【急！！】双曲线x^2/a^2-y^2-b^2=1一条渐近线的倾斜角为3/π,离心率为e,则(a^2+e)/b的最小值

RT，答案应为（2√6）/3，求过程

啊哦，题弄错了，应该是x^2/a^2-y^2/b^2=1

一条渐近线的倾斜角为π/3==>b/a=√3==>b=√3a
c^2=a^2+b^2==>c^2=4a^2==>c=2a==>e=2
[(a^2+e)/b]^2
=(a^2+2)^2/(3a^2)
=(a^4+4a^2+4)/(3a^2)
=1/3(a^2+4/a^2+4)
≥1/3[2√(a^2*4/a^2)+4]
=1/3(4+4)=8/3
当a^2=4/a^2时取等号

∴(a^2+e)/b≥（2√6）/3

渐近线斜率k=b/a k²=b²/a²=(c²-a²)/a² =c²/a²-1 =e²-1 √2<=e<2 2<=e²<=4 1<=k²<=3 我们取k>0 则1<=k<=√3 所以夹角范围是[π/4，π/3]