摘要：26．解:(1)由已知.得.. . ． ．············································································································ 设过点的抛物线的解析式为． 将点的坐标代入.得． 将和点的坐标分别代入.得 ··································································································· 解这个方程组.得 故抛物线的解析式为．··························································· (2)成立．························································································· 点在该抛物线上.且它的横坐标为. 点的纵坐标为．······················································································· 设的解析式为. 将点的坐标分别代入.得 解得 的解析式为．········································································ .．··························································································· 过点作于点. 则． . ． 又. ． ． ．··········································································································· ． (3)点在上...则设． ..． ①若.则. 解得．.此时点与点重合． ．··········································································································· ②若.则. 解得 ..此时轴． 与该抛物线在第一象限内的交点的横坐标为1. 点的纵坐标为． ．······································································································· ③若.则. 解得..此时.是等腰直角三角形． 过点作轴于点. 则.设. ． ． 解得． ．··········································· 综上所述.存在三个满足条件的点. 即或或． 26．如图.已知抛物线经过点.抛物线的顶点为.过作射线．过顶点平行于轴的直线交射线于点.在轴正半轴上.连结． (1)求该抛物线的解析式, (2)若动点从点出发.以每秒1个长度单位的速度沿射线运动.设点运动的时间为．问当为何值时.四边形分别为平行四边形?直角梯形?等腰梯形? (3)若.动点和动点分别从点和点同时出发.分别以每秒1个长度单位和2个长度单位的速度沿和运动.当其中一个点停止运动时另一个点也随之停止运动．设它们的运动的时间为.连接.当为何值时.四边形的面积最小?并求出最小值及此时的长． *26．解:(1)抛物线经过点. ·························································································· 1分 二次函数的解析式为:·················································· 3分 (2)为抛物线的顶点过作于.则. ··················································· 4分 当时.四边形是平行四边形 ················································ 5分 当时.四边形是直角梯形 过作于.则 (如果没求出可由求) ····························································································· 6分 当时.四边形是等腰梯形 综上所述:当.5.4时.对应四边形分别是平行四边形.直角梯形.等腰梯形．·· 7分 及已知.是等边三角形 则 过作于.则········································································· 8分 =·································································································· 9分 当时.的面积最小值为··································································· 10分 此时 ······················································ 11分

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