摘要：解:(1)解方程.得．------1分 ∴点.点． ∴ 解.得 ∴抛物线的解析式为．·················································· 2分 (2)∵抛物线与y轴交于点C． ∴点C的坐标为(0.2)． 又点.可求直线BC的解析式为． ∵AD∥CB.∴设直线AD的解析式为． 又点.∴.直线AD的解析式为． 解.得. ∴点D的坐标为(4.)．·········································································· 4分 过点D作DD’轴于D’. DD’＝.则又AB＝4． ∴四边形ACBD的面积＝AB•OC+AB•DD’＝······························ 5分 (3)假设存在满足条件的点R.设直线l交y轴于点E(0.m). ∵点P不与点A.C重合.∴0< m <2.∵点.点. ∴可求直线AC的解析式为.∴点． ∵直线BC的解析式为.∴点． ∴．在△PQR中. ①当RQ为底时.过点P作PR1⊥x轴于点R1.则∠R1PQ＝90°.PQ＝PR1＝m． ∴.解得.∴点. ∴点R1坐标为(.0)．··········································································· 6分 ②当RP为底时.过点Q作Q R2⊥x轴于点R2. 同理可求.点R2坐标为(1.0)．································································ 7分 ③当PQ为底时.取PQ中点S.过S作SR3⊥PQ交x轴于点R3.则PR3＝QR3.∠PR3Q＝90°．∴PQ＝2R3S＝2m．∴.解.得. ∴点.点.可求点R3坐标为(.0)． -------8分 经检验.点R1.点R2.点R3都满足条件． 综上所述.存在满足条件的点R.它们分别是R1(.0).R2(1.0)和点R3(.0)．

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