分析:A(-a,0),B(a,0),P(x,y),tanα= ,-tanβ= ,由x2-y2=a2得 =1,所以-tanαtanβ=1,tanγ=-tan(α+β)="-" ="-" (tanα+tanβ),故tanα+tanβ+2tanγ=0. 解:A(-a,0),B(a,0),P(x,y), PA的斜率tanα= ,① PB的斜率-tanβ= ,∴tanβ=- ,② 由x2-y2=a2得 =1, ①×②,得-tanαtanβ=1, tanγ=tan[π-(β+α)]=-tan(α+β)=- =- (tanα+tanβ), ∴tanα+tanβ+2tanγ=0. 故选C. |