如图10.平行四边形ABCD中.AB＝5.BC＝10.BC边上的高AM=4.E为 BC边上的一个动点(不与B.C重合)．过E作直线AB的垂线.垂足为F． FE与DC的延长线相交于点G.连结DE.DF．——青夏教育精英家教网—

如图10.平行四边形ABCD中.AB＝5.BC＝10.BC边上的高AM=4.E为 BC边上的一个动点(不与B.C重合)．过E作直线AB的垂线.垂足为F． FE与DC的延长线相交于点G.连结DE.DF．． (1) 求证:ΔBEF ∽ΔCEG． (2) 当点E在线段BC上运动时.△BEF和△CEG的周长之间有什么关系?并说明你的理由． (3)设BE＝x.△DEF的面积为 y.请你求出y和x之间的函数关系式.并求出当x为何值时,y有最大值.最大值是多少? (1) 因为四边形ABCD是平行四边形. 所以 1分所以所以 ···························································································· 3分 (2)的周长之和为定值．····························································· 4分理由一: 过点C作FG的平行线交直线AB于H . 因为GF⊥AB.所以四边形FHCG为矩形．所以 FH＝CG.FG＝CH 因此.的周长之和等于BC+CH+BH 由 BC＝10.AB＝5.AM＝4.可得CH＝8.BH＝6. 所以BC+CH+BH＝24 ···························································································· 6分理由二: 由AB＝5.AM＝4.可知在Rt△BEF与Rt△GCE中.有: . 所以.△BEF的周长是. △ECG的周长是又BE+CE＝10.因此的周长之和是24．·········································· 6分 (3)设BE＝x.则所以 ···································· 8分配方得:．所以.当时.y有最大值．············································································· 9分最大值为．··········································································································· 10分 52 如图13.在平面直角坐标系中.圆M经过原点O.且与轴.轴分别相交于两点． (1)求出直线AB的函数解析式, (2)若有一抛物线的对称轴平行于轴且经过点M.顶点C在⊙M上.开口向下.且经过点B.求此抛物线的函数解析式, 中的抛物线交轴于D.E两点.在抛物线上是否存在点P.使得?若存在.请求出点P的坐标,若不存在.请说明理由．解:(1)设AB的函数表达式为 ∵∴∴ ∴直线AB的函数表达式为．··································································· 3分 (2)设抛物线的对称轴与⊙M相交于一点.依题意知这一点就是抛物线的顶点C.又设对称轴与轴相交于点N.在直角三角形AOB中. 因为⊙M经过O.A.B三点.且⊙M的直径.∴半径MA=5.∴N为AO的中点AN=NO=4.∴MN=3∴CN=MC-MN=5-3=2.∴C点的坐标为．设所求的抛物线为则 ∴所求抛物线为 ············································································· 7分 (3)令得D.E两点的坐标为D.所以DE=4．又AC=直角三角形的面积假设抛物线上存在点．当故满足条件的存在．它们是． ························· 10分 53 已知抛物线经过点A(5.0).B和原点. (1)求抛物线的函数关系式, (2)若过点B的直线与抛物线相交于点C(2.m).请求出OBC的面积S的值. (3)过点C作平行于x轴的直线交y轴于点D.在抛物线对称轴右侧位于直线DC下方的抛物线上.任取一点P.过点P作直线PF平行于y轴交x轴于点F.交直线DC于点E. 直线PF与直线DC及两坐标轴围成矩形OFED.是否存在点P.使得OCD与CPE相似?若存在.求出点P的坐标,若不存在.请说明理由. 解:(1)由题意得: 2分解得 ·················································· 3分故抛物线的函数关系式为·············· 4分 (2)在抛物线上.·· 5分点坐标为(2.6)..C在直线上解得直线BC的解析式为······································································· 6分设BC与x轴交于点G.则G的坐标为(4.0) ·································································· 7分 (3)存在P.使得∽·············································································· 8分设P. 故若要∽.则要或即或解得或又在抛物线上.或解得或故P点坐标为和········································································· 10分 (只写出一个点的坐标记9分) 【查看更多】

题目列表(包括答案和解析)

16、如图，在平行四边形ABCD中，E是对角线BD上的点，且EF∥AB，DE：EB=2：3，EF=4，则CD的长为

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