题目列表(包括答案和解析)
76.(08天津市卷26题)(本小题10分)
已知抛物线,
(Ⅰ)若,,求该抛物线与轴公共点的坐标;
(Ⅱ)若,且当时,抛物线与轴有且只有一个公共点,求的取值范围;
(Ⅲ)若,且时,对应的;时,对应的,试判断当时,抛物线与轴是否有公共点?若有,请证明你的结论;若没有,阐述理由.
(08天津市卷26题解析)解(Ⅰ)当,时,抛物线为,
方程的两个根为,.
∴该抛物线与轴公共点的坐标是和. ················································ 2分
(Ⅱ)当时,抛物线为,且与轴有公共点.
对于方程,判别式≥0,有≤. ········································ 3分
①当时,由方程,解得.
此时抛物线为与轴只有一个公共点.································· 4分
②当时,
时,,
时,.
由已知时,该抛物线与轴有且只有一个公共点,考虑其对称轴为,
应有 即
解得.
综上,或. ················································································ 6分
(Ⅲ)对于二次函数,
由已知时,;时,,
又,∴.
于是.而,∴,即.
∴. ············································································································ 7分
∵关于的一元二次方程的判别式
,
∴抛物线与轴有两个公共点,顶点在轴下方.····························· 8分
又该抛物线的对称轴,
由,,,
得,
∴.
又由已知时,;时,,观察图象,
可知在范围内,该抛物线与轴有两个公共点. ············································ 10分
77(08湖北宜昌25题)如图1,已知四边形OABC中的三个顶点坐标为O(0,0),A(0,n),C(m,0).动点P从点O出发依次沿线段OA,AB,BC向点C移动,设移动路程为z,△OPC的面积S随着z的变化而变化的图象如图2所示.m,n是常数, m>1,n>0.
(1)请你确定n的值和点B的坐标;
(2)当动点P是经过点O,C的抛物线y=ax+bx+c的顶点,且在双曲线y=上时,求这时四边形OABC的面积.
(08湖北宜昌25题解析)解:(1) 从图中可知,当P从O向A运动时,△POC的面积S=mz, z由0逐步增大到2,则S由0逐步增大到m,故OA=2,n=2 . (1分)
同理,AB=1,故点B的坐标是(1,2).(2分)
(2)解法一:
∵抛物线y=ax+bx+c经过点O(0,0),C(m ,0),∴c=0,b=-am,(3分)
∴抛物线为y=ax-amx,顶点坐标为(,-am2).(4分)
如图1,设经过点O,C,P的抛物线为l.
当P在OA上运动时,O,P都在y轴上,
这时P,O,C三点不可能同在一条抛物线上,
∴这时抛物线l不存在, 故不存在m的值..①
当点P与C重合时,双曲线y=不可能经过P,
故也不存在m的值.②(5分)
(说明:①②任做对一处评1分,两处全对也只评一分)
当P在AB上运动时,即当0<x≤1时,y=2,
抛物线l的顶点为P(,2).
∵P在双曲线y=上,可得 m=,∵>2,与 x=≤1不合,舍去.(6分)③
容易求得直线BC的解析式是:,(7分)
当P在BC上运动,设P的坐标为 (x,y),当P是顶点时 x=,
故得y==,顶点P为(,),
∵1< x=<m,∴m>2,又∵P在双曲线y=上,
于是,×=,化简后得5m-22m+22=0,
解得,,(8分)
与题意2<x=<m不合,舍去.④(9分)
故由①②③④,满足条件的只有一个值:.
这时四边形OABC的面积==.(10分)
(2)解法二:
∵抛物线y=ax+bx+c经过点O(0,0),C(m ,0)
∴c=0,b=-am,(3分)
∴抛物线为y=ax-amx,顶点坐标P为(,-am2). (4分)
∵m>1,∴>0,且≠m,
∴P不在边OA上且不与C重合. (5分)
∵P在双曲线y=上,∴×(- am2)=即a=- .
.①当1<m≤2时,<≤1,如图2,分别过B,P作x轴的垂线,
M,N为垂足,此时点P在线段AB上,且纵坐标为2,
∴-am2=2,即a=-.
