10. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, the upper and lower endpoints of the ellipse's minor axis are $A, B$ respectively. A circle centered at $A$ with the semi-major axis length of the ellipse as its radius intersects the ellipse at points $C, D$. The y-coordinate of the midpoint of $C D$ is $6-3 \sqrt{3}$.
(1) Find the equation of the ellipse;
(2) A line $l$ passes through the right focus $F$ of the ellipse and is not perpendicular to the $x$-axis. $l$ intersects the ellipse at points $M, N$. Let the point $N$'s reflection over the $x$-axis be $N^{\prime}$. Determine whether the line $M N^{\prime}$ passes through a fixed point. If it does, find this fixed point; otherwise, explain the reason.
(1) Since $\frac{b^{2}}{a^{2}}=1-e^{2}=\frac{3}{4}$, let the equation of the ellipse be $\frac{x^{2}}{4}+\frac{y^{2}}{3}=k^{2}(k>0)$,

Solving the system of equations $\left\{\begin{array}{l}3 x^{2}+4 y^{2}=12 k^{2}, \\ x^{2}+(y-\sqrt{3} k)^{2}=4 k^{2}\end{array} \Rightarrow 3(y-\sqrt{3} k)^{2}=4 y^{2} \Rightarrow \sqrt{3}(\sqrt{3} k-y)=2 y\right.$ $\Rightarrow 3 k=(2+\sqrt{3})(6-3 \sqrt{3}) \Rightarrow k=1$. Therefore, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let the line $l: x=n y+1, M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right), N^{\prime}\left(x_{2},-y_{2}\right)$,

Solving the system of equations $\left\{\begin{array}{l}3 x^{2}+4 y^{2}=12, \\ x=n y+1\end{array} \Rightarrow\left(3 n^{2}+4\right) y^{2}+6 n y-9=0\right.$.
Then the line $M N^{\prime}: y-y_{1}=\frac{y_{1}+y_{2}}{x_{1}-x_{2}}\left(x-x_{1}\right)$, setting $y=0 \Rightarrow x=x_{1}-\frac{\left(x_{1}-x_{2}\right) y_{1}}{y_{1}+y_{2}}$ $=\frac{x_{1} y_{2}+x_{2} y_{1}}{y_{1}+y_{2}}=\frac{\left(n y_{1}+1\right) y_{2}+\left(n y_{2}+1\right) y_{1}}{y_{1}+y_{2}}=\frac{2 n y_{1} y_{2}}{y_{1}+y_{2}}+1=4$.
Therefore, the line $M N^{\prime}$ passes through the fixed point $P(4,0)$.