椭圆x2a2+y2b2=1(a>b>0)且满足a≤3b,若离心率为e,则e2+1e2的最小值为______.-数学

题目简介

椭圆x2a2+y2b2=1(a>b>0)且满足a≤3b,若离心率为e,则e2+1e2的最小值为______.-数学

题目详情

椭圆
x2
a2
+
y2
b2
=1
(a>b>0)且满足a≤
3
b
,若离心率为e,则e2+
1
e2
的最小值为______.
题型:填空题难度:中档来源:不详

答案

∵a≤
3
b

e2+class="stub"1
e2
=
c2
a2
+
a2
c2

=
a2-b2
a2
+
a2
a2-b2

=2+
b2
a2
b2
a2-b2

∵a≤
3
b
,,∴a2≤3b2,
b2
a2
class="stub"1
3
,且
b2
a2-b2
b2
3b2-b2
=class="stub"1
2

b2
a2
b2
a2-b2
class="stub"1
3
×class="stub"1
2
=class="stub"1
6

∴e2+class="stub"1
e2
class="stub"13
6

故答案为:class="stub"13
6

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