而a=- ,∴- =-,m=>2,而1<m≤2,不合题意,舍去.(6分)
②当m≥2时,>1,如图3,分别过B,P作x轴的垂线,M,N为垂足,ON>OM,
此时点P在线段CB上,易证Rt△BMC∽Rt△PNC,
∴BM∶PN=MC∶NC,即: 2∶PN=(m-1)∶,∴PN=(7分)
而P的纵坐标为- am2,∴=- am2,即a=
而a=-,∴- =
化简得:5m2-22m+22=0.解得:m= ,(8分)
但m≥2,所以m=舍去,(9分)
取m = .
由以上,这时四边形OABC的面积为:
(AB+OC) ×OA=(1+m) ×2=. (10分)
74.(08广东东莞22题)(本题满分9分)将两块大小一样含30°角的直角三角板,叠放在一起,使得它们的斜边
AB重合,直角边不重合,已知AB=8,BC=AD=4,AC与BD相交于点E,连结CD.
(1)填空:如图9,AC= ,BD= ;四边形ABCD是 梯形.
(2)请写出图9中所有的相似三角形(不含全等三角形).
(3)如图10,若以AB所在直线为轴,过点A垂直于AB的直线为轴建立如图10的平面直角坐标系,保持ΔABD不动,将ΔABC向轴的正方向平移到ΔFGH的位置,FH与BD相交于点P,设AF=t,ΔFBP面积为S,求S与t之间的函数关系式,并写出t的取值值范围.
(08广东东莞22题解析)解:(1),,…………………………1分
等腰;…………………………2分
(2)共有9对相似三角形.(写对3-5对得1分,写对6-8对得2分,写对9对得3分)
①△DCE、△ABE与△ACD或△BDC两两相似,分别是:△DCE∽△ABE,△DCE∽△ACD,△DCE∽△BDC,△ABE∽△ACD,△ABE∽△BDC;(有5对)
②△ABD∽△EAD,△ABD∽△EBC;(有2对)
③△BAC∽△EAD,△BAC∽△EBC;(有2对)
所以,一共有9对相似三角形.…………………………………………5分
(3)由题意知,FP∥AE,
∴ ∠1=∠PFB,
又∵ ∠1=∠2=30°,
∴ ∠PFB=∠2=30°,
∴ FP=BP.…………………………6分
过点P作PK⊥FB于点K,则.
∵ AF=t,AB=8,
∴ FB=8-t,.
在Rt△BPK中,. ……………………7分
∴ △FBP的面积,
∴ S与t之间的函数关系式为:
,或. …………………………………8分
t的取值范围为:. …………………………………………………………9分
75(08甘肃兰州28题)(本题满分12分)如图19-1,是一张放在平面直角坐标系中的矩形纸片,为原点,点在轴的正半轴上,点在轴的正半轴上,,.
(1)在边上取一点,将纸片沿翻折,使点落在边上的点处,求两点的坐标;
(2)如图19-2,若上有一动点(不与重合)自点沿方向向点匀速运动,运动的速度为每秒1个单位长度,设运动的时间为秒(),过点作的平行线交于点,过点作的平行线交于点.求四边形的面积与时间之间的函数关系式;当取何值时,有最大值?最大值是多少?
(3)在(2)的条件下,当为何值时,以为顶点的三角形为等腰三角形,并求出相应的时刻点的坐标.
(08甘肃兰州28题解析)(本题满分12分)
解:(1)依题意可知,折痕是四边形的对称轴,
在中,,.
..
点坐标为(2,4).································································································ 2分
在中,, 又.
. 解得:.
点坐标为···································································································· 3分
(2)如图①,.
,又知,,
, 又.
而显然四边形为矩形.
·························································· 5分
,又
当时,有最大值.········································································ 6分
(3)(i)若以为等腰三角形的底,则(如图①)
在中,,,为的中点,
.
又,为的中点.
过点作,垂足为,则是的中位线,
,,
当时,,为等腰三角形.
此时点坐标为.··························································································· 8分
(ii)若以为等腰三角形的腰,则(如图②)
在中,.
过点作,垂足为.
,.
.
,.
,,
当时,(),此时点坐标为.·························· 11分
综合(i)(ii)可知,或时,以为顶点的三角形为等腰三角形,相应点的坐标为或.·········································································································· 12分
71.(08江苏镇江28题)(本小题满分8分)探索研究
如图,在直角坐标系中,点为函数在第一象限内的图象上的任一点,点的坐标为,直线过且与轴平行,过作轴的平行线分别交轴,于,连结交轴于,直线交轴于.
