14. Let the curve $C_{1}: \frac{x^{2}}{a^{2}}+y^{2}=1$ (where $a$ is a positive constant) intersect with $C_{2}: y^{2}=2(x+m)$ at a single common point $P$ above the $x$-axis.
(1) Find the range of real numbers $m$ (expressed in terms of $a$);
(2) $O$ is the origin, and if $C$ intersects the negative half of the $x$-axis at point $A$, when $0<a<\frac{1}{2}$, try to find the maximum value of the area of $\triangle O A P$ (expressed in terms of $a$).
(1) $-a0$, thus $y_{p}=\sqrt{1-\frac{x_{p}^{2}}{a^{2}}}$.
To maximize $y_{p}$, $x_{p}$ should be minimized. It is easy to see that when $m=a$, $\left(x_{p}\right)_{\text {min }}=a-2 a^{2}$. Thus, $\left(y_{p}\right)_{\text {max }}=2 \sqrt{a-a^{2}}$. Therefore, $\left(S_{\triangle O A P}\right)_{\max }=a \sqrt{a-a^{2}}$.
When $m=\frac{a^{2}+1}{2}$, $x_{p}=-a^{2}$. Thus, $y_{p}=\sqrt{1-a^{2}}$, hence $S_{\triangle O A P}=\frac{1}{2} a \sqrt{1-a^{2}}$.
Now, let's compare $a \sqrt{a-a^{2}}$ with $\frac{1}{2} a \sqrt{1-a^{2}}$.
$$
\begin{array}{l}
\because\left(\sqrt{a-a^{2}}\right)^{2}-\left(\frac{1}{2} \sqrt{1-a^{2}}\right)^{2} \\
=\cdots=-\frac{1}{4}(3 a-1)(a-1) .
\end{array}
$$
$\therefore$ When $0\frac{1}{2} a \sqrt{1-a^{2}}$.
$$
\left(S_{\triangle D A P}\right)_{\max }=\left\{\begin{array}{l}
\frac{1}{2} a \sqrt{1-a^{2}}\left(0<a \leqslant \frac{1}{3}\right) \\
a \sqrt{a-a^{2}}\left(\frac{1}{3}<a<\frac{1}{2}\right)
\end{array}\right.
$$