摘要： 解:(1)直线与轴交于点.与轴交于点． .······························································································· 1分 点都在抛物线上. 抛物线的解析式为······························································ 3分 顶点····································································································· 4分 (2)存在····················································································································· 5分 ··················································································································· 7分 ·················································································································· 9分 (3)存在···················································································································· 10分 理由: 解法一: 延长到点.使.连接交直线于点.则点就是所求的点． ··························································································· 11分 过点作于点． 点在抛物线上. 在中.. .. 在中.. ..····················································· 12分 设直线的解析式为 解得 ······································································································ 13分 解得 在直线上存在点.使得的周长最小.此时．········· 14分 解法二: 过点作的垂线交轴于点.则点为点关于直线的对称点．连接交于点.则点即为所求．···················································································· 11分 过点作轴于点.则.． . 同方法一可求得． 在中...可求得. 为线段的垂直平分线.可证得为等边三角形. 垂直平分． 即点为点关于的对称点．··················································· 12分 设直线的解析式为.由题意得 解得 ······································································································ 13分 解得 在直线上存在点.使得的周长最小.此时． 1

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