摘要：053]解:(1)设.把代入.得.··························· 2分 ∴抛物线的解析式为:．顶点的坐标为．······························· 5分 (2)设直线解析式为:().把两点坐标代入. 得解得．∴直线解析式为．······················ 7分 .∴················ 9分 ．···························································· 10分 ∴当时.取得最大值.最大值为．······························································ 11分 (3)当取得最大值...∴．∴四边形是矩形． 作点关于直线的对称点.连接． 法一:过作轴于.交轴于点． 设.则． 在中.由勾股定理.． 解得．∵.∴． 由.可得.．∴． ∴坐标．································································································ 13分 法二:连接.交于点.分别过点作的垂线.垂足为． 易证．∴． 设.则．∴.． 由三角形中位线定理.． ∴.即． ∴坐标．······················································· 13分 把坐标代入抛物线解析式.不成立.所以不在抛物线上．·················· 14分

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