M点的切线方程Y-y=f'(x)(X-x)
令X=0得到交点的纵坐标
Y|M=y-xf'(x)
所以根据题意
x^2=Y|M=y-xf'(x)=y-xy'
所以xy'-y=-x^2
(xy'-y)/x^2=-1
所以(y/x)'=-1
y/x=-x+C
y= -x^2+Cx