Question

已知双曲线x^2/m^2-y^2/n^2=1(m>o,n>0)的离心率为4/3,则双曲线-x^2/m^2+y^2/n^2

Answer 1

双曲线x^2.m^2-y^2/n^2=1,(m>0,n>0)

半焦距c=√(m^2+n^2).a=m.

离心率e=c/a=√(m^2+n^2)/m=4/3.

√(m^2+n^2)=4/3m.

(m^2+n^2=(16/9)m^2.

n^2=(16/9)m^2-m^2=(7/9)m^2.

椭圆：x^2/m^2+y^2/n^2=1.

半焦距c=√(m^2-n^2)=√[ m^2-(7/9)m^2],

c=(√2/3)m,a=m.

其离心率e=c/a=√2/3m/m=√2/3.---=即为所求.