Question

已知直线y=x+b与抛物线x^2=2y交于A,B两点，且OA垂直于OB（O为坐标原点），求b的取值范围请写清楚过程谢谢

Answer 1

y = x + b = x²/2
x² - 2x -2b = 0
x1 = 1+ √(1+2b), y1 = 1+b + √(1+2b), OA的斜率为k1 = [1+b + √(1+2b)]/[1+ √(1+2b)]
x2 = 1- √(1+2b), y2 = 1+b -√(1+2b), OB的斜率为k2 = [1+b - √(1+2b)]/[1- √(1+2b)]
OA垂直于OB, k1*k2 = -1
[(1+b)² - 1-2b]/[1 - 1 -2b] = b²/(-2b) = -1
b² = 2b
b = 0 (A,B重合，舍去）
b = 2