交点A(x1,y1),B(x2,y2), y1=ax1+1, y2=ax2+1 y1y2 =a^2x1x2+a(x1+x2)+1 ...(1) y=ax+1代入3x^2-y^2=1: (a^2-3)x^2+2ax+2=0 x1+x2=-2a/(a^2-3), x1x2=2/(a^2-3) ...(2) 以线段AB为直径的圆过坐标原点: OA垂直OB ==> (y1/x1)(y2/x2) =-1, x1x2+y1y2=0 ...(3) (1)(2)(3) ==> a =1,-1
把直线L:y=ax+1带入双曲线C:3x^2-y^2=1中得(3-a^2)x^2-2ax-2=0; A(x1,y1),B(x2,y2),原点O(0,0).OA垂直于OB. 所以x1x2+y1y2=0; x1x2=2/(a^2-3);y1y2=(a^2)x1x2+a(x1+x2)+1=[5(a^2)-3]/(a^2-3); 所以[5(a^2)-1]/(a^2-3); 即a=1/(√5)
把y=ax+1代入3x²-y²=1，得(3-a²)x²-2ax-3=0, 设A(x1,y1),B(x2,y2) ∴ x1x2=-3/(3-a²),x1+x2=2a/(3-a²), y1y2=(ax1+1)(ax2+1)=a²x1x2+a(x1+x2)+1 ∵ AO⊥BO , ∴ x1x2+y1y2=0,即(1+a²)x1x2+a(x1+x2)+1=0 ∴ -3(1+a²)/(3-a²)+2a/(3-a²)+1=0,解得a=1/2
答：y=ax+1代入3x^2-y^2=1，得：(3-a^2)x^2 -2ax -2 =0 1. 交于A,B两点 ==> 判别式 =(2a)^2 -4*(3-a^2)...详情>>
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