-x^2-4y^2+4y=0
4y^2-4y+x^2=0
把这个方程看作是关于y的一元二次方程,x是常数
16-16x^2>=0
1-x^2>=0
x^2<=1
-1<=x<=1
y=(4+-4(1-x^2)^1/2)/8=1/2+-1/2(1-x^2)^1/2
(y-y1)(y-y2)
=(y-(1/2+1/2(1-x^2)^1/2))(y-(1/2-1/2(1-x^2)^1/2))
-x^2-4y^2+4y????
应是-x^2-4y^2+4xy
-x^2-4y^2+4xy
=-(x^2+4y^2-4xy)
=-(x-2y)^2
-x^2-4y^2+4y=-(x-2y)^2
-x^2-(4y^2-4y+1)+1
=-x^2-(2y-1)^2+1
-(x-1)(x+1)-(2y-1)(2y+1)