20.(1) 设动点为
![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023171335-37190.jpg) 依据题意,有 ,化简得 . 即为动点P所在曲线C的方程。·························································· 3分 (2) 点F在以MN为直径的圆的外部. 理由:由题意可知,当过点F的直线 的斜率为0时,不合题意,故可设直线 : ,如图所示.联立方程组 ,可化为 ,则点 、 的坐标满足 . 又 、 ,可得点 、 . 因 , ,则 = . 于是, 为锐角,即点F在以MN为直径的圆的外部.······················· 10分 (3) 依据 (2) 可算出 , , 则![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023171338-13747.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023171338-50189.png) ,
![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023171339-38149.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023171339-16330.png) . 所以, ,即存在实数 使得结论成立.······························· 12分 |