摘要：25．解:(1)依条件有.． 由知． ∴由得． ∴． 将的坐标代入抛物线方程. 得． ∴抛物线的解析式为．····································································· 3分 (2)设..． ∴ 设.则 ∴.(舍去) 此时点与点重合.... 则为等腰梯形．······························································································· 3分 (3)在射线上存在一点.在射线上存在一点． 使得.且成立.证明如下: 当点如图①所示位置时.不妨设.过点作...垂足分别为． 若．由得: . ． 又 ．············································································································· 2分 当点在如图②所示位置时. 过点作.. 垂足分别为． 同理可证． ． 又. . ．············································································································· 1分 当在如图③所示位置时.过点作.垂足为.延长线.垂足为． 同理可证． ．············································································································· 1分 注意:分三种情况讨论.作图正确并给出一种情况证明正确的.同理可证出其他两种情况的给予4分,若只给出一种正确证明.其他两种情况未作出说明.可给2分.若用四点共圆知识证明且证明过程正确的也没有讨论三种情况的．只给2分．

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