(1)求证:点为线段的中点;
(2)求证:①四边形为平行四边形;
②平行四边形为菱形;
(3)除点外,直线与抛物线有无其它公共点?并说明理由.
(08江苏镇江28题解析)(1)法一:由题可知.
,,
.························································································· (1分)
,即为的中点.····································································· (2分)
法二:,,.·························································· (1分)
又轴,.··············································································· (2分)
(2)①由(1)可知,,
,,
.·························································································· (3分)
,
又,四边形为平行四边形.···················································· (4分)
②设,轴,则,则.
过作轴,垂足为,在中,
.
平行四边形为菱形.··············································································· (6分)
(3)设直线为,由,得,代入得:
直线为.························ (7分)
设直线与抛物线的公共点为,代入直线关系式得:
,,解得.得公共点为.
所以直线与抛物线只有一个公共点.············································· (8分)
72(08黑龙江齐齐哈尔28题)(本小题满分10分)
如图,在平面直角坐标系中,点,点分别在轴,轴的正半轴上,且满足.
(1)求点,点的坐标.
(2)若点从点出发,以每秒1个单位的速度沿射线运动,连结.设的面积为,点的运动时间为秒,求与的函数关系式,并写出自变量的取值范围.
(3)在(2)的条件下,是否存在点,使以点为顶点的三角形与相似?若存在,请直接写出点的坐标;若不存在,请说明理由.
(08黑龙江齐齐哈尔28题解析)解:(1)
,··················································································· (1分)
,
点,点分别在轴,轴的正半轴上
······························································································· (2分)
(2)求得························································································· (3分)
(每个解析式各1分,两个取值范围共1分)························································· (6分)
(3);;;(每个1分,计4分)
···························································································································· (10分)
注:本卷中所有题目,若由其它方法得出正确结论,酌情给分.
73(08海南省卷24题)(本题满分14分)如图13,已知抛物线经过原点O和x轴上另一点A,它的对称轴x=2 与x轴交于点C,直线y=-2x-1经过抛物线上一点B(-2,m),且与y轴、直线x=2分别交于点D、E.
(1)求m的值及该抛物线对应的函数关系式;
(2)求证:① CB=CE ;② D是BE的中点;
(3)若P(x,y)是该抛物线上的一个动点,是否存在这样的点P,使得PB=PE,若存在,试求出所有符合条件的点P的坐标;若不存在,请说明理由.
(08海南省卷24题解析)(1)∵ 点B(-2,m)在直线y=-2x-1上,
∴ m=-2×(-2)-1=3. ………………………………(2分)
∴ B(-2,3)
∵ 抛物线经过原点O和点A,对称轴为x=2,
∴ 点A的坐标为(4,0) .
设所求的抛物线对应函数关系式为y=a(x-0)(x-4). ……………………(3分)
将点B(-2,3)代入上式,得3=a(-2-0)(-2-4),∴ .
∴ 所求的抛物线对应的函数关系式为,即. (6分)
(2)①直线y=-2x-1与y轴、直线x=2的交点坐标分别为D(0,-1) E(2,-5).
过点B作BG∥x轴,与y轴交于F、直线x=2交于G,
则BG⊥直线x=2,BG=4.
在Rt△BGC中,BC=.
∵ CE=5,
∴ CB=CE=5. ……………………(9分)
②过点E作EH∥x轴,交y轴于H,
则点H的坐标为H(0,-5).
又点F、D的坐标为F(0,3)、D(0,-1),
∴ FD=DH=4,BF=EH=2,∠BFD=∠EHD=90°.
∴ △DFB≌△DHE (SAS),
∴ BD=DE.
即D是BE的中点. ………………………………(11分)
(3) 存在. ………………………………(12分)
由于PB=PE,∴ 点P在直线CD上,
∴ 符合条件的点P是直线CD与该抛物线的交点.
设直线CD对应的函数关系式为y=kx+b.
将D(0,-1) C(2,0)代入,得. 解得 .
∴ 直线CD对应的函数关系式为y=x-1.
∵ 动点P的坐标为(x,),
∴ x-1=. ………………………………(13分)
解得 ,. ∴ ,.
