设PF1:y=k1(x+1),PF2=k2(x-1)
分别与椭圆联立方程
→(1+2k1²)x²+4k1²x+2k1²-2=0,(所以设A(x1,y1),B(x2,y2))
→x1+x2=-4k1²/(1+2k1²)①,x1x2=(2k1²-2)/(1+2k1²)②
同理,设C(x3,y3),D(x4,y4)
→(1+2k2²)x²-4k2²x+2k2²-2=0
→x3+x4=4k2²/(1+2k2²)③,x3x4=(2k2²-2)/(1+2k2²)④
根据kOA+kOB+kOC+kOD=0
→y1/x1+y2/x2+y3/x3+y4/x4=0
根据y=k1(x+1)→y1=k1(x1+1),y2~
根据y=k2(x-1)→y3=k2(x3-1),y4~
代入进行化简
→k1(2x1x2+x1+x2)/(x1x2)+k2[2x3x4-(x3+x4)]/x3x4=0
由①②③④→-2k1/(k1²-1)-2k2/(k2²-1)=0⑤
设P(n,2-n)→k1=(2-n-0)/(n+1)=(2-n)/(n+1),k2=(2-n)/(n-1)
代⑤→k1²k2+k1k2²=k1+k2
→k1k2(k2+k1)=k1+k2
→k1k2=1或者k1k2=0或者(k1+k2)=0
均成立
→n=5/4,n=2,n=0均可
→P(5/4,3/4),P(2,0),P(0,2)