Question

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

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

Answer 1

A

设A(x 1 ,y 1 )、B(x 2 ,y 2 )，将y=x+b代入x 2 =2y得x 2 -2x-2b=0.则x 1 x 2 =?-2b?,x 1 +x 2 =2. ∴y 1 y 2 =(x 1 +b)(x 2 +b)=x 1 x 2 +b(x 1 +x 2 )+b 2 =b 2 . 由OA⊥OB，得x 1 x 2 +y 1 y 2 =0,即-2b+b 2 =0.∴b=0(舍去)或2. 又Δ=2 2 +8b>0,得b>- .∴b=2符合.