中心在原点的椭圆的右焦点为(3,0)

右焦点F(3,0),c=3
右准线L的方程为：x=12,a^2/c=12,a^2=12c=36,a=6,b^2=36-9=27
e=c/a=3/6=0.5
椭圆O:x^2/36+y^2/27=1
∠P1FP2=∠P2FP3=∠P3FP1=∠120°
用椭圆第二定义:
12-3=9
过F作垂直X轴的直线L,设P1F在X轴的上方,在L的左边;P2在X轴的上方,在L的右边,椭圆的原点为O,右准线与X轴交于点K,并且设M=∠P1FO,N=∠P2FK,H=∠P3FO,则
M+H=120°,M+N+120°=180°
H=120°-M,N=60°-M
cosM+cosH-cosN
=cosM+cos(120°-M)-cos(60°-M)
=cosM-sin(30°-M)-cos(60°-M)
=0
P1F/(9+P1F*cosM)=e=0.5
1/P1F=(1-0.5cosM)/4.5
同理
1/P2F=(1+0.5cosN)/4.5
1/P3F=(1-0.5cosH)/4.5
1/FP1+1/FP2+1/FP3
=(1-0.5cosM)/4.5+(1+0.5cosN)/4.5+(1-0.5cosH)/4.5
=[3-0.5*(cosM+cosH-cosN)]/4.5
=3/4.5
=2/3
∴1/FP1+1/FP2+1/FP3为定值

平面几何不是很显然。用极坐标做，以F为原点。这样的话就几乎没有计算量了：求和的时候直接看出那三个cos的和是零。于是定值就是2/3
给分吧~~