动直线y=x+b和抛物线y^2=-4x交于两个相异的点A,B求是否存在定点M,使k(MA)+k(MB)为常数?

1个回答

  • y=x+b

    y^2=-4x

    (x+b)^2=-4x

    x^2+(2b+4)x+b^2=0

    (2b+4)^2-4b^2>0

    16b+16>0

    b>-1

    x1+x2=-2(b+2),x1x2=b^2

    M(x0,y0)A(x1,x1+B),b(x2,x2+b)

    k1=(x1+b-y0)/(x1-x0),k2=(x2+b-y0)/(x2-x0)

    k1+k2=((x1+b-y0)(x2-x0)+(x2+b-y0)(x1-x0))/(x1-x0)(x2-x0)

    =(x1x2-x0x1+bx2-bx0-y0x2+x0y0+x1x2-x0x2+bx1-bx0-y0x1+x0y0)/(x1-x0)(x2-x0)

    =(2x1x2+2x0y0-x0(x1+x2)+b(x1+x2)-2bx0-y0(x1+x2))/(x1-x0)(x2-x0)

    =(2b^2+2x0y0+(b-x0-y0)*(-2(b+2))-2bx0)/(x1x2-x0(x1+x2)+x0^2)

    =(2b^2+2x0y0+(b-(x0+y0))*(-2(b+2)-2bx0)/(b^2-x0(-2(b+2)+x0^2)

    =k

    解出x0,y0

    x0=4+-2*6^1/2,y0=2