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$overrightarrow{AP}=lambdaoverrightarrow{PC},overrightarrow{BP}=lambdaoverrightarrow{PD}$, 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}acksimDelta{MNP} $故$K_{AB}=K_{MN}=-dfrac{3}{4}$