10. (15 points) Given the curve $C_{1}: \frac{(x-2)^{2}}{2}+y^{2}=$ 1 , and the line $l: x=-a-\frac{1}{8}$, if there exist four points on the curve $C_{1}$ such that their distances to the point $\left(-a+\frac{1}{8}, 0\right)$ and to the line $l$ are equal, find the range of values for $a$.
10. Solution: The points equidistant from the point $\left(-a+\frac{1}{8}, 0\right)$ and the line $l: x=-a-\frac{1}{8}$ lie on the parabola $C_{2}: y^{2}=\frac{1}{2}(x+a)$. The necessary and sufficient condition for curves $C_{1}, C_{2}$ to have four common points is:
$$
\text { The system of equations }(I)\left\{\begin{array}{l}
\frac{(x-a)^{2}}{2}+y^{2}=1 \\
y^{2}=\frac{1}{2}(x+a)
\end{array}\right.
$$

has four distinct

real solutions.
Substituting (2) into (1) and simplifying, we get $x^{2}-(2 a-1) x+a^{2}+a-2$ $=0 \quad$ (3)
Let the roots of (3) be $x_{1}, x_{2}$,
then $\left\{\begin{array}{l}x_{1}+x_{2}=2 a-1 \\ x_{1} x_{2}=a^{2}+a-2\end{array}\right.$
The necessary and sufficient condition for the system of equations $(I)$ to have 4 distinct real solutions is
$$
\left\{\begin{array}{l}
\Delta=(2 a-1)^{2}-4\left(a^{2}+a-2\right)>0 \\
\frac{1}{2}\left(x_{1}+a\right)+\frac{1}{2}\left(x_{2}+a\right)>0 \\
\frac{1}{2}\left(x_{1}+a\right) \cdot \frac{1}{2}\left(x_{2}+a\right)>0
\end{array}\right.
$$

From (6) we get $a0 \\
\left(a^{2}+a-2\right)^{2}+a(2 a-1)+a^{2}>0
\end{array}\right.
$$

From (9) we get, $a>\frac{1}{4}$.
From (10) we get, $|a|>\frac{\sqrt{2}}{2}$.
In summary, $a \in\left(\frac{\sqrt{2}}{2}, \frac{9}{8}\right)$.