1)上一点M满足.角F1PF2=α,其中F1F2为椭圆的两焦点求证△F1PF2的面积是b^2tanα

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证明：记PF1=r1,PF2=r2,角[email protected],
F1F2=2c,PF1+PF2=2a,a^2-b^2=c^2,
易知三角形PF1F2面积为S=(1/2)[email protected],
在三角形PF1F2中由余弦定理得
[email protected]=(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/([email protected]),
从而S=(1/2)r1r2sin
@=b^2[[email protected]/([email protected])]
=b^2[2sin(@/2)cos(@/2)]/[2cos^2(@/2)]
=b^2tan(@/2),进而命题得证.