10. (20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b>0), c^{2}+b^{2}=a^{2}$, a line passing through the focus intersects the ellipse at points $M, N$, and the length of segment $\mathrm{MN}$ is in the range $[3,4]$.
(1) Find the equation of the ellipse.
(2) Let $\mathrm{D}, \mathrm{E}$ be two points on the ellipse that are symmetric with respect to the $\mathrm{x}$-axis. For any two points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$ on the $\mathrm{x}$-axis, when $D A$ is not tangent to the ellipse, it always makes the intersection point $C$ of lines $D A, B E$ lie on the ellipse. Find the relationship between $x_{1}, x_{2}$.
(1) From the knowledge of ellipses, it is easy to know that the shortest length is when the line is perpendicular to the $x$-axis, which is $\frac{2 b^{2}}{a}$. When the line is the $x$-axis, it is the longest, which is $2a$. Solving for $a=2, b=\sqrt{3}, c=1$, the equation is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) $x_{1} x_{2}=a^{2}$.

Let $D(m, n), E(m,-n)$, then the equation of line $DA$ is $y=\frac{n}{m-x_{1}}\left(x-x_{1}\right)$, and combining it with the ellipse equation and eliminating $y$ gives $\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right] x^{2}-2 a^{2} n^{2} x_{1} x+a^{2} n^{2} x_{1}^{2}-a^{2} b^{2}\left(m-x_{1}\right)^{2}=0$.
Let the other intersection point of line $DA$ with the ellipse be $C_{1}\left(x_{3}, y_{3}\right)$, then $x_{3}=\frac{2 a^{2} n^{2} x_{1}}{a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}}-m=\frac{2 a^{2} n^{2}}{b^{2}\left(x_{1}-2 m+x_{2}\right)}-m$. Similarly, let the other intersection point of line $BE$ with the ellipse be $C_{2}\left(x_{4}, y_{4}\right)$, $x_{4}=\frac{2 a^{2} n^{2}}{b^{2}\left(x_{1}-2 m+x_{2}\right)}-m$, so $x_{3}=x_{4}$.
Combining the equation of line $DA$ with the ellipse equation and eliminating $x$ gives
$$
\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right] y^{2}+2 b^{2} n x_{1}\left(m-x_{1}\right) y+b^{2} n^{2} x_{1}^{2}-a^{2} b^{2} n^{2}=0
$$
$y_{3}=\frac{2 b^{2} n^{2} x_{1}^{2}-a^{2} b^{2} n^{2}}{n\left[a^{2} n^{2}+b^{2}\left(m-x_{1}\right)^{2}\right]}=\frac{n\left(x_{1}-x_{2}\right)}{x_{1}-2 m+x_{2}}$, similarly, $y_{4}=y_{3}$. Therefore, $C_{1}$ and $C_{2}$ coincide. Thus, the intersection point $C$ of lines $AD$ and $BE$ is still on the ellipse.