∴ 符合条件的点P的坐标为(,)或(,).…(14分)
(注:用其它方法求解参照以上标准给分.)
91.(08新疆自治区24题)(10分)某工厂要赶制一批抗震救灾用的大型活动板房.如图,板房一面的形状是由矩形和抛物线的一部分组成,矩形长为12m,抛物线拱高为5.6m.
(1)在如图所示的平面直角坐标系中,求抛物线的表达式.
(2)现需在抛物线AOB的区域内安装几扇窗户,窗户的底边在AB上,每扇窗户宽1.5m,高1.6m,相邻窗户之间的间距均为0.8m,左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m.请计算最多可安装几扇这样的窗户?
(08新疆自治区24题解析)24.(10分)解:(1)设抛物线的表达式为 1分
点在抛物线的图象上.
∴
······························································ 3分
∴抛物线的表达式为············································································· 4分
(2)设窗户上边所在直线交抛物线于C、D两点,D点坐标为(k,t)
已知窗户高1.6m,∴··························································· 5分
(舍去)············································································ 6分
∴(m)·············································································· 7分
又设最多可安装n扇窗户
∴····················································································· 9分
.
答:最多可安装4扇窗户.···················································································· 10分
(本题不要求学生画出4个表示窗户的小矩形)
90.(08四川自贡26题)抛物线的顶点为M,与轴的交点为A、B(点B在点A的右侧),△ABM的三个内角∠M、∠A、∠B所对的边分别为m、a、b。若关于的一元二次方程有两个相等的实数根。
(1)判断△ABM的形状,并说明理由。
(2)当顶点M的坐标为(-2,-1)时,求抛物线的解析式,并画出该抛物线的大致图形。
(3)若平行于轴的直线与抛物线交于C、D两点,以CD为直径的圆恰好与轴相切,求该圆的圆心坐标。
(08四川自贡26题解析)解:(1)令
得
由勾股定理的逆定理和抛物线的对称性知
△ABM是一个以、为直角边的等腰直角三角形
(2)设
∵△ABM是等腰直角三角形
∴斜边上的中线等于斜边的一半
又顶点M(-2,-1)
∴,即AB=2
∴A(-3,0),B(-1,0)
将B(-1,0) 代入中得
∴抛物线的解析式为,即
图略
(3)设平行于轴的直线为
解方程组错误!不能通过编辑域代码创建对象。
得, (
∴线段CD的长为
∵以CD为直径的圆与轴相切
据题意得
∴
解得
∴圆心坐标为和
89.(08四川巴中30题)(12分)30.已知:如图14,抛物线与轴交于点,点,与直线相交于点,点,直线与轴交于点.
(1)写出直线的解析式.
(2)求的面积.
(3)若点在线段上以每秒1个单位长度的速度从向运动(不与重合),同时,点在射线上以每秒2个单位长度的速度从向运动.设运动时间为秒,请写出的面积与的函数关系式,并求出点运动多少时间时,的面积最大,最大面积是多少?
(08四川巴中30题解析)解:(1)在中,令
,
,··············································· 1分
又点在上
的解析式为·············································································· 2分
(2)由,得 ···················································· 4分
,
,······························································································· 5分
························································································· 6分
(3)过点作于点
······························································································· 7分
·········································································································· 8分
由直线可得:
在中,,,则
,······················································································· 9分
···················································································· 10分
····························································································· 11分
此抛物线开口向下,当时,
当点运动2秒时,的面积达到最大,最大为.···························· 12分
88.(08山东济宁26题)(12分)
中,,,cm.长为1cm的线段在的边上沿方向以1cm/s的速度向点运动(运动前点与点重合).过分别作的垂线交直角边于两点,线段运动的时间为s.
(1)若的面积为,写出与的函数关系式(写出自变量的取值范围);
(2)线段运动过程中,四边形有可能成为矩形吗?若有可能,求出此时的值;若不可能,说明理由;
(3)为何值时,以为顶点的三角形与相似?
(08山东济宁26题解析)解:(1)当点在上时,,.
.········································································ 2分
当点在上时,.
.·················································· 4分
(2),..
.········································································ 6分
由条件知,若四边形为矩形,需,即,
.
