13 (14 points) Let the equation of the ellipse $C_{1}$ be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and the equation of the curve $C_{2}$ be $y=\frac{1}{x}$. Suppose $C_{1}$ and $C_{2}$ have only one common point $P$ in the first quadrant.
(1) Try to express the coordinates of point $P$ in terms of $a$;
(2) Let $F_{1}$ and $F_{2}$ be the two foci of $C_{1}$. Find the range of the area function $f(a)$ of $\triangle P F_{1} F_{2}$ as $a$ varies;
(3) Let $g(a)$ be the area of a square with side length equal to the semi-focal distance of $C_{1}$. Find the expression for the function $y=\min \{f(a), g(a)\}$.
13 (1) Substituting $y=\frac{1}{x}$ into the equation of the ellipse $C_{1}$, we get $\frac{x^{2}}{a^{2}}+\frac{1}{b^{2} x^{2}}=1$, simplifying to $b^{2} x^{4}-a^{2} b^{2} x^{2}+a^{2}=0$. Given $\Delta=a^{4} b^{4}-4 a^{2} b^{2}=0$, we get $a b=2$, solving for $x=\frac{\sqrt{2}}{2} a, x=-\frac{\sqrt{2}}{2} a$ (discard), hence the coordinates of point $P$ are $\left(\frac{\sqrt{2}}{2} a, \frac{\sqrt{2}}{a}\right)$.
(2) Given $\left|F_{1} F_{2}\right|=2 \sqrt{a^{2}-b^{2}}$, and the height of $\triangle P F_{1} F_{2}$ is $\frac{\sqrt{2}}{a}$, we have
$$
f(a)=\frac{1}{2} \cdot 2 \sqrt{a^{2}-b^{2}} \cdot \frac{\sqrt{2}}{a}=\sqrt{2\left(1-\frac{4}{a^{4}}\right)} .
$$

Also, $a>b>0, b=\frac{2}{a}$ gives $a>\frac{2}{a}$, i.e., $a>\sqrt{2}$, leading to $0\sqrt{2}$, solving for $a \geqslant \sqrt[4]{6}$. Therefore,
$$
y=\min \{f(a), g(a)\}=\left\{\begin{array}{ll}
a^{2}-\frac{4}{a^{2}}, & \sqrt{2}\sqrt[4]{6} .
\end{array}\right.
$$