关于抛物线的难题过抛物线y^2=2px的焦点的一条直线与它交于P,Q两点,过点P和此抛物线顶点的直线与准线交于M,求证直

4个回答

  • 答:证明M点和Q点纵坐标相同即是命题.

    ① 求M点纵坐标:

    设P点坐标是(t²/2p,t),t是参数.

    PM斜率 k=t²/2pt = t/[t²/2p]=2p/t

    PM直线方程:y = 2p/t x

    准线x=-p/2

    y = 2p/t * (-p/2) = -p²/t

    ② 求Q点纵坐标y`:

    通过焦点的直线与抛物线交点y1和y2满足:

    y1y2 = - p²

    所以:

    y`*t = -p²

    y` = -p²/t

    M点和Q点纵坐标相等.

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    /***

    y1y2 = -p² 的证明:

    y² = 2px

    y = k(x - p/2)

    (x - p/2)² = 2px/k²

    x² -(p + 2p/k²)x + p²/4 = 0

    x1x2 = p²/4

    (2px1)(2px2)=p^4

    y1²y2² =p^4

    y1y2 = + - p² (y1、y2符号相反)

    y1y2 = -p²

    /***