Question

已知双曲线C:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,实轴长为2

Answer 1

(1)

e=c/a=√3

设c=√3k a=k

2a=2

a=1

k=1

c=√3

b^2=c^2-a^2=3-1=2

∴双曲线C的方程是

x^2-y^2/2=1

(2)

把2x-y-m=0 -->y=2x-m代入双曲线方程得

x^2-(2x-m)^2/2=1

2x^2-(4x^2-4mx+m^2)=4

2x^2-4mx+4+m^2=0

xa+xb=4m/2=2m

(xa+xb)/2=m

(ya+yb)/2=(2xa-m+2xb-m)/2

=xa+xb-m

=2m-m

=m

∵线段AB的中点在圆x^2=(y-1)^2=5上

∴m^2+(m-1)^2=5

m^2+m^2-2m+1=5

m^2-m-2=0

(m-2)(m+1)=0

m=2或m=-1