10. (20 points) As shown in Figure 1, given $A(-1,0), B(1,0)$, $Q$ and $G$ are the circumcenter and centroid of $\triangle ABC$ respectively, and $QG \parallel AB$.
(1) Find the equation of the locus $E$ of point $C$.
(2) Let the two intersection points of the locus $E$ with the $y$-axis be $A_{1}$ and $A_{2}$ ($A_{1}$ is below $A_{2}$). Suppose moving points $M$ and $N$ are both on the locus $E$, and satisfy $A_{1} M \perp A_{1} N$. Is the intersection point $P$ of line $A_{1} N$ and $A_{2} M$ always on a fixed line $l$? If so, find the equation of the line $l$; if not, explain the reason.
10. (1) Let $C(x, y)(y \neq 0)$. Then
$$
G\left(\frac{x}{3}, \frac{y}{3}\right), Q\left(0, \frac{y}{3}\right) \text {. }
$$

From $|Q A|=|Q C| \Rightarrow 1+\frac{y^{2}}{9}=x^{2}+\frac{4 y^{2}}{9}$
$$
\Rightarrow x^{2}+\frac{y^{2}}{3}=1 \text {. }
$$

Thus, the equation of the trajectory $E$ is $x^{2}+\frac{y^{2}}{3}=1(y \neq 0)$.
(2) From (1), we know $A_{1}(0,-\sqrt{3}), A_{2}(0, \sqrt{3})$.

It is easy to see that the slope of the line $A_{1} N$ exists and is not 0.
Let $l_{A, N}: y=k x-\sqrt{3}(k \neq 0)$.
Then $l_{A, M}: y=-\frac{1}{k} x-\sqrt{3}$.
Combining equations (1) and (3), we get
$$
\begin{array}{l}
\left(3 k^{2}+1\right) x^{2}+2 \sqrt{3} k x=0 \\
\Rightarrow x_{1}=0, x_{2}=\frac{-2 \sqrt{3} k}{3 k^{2}+1} .
\end{array}
$$

Thus, the point $M\left(\frac{-2 \sqrt{3} k}{3 k^{2}+1}, \frac{\sqrt{3}-3 \sqrt{3} k^{2}}{3 k^{2}+1}\right)$,
$$
k_{1, M}=\frac{\frac{\sqrt{3}-3 \sqrt{3} k^{2}}{3 k^{2}+1}-\sqrt{3}}{\frac{-2 \sqrt{3} k}{3 k^{2}+1}}=3 k \text {. }
$$

Therefore, $l_{A, M}: y=3 k x+\sqrt{3}$.
Combining equations (2) and (4), we get $x_{P}=-\frac{\sqrt{3}}{k}, y_{P}=-2 \sqrt{3}$. Hence, point $P$ is always on the fixed line $l: y=-2 \sqrt{3}$.