3. Let the curve $C_{1}: \frac{x^{2}}{a^{2}}+y^{2}=1$ (where $a$ is a positive constant), $C_{2}: y^{2}=2(x+m)$ intersect at only one point $P$ above the $x$-axis, find the range of the real number $m$.

The curves $C_{1}: \frac{x^{2}}{a^{2}}+y^{2}=1$ (where $a$ is a positive constant), and $C_{2}: y^{2}=2(x+m)$ intersect at only one point $P$ above the $x$-axis. Determine the range of the real number $m$.
3. From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1 \\ y^{2}=2(x+m)\end{array}\right.$ we get: $x^{2}+2 a^{2} x+2 a^{2} m-a^{2}=0$ which can be transformed into the equation $f(x)=x^{2}+2 a^{2} x+2 a^{2} m-a^{2}=0$ having a unique or equal root in $(-u, a)$.
1) If $\Delta=0$ then $m=\frac{a^{2}+1}{2}, x_{P}=-a^{2}$. This is valid if and only if $-a<-a^{2}<a \Rightarrow 0<a<1$:
2) $f(a) f(-a)<0$ if and only if $-a<m<a$;
3) $f(-a)=0$ gives: $m=a$, in which case $x_{P}=a-2 a^{2}$, valid if and only if $-a<a-2 a^{2}<a$ i.e., $0<a<1$; $f(a)=0$ gives: $m=-a$, in which case $x_{P}=-a-2 a^{2}$, but since $-a-2 a^{2}<-a$, thus $m \neq-a$. In summary: When $0<a<1$, $m=\frac{a^{2}+1}{2}$ or $-a<m \leqslant a$; When $a \geqslant 1$, $-a<m<a$.