如图：
B1F1JF2正方形
设正方形的边长为1
则b=1 OF1=1=c
a^2=b^2+c^2=2
所以椭圆为x^2/2+y^2=1
令A(根号2cost1,sint1) B(根号2 cost2,sint2) M(根号2 cost3,sint3)
OM=(根号2 cost3,sint3)
OA=(根号2cost1,sint1) OB=(根号2 cost2,sint2)
所以(根号2 cost3,sint3) =cosθ(根号2cost1,sint1) +sinθ(根号2 cost2,sint2)
得cost3=cost1 cosθ +cost2 sinθ sint3=sint1 cosθ+sint2sinθ
cos^2 t3+sin^2 t3=(cost1cosθ+cost2sinθ)^2+(sint1cosθ+sint2sinθ)^2 =1
因为A,B,M异于椭圆顶点 所以 cosθ sinθ 是不可能=0的
cos^2 t1 cos^2 θ +cos^2 t2 sin^2 θ +sin^2 t1cos^2 θ +sin^2 t2 sin^2 θ
+2cost1cosθcost2sinθ +2sint1cosθ sint2 sinθ =1
cos^2 θ +sin^2 θ +2cost1cosθcost2sinθ +2sint1cosθ sint2 sinθ =1
cost1cosθcost2sinθ +sint1cosθ sint2 sinθ =0
cost1 cost2+ sint1 sint2 =0
cost1cost2 =-sint1sint2
sint1/cost1 *sint2/cost2=-1
由于OA,OB斜率为k1=sint1/根号2cost1 k2=sint2/根号2cost2
所以k1*k2=-1 /2