13. 2. 23 ** Given the ray $O A: y=k x(k>0, x>0)$, the ray $O B: y=-k x(x>0)$, a moving point $P(x, y)$ is inside $\angle A O x$, $P M \perp O A, P N \perp O B$, with the feet of the perpendiculars being $M$ and $N$ respectively, and the area of quadrilateral $O M P N$ is exactly $k$.
(1) When $k$ is a constant, find the function $y=f(x)$ expressing the ordinate $y$ of the moving point $P$ in terms of its abscissa $x$;
(2) Based on the range of $k$, determine the domain of the function $y=f(x)$.
(1) Let the coordinates of point $P$ be $\left(x_{0}, y_{0}\right)$, then $y_{0}>$
0. From
$$
\left\{\begin{array}{l}
l_{O A}: y=k x, \\
l_{F M}: y-y_{0}=-\frac{1}{k}\left(x-x_{0}\right),
\end{array}\right.
$$

By solving (1) and (2) simultaneously, we can find the coordinates of point $M$ to be $\left(\frac{x_{0}+k y_{0}}{1+k^{2}}, \frac{k\left(x_{0}+k y_{0}\right)}{1+k^{2}}\right)$. Therefore, $|O M|=$ $\sqrt{1+k^{2}} \frac{\left|x_{0}+k y_{0}\right|}{1+k^{2}}=\frac{\left|x_{0}+k y_{0}\right|}{\sqrt{1+k^{2}}}$, and $|P M|=\frac{\left|k x_{0}-y_{0}\right|}{\sqrt{1+k^{2}}}$. Thus, $S_{\triangle O M P}=$ $\frac{1}{2}|O M| \cdot|P M|=\frac{1}{2} \frac{\left|x_{0}+k y_{0}\right| \cdot\left|k x_{0}-y_{0}\right|}{1+k^{2}}$. Since $P$ is in $\angle \Lambda O x$, we have $k x_{0}-$ $y_{0}>0, x_{0}+k y_{0}>0$. Therefore, $S_{\triangle O M P}=\frac{1}{2} \frac{\left(x_{0}+k y_{0}\right)\left(k x_{0}-y_{0}\right)}{1+k^{2}}$. By substituting $-k$ for $k$, similar analysis yields $S_{\triangle O N P} \doteq \frac{1}{2} \frac{\left(k x_{0}+y_{0}\right)\left(x_{0}-k y_{0}\right)}{1+k^{2}}$, so
$$
\begin{aligned}
k & =S_{\text {quadrilateral OMPN }}=S_{\triangle O M P}+S_{\triangle O N P} \\
& =\frac{1}{2} \cdot \frac{1}{1+k^{2}}\left[\left(x_{0}+k y_{0}\right)\left(k x_{0}-y_{0}\right)+\left(k x_{0}+y_{0}\right)\left(x_{0}-k y_{0}\right)\right] \\
& =\frac{k}{1+k^{2}}\left(x_{0}^{2}-y_{0}^{2}\right),
\end{aligned}
$$

Thus, $y_{0}^{2}=x_{0}^{2}-1-k^{2}$, and since $y_{0}>0$, we have $y_{0}=\sqrt{x_{0}^{2}-1-k^{2}}$. Therefore, the function $y=f(x)$ $=\sqrt{x^{2}-1-k^{2}}$.
(2) From (1), we know that point $P$ is above the $x$-axis, below the lines $y=k x$ and $y=\frac{1}{k} x$. Therefore,
$$
\left\{\begin{array}{l}
0 < k x < y_{0} < \frac{1}{k} x, \\
y_{0}=\sqrt{x_{0}^{2}-1-k^{2}}, \\
\left(1-k^{2}\right) x^{2} < 1.
\end{array}\right.
$$

When $k > 1$, the domain is $\left(\sqrt{k^{2}+1}, k \sqrt{\frac{k^{2}+1}{k^{2}-1}}\right)$.