上海市上宝中学数学圆 几何综合章末训练(Word版 含解析)

上海市上宝中学数学圆 几何综合章末训练(Word版 含解析)
上海市上宝中学数学圆 几何综合章末训练(Word版 含解析)

上海市上宝中学数学圆几何综合章末训练(Word版含解析)

一、初三数学圆易错题压轴题(难)

1.如图所示,CD为⊙O的直径,点B在⊙O上,连接BC、BD,过点B的切线AE与CD 的延长线交于点A,OE//BD,交BC于点F,交AB于点E.

(1)求证:∠E=∠C;

(2)若⊙O的半径为3,AD=2,试求AE的长;

(3)在(2)的条件下,求△ABC的面积.

【答案】(1)证明见解析;(2)10;(3)48 5

.

【解析】

试题分析:(1)连接OB,利用已知条件和切线的性质证明:OE∥BD,即可证明:∠E=∠C;

(2)根据题意求出AB的长,然后根据平行线分线段定理,可求解;

(3)根据相似三角形的面积比等于相似比的平方可求解.

试题解析:(1)如解图,连接OB,

∵CD为⊙O的直径,

∴∠CBD=∠CBO+∠OBD=90°,

∵AB是⊙O的切线,

∴∠ABO=∠ABD+∠OBD=90°,

∴∠ABD=∠CBO.

∵OB、OC是⊙O的半径,

∴OB=OC,∴∠C=∠CBO.

∵OE∥BD,∴∠E=∠ABD,

∴∠E=∠C;

(2)∵⊙O的半径为3,AD=2,

∴AO=5,∴AB=4.

∵BD∥OE,

∴=,

∴=,

∴BE=6,AE=6+4=10

(3)S △AOE==15,然后根据相似三角形面积比等于相似比的平方可得

S △ABC = S △AOE ==

2.已知:

图1 图2 图3 (1)初步思考:

如图1, 在PCB ?中,已知2PB =,BC=4,N 为BC 上一点且1BN =,试说明:

1

2

PN PC =

(2)问题提出:

如图2,已知正方形ABCD 的边长为4,圆B 的半径为2,点P 是圆B 上的一个动点,求

1

2

PD PC +的最小值.

(3)推广运用:

如图3,已知菱形ABCD 的边长为4,∠B ﹦60°,圆B 的半径为2,点P 是圆B 上的一个动点,求1

2

PD PC -的最大值.

【答案】(1)详见解析;(2)5;(3)最大值37DG =【解析】 【分析】

(1)利用两边成比例,夹角相等,证明BPN ?∽BCP ?,得到PN BN

PC BP

=,即可得到结论成立;

(2)在BC 上取一点G ,使得BG=1,由△PBG ∽△CBP ,得到1

2

PG PC =,当D 、P 、G 共线时,1

2

PD PC +

的值最小,即可得到答案; (3)在BC 上取一点G ,使得BG=1,作DF ⊥BC 于F ,与(2)同理得到1

2

PG PC =,当点P 在DG 的延长线上时,1

2

PD PC -的值最大,即可得到答案. 【详解】

(1)证明:∵2,1,4PB BN BC ===,

∴24,4PB BN BC =?=, ∴2PB BN BC =?, ∴

BN BP

BP BC

=, ∵B B ∠=∠, ∴BPN BCP ??∽, ∴

1

2

PN BN PC BP ==, ∴1

2

PN PC =

; (2)解:如图,在BC 上取一点G ,使得BG=1,

∵24

2,212PB BC BG PB ====, ∴

,PB BC

PBG PBC BG PB

=∠=∠, ∴PBG CBP ??∽, ∴

1

2

PG BG PC PB ==, ∴1

2

PG PC =, ∴1

2

PD PC DP PG +

=+; ∵DP PG DG +≥, ∴当D 、P 、G 共线时,1

2

PD PC +

的值最小, ∴最小值为:22435DG =+=;

(3)如图,在BC 上取一点G ,使得BG=1,作DF ⊥BC 于F ,

与(2)同理,可证

1

2

PG PC

=,

在Rt△CDF中,∠DCF=60°,CD=4,

∴DF=CD?sin60°=23,CF=2,

在Rt△GDF中,DG=22

(23)537

+=,

1

2

PD PC PD PG DG -=-≤,

当点P在DG的延长线上时,

1

2

PD PC

-的值最大,

∴最大值为:37

DG=.

