设直线 y = kx + b
∵ 倾斜角是30°
∴ 斜率 k = tan30° = 1/√3 = √3/3
∴直线 y = √3x/3 + b 经过点P(1,-2)
∴ -2 = √3/3+b => b = -2-√3/3
y = √3x/3 -2 - √3/3
将直线方程代入抛物线方程x²=y+m 得到
x² = √3x/3 - 2 - √3/3+m
即 x²-√3x/3 + 2 + √3/3 - m = 0
∵方程有两个解
∴ (√3/3)²-4(2-m+√3/3) >0
1/3 - 8 + 4m - 4√3/3 > 0
∴m > 23/12 + √3/3