设A(a,0)
M是(acosm,bsinm)
所以MA,MO斜率是bsinm/(acosm-a),bsinm/acosm
垂直则b²sin²m/[(acosm-a)acosm]=-1
b²sin²m=a²cosm-a²cos²m
(a²-c²)sin²m=a²cosm-a²cos²m
a²sin²m-c²sin²m=a²cosm-a²cos²m
a²(sin²m+cos²m)-c²sin²m=a²cosm
a²-a²cosm=c²sin²m
c²/a²=(1-cosm)/sin²m
=(1-cosm)/(1-cos²m)
=(1-cosm)/(1-cosm)(1+cosm)
=1/(1+cosm)
0=1/2
所以√2/2