题目内容
(本小题满分12分)
如图,圆
与
轴相切于点
,与
轴正半轴相交于两点
(点
在点
的左侧),且
.
![]()
(Ⅰ)求圆
的方程;
(Ⅱ)过点
任作一条直线与椭圆
相交于
两点,连接
,求证:
.
(Ⅰ)
.(Ⅱ)见解析。
【解析】(I)由于圆
与
轴相切于点
, 所以圆心坐标为
,然后根据![]()
建立关于r的方程求出r值,圆的标准确定.
(2)将y=0代入圆的方程求出M,N的坐标,然后再分两种情况证明.
(i) 当
轴时,由椭圆对称性可知
.
当
与
轴不垂直时,可设直线
的方程为
.证明
,然后直线方程与椭圆方程联立借助韦达定理来解决即可.
(Ⅰ)设圆
的半径为
(
),依题意,圆心坐标为
.································ 1分
∵ ![]()
∴
,解得
.····································································· 3分
∴ 圆
的方程为
.······················································· 5分
(Ⅱ)把
代入方程
,解得
,或
,
即点
,
.····················································································· 6分
(1)当
轴时,由椭圆对称性可知
.······························· 7分
(2)当
与
轴不垂直时,可设直线
的方程为
.
联立方程
,消去
得,
.······················ 8分
设直线
交椭圆
于
两点,则
,
.······································································· 9分
∵
,
∴ ![]()
.······························································· 10分
∵
,
11分
∴
,
.····························································· 12分
综上所述,
. 13分