【点睛】

本题考查圆综合题、正方形的性质、菱形的性质、相似三角形的判定和性质、两点之间线段最短等知识,解题的关键是学会构建相似三角形解决问题,学会用转化的思想思考问题,把问题转化为两点之间线段最短解决,题目比较难,属于中考压轴题.

3.在直角坐标系中,⊙C过原点O,交x轴于点A(2,0),交y轴于点B(0,).(1)求圆心C的坐标.

(2)抛物线y=ax2+bx+c过O,A两点,且顶点在正比例函数y=-的图象上,求抛物线的解析式.

(3)过圆心C作平行于x轴的直线DE,交⊙C于D,E两点,试判断D,E两点是否在(2)中的抛物线上.

(4)若(2)中的抛物线上存在点P(x0,y0),满足∠APB为钝角,求x0的取值范围.

【答案】(1)圆心C的坐标为(1,);

(2)抛物线的解析式为y=x2﹣x;

(3)点D、E均在抛物线上;

(4)﹣1<x0<0,或2<x0<3.

【解析】

试题分析:(1)如图线段AB是圆C的直径,因为点A、B的坐标已知,根据平行线的性

质即可求得点C的坐标;

(2)因为抛物线过点A、O,所以可求得对称轴,即可求得与直线y=﹣x的交点,即是二次函数的顶点坐标,利用顶点式或者一般式,采用待定系数法即可求得抛物线的解析式;

(3)因为DE∥x轴,且过点C,所以可得D、E的纵坐标为,求得直径AB的长,可得D、E的横坐标,代入解析式即可判断;

(4)因为AB为直径,所以当抛物线上的点P在⊙C的内部时,满足∠APB为钝角,所以﹣1<x0<0,或2<x0<3.

试题分析:(1)∵⊙C经过原点O

∴AB为⊙C的直径

∴C为AB的中点

过点C作CH垂直x轴于点H,则有CH=OB=,OH=OA=1

∴圆心C的坐标为(1,).

(2)∵抛物线过O、A两点,

∴抛物线的对称轴为x=1,

∵抛物线的顶点在直线y=﹣x上,

∴顶点坐标为(1,﹣).

把这三点的坐标代入抛物线y=ax2+bx+c,得,

解得,

∴抛物线的解析式为y=x2﹣x.

(3)∵OA=2,OB=2,

∴AB==4,即⊙C的半径r=2,

∴D(3,),E(﹣1,),

代入y=x2﹣x检验,知点D、E均在抛物线上.

(4)∵AB为直径,

∴当抛物线上的点P在⊙C的内部时,满足∠APB为钝角,

∴﹣1<x0<0,或2<x0<3.

考点:二次函数综合题.

4.在△ABC中,∠A=90°,AB=4,AC=3,M是AB上的动点(不与A,B重合),过M点作MN∥BC交AC于点N.

(1)如图1,把△AMN沿直线MN折叠得到△PMN,设AM=x.

i.若点P正好在边BC上,求x的值;

ii.在M的运动过程中,记△MNP与梯形BCNM重合的面积为y,试求y关于x的函数关系式,并求y的最大值.

(2)如图2,以MN为直径作⊙O,并在⊙O内作内接矩形AMQN.试判断直线BC与⊙O的位置关系,并说明理由.

【答案】(1)i.当x=2时,点P恰好落在边BC上;ii. y=,

当x=时,重叠部分的面积最大,其值为2;(2)当x=时,⊙O与直线BC相切;当x<

时,⊙O与直线BC相离;x>时,⊙O与直线BC相交.

