26．如图(9)-1.抛物线经过A(.0).C(3.)两点.与轴交于点D.与轴交于另一点B． (1)求此抛物线的解析式, (2)若直线将四边形ABCD面积二等分.求的值, -2.过点E(1.——青夏教育精英家教网—

26．如图(9)-1.抛物线经过A(.0).C(3.)两点.与轴交于点D.与轴交于另一点B． (1)求此抛物线的解析式, (2)若直线将四边形ABCD面积二等分.求的值, -2.过点E(1.1)作EF⊥轴于点F.将△AEF绕平面内某点旋转180°得△MNQ(点M.N.Q分别与点A.E.F对应).使点M.N在抛物线上.作MG⊥轴于点G.若线段MG︰AG＝1︰2.求点M.N的坐标． (1)解:把A(.0).C(3.)代入抛物线得 ················································································· 1分整理得 ------ 2分解得------3分 ∴抛物线的解析式为 ··································································· 4分 (2)令解得 ∴ B点坐标为(4.0) 又∵D点坐标为(0.) ∴AB∥CD ∴四边形ABCD是梯形． ∴S梯形ABCD ＝···························· 5分设直线与x轴的交点为H. 与CD的交点为T. 则H(.0). T(.)··················· 6分 ∵直线将四边形ABCD面积二等分 ∴S梯形AHTD ＝S梯形ABCD＝4 ∴····································· 7分 ∴···························································· 8分 (3)∵MG⊥轴于点G.线段MG︰AG＝1︰2 ∴设M(m.).········································ 9分 ∵点M在抛物线上 ∴ 解得 ···························· 10分 ∴M点坐标为(3.)····················································································· 11分根据中心对称图形性质知.MQ∥AF.MQ＝AF.NQ＝EF. ∴N点坐标为(1.) ···················································································· 12分【查看更多】

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