设 $M(x_{1},y_{1}),N(x_{2},y_{2})$，$\overrightarrow{MP}=\lambda\overrightarrow{PA}$，$\overrightarrow{NP}=\mu\overrightarrow{PB}$，则\[\left(\dfrac{x_{1}-2\lambda}{1+\lambda},\dfrac{y_{1}}{1+\lambda}\right)=\left(\dfrac{x_{2}+2\mu}{1+\mu},\dfrac{y_{2}}{1+\mu}\right)=(4,t)\]从而\[ x_{1}= 4+6\lambda,y_{1}=t(1+\lambda),x_{2}= 4+2\mu,y_{2}=t(1+\mu) \cdots\cdots (1) \]$\dfrac{4}{4}\cdot\dfrac{x_{1}+2\lambda}{1-\lambda}+\dfrac{t}{3}\cdot\dfrac{y_{2}}{1-\lambda}=1$，即\[x_{1}+\dfrac{t}{3}y_{1}=1-3\lambda\cdots\cdots(2),\]$\dfrac{4}{4}\cdot\dfrac{x_{2}-2\mu}{1-\mu}+\dfrac{t}{3}\cdot\dfrac{y_{2}}{1-\mu}=1$，即\[x_{2}+\dfrac{t}{3}y_{2}=1+\mu\cdots\cdots(3),\]将 $(1)$ 代入 $(2)(3)$ 中，有\[\dfrac{t}{3}y_{1}=-3-9\lambda,\dfrac{t}{3}y_{2}=-3-\mu,\]于是\[\dfrac{y_{1}}{y_{2}}=\dfrac{1+\lambda}{1+\mu}=\dfrac{-3-9\lambda}{-3-\mu},\]得\[4\lambda\mu=-3\lambda-\mu\cdots\cdots (4)\]直线 $MN$ 与 $x$ 轴的交点为 $\left(\dfrac{y_{1}x_{2}-y_{2}x_{1}}{y_{1}-y_{2}},0\right)$，\[\dfrac{y_{1}x_{2}-y_{2}x_{1}}{y_{1}-y_{2}}=\dfrac{t(1+\lambda)(4+2\mu)-t(1+\mu)(4+6\lambda)}{t(\lambda-\mu)}=\dfrac{-2\lambda-2\mu-4\lambda\mu}{\lambda-\mu}=1,\]因此 $MN$ 过定点 $(1,0)$．