试在曲线y=x^2-x上求一点P的坐标使P点与定点A(0,1)的距离最近

3个回答

  • 设P(x,y),要使P点与定点A(0,1)的距离最近,那意思就是P点与A点的连线垂直于过P点的切线呀,那就是这两条线的斜率相乘等于-1.

    y'=2x-1

    即切线的斜率k=y'=2x-1

    k(PA)=(y-1)/(x-0)

    k(PA)*k=(y-1)/x*(2x-1)=-1

    (y-1)*(2x-1)=-x

    (x^2-x-1)*(2x-1)=-x

    2x^3-x^2-2x^2+x-2x+1+x=0

    2x^3-3x^2+1=0

    2x^2(x-1)-(x+1)(x-1)=0

    (x-1)(2x^2-x-1)=0

    (x-1)(2x+1)(x-1)=0

    x1=-1/2,x2=1.

    y1=1/4+1/2=3/4

    y2=0

    即P坐标是P(1,0)或(-1/2,3/4)

    经检验,P(-1/2,3/4)是最近的点.