Question

过抛物线y^2=4px(p>0)的顶点作互相垂直的两弦OA.OB,求AB中点P的轨迹方程

Answer 1

1. 设A(pm^2,2pm),B(pn^2,2pn), ( m与n都不为0，且不相等)  AB中点P(x,y)

2. 两弦OA.OB互相垂直得 (pmn)^2+4P^2mn=0

   mn=-4  (1)

3. pm^2+pn^2=2x  (2)

   pm+pn=y           (3)

       由3个方程消去m,n得  (y/p)^2=2x/p-8

       即y^2=2p(x-4p)

所以 AB中点P的轨迹方程是 y^2=2p(x-4p)

 

希望对你有点帮助！

Answer 2

设A(x1,y1),B(x2,y2) x1,x2,y1,y2均不等于0
∴y²1=4px1,y²2=4px2
x1=y²1/(4p),x2=y²2/(4p)
∵OA⊥OB
∴OA●OB=0
∴x1x2+y1y2=0
∴(y1y2)²/(16p²)+y1y2=0
∵y1y2≠0
∴y1y2=-16p²
x1x2=16p²
设AB与x轴交于M(m,0)
则AB:x=ty+m代入y²=4px
得y²=4p(ty+m)
即y²-4pty-4pm=0
∴y1y2=-4pm=-16p²
∴m=4p
即M(4p,0)
设AB中点为(x,y)
则 y=(y1+y2)/2=2pt ==>t=y/(2p)
x=ty+4p=2pt²+4p
∴x=2p*y²/(4p²)+4p
x=y²/(2p)+4p
∴AB中点P的轨迹方程为
y²=2p(x-4p)