摘要：已知与是反比例函数图象上的两个点． (1)求的值, (2)若点.则在反比例函数图象上是否存在点.使得以四点为顶点的四边形为梯形?若存在.求出点的坐标,若不存在.请说明理由． 解:(1)由.得.因此．···························· 2分 (2)如图1.作轴.为垂足.则...因此． 由于点与点的横坐标相同.因此轴.从而． 当为底时.由于过点且平行于的直线与双曲线只有一个公共点. 故不符题意．················································································································ 3分 当为底时.过点作的平行线.交双曲线于点. 过点分别作轴.轴的平行线.交于点． 由于.设.则.. 由点.得点． 因此. 解之得(舍去).因此点． 此时.与的长度不等.故四边形是梯形．······························· 5分 如图2.当为底时.过点作的平行线.与双曲线在第一象限内的交点为． 由于.因此.从而．作轴.为垂足. 则.设.则. 由点.得点. 因此． 解之得(舍去).因此点． 此时.与的长度不相等.故四边形是梯形．··································· 7分 如图3.当过点作的平行线.与双曲线在第三象限内的交点为时. 同理可得.点.四边形是梯形．··················································· 9分 综上所述.函数图象上存在点.使得以四点为顶点的四边形为梯形.点的坐标为:或或．···················································································· 10分

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