过抛物线y^2=2x的顶点做互相垂直的两弦OA,OB,

1个回答

  • 1.设A(x1,y1),B(x2,y2),AB中点M(x,y),则 y=(y1+y2)/2

    x=(x1+x2)/2=2(x1+x2)/4=(y1^2+y2^2)/4

    =[(y1+y2)^2-2y1y2]/4

    =[(y1+y2)^2-2*4]/4

    =[(y1+y2)/2]^2-2=y^2-2

    故AB中点M的轨迹方程x+2=y^2

    2.OA方程y=kx 则OB方程为y=-x/k

    则,k^2*x1^2=2x1 x1=2/k^2

    (x2)^2/k^2=2x2 x2=2k^2

    则x1x2=4=定值

    y1y2=根号(2x1*x2)=4=定值

    直线AB方程为(x-x1)/(y-y1)=(x-x2)/(y-y2)

    设与x轴交点为(t,0)

    则:(t-x1)/y1=(t-x2)/y2

    y2t-y2x1=y1t-x2y1

    t=-2(y2x1-y1x2)/2(y1-y2)

    =-(y2*y1^2-y1*y2^2)/2(y1-y2)

    =y1y2((y1-y2))/2(y1-y2)=y1y2/2=2

    即为定点