20.(1) 设动点为
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, . 所以,,即存在实数使得结论成立.······························· 12分 |