aligned}
\]

The minimum value of \(c^{2} \sin \alpha + \frac{b^{2}}{\sin \alpha}\) is \(a^{2}\) when and only when \(\sin \alpha = 1\). Therefore, \(S_{\triangle F_{1} A B} \leqslant c \cdot \frac{2 b^{2}}{a} = \frac{2 b^{2} c}{a}\), i.e., the maximum value of \(S_{\triangle F_{1} A B}\) is \(\frac{2 b^{2} c}{a}\).

Thus, the maximum area of \(\triangle F_{1} A B\) is \(a b\) when \(b \leqslant c\) and \(\frac{2 b^{2} c}{a}\) when \(b > c\).