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1、学习好资料欢迎下载2007 年各地中考压轴题汇编(1)1、(安徽)按右图所示的流程,输入一个数据x,根据 y 与 x 的关系式就输出一个数据 y,这样可以将一组数据变换成另一组新的数据,要使任意一组都在20100(含 20 和 100)之间的数据,变换成一组新数据后能满足下列两个要求:()新数据都在 60100(含 60 和 100)之间;()新数据之间的大小关系与原数据之间的大小关系一致,即原数据大的对应的新数据也较大。(1)若 y 与 x 的关系是yxp(100 x),请说明:当p12时,这种变换满足上述两个要求;(2)若按关系式y=a(xh)2k(a0)将数据进行变换,请写出一个满足上述
2、要求的这种关系式。(不要求对关系式符合题意作说明,但要写出关系式得出的主要过程)【解】(1)当 P=12时,y=x11002x,即 y=1502x。y 随着 x 的增大而增大,即P=12时,满足条件()3 分又当 x=20 时,y=1100502=100。而原数据都在20100 之间,所以新数据都在60100 之间,即满足条件(),综上可知,当P=12时,这种变换满足要求;6 分(2)本题是开放性问题,答案不唯一。若所给出的关系式满足:(a)h20;(b)若 x=20,100 时,y 的对应值m,n 能落在 60100 之间,则这样的关系式都符合要求。如取 h=20,y=220a xk,8 分
3、a0,当 20 x100 时,y 随着 x 的增大 10 分令 x=20,y=60,得 k=60 令 x=100,y=100,得 a802k=100 由解得116060ak,212060160yx。14 分2、(常州)已知(1)Am,与(23 3)Bm,是反比例函数kyx图象上的两个点(1)求k的值;(2)若点(1 0)C,则在反比例函数kyx图象上是否存在点D,使得以ABCD,四点为顶点的四边形为梯形?若存在,求出点D的坐标;若不存在,请说明理由解:(1)由(1)2(3 3)mm,得2 3m,因此2 3k 2 分BCxy1111O开始y 与 x 的关系式结束输入 x输出 y精品资料-欢迎下载
4、-欢迎下载 名师归纳-第 1 页,共 10 页 -学习好资料欢迎下载(2)如图 1,作BEx轴,E为垂足,则3CE,3BE,2 3BC,因此30BCE由于点C与点A的横坐标相同,因此CAx轴,从而120ACB当AC为底时,由于过点B且平行于AC的直线与双曲线只有一个公共点B,故不符题意 3 分当BC为底时,过点A作BC的平行线,交双曲线于点D,过点AD,分别作x轴,y轴的平行线,交于点F由于30DAF,设11(0)DFm m,则13AFm,12ADm,由点(12 3)A,得点11(132 3)Dmm,因此11(13)(2 3)2 3mm,解之得1733m(10m舍去),因此点363D,此时14
5、33AD,与BC的长度不等,故四边形ADBC是梯形 5 分如图 2,当AB为底时,过点C作AB的平行线,与双曲线在第一象限内的交点为D由于ACBC,因此30CAB,从而150ACD作DHx轴,H为垂足,则60DCH,设22(0)CHm m,则23DHm,22CDm由点(1 0)C,得点22(13)Dmm,因此22(1)32 3mm解之得22m(21m舍去),因此点(12 3)D,此时4CD,与AB的长度不相等,故四边形ABDC是梯形 7 分如图 3,当过点C作AB的平行线,与双曲线在第三象限内的交点为D时,同理可得,点(23)D,四边形ABCD是梯形 9 分图 1 ABCxyOFDE图 2 A
6、BCxyODH精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文
7、档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4
8、文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O
9、4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6
10、O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F
11、6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9
12、F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P
13、9F6O4学习好资料欢迎下载综上所述,函数2 3yx图象上存在点D,使得以ABCD,四点为顶点的四边形为梯形,点D的坐标为:363D,或(12 3)D,或(23)D,10 分3、(福建龙岩)如图,抛物线254yaxax经过ABC的三个顶点,已知BCx轴,点A在x轴上,点C在y轴上,且ACBC(1)求抛物线的对称轴;(2)写出ABC,三点的坐标并求抛物线的解析式;(3)探究:若点P是抛物线对称轴上且在x轴下方的动点,是否存在PAB是等腰三角形若存在,求出所有符合条件的点P坐标;不存在,请说明理由解:(1)抛物线的对称轴5522axa 2 分(2)(3 0)A,(5 4)B,(0 4)C,5 分把
14、点A坐标代入254yaxax中,解得16a 6 分215466yxx7 分(3)存在符合条件的点P共有 3 个以下分三类情形探索设抛物线对称轴与x轴交于N,与CB交于M过点B作BQx轴于Q,易得4BQ,8AQ,5.5AN,52BM以AB为腰且顶角为角A的PAB有 1个:1P AB图 3 ABCxyODA C B y x 0 1 1 A B C x 0 1 1 2P1P3Py 精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP
15、4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 Z
16、P4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1
17、ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1
18、 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H
19、1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9
20、H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K
21、9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4学习好资料欢迎下载222228480ABAQBQ 8 分在1RtANP中,222221119980(5.5)2PNAPANABAN1519922P,9 分以AB为腰且顶角为角B的PAB有 1 个:2P AB在2RtBMP中,2
22、22222252958042MPBPBMABBM 10 分25 829522P,11 分以AB为底,顶角为角P的PAB有 1 个,即3P AB画AB的垂直平分线交抛物线对称轴于3P,此时平分线必过等腰ABC的顶点C过点3P作3P K垂直y轴,垂足为K,显然3RtRtPCKBAQ312P KBQCKAQ32.5P K5CK于是1OK 13 分3(2.51)P,14 分注:第(3)小题中,只写出点P的坐标,无任何说明者不得分4、(福州)如图 12,已知直线12yx与双曲线(0)kykx交于AB,两点,且点A的横坐标为4(1)求k的值;(2)若双曲线(0)kykx上一点C的纵坐标为8,求AOC的面积
