题目内容

(本小题满分12分)

如图,圆轴相切于点,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:

 

【答案】

(Ⅰ).(Ⅱ)见解析。

【解析】(I)由于圆轴相切于点, 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.

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

设直线交椭圆两点,则

.······································································· 9分

∵ 

∴ 

.······························································· 10分

11分

∴ .····························································· 12分

综上所述,.   13分

 

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