摘要：23．本题满分11分． 如图11所示.在梯形ABCD中.已知AB∥CD. AD⊥DB.AD=DC=CB.AB=4．以AB所在直线为轴.过D且垂直于AB的直线为轴建立平面直角坐标系． (1)求∠DAB的度数及A.D.C三点的坐标, (2)求过A.D.C三点的抛物线的解析式及其对称轴L． (3)若P是抛物线的对称轴L上的点.那么使PDB为等腰三角形的点P有几个?(不必求点P的坐标.只需说明理由) 解: (1) DC∥AB.AD=DC=CB. ∠CDB=∠CBD=∠DBA.··············································································· 0.5分 ∠DAB=∠CBA. ∠DAB=2∠DBA. ············ 1分 ∠DAB+∠DBA=90. ∠DAB=60. ·········· 1.5分 ∠DBA=30.AB=4. DC=AD=2. ········· 2分 RtAOD.OA=1.OD=.··························· 2.5分 A.D(0. ).C(2. )． · 4分 (2)根据抛物线和等腰梯形的对称性知.满足条件的抛物线必过点A.B(3.0). 故可设所求为 = (+1)( -3) ······························································ 6分 将点D(0. )的坐标代入上式得. =． 所求抛物线的解析式为 = ·········································· 7分 其对称轴L为直线=1．······················································································ 8分 (3) PDB为等腰三角形.有以下三种情况: ①因直线L与DB不平行.DB的垂直平分线与L仅有一个交点P1.P1D=P1B. P1DB为等腰三角形, ················································································· 9分 ②因为以D为圆心.DB为半径的圆与直线L有两个交点P2.P3.DB=DP2.DB=DP3. P2DB. P3DB为等腰三角形, ③与②同理.L上也有两个点P4.P5.使得 BD=BP4.BD=BP5． ···················· 10分 由于以上各点互不重合.所以在直线L上.使PDB为等腰三角形的点P有5个．

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