已知椭圆方程:$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,过点$P(1,1)$的两条直线分别与椭圆交于点$A,C$和$B,D$,且满足$\overrightarrow{AP}=\lambda\overrightarrow{PC},\overrightarrow{BP}=\lambda\overrightarrow{PD}$, 当$\lambda$变化时,直线$AB$的斜率是否为定值?若是求此定值.

分析:设点$M$在直线$AC$上,$N$在直线$BD$上且满足$\dfrac{PA}{PC}=\dfrac{MA}{MC}=\lambda$且$\dfrac{PB}{PD}=\dfrac{NB}{ND}=\lambda$

则$M,P$调和分割$A,C$;$N,P$调和分割$B,D$,故由极线知识知道$M,N$在$P$ 所对应的极线上,故$MN:\dfrac{x}{4}+\dfrac{y}{3}=1$又由于$\Delta{ABP}\backsim\Delta{MNP} $故$K_{AB}=K_{MN}=-\dfrac{3}{4}$