当s时,四边形为矩形.································································· 8分
(3)由(2)知,当s时,四边形为矩形,此时,
.··························································································· 9分
除此之外,当时,,此时.
,..····························· 10分
,.
又,.········································ 11分
,.
当s或s时,以为顶点的三角形与相似.··················· 12分
87.(08青海省卷28题)王亮同学善于改进学习方法,他发现对解题过程进行回顾反思,效果会更好.某一天他利用30分钟时间进行自主学习.假设他用于解题的时间(单位:分钟)与学习收益量的关系如图甲所示,用于回顾反思的时间(单位:分钟)与学习收益量的关系如图乙所示(其中是抛物线的一部分,为抛物线的顶点),且用于回顾反思的时间不超过用于解题的时间.
(1)求王亮解题的学习收益量与用于解题的时间之间的函数关系式,并写出自变量的取值范围;
(2)求王亮回顾反思的学习收益量与用于回顾反思的时间之间的函数关系式;
(3)王亮如何分配解题和回顾反思的时间,才能使这30分钟的学习收益总量最大?
(学习收益总量解题的学习收益量回顾反思的学习收益量)
(08青海省卷28题解析)解:(1)设,
把代入,得.
.······································································································ (1分)
自变量的取值范围是:.···························································· (2分)
(2)当时,
设,···················································································· (3分)
把代入,得,.
.································································· (5分)
当时,
············································································································· (6分)
即.
(3)设王亮用于回顾反思的时间为分钟,学习效益总量为,
则他用于解题的时间为分钟.
当时,
.························· (7分)
当时,.············································································ (8分)
当时,
.····································································· (9分)
随的增大而减小,
当时,.
综合所述,当时,,此时.································ (10分)
即王亮用于解题的时间为26分钟,用于回顾反思的时间为4分钟时,学习收益总量最大.
······················································································································· (11分)
86.(08青海西宁28题)如图14,已知半径为1的与轴交于两点,为的切线,切点为,圆心的坐标为,二次函数的图象经过两点.
(1)求二次函数的解析式;
(2)求切线的函数解析式;
(3)线段上是否存在一点,使得以为顶点的三角形与相似.若存在,请求出所有符合条件的点的坐标;若不存在,请说明理由.
(08青海西宁28题解析)解:(1)圆心的坐标为,半径为1,,……1分
二次函数的图象经过点,
可得方程组················································································ 2分
解得:二次函数解析式为············································· 3分
(2)过点作轴,垂足为.······························································· 4分
是的切线,为切点,(圆的切线垂直于经过切点的半径).
在中,
为锐角,···························· 5分
,
在中,.
.
点坐标为························································································· 6分
设切线的函数解析式为,由题意可知,······ 7分
切线的函数解析式为···································································· 8分
(3)存在.············································································································ 9分
①过点作轴,与交于点.可得(两角对应相等两三角形相似)
,············································ 10分
②过点作,垂足为,过点作,垂足为.
可得(两角对应相等两三角开相似)
在中,,,
在中,,
,······································· 11分
符合条件的点坐标有,······················································ 12分
85.(08内蒙古赤峰25题)(本题满分14分)
在平面直角坐标系中给定以下五个点.
(1)请从五点中任选三点,求一条以平行于轴的直线为对称轴的抛物线的解析式;
(2)求该抛物线的顶点坐标和对称轴,并画出草图;
(3)已知点在抛物线的对称轴上,直线过点且垂直于对称轴.验证:以为圆心,为半径的圆与直线相切.请你进一步验证,以抛物线上的点为圆心为半径的圆也与直线相切.由此你能猜想到怎样的结论.
(08内蒙古赤峰25题解析)25.解:(1)设抛物线的解析式为,
且过点,
由在H .
则.········································································································ (2分)
得方程组,
解得.
抛物线的解析式为················ (4分)
(2)由············· (6分)
得顶点坐标为,对称轴为.·········· (8分)
(3)①连结,过点作直线的垂线,垂足为,
则.
在中,,,
,
,
以点为圆心,为半径的与直线相切.····························· (10分)
②连结过点作直线的垂线,垂足为.过点作垂足为,
则.
在中,,.
.
以点为圆心为半径的与直线相切.································ (12分)
③以抛物线上任意一点为圆心,以为半径的圆与直线相切.····· (14分)
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