摘要：24． 解:(1)设抛物线的解析式为．······················································· 1分 将代入上式.得． 解.得．············································································································ 2分 抛物线的解析式为． 即．··································································································· 3分 (2)连接.交直线于点． 点与点关于直线 对称. ．······························································ 4分 ． 由“两点之间.线段最短 的原理可知: 此时最小.点的位置即为所求．················ 5分 设直线的解析式为. 由直线过点..得 解这个方程组.得 直线的解析式为．············································································· 6分 由(1)知:对称轴为.即． 将代入.得． 点的坐标为(1.2)．···························································································· 7分 说明:用相似三角形或三角函数求点的坐标也可.答案正确给2分． (3)①连接．设直线与轴的交点记为点． 由(1)知:当最小时.点的坐标为(1.2)． ． ．·························································································· 8分 ． ． 与相切．······································································································ 9分 ②．················································································································· 11分

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