Question

已知双曲线x2/a2-y2/b2=1(a>0,b>0)和x2/a2-y2/b2=-1的离心率分别为e1和e2

Answer 1

x^2/a^2-y^2/b^2=1和y^2/b^2-x^2/a^2=1互为共扼
双曲线
,
离心率
e1=c/a,e2=c/b,
∴
1/(e1)^2+1/(e2)^2=(a/c)^2+(b/c)^2=(a^2+b^2)/(c^2)=c^2/c^2=1.
(2)
由1=1/(e1)^2+1/(e2)^2=[(e1)^2+(e2)^2]/(e1·e2)^2≥2(e1·e2)/(e1·e2憨花封拘莩饺凤邪脯矛)^2=2/(e1·e2),
∴
e1·e2≥2.
∴
e1+e2≥2√(e1·e2)≥2√2,
当且仅当e1=e2,即a=b时,取

Answer 2

设双曲线左焦点f(-c,0)
,m(-c,
y1),
n(-c,
-y1)
,(y1
>0)
∵点m(-c,
y1)在双曲线上
∴
(-c)²/a²
-
(y1)²/b²
=
1
∴
y1
=
b²
/a
∴fm
=
b²
/a
,
又∵fm
=
c
+
a
∴
b²
/a
=c
+
a
∴
(c²
-
a²
)
/a
=c
+
a
两边同除以a，得
c/a
=
2
即
离心率
e
=
2