证明：方程x方+6xy+9y方+3x+9y-4=0的曲线是两条平行直线.

x^2 + 6xy + 9y^2 + 3x + 9y - 4 = 0
x^2 + 6xy + 9y^2 + 3x + 9y - 4
= (x + 3y)^2 + 3*(x + 3y) - 4
= (x + 3y - 1)(x + 3y + 4)
= 0
x + 3y - 1 = 0
x + 3y + 4 = 0
两条直线,斜率相同而且不重合,自然是平行的了