Question

（本小題滿分12分）

如圖，圓與軸相切于點，與軸正半軸相交于兩點（點在點的左側(cè)），且．

         （Ⅰ）求圓的方程；

         （Ⅱ）過點任作一條直線與橢圓相交于兩點，連接，求證：．

 

Answer 1

【答案】

（Ⅰ）．（Ⅱ）見解析。

【解析】(I)由于圓與軸相切于點, 所以圓心坐標為,然后根據(jù)

建立關(guān)于r的方程求出r值，圓的標準確定.

(2)將y=0代入圓的方程求出M,N的坐標，然后再分兩種情況證明.

(i) 當軸時，由橢圓對稱性可知.

當與軸不垂直時，可設直線的方程為．證明,然后直線方程與橢圓方程聯(lián)立借助韋達定理來解決即可.

（Ⅰ）設圓的半徑為（），依題意，圓心坐標為．································ 1分

∵　

∴　，解得．····································································· 3分

∴　圓的方程為．······················································· 5分

（Ⅱ）把代入方程，解得，或，

即點，．····················································································· 6分

（1）當軸時，由橢圓對稱性可知．······························· 7分

（2）當與軸不垂直時，可設直線的方程為．

聯(lián)立方程，消去得，．······················ 8分

設直線交橢圓于兩點，則

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

∵　，

∴　

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

∵，

11分

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

綜上所述，．   13分