设点为(x,y)
则距离d=|4x+3y-8|/5
=|4x-3x²-8|/5
=|3x²-4x+8|/5
因为3x²-4x+8=3(x-2/3)²+20/3
所以 当x=2/3时,3x²-4x+8有最小值20/3
所以 抛物线y=-x^2上的点到直线4x+3y-8=0距离的最小值是4/33(x-2/3)²+20/3是怎么来的啊？配方 3x²-4x+8 =3(x²-4x/3)+8 =3[x²-(4/3)x+(2/3)²]+8-3*(2/3)² =3(x-2/3)²+20/3