10. (20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the fixed point $A(1,0)$, and the foci are on the $x$-axis, the ellipse intersects the curve $|y|=x$ at points $B$ and $C$. A parabola with focus at $A$ and passing through $B$ and $C$, opening to the left, has its vertex at $M(m, 0)$. When the eccentricity of the ellipse satisfies $\sqrt{\frac{2}{3}}<e<1$, find the range of the real number $m$.
10. The ellipse passes through the fixed point $A(1,0)$. Then
$a=1, c=\sqrt{1-b^{2}}$,
$e=\sqrt{1-b^{2}} \in\left(\sqrt{\frac{2}{3}}, 1\right)$.
Therefore $00, m>1) \text {. }
$

Since $\frac{p}{2}=m-1$, we have
$
y^{2}=(1-m)(x-m) \text {. }
$

Combining with equation (1), we get
$
\begin{array}{l}
x^{2}+4(m-1) x-4 m(m-1)=0 . \\
\text { Let } f(x)=x^{2}+4(m-1) x-4 m(m-1) .
\end{array}
$

Since $f(x)$ has a root in the interval $\left(0, \frac{1}{2}\right)$ and is monotonically increasing, we have
$
\left\{\begin{array}{l}
f(0)=-4 m(m-1)0 .
\end{array}\right.
$

Solving this, we get $1<m<\frac{3+\sqrt{2}}{4}$.