Question

点p(X,Y)是椭圆X^2/a^2+y^/b^2=1(a>b>1)上一点M满足.角F1PF2=α,其中F1F2为椭圆的两

Answer 1

证明：记PF1=r1,PF2=r2,角F1PF2=@,

F1F2=2c,PF1+PF2=2a,a^2-b^2=c^2,

易知三角形PF1F2面积为S=(1/2)r1r2sin@,

在三角形PF1F2中由余弦定理得

cos@=(r1^2+r2^2-4c^2)/2r1r2

=[(r1+r2)^2-2r1r2-4c^2]/(2r1r2)

=(4a^2-4c^2-2r1r2)/(2r1r2)

=2b^2/(r1r2)-1,

于是得到

r1r2=2b^2/(1+cos@),

从而

S=(1/2)r1r2sin@

=b^2[sin@/(1+cos@)]

=b^2[2sin(@/2)cos(@/2)]/[2cos^2(@/2)]

=b^2tan(@/2),进而命题得证.