【解析】

试题分析:(1)i.根据轴对称的性质,可求得相等的线段与角,可得点M是AB中点,即当x=AB=2时,点P恰好落在边BC上;

ii.分两种情况讨论:①当0<x≤2时,△MNP与梯形BCNM重合的面积为△MNP的面积,根据轴对称的性质△MNP的面积等于△AMN的面积,易见y=x2

②当2<x<4时,如图2,设PM,PN分别交BC于E,F,由i.知ME=MB=4-x∴PE=PM-ME=x-(4-x)=2x-4,由题意知△PEF∽△ABC,利用相似三角形的性质即可求得.

(2)利用分类讨论的思想,先求的直线BC与⊙O相切时,x的值,然后得到相交,相离时x的取值范围.

试题解析:(1)i.如图1,

由轴对称性质知:AM=PM,∠AMN=∠PMN,

又MN∥BC,

∴∠PMN=∠BPM,∠AMN=∠B,

∴∠B=∠BPM,

∴AM=PM=BM,

∴点M是AB中点,即当x=AB=2时,点P恰好落在边BC上.

ii.以下分两种情况讨论:

①当0<x≤2时,

∵MN∥BC,

∴△AMN∽△ABC,

∴,

∴,

∴AN=,

△MNP与梯形BCNM重合的面积为△MNP的面积,∴,

②当2<x<4时,如图2,

设PM,PN分别交BC于E,F,

由(2)知ME=MB=4-x,

∴PE=PM-ME=x-(4-x)=2x-4,

由题意知△PEF∽△ABC,

∴,

∴S△PEF=(x-2)2,

∴y=S△PMN-S△PEF=,

∵当0<x≤2时,y=x2,

∴易知y最大=,

又∵当2<x<4时,y=,∴当x=时(符合2<x<4),y最大=2,

综上所述,当x=时,重叠部分的面积最大,其值为2.(2))如图3,

设直线BC与⊙O相切于点D,连接AO,OD,则AO=OD=MN.

在Rt△ABC中,BC==5;

由(1)知△AMN∽△ABC,

∴,即,

∴MN=x

∴OD=x,

过M点作MQ⊥BC于Q,则MQ=OD=x,

在Rt△BMQ与Rt△BCA中,∠B是公共角,

∴△BMQ∽△BCA,

∴,

∴BM=,AB=BM+MA=x+x=4

∴x=,

∴当x=时,⊙O与直线BC相切;

当x<时,⊙O与直线BC相离;

x>时,⊙O与直线BC相交.

考点:圆的综合题.

5.四边形ABCD内接于⊙O,连接AC、BD,2∠BDC+∠ADB=180°.

(1)如图1,求证:AC=BC;

(2)如图2,E为⊙O上一点,AE=BE,F为AC上一点,DE与BF相交于点T,连接

AT,若∠BFC=∠BDC+1

2

∠ABD,求证:AT平分∠DAB;

(3)在(2)的条件下,DT=TE,AD=8,BD=12,求DE的长.【答案】(1)见解析;(2)见解析;(3)2

【解析】

【分析】

(1)只要证明∠CAB=∠CBA即可.

(2)如图2中,作TH⊥AD于H,TR⊥BD于R,TL⊥AB于L.想办法证明TL=TH即可解决问题.

(3)如图3中,连接EA,EB,作EG⊥AB,TH⊥AD于H,TR⊥BD于R,TL⊥AB于L,AQ⊥BD于Q.证明△EAG≌△TDH(AAS),推出AG=DH,证明

Rt△TDR≌Rt△TDH(HL),推出DH=DR,同理可得AL=AH,BR=BL,设DH=x,则AB=2x,

由S△ADB=1

2

?BD?AQ=

1

2

?AD?h+

1

2

?AB?h+

1

2

?DB?h,可得AQ=

5

2

h,再根据

sin∠BDE=sin∠ADE,sin∠AED=sin∠ABD,构建方程组求出m即可解决问题.【详解】

解:(1)如图1中,

∵四边形ABCD内接于⊙O,

∴∠ADC+∠ABC=180°,

即∠ADB+∠BDC+∠ABC=180°,

∵2∠BDC+∠ADB=180°,

∴∠ABC=∠BDC,

∵∠BAC=∠BDC,

∴∠BAC=∠ABC,

∴AC=BC.

