过抛物线y^2=2x的顶点作互相垂直的弦OA,OB,其中A(x1,y1)B(x2,y2)

题目

过抛物线y^2=2x的顶点作互相垂直的弦OA,OB,其中A(x1,y1)B(x2,y2)
①求证：x1x2,y1y2均为定值
②证明直线AB与x轴交点为定点
3,求AB中点M的轨迹方程

答案

1.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
2.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
3.设AB中点M（x,y)
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