Question

双曲线 (10 12:37:16)

Answer 1

y=ax+2带入x2/3a2+y2/a2=1

得到

x^2/3a^2+(ax+2)^2/a^2=1

x^2+3(a^2*x^2+4ax+4)=3a^2

(1+3a^2)x^2+12ax+12-3a^2=0

(12a)^2-4(1+3a^2)(12-3a^2)>0才会有两个交点

从而推出a的取值范围 -1