已知椭圆 $C:\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1$ 的左、右顶点分别为 $A_{1},A_{2}$，$F_{2}$ 为椭圆 $C$ 的右焦点．若点 $P$ 是椭圆 $C$ 上异于 $A_{1},A_{2}$ 的任意一点，直线 $A_{1}P,A_{2}P$ 与直线 $x=4$ 分别交于 $M,N$ 两点，证明：以 $MN$ 为直径的圆与直线 $PF_{2}$ 相切于点 $F_{2}$．
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【解析】
设线段 $MN$ 的中点为 $Q$，则题意即需要证明 $F_{2}M\perp F_{2}N$ 且 $F_{2}Q\perp PF_{2}$．
设 $P(x_{0},y_{0})$，$M(4,y_{1})$，$N(4,y_{2})$，则$$\dfrac{y_{0}^2}{x_{0}^{2}-4}=-\dfrac{3}{4},$$且\[A_{1}P:y=\dfrac{y_{0}}{x_{0}+2}(x+2), A_{2}P:y=\dfrac{y_{0}}{x_{0}-2}(x-2).\]所以 $y_{1}=\dfrac{6y_{0}}{x_{0}+2}$，$y_{2}=\dfrac{2y_{0}}{x_{0}-2}$．于是\[\overrightarrow{F_{2}M}\cdot \overrightarrow{F_{2}N}=\left(3,\dfrac{6y_{0}}{x_{0}+2}\right)\cdot \left(3,\dfrac{2y_{0}}{x_{0}-2}\right)=9+\dfrac{12y_{0}^{2}}{x_{0}^{2}-4}=0.\]所以 $\overrightarrow{F_{2}M}\perp \overrightarrow{F_{2}N}$．\[\overrightarrow{PF_{2}}\cdot \overrightarrow{F_{2}Q}=(x_{0}-1,y_{0})\cdot\left(3, \dfrac{1}{2}\left(\dfrac{6y_{0}}{x_{0}+2}+\dfrac{2y_{0}}{x_{0}-2}\right)\right)=3(x_{0}-1)+4(x_{0}-1)\cdot \dfrac{y_{0}^{2}}{x_{0}^{2}-4}=0,\]所以 $\overrightarrow{F_{2}Q}\perp \overrightarrow{PF_{2}}$．
综上，原命题得证．
设 $P(x_{0},y_{0})$，$M(4,y_{1})$，$N(4,y_{2})$，则$$\dfrac{y_{0}^2}{x_{0}^{2}-4}=-\dfrac{3}{4},$$且\[A_{1}P:y=\dfrac{y_{0}}{x_{0}+2}(x+2), A_{2}P:y=\dfrac{y_{0}}{x_{0}-2}(x-2).\]所以 $y_{1}=\dfrac{6y_{0}}{x_{0}+2}$，$y_{2}=\dfrac{2y_{0}}{x_{0}-2}$．于是\[\overrightarrow{F_{2}M}\cdot \overrightarrow{F_{2}N}=\left(3,\dfrac{6y_{0}}{x_{0}+2}\right)\cdot \left(3,\dfrac{2y_{0}}{x_{0}-2}\right)=9+\dfrac{12y_{0}^{2}}{x_{0}^{2}-4}=0.\]所以 $\overrightarrow{F_{2}M}\perp \overrightarrow{F_{2}N}$．\[\overrightarrow{PF_{2}}\cdot \overrightarrow{F_{2}Q}=(x_{0}-1,y_{0})\cdot\left(3, \dfrac{1}{2}\left(\dfrac{6y_{0}}{x_{0}+2}+\dfrac{2y_{0}}{x_{0}-2}\right)\right)=3(x_{0}-1)+4(x_{0}-1)\cdot \dfrac{y_{0}^{2}}{x_{0}^{2}-4}=0,\]所以 $\overrightarrow{F_{2}Q}\perp \overrightarrow{PF_{2}}$．综上，原命题得证．
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