已知双曲线X2/A2-Y2/2=1(A＞根号2）的两条渐近线的夹角为π/3,则双曲线的离心率

Question

已知双曲线X2/A2-Y2/2=1(A＞根号2）的两条渐近线的夹角为π/3,则双曲线的离心率

Answer 1

双曲线X^2/A2-Y^2/2=1(A＞根号2）
其渐近线为X^2/A2-Y^2/2=0,即Y=±√2/A X
A＞√2
Y=-√2/A X的斜率K1=-√2/A,-1＜K1＜0,令其与x轴夹角α
Y=√2/A X的斜率K2=√2/A,0＜K2＜1,令其与x轴夹角β
相当每条渐近线与X轴的夹角都小于π/4
两条渐近线的夹角为π/3
则β-α=π/3
tan(β-α) = tan(π/3)
(tanβ-tanα)/(1+tanβtanα) = tan(π/3)
(K2-K1)/(1+K2K1) = √3
(√2/A+√2/A)/(1-√2/A√2/A) = √3
2√2A/(A^2-2)=√3
√3A^2-2√2A-2√3=0
(√3A+√2)(A-√6)=0
A=√6
离心率e = √(A^2+B&2)/A = √(6+2)/√6 = 2√3/3