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 only one 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 $C_{1}$ intersects the negative half of the $x$-axis at point $A$. When $0<a<\frac{1}{2}$, find the maximum value of the area of $\triangle O A P$ (expressed in terms of $a$).
14. (1) From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y^{2}=2(x+m)\end{array}\right.$ eliminating $y$, we get
$$
x^{2}+2 a^{2} x+2 a^{2} m-a^{2}=0 \text {. }
$$

Let $f(x)=x^{2}+2 a^{2} x+2 a^{2} m-a^{2}$, problem (1) is transformed into the equation (1) having a unique solution or equal roots in $x \in(-a, a)$.
It is only necessary to discuss the following three cases:
$1^{\circ} \Delta=0$ gives $m=\frac{a^{2}+1}{2}$. At this time $a_{P}=\cdots a^{2}$, if and only if $-a < 0$, thus $y_{P}=\sqrt{1-\frac{x_{P}^{2}}{a^{2}}}$ takes the maximum value, at this time $y_{P}=2 \sqrt{a-a^{2}}$. Hence $S=a \sqrt{a-a^{2}}$.

When $m=\frac{a^{2}+1}{2}$, $x_{P}=-a^{2}, y_{P}=\sqrt{1-a^{2}}$, at this time $S=\frac{1}{2} a \sqrt{1-a^{2}}$.
Now compare $a \sqrt{a-a^{2}}$ with $\frac{1}{2} a \sqrt{1-a^{2}}$.
Let $a \sqrt{a-a^{2}}=\frac{1}{2} a \sqrt{1-a^{2}}$, we get $a=\frac{1}{3}$.
Therefore, when $0 < a < \frac{1}{3}$, $a \sqrt{a-a^{2}} < \frac{1}{2} a \sqrt{1-a^{2}}$; when $a > \frac{1}{3}$, $a \sqrt{a-a^{2}} > \frac{1}{2} a \sqrt{1-a^{2}}$.
At this time, $S_{\text {nux }}=a \sqrt{a-a^{2}}$.