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设双曲线的起点在x轴上,两条渐近线方程为y=正负2x,则该双曲线的离心率为
人气:371 ℃ 时间:2019-10-18 08:29:08
解答
渐近线方程是:y=b/a*x
所以b/a=2;
e^2=(c/a)^2 =c^2/a^2=(a^2+b^2)/a^2=1+b^2/a^2=c^2/a^2=5;
所以离心率e=c/a=根号5
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