设A(X1,Y1),B(X2,Y2),AB中点P(X,Y)
则有:2X=X1+X2,2Y=Y1+Y2
X1^2/2+Y1^2=1
X2^2/2+Y2^2=1
二式相减得:(x1+x2)(x1-x2)/2+(y1+y2)(y1-y2)=0
即:2x(x1-x2)=-2y(y1-y2)
(y1-y2)/(x1-x2)=-x/y
设A(0,2),则AP的斜率是:k=(y-2)/(x-0)
又k=(y2-y1)/(x2-x1)
所以:-x/y=(y-2)/x
y(y-2)=-x^2
化简得:x^2+y^2-2y=0