Question

過拋物線y²=4x的頂點O作兩條互相垂直的直線分別交拋物線于A、B兩點,則線段AB的中點P(x,y)的軌跡方程是(    )
A.y²="-2x-8                               " B.y²=2x-8
C.y²="2x+8                                " D.y²=-2x+8

Answer 1

C

設(shè)OA:y=kx,代入y²=4x得k²x²=4x,解得A(,).
∵OB⊥OA,則OB:y=-x,
用-代替A點坐標中的k得B(4k²,-4k).
又AB中點P(x,y),
∴
消去參數(shù)k得y²-2x-8=0.