(2)如图2中,作TH⊥AD于H,TR⊥BD于R,TL⊥AB于L.

∵∠BFC=∠BAC+∠ABF,∠BAC=∠BDC,

∴∠BFC=∠BDC+∠ABF,

∵∠BFC=∠BDC+1

2

∠ABD,

∴∠ABF=1

2

∠ABD,

∴BT平分∠ABD,

∵AE=BE

∴∠ADE=∠BDE,

∴DT平分∠ADB,

∵TH⊥AD于H,TR⊥BD于R,TL⊥AB于L.

∴TR=TL,TR=TH,

∴TL=TH,

∴AT平分∠DAB.

(3)如图3中,连接EA,EB,作EG⊥AB,TH⊥AD于H,TR⊥BD于R,TL⊥AB于L,AQ⊥BD于Q.

∵AE=BE

∴∠EAB=∠EDB=∠EDA,AE=BE,

∵∠TAE=∠EAB+∠TAB,∠ATE=∠EDA+∠DAT,

∴∠TAE=∠ATE,

∴AE=TE,

∵DT=TE,

∴AE=DT,

∵∠AGE=∠DHT=90°,

∴△EAG≌△TDH(AAS),

∴AG=DH,

∵AE=EB,EG⊥AB,

∴AG=BG,

∴2DH=AB,

∵Rt△TDR≌Rt△TDH(HL),

∴DH=DR,同理可得AL=AH,BR=BL,

设DH=x,则AB=2x,

∵AD=8,DB=12,

∴AL=AH=8﹣x,BR=12﹣x,AB=2

x=8﹣x+12﹣x,

∴x=5,

∴DH=5,AB=10,

设TR=TL=TH=h,DT=m,

∵S△ADB=1

2

?BD?AQ=

1

2

?AD?h+

1

2

?AB?h+

1

2

?DB?h,

∴12AQ=(8+12+10)h,

∴AQ=5

2 h,

∵sin∠BDE=sin∠ADE,可得h

m

AP

AD

AP

8

sin∠AED=sin∠ABD,可得AP

m

AQ

AB

AQ

10

5

2

10

h

∴AP

m

5

28

10

mAP

解得m=42或﹣42(舍弃),

∴DE=2m=82.

【点睛】

本题属于圆综合题,考查了圆内接四边形的性质,圆周角定理,锐角三角函数,全等三角形的判定和性质,角平分线的性质定理和判定定理等知识,解题的关键是学会添加常用辅助线,学会利用参数构建方程组解决问题,属于中考压轴题.

6.我们把“有两条边和其中一边的对角对应相等的两个三角形”叫做“同族三角形”,如图1,在△ABC和△ABD中,AB=AB,AC=AD,∠B=∠B,则△ABC和△ABD是“同族三角形”.

(1)如图2,四边形ABCD内接于圆,点C是弧BD的中点,求证:△ABC和△ACD是同族三角形;

(2)如图3,△ABC内接于⊙O,⊙O的半径为32AB=6,∠BAC=30°,求AC的长;(3)如图3,在(2)的条件下,若点D在⊙O上,△ADC与△ABC是非全等的同族三角

形,AD>CD,求AD

CD

的值.

【答案】(1)详见解析;(2)33+3;(3)AD

CD

=

62

+

6

【解析】

【分析】

(1)由点C是弧BD的中点,根据弧与弦的关系,易得BC=CD,∠BAC=∠DAC,又由公共边AC,可证得:△ABC和△ACD是同族三角形;

(2)首先连接0A,OB,作点B作BE⊥AC于点E,易得△AOB是等腰直角三角形,继而求得答案;

(3)分别从当CD=CB时与当CD=AB时进行分析求解即可求得答案.

