双曲线x^2-y^2=1,A为右焦点,BC在左支上,三角形ABC为正三角形,
设B(x1,y1),则C(x1,-y1) A(√2,0) x10 y1=√(x1^2-1)
BC中点D在x轴上,
|AD|=√2-x1
|BD|=y1 |AD|=√3*|BD|
√3*√(x1^2-1)=(√2-x1)
2x1^2+2√2x1-5=0 x1=-[√2/2]±√3 x1