Question

高二数学圆锥曲线（椭圆）

设F1，F2是椭圆x^2/a^2+y^2/b^2=1(a>b>0)的两个焦点，P是椭圆上任意点，若角F1PF2=2θ，求证：PF1*PF2*(cosθ)^2为定值

Answer 1

cos2θ=（PF1^2+PF2^2-4c^2）/2PF1PF2
cos2θ=2cos^2θ-1
PF1PF2cos^2θ=(PF1^2+PF2^2-4c^2)/4+1/2PF1PF2
=(PF1^2+PF2^2+2PF1PF2)/4-c^2
=(PF1+PF2)^2/4-c^2
=4a^2/4-c^2
=b^2
所以PF1*PF2*(cosθ)^2为定值

Answer 2

cos2θ=（PF1^2+PF2^2-4c^2）/2PF1PF2
cos2θ=2cos^2θ-1
PF1PF2cos^2θ=(PF1^2+PF2^2-4c^2)/4+1/2PF1PF2
=(PF1^2+PF2^2+2PF1PF2)/4-c^2
=(PF1+PF2)^2/4-c^2
=4a^2/4-c^2
=b^2