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    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． 【查看更多】

     

    题目列表(包括答案和解析)

    已知抛物线y＝x2－(2m＋4)x＋m2－10与x轴交于A、B两点，C是抛物线的顶点．

    (1)用配方法求顶点C的坐标(用含有m的代数式表示)；

    (2)“若AB的长为2，求抛物线的解析式”的解法如下：

    由(1)知，对称轴与x轴交于点D(________，0)．

    ∵抛物线具有对称性，且AB＝2，

    ∴AD＝DB＝|xA－xD|＝．

    ∵A(xA，0)在抛物线y＝(x－h)2＋k上，

    ∴(xA－h)2＋k＝0．　　　　①

    ∵h＝xC＝xD，

    ∴将|xA－xD|＝代入①，得到关于m的方程0＝()2＋(________)．　　②

    补全解题过程，并简述步骤①的解题依据，步骤②的解题方法．

    (3)将(2)中条件“AB的长为2”改为“△ABC为等边三角形”，用类似的方法求出抛物线的解析式．

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