Question

设点p（p,p^3）是曲线C:y＝x^3的一点,过点p引曲线C的切线,将切线以P为中心逆时针方向旋转45°,得到直线l.

Answer 1

y'=3x²

过P点的C的切线斜率为3p²

设tanθ=3p²

旋转后,斜率为tan(θ+45度)=(1+3p²)/(1-3p²)

L的方程为(斜截式)

y-p³=[(1+3p²)/(1-3p²)](x-p)

将y=x³带入L的方程可得

x³-p³=[(1+3p²)/(1-3p²)](x-p)

(x-p)(x²+px+p²)=[(1+3p²)/(1-3p²)](x-p)

若直线L与曲线C相交于相异的三点,则除去P点,还有两个相异的点

所以

x²+px+p²=(1+3p²)/(1-3p²)有两相异的根

则Δ>0

p²-4[p²-(1+3p²)/(1-3p²)]>0

(4+12p²)/(1-3p²)-3p²>0

1-3p²>0时：

4+12p²-3p²+9p^4>0

9p^4+9p²+4>0

9(p^4+p²+1/4)>-7/4

所以

-√3/3