(10 (20 points) Let $M$ be a point on the ellipse $C: \frac{x^{2}}{48}+\frac{y^{2}}{16}=1$, and let $P$, $Q$, and $T$ be the points symmetric to $M$ with respect to the $y$-axis, the origin, and the $x$-axis, respectively. Let $N$ be a point on the ellipse $C$ different from $M$, and suppose that $M N \perp M Q$. The intersection of $Q N$ and $P T$ is $E$. Find the equation of the trajectory of the moving point $E$ as $M$ moves along the ellipse $C$.
)
10. Let \( M\left(x_{1}, y_{1}\right) \) be any point on the ellipse \( C \) \(\left(x_{1} y_{1} \neq 0\right)\), \( N\left(x_{2}, y_{2}\right) \), and the coordinates of the moving point \( E \) be \((x, y)\). Then \( P\left(-x_{1}, y_{1}\right) \), \( Q\left(-x_{1},-y_{1}\right) \), \( T\left(x_{1},-y_{1}\right) \). Therefore,
\[
\begin{array}{l}
\frac{x_{1}^{2}}{48}+\frac{y_{1}^{2}}{16}=1 \\
\frac{x_{2}^{2}}{48}+\frac{y_{2}^{2}}{16}=1
\end{array}
\]
Subtracting (2) from (1), we get
\[
\frac{\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{48}+\frac{\left(y_{1}-y_{2}\right)\left(y_{1}+y_{2}\right)}{16}=0 .
\]

Thus,
\[
\frac{\left(y_{1}-y_{2}\right)\left(y_{1}+y_{2}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}=-\frac{1}{3},
\]

Therefore,
\[
k_{M N} \cdot k_{Q V}=-\frac{1}{3} .
\]

Since \( M N \perp M Q \), we have \( k_{M N} \cdot k_{Q M}=-1 \).
Also, \( k_{Q M}=\frac{y_{1}}{x_{1}} \), so \( k_{M N}=-\frac{x_{1}}{y_{1}} \), and thus \( k_{Q N}=\frac{y_{1}}{3 x_{1}} \). Therefore, the equation of the line \( Q N \) is
\[
y=\frac{y_{1}}{3 x_{1}}\left(x+x_{1}\right)-y_{1} .
\]

Since the equation of the line \( P T \) is \( y=-\frac{y_{1}}{x_{1}} x \), solving the system
\[
\left\{\begin{array}{l}
y=\frac{y_{1}}{3 x_{1}}\left(x+x_{1}\right)-y_{1}, \\
y=-\frac{y_{1}}{x_{1}} x,
\end{array}\right.
\]
we get \(\left\{\begin{array}{l}x=\frac{1}{2} x_{1}, \\ y=-\frac{1}{2} y_{1},\end{array}\right.\) so \( x_{1}=2 x \) and \( y_{1}=-2 y \).
Since the point \(\left(x_{1}, y_{1}\right)\) lies on the ellipse \( C: \frac{x^{2}}{48}+\frac{y^{2}}{16}=1 \), we have \(\frac{x_{1}^{2}}{48}+\frac{y_{1}^{2}}{16}=1\), so \(\frac{4 x^{2}}{48}+\frac{4 y^{2}}{16}=1\), which simplifies to \(\frac{x^{2}}{12}+\frac{y^{2}}{4}=1(x y \neq 0)\). This is the equation of the trajectory of point \( E \).