y=k(x+1)代入y²=4x[k(x+1)]²=4x,整理,得k²(x²+2x+1)-4x=0k²x²+2(k²-2)x+k²=0判别式△=[2(k²-2)]²-4k⁴=16(1-k²)令△1或k0 -1