P为椭圆X^2/a^2+y^2/b^2=1(a>b>0)上一点，F1、F2为左、右焦点，∠F1PF2=2θ，则|PF1||PF2| cos2θ值为
热心网友
设：|PF1|=x，|PF2|=y三角形F1PF2中，余弦定理：|F1F2|^2 = |PF1|^2 +|PF2|^2 -2*|PF1||PF2|*cos2θ= (|PF1| +|PF2|)^2 - 2*|PF1||PF2|*(1 + cos2θ)|F1F2| = 2c, |PF1| +|PF2| = 2a=== 4*c^2 = 4*a^2 -2*|PF1||PF2|*(1 +cos2θ)== |PF1||PF2| = 2*(a^2 -c^2)/(1+cos2θ) = 2*b^2/(1+cos2θ)因此：|PF1||PF2|*cos2θ = 2*b^2*cos2θ/(1+cos2θ)