Question

A(x1,y1),B(x2,y2)是抛物线y平方=4x上异于原点O的两点,且向量OA*向量OB=0,OM⊥AB于M.

Answer 1

(1)设kOA=k kOB=-1/k

则A(2P/k^2,2P/k) B(2Pk^2,-2Pk)

kAB=k/(1-k^2)

AB:y+2Pk=[k/(1-k^2)](x-2Pk^2)

即y=[k/(1-k^2)](x-2P)

∴AB经过定点(2P,0)

(2)①交轨法

AB:y=[k/(1-k^2)](x-2P)①

OD:y=[-(1-k^2)/k]x②

两式相乘得x^2+y^2-2Px=0

即(x-P)^2+y^2=P^2 (x≠0)

②记M的坐标为(x,y)

由OM⊥PM得[(y-0)/(x-0)][(y-0)/(x-2P)]=-1

同样得x^2+y^2-2Px=0