抛物线切线交点的轨迹问题y=x^2 过(0,2)作直线交抛物线于M N点,过M N分别作抛物线的切线交于Q点 求Q的轨迹

1个回答

  • y=kx+2

    与y=x^2 联立

    x^2-kx-2=0

    △=k^2+8>0 k2√2

    x1=0.5k-√2 x2=0.5k+√2

    y1=0.5k^2-√2k+2 y2=0.5k^2+√2k+2

    M(0.5k-√2,0.5k^2-√2k+2 )

    N(0.5k+√2,0.5k^2+√2k+2)

    y=x^2求导 y'=2x

    过M N分别作抛物线的切线

    M处切线斜率=2(0.5k-√2)=k-2√2

    N处切线斜率=2(0.5k+√2)=k+2√2

    过M点L1:y-(0.5k^2-√2k+2)=(k-2√2) (x-0.5k+√2)

    过N点L2:y-(0.5k^2+√2k+2)=(k+2√2) (x-0.5k-√2)

    二者交点 (k/2,0.5k^2+2√2k-6)

    轨迹为 y=0.5(2x)^2+2√2*2x-6=2x^2+4√2x-6

    ∵k2√2

    x4√2