23、;(3)过原点O的另一条直线l交双曲线(0)kykx于PQ,两点(P点在第一象限),若由点ABPQ,为顶点组成的四边形面积为24,求点P的坐标解:(1)点A横坐标为 4,当x=4 时,y=2.点A的坐标为(4,2).点A是直线与双曲线(k0)的交点 ,k=4 2=8.图 12 OxAyBxy21xy8精品资料-欢迎下载-欢迎下载 名师归纳-第 4 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9
24、H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K
25、9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10
26、K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N1
27、0K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N
28、10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7
29、N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X
30、7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4学习好资料欢迎下载(2)解法一:如图12-1,点C在双曲线上,当y=8时,x=1 点C的坐标为 (1,8).过点A、C分别做x轴、y轴的垂线,垂足为M、N,得矩形DMON.S矩形 ONDM=32,SONC=4,SCDA=9,SOAM=4.SAOC=S矩形 ONDM-SONC-SCDA-SOAM=32-
31、4-9-4=15.解法二:如图12-2,过点C、A分别做x轴的垂线,垂足为E、F,点C在双曲线8yx上,当y=8时,x=1.点C的坐标为 (1,8).点C、A都在双曲线8yx上,SCOE=SAOF =4。SCOE+S梯形 CEFA=SCOA+SAOF.SCOA=S梯形 CEFA .S梯形 CEFA=12(2+8)3=15,SCOA=15.(3)反比例函数图象是关于原点O的中心对称图形,OP=OQ,OA=OB.四边形APBQ是平行四边形 .SPOA=S平行四边形APBQ=24=6 .设点P的横坐标为m(m 0 且4m),得P(m,).过点 P、A 分别做x轴的垂线,垂足为E、F,点P、A在双曲线
32、上,SPOE=SAOF =4.若 0m4,如图 12-3,SPOE+S梯形PEFA=SPOA+S AOF,4141m8精品资料-欢迎下载-欢迎下载 名师归纳-第 5 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4
33、G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP
34、4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 Z
35、P4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1
36、ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1
37、 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H
38、1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9
39、H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4学习好资料欢迎下载 S梯形 PEFA=SPOA=6.18(2)(4)62mm.解得m=2,m=-8(舍去).P(2,4).若m 4,如图 12-4,SAOF+S梯形AFEP=SAOP+S POE,S梯形PEFA=SPOA=6.18(2)(4)62mm,解得m=8,m=-2(舍去).P(8,1).点P的坐标是P(2,4)或P(8,1).5、(甘肃陇南)如图,抛物线212yxmxn交x轴于 A、B 两点,交 y 轴于点 C,点 P 是它的顶点,点 A 的横坐标是3,点 B 的横坐标是1
40、(1)求m、n的值;(2)求直线 PC 的解析式;(3)请探究以点A 为圆心、直径为5 的圆与直线PC 的位置关系,并说明理由(参考数:21.41,31.73,52.24)解:(1)由已知条件可知:抛物线212yxmxn经过 A(-3,0)、B(1,0)两点903,210.2mnmn2分解得31,2mn3分(2)21322yxx,P(-1,-2),C3(0,)24分精品资料-欢迎下载-欢迎下载 名师归纳-第 6 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9
41、F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P
42、9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7
43、P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G
44、7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4
45、G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP
46、4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 Z
47、P4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4学习好资料欢迎下载设直线 PC 的解析式是ykxb,则2,3.2kbb解得13,22kb 直线 PC 的解析式是1322yx6分说明:只要求对1322kb,不写最后一步,不扣分(3)如图,过点A 作 AEPC,垂足为 E设直线 PC 与
48、x轴交于点 D,则点 D 的坐标为(3,0)7分在 RtOCD 中,OC=32,3OD,2233()3522CD8 分 OA=3,3OD,AD=69 分 COD=AED=90o,CDO 公用,COD AED 10 分OCCDAEAD,即335226AE 655AE11 分652.6882.55,以点 A 为圆心、直径为5 的圆与直线PC 相离 12 分6、(贵阳)如图14,从一个直径是2 的圆形铁皮中剪下一个圆心角为90的扇形(1)求这个扇形的面积(结果保留)(3 分)(2)在剩下的三块余料中,能否从第块余料中剪出一个圆作为底面与此扇形围成一个圆锥?请说明理由(4 分)(3)当O的半径(0)R
49、 R为任意值时,(2)中的结论是否仍然成立?请说明理由(5 分)解:(1)连接BC,由勾股定理求得:2ABAC 1 分213602n RS 2 分(2)连接AO并延长,与弧BC和O交于EF,22EFAFAE 1 分弧BC的长:21802n Rl 2 分222rABCOEF精品资料-欢迎下载-欢迎下载 名师归纳-第 7 页,共 10 页 -文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文
50、档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4文档编码:CD10V2M3T10T7 HH4X7N10K9H1 ZP4G7P9F6O4