28．如图14(1).抛物线与x轴交于A.B两点.与y轴交于点C(0.)．［图14为解答备用图］ (1) .点A的坐标为 .点B的坐标为 , (2)设抛物线的顶点为M.求四边形ABMC的面积——青夏教育精英家教网—

28．如图14(1).抛物线与x轴交于A.B两点.与y轴交于点C(0.)．［图14为解答备用图］ (1) .点A的坐标为 .点B的坐标为 , (2)设抛物线的顶点为M.求四边形ABMC的面积, (3)在x轴下方的抛物线上是否存在一点D.使四边形ABDC的面积最大?若存在.请求出点D的坐标,若不存在.请说明理由, (4)在抛物线上求点Q.使△BCQ是以BC为直角边的直角三角形．解:(1).··················· 1分 A.····························································· 2分 B(3.0)．······························································· 3分 .抛物线的顶点为M.连结OM． ······································································ 4分则 △AOC的面积=.△MOC的面积=. △MOB的面积=6.····················································· 5分 ∴ 四边形 ABMC的面积 =△AOC的面积+△MOC的面积+△MOB的面积=9．········································· 6分说明:也可过点M作抛物线的对称轴.将四边形ABMC的面积转化为求1个梯形与2个直角三角形面积的和． .设D(m.).连结OD．则 0＜m＜3. ＜0．且 △AOC的面积=.△DOC的面积=. △DOB的面积=-().····························································· 8分 ∴ 四边形 ABDC的面积=△AOC的面积+△DOC的面积+△DOB的面积 = =．···················································································· 9分 ∴ 存在点D.使四边形ABDC的面积最大为．······························ 10分 (4)有两种情况: 如图14(3).过点B作BQ1⊥BC.交抛物线于点Q1.交y轴于点E.连接Q1C． ∵ ∠CBO=45°.∴∠EBO=45°.BO=OE=3． ∴ 点E的坐标为(0.3)． ∴ 直线BE的解析式为．···································································· 12分由解得 ∴ 点Q1的坐标为．··············································································· 13分如图14(4).过点C作CF⊥CB.交抛物线于点Q2.交x轴于点F.连接BQ2． ∵ ∠CBO=45°.∴∠CFB=45°.OF=OC=3． ∴ 点F的坐标为． ∴ 直线CF的解析式为．···································································· 14分由解得 ∴点Q2的坐标为．················································································· 15分综上.在抛物线上存在点Q1.Q2.使△BCQ1.△BCQ2是以BC为直角边的直角三角形． 16分说明:如图14(4).点Q2即抛物线顶点M.直接证明△BCM为直角三角形同样得2分．【查看更多】

题目列表(包括答案和解析)

(2009年甘肃定西)计算：