不妨设M(a^2/c,y1),N(a^2/c,y2),
由F2(c,0)且F2M丄F2N可得F2M^2+F2N^2=MN^2,
即 (a^2/c-c)^2+y1^2+(a^2/c-c)^2+y2^2=(y1-y2)^2,
化简得 b^4/c^2=-y1*y2.
而C(a^2/c,(y1+y2)/2),r=|y1-y2|/2,
所以,由
OC^2-r^2
=[(a^2/c)^2+(y1+y2)^2/4]-(y1-y2)^2/4
=a^4/c^2+(y1^2+y2^2+2y1*y2)/4-(y1^2-2y1*y2+y2^2)/4
=a^4/c^2+y1*y2
=a^4/c^2-b^4/c^2
=a^2+b^2
>0,
可得 OC>r,
即原点O在圆C外.
(事实上,圆C恒过F2,而直线x=a^2/c是椭圆的右准线,O在F2的左侧,因此O在圆C外）