已知抛物线C:y2(方)=4x的焦点为F,过点K(-1,0)的直线L与C相交于A.B两点,点A关于X轴的对称点为D.证明

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  • 抛物线C:y^2=4x①的焦点为F(1,0),

    设过点K(-1,0)的直线L:x=my-1,

    代入①,整理得

    y^2-4my+4=0,

    设L与C 的交点A(x1,y1),B(x2,y2),则

    y1+y2=4m,y1y2=4,

    点A关于X轴的对称点D为(x1,-y1).

    1.BD的斜率k1=(y2+y1)/(x2-x1)=4m/[m(y2-y1)]=4/(y2-y1),

    BF的斜率k2=y2/(x2-1).

    k1=k24(x2-1)=y2(y2-y1),4x2=y2^2,

    上式成立,∴k1=k2,∴点F在直线BD上.

    2.向量FA*FB=(x1-1,y1)*(x2-1,y2)=(x1-1)(x2-1)+y1y2=(my1-2)(my2-2)+y1y2

    =(m^2+1)y1y2-2m(y1+y2)+4=4(m^2+1)-8m^2+4=8-4m^2=8/9,

    ∴m^2=16/9,m=土4/3.

    取y2-y1=√(16m^2-16)=4√(m^2-1)=(4/3)√7,

    ∴k1=3/√7,BD:y=(3/√7)(x-1).

    易知圆心M在x轴上,设为(a,0),M到x=(4/3)y-1和到BD的距离相等,即

    |a+1|/(5/3)=|(3/√7)(a-1)|/(4/√7),

    ∴4|a+1|=5|a-1|,-1