直线y=ax+1和双曲线3x^2-y^2=1交于Ａ，Ｂ两点，a为何值时，以Ａ，Ｂ为直径的圆经过原点？
热心网友
OA、OB垂直时，以A(x1,y1)、B(x2,y2)为直径的圆经过原点即：(y1/x1)*(y2/x2) = -1 === x1*x2 +y1*y2 = 0将y=ax+1代入3x^2-y^2=1：(a^2-3)x^2 +2ax+2 = 0x1*x2 +y1*y2 = x1*x2 +(a*x1+1)(a*x2+1) = (1 +a^2)x1*x2 +2a*(x1+x2)+1 = 0== (1 +a^2)*[2/(a^2 -3)] + a*[(-2a)/(a^2 -3)]x1 + 1 = 0== a = 1 or -1