由y=ax+1; 3x^2-y^2=1消去y得到
(3-a^2)x^2-2ax-2=0
--->x1+x2=-2a/(a^2-3); x1x2=2/(a^2-3)
y1=ax1+1; y2=ax2+1
--->y1y2=(ax1+1)(ax2+1)
=a^2*x1x2+a(x1+x2)+1
=2a^2/(a^2-3)-2a^2/(a^2-3)+1
=1
以AB为直径的圆经过原点O,意味着线段OA垂直于OB,于是k(OA)*k(OB)=-1
--->y1/x1*y2/x2=-1
--->x1x2+y1y2=0
--->2/(a^2-3)+1=0
--->2+(a^2-3)=0
--->a^2=1
--->a=+'-1.