14. Let the curve $C_{1}: \frac{x^{2}}{a^{2}}+y^{2}=1\left(a\right.$ be a positive constant) intersect with $C_{2}: y^{2}=2(x+m)$ at only one common point $P$ above the $x$-axis.
(1) Find the range of real number $m$ (expressed in terms of $\mu$);
(2) $O$ is the origin, and if $C_{1}$ 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 $a$).
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
$$
x^{2}+2 a^{2} x+2 a^{2} m-a^{2}=0 .
$$

Let $f(x)=x^{2}+2 u^{2} x+2 a^{2} m-a^{2}$, problem (1) is transformed into the equation (1) having a unique solution or equal roots in $x \in(-a, a)$.
We only need to discuss the following 3 cases:
a. $\Delta=0$ gives $m=\frac{a^{2}+1}{2}$. At this time, $x_{P}=-a^{2}$, if and only if $-a < -a^{2} < a$, which is equivalent to $0 < a < 1$.
b. $\Delta>0$ and $f(-a)=0$, we get $m=-a$. At this time, $x_{P}=-a$, if and only if $-a < -a < a$, which is equivalent to $0 < a < 1$.
c. $\Delta>0$ and $f(a)=0$, we get $m=-a-1$. At this time, $x_{P}=a$, if and only if $-a < a < a$, which is equivalent to $0 < a < \frac{1}{2} a \sqrt{1-a^{2}}$.

At this time, $S_{\operatorname{mex}}=a \sqrt{a-a^{2}}$.