已知过点(1.0)直线L与抛物线y²=4x相交于A·B两点,若|AB|=16/3则直线的斜率

1个回答

  • 设AB直线方程:y=k(x-1)

    与y²=4x联立:

    k²(x-1)²=4x

    k²x²-2k²x+k²=4x

    k²x²-(2k²+4)x+k²=0

    x1+x2=(2k²+4)/k²

    x1x2=1

    (x1-x2)²=x1²-2x1x2+x2²=(x1+x2)²-4x1x2=(x1+x2)²-4

    (y1-y2)²=y1²+y2²-2y1y2=4x1+4x2-2√4x1×√4x2=4(x1+x2)-8

    ∴|AB|=√(x1-x2)²+(y1-y2)²=√[(x1+x2)²+4(x1+x2)-12]=16/3

    整理:9(x1+x2)²+36(x1+x2)-364=0

    x1+x2=14/3

    x1+x2=-26/3(舍去)

    ∴(2k²+4)/k²=14/3

    整理:2k²=3

    k=±√6/2