Question

直线y=x+b与抛物线x²=2y交于A、B两点，O为坐标原点，且OA⊥OB，则b的值是(    )
A.2                                B.-2
C.1                                D.-1

Answer 1

A

设A(x₁,y₁)、B(x₂,y₂)，将y=x+b代入x²=2y得x²-2x-2b=0.则x₁x₂=?-2b?,x₁+x₂=2.
∴y₁y₂=(x₁+b)(x₂+b)=x₁x₂+b(x₁+x₂)+b²=b².
由OA⊥OB，得x₁x₂+y₁y₂=0,即-2b+b²=0.∴b=0(舍去)或2.
又Δ=2²+8b>0,得b>-.∴b=2符合.