Solution: The ellipse is: $3 x^{2}+8 y^{2}=48$, let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, from $A(-4,0), B(4,0)$ we get the lines $A P, Q B$ as $\left\{\begin{array}{l}y\left(x_{1}+4\right)=y_{1}(x+4) \\ y\left(x_{2}-4\right)=y_{2}(x-4)\end{array}\right.$, eliminating $\mathrm{y}$ we get $x_{M}\left(x_{1} y_{2}-x_{2} y_{1}+4\left(y_{1}+y_{2}\right)\right)=4\left(x_{1} y_{2}+x_{2} y_{1}+4\left(y_{2}-y_{1}\right)\right)$ (1), by $\mathrm{P}, \mathrm{C}, \mathrm{Q}$ being collinear, i.e., $\overrightarrow{C P}=\left(x_{1}-2, y_{1}\right), \overrightarrow{C Q}=\left(x_{2}-2, y_{2}\right)$ being collinear, we get $x_{1} y_{2}-x_{2} y_{1}=2\left(y_{2}-y_{1}\right)$ (3), combining (2) and (3) we get $x_{1} y_{2}+x_{2} y_{1}=8\left(y_{2}+y_{1}\right)$ (4), substituting (3) and (4) into (1) we get $x_{M}\left(y_{1}+3 y_{2}\right)=8\left(y_{1}+3 y_{2}\right)$, since $y_{1}+3 y_{2}$ is not always 0 during the movement of $\mathrm{PQ}$, hence $x_{M}=8$, similarly $x_{N}=8$, the equation of line $\mathrm{MN}$ is $x=8$.