若直线ax+2by-2=0(a>0,b>0)始终平分圆x2+y2-4x-2y-8=0的周长,则1a+2b的最小值为______,ab的取值范围是______.-数学

题目简介

若直线ax+2by-2=0(a>0,b>0)始终平分圆x2+y2-4x-2y-8=0的周长,则1a+2b的最小值为______,ab的取值范围是______.-数学

题目详情

若直线ax+2by-2=0(a>0,b>0)始终平分圆x2+y2-4x-2y-8=0的周长,则
1
a
+
2
b
的最小值为______,ab的取值范围是______.
题型:填空题难度:中档来源:不详

答案

x2+y2-4x-2y-8=0可化为:(x-2)2+(y-1)2=13,∴圆的圆心是(2,1)
∵直线平分圆的周长,所以直线恒过圆心(2,1)
把(2,1)代入直线ax+2by-2=0,得a+b=1
class="stub"1
a
+class="stub"2
b
=(class="stub"1
a
+class="stub"2
b
)(a+b)=3+class="stub"b
a
+class="stub"2a
b

∵a>0,b>0,
class="stub"1
a
+class="stub"2
b
=(class="stub"1
a
+class="stub"2
b
)(a+b)=3+class="stub"b
a
+class="stub"2a
b
≥3+2
2

0≤ab≤(class="stub"a+b
2
)
2
=class="stub"1
4

故答案为:3+2
2
(0,class="stub"1
4
]

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