【详解】

(1)证明:∵点C是弧BD的中点,即BC CD

=,

∴BC=CD,∠BAC=∠DAC,

∵AC=AC,

∴△ABC和△ACD是同族三角形.

(2)解:如图1,连接OA,OB,作点B作BE⊥AC于点E,

∵2,AB=6,

∴OA2+OB2=AB2,

∴△AOB是等腰直角三角形,且∠AOB=90°,

∴∠C=∠AOB=45°,

∵∠BAC=30°,

∴BE=AB=3,

∴22

AB BE

-3,

∵CE=BE=3,

∴3

(3)解:∵∠B=180°﹣∠BAC﹣∠ACB=180°﹣30°﹣45°=105°,

∴∠ADC=180°﹣∠B=75°,

如图2,当CD=CB时,∠DAC=∠BAC=30°,

∴∠ACD=75°,

∴AD=AC=33+3,CD=BC=2BE=32, ∴

AD 333CD 32

+=

=62

2+; 如图3,当CD=AB 时,过点D 作DF ⊥AC ,交AC 于点F ,

则∠DAC=∠ACB=45°,

∴∠ACD=180°﹣∠DAC ﹣∠ADC=60°, ∴DF=CD?sin60°=6×

3

2

=33, ∴AD=2DF=36, ∴

AD 36CD =

=62

. 综上所述:AD CD =62

2+或62

. 【点睛】

本题考查圆的综合应用问题,综合运用弧与弦的关系,等腰三角形的性质结合图形作辅助线进行分析证明以及求解,难度较大.

7.已知ABD △内接于圆O ,点C 为弧BD 上一点,连接BC AC AC 、,交BD 于点E ,CED ABC ∠=∠.

(1)如图1,求证:弧AB =弧AD ;

(2)如图2,过B 作BF AC ⊥于点F ,交圆O 点G ,连接AG 交BD 于点H ,且

∠的度数;

222

EH BE DH

=+,求CAG

(3)如图3,在(2)的条件下,圆O上一点M与点C关于BD对称,连接ME,交

∥交AD于点Q,交BD的延长线于点R,AB于点N,点P为弧AD上一点,PQ BG

=,ANE的周长为20,52

AQ BN

DR=,求圆O半径.

【答案】(1)见解析;(2)∠CAG=45°;(3)r=62

【解析】

【分析】

(1)证∠ABD=∠ACB可得;

(2)如下图,△AHD绕点A旋转至△ALE处,使得点D与点B重合,证△ALE≌△AHE,利用勾股定理逆定理推导角度;

(3)如下图,延长QR交AB于点T,分别过点N、Q作BD的垂线,交于点V,I,取QU=AE,过点U作UK垂直BD.先证△AEN≌△QUD,再证△NVE≌△RKU,可得到

NV=KR=DK,进而求得OB的长.

