过抛物线的对称轴上一点的直线与抛物线相交于M、N两点,自M、N向

过抛物线的对称轴上一点的直线与抛物线相交于M、N两点,自M、N向直线作垂线,垂足分别为

(Ⅰ)当时,求证:

(Ⅱ)记 、的面积分别为,是否存在,使得对任意的,都有成立。若存在,求出的值;若不存在,说明理由。

答案


解析:

本小题主要考察抛物线的定义和几何性质等平面解析几何的基础知识,考查综合运用数学知识进行推理运算的能力。(14分)

解:依题意,可设直线MN的方程为,则有

消去x可得

从而有                                             ①

于是                                    ②

又由可得                  ③

(Ⅰ)如图1,当时,点即为抛物线的焦点,为其准线

此时 ①可得

证法1:

证法2:

(Ⅱ)存在,使得对任意的,都有成立,证明如下:

证法1:记直线与x轴的交点为,则。于是有

将①、②、③代入上式化简可得

上式恒成立,即对任意成立

证法2:如图2,连接,则由可得

,所以直线经过原点O,

同理可证直线也经过原点O

(2)当得对称轴x=b位于区间之外

此时

①         若

于是

②         若,则

于是

综上,对任意的b、c都有

而当,时,在区间上的最大值

对任意的b,c恒成立的k的最大值为

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