直线y=ax+1与双曲线3x^2-y^2=1相交A、B两点,O为坐标原点

原点对AB张直角
<==>OA垂直OB y-ax = 1 3x^2 - y^2 = 1
==> (y-ax)^2 = 3x^2-y^2 两边同时除以x^2,
==> 2(y/x)^2 - 2a(y/x) + (a^2 -3) = 0 (注意,要补充检查x是否为0) OA垂直OB
==> (a^2-3)/2 = -1 (两根之积为-1,垂直)
==> a = ...