【详解】

(1)∵∠CED是△BEC的外角,∴∠CED=∠EBC+∠BCA

∵∠ABC=∠ABD+∠EBC

又∵∠CED=∠ABC

∴∠ABD=∠ACB

∴弧AB=弧AD

(2)如下图,△AHD绕点A旋转至△ALE处,使得点D与点B重合

∵△ALB是△AHD旋转所得

∴∠ABL=∠ADB,AL=AH

设∠CAG=a,则∠CBG=a

∵BG⊥AC

∴∠BCA=90°-a,∴∠ADB=∠ABD=90°-a

∴在△BAD中,BAE+∠HAD=180-a-(90°-a)-(90°-a)=a

∴∠LAE=∠EAH=a

∵LA=AH,AE=AE

∴△ALE≌△AHE,∴LE=EH

∵HD=LB,222

=+

EH BE DH

∴△LBE为直角三角形

∴∠LBE=(90°-a)+(90°-a)=90°,解得:a=45°

∴∠CAG=45°

(3)如下图,延长QR交AB于点T,分别过点N、Q作BD的垂线,交于点V,I,取

QU=AE,过点U作UK垂直BD

由(2)得∠BAD=90°

∴点O在BD上

设∠R=n,则∠SER=∠BEC=∠MEB=90°-n

∴∠AEN=2n

∵SQ⊥AC

∴∠TAS=∠AQS=∠DQR,AN=QD

∵QU=AE

∴△AEN≌△QUD

∴∠QUD=∠AEN=2n

∴UD=UR=NE,

∵△ANE的周长为20

∴QD+QR=20

在△DQR中,QD=7

∵∠ENR=∠UDK=∠R=n

∴△NVE≌△RKU

52

∴BN=5

∴22r

【点睛】

本题考查了圆的证明,涉及到全等、旋转和勾股定理,解题关键是结合图形特点,适当构造全等三角形

8.如图,PA,PB分别与O相切于点A和点B,点C为弧AB上一点,连接PC并延长交O于点F,D为弧AF上的一点,连接BD交FC于点E,连接AD,且

2180APB PEB ∠+∠=?.

(1)如图1,求证://PF AD ;

(2)如图2,连接AE ,若90APB ∠=?,求证:PE 平分AEB ∠; (3)如图3,在(2)的条件下,连接AB 交PE 于点H ,连接OE ,8AD =,

4

sin 5

ABD ∠=

,求PH 的长. 【答案】(1)见解析;(2)见解析;(3)257

【解析】 【分析】

(1)连接OA 、OB ,由切线的性质可得90OAP OBP ∠=∠=?,由四边形内角和是

360?,得180∠+∠=?P AOB ,由同弧所对的圆心角是圆周角的一半,得到

2AOB ADB ∠=∠,等量代换得到ADB PEB ∠=∠,由同位角相等两直线平行,得到//PF AD ;

(2)过点P 做PK PF ⊥交EB 延长线于点K ,由90APB ∠=?得290PEB ∠=?,从而45PEB ∠=?,由切线的性质,得PA PB =,由PK PE ⊥,45PEK ∠=?,得

PE PK =,从而90APE EPB ?∠=-∠,进而APE BPK ∠=∠,即可证得

APE BPK ??≌由此45K AEP ∠=∠=?,得到AEP PEB ∠=∠,即可证得PE 平分AEB ∠;

(3)连接AO 并延长交圆O 于点M ,连接OB 、OH 、OP 、OD 、DM ,由

45ADE ∠=?,90AED ∠=?,可得DE AE =,由OA 、OD 为半径,可得OA OD =,即可证出DEO AEO ??≌,由直径所对的圆周角是直角,可得90ADM ∠=?,在

Rt ADM ?中,由正弦定义可得10AM =,由此5OA OB ==,由OAPB 为正方形,对

角线AB 垂直平分OP ,从而,OH PH =.在Rt OAP ?中,252OP OA =

=延长EO

交AD 于K ,在Rt OEP ?中,由勾股定理得7PE =,在Rt OEH ?中,由勾股定理得

257PH =

. 【详解】 (1)连接OA 、OB

∵PA 、PB 与圆O 相切于点A 、B ,且OA 、OB 为半径, ∴OA AP ⊥,OB BP ⊥, ∴90OAP OBP ∠=∠=?,

∴在四边形AOBP 中,360180180P AOB ∠+∠=?-?=?, ∵AB AB =, ∴2AOB ADB ∠=∠, ∴2180P ADB ∠+∠=?, ∵2180P PEB ∠+∠=?, ∴ADB PEB ∠=∠, ∴//PF AD

(2)过点P 做PK PF ⊥交EB 延长线于点K

∵90APB ∠=?,

∴21809090PEB ∠=?-?=?, ∴45PEB ∠=?,

∵PA 、PB 为圆O 的切线, ∴PA PB =,

∵PK PE ⊥,45PEK ∠=?, ∴PE PK = ,

∵9090APE EPB KPB EPB ??∠=-∠=∠=-∠, ∴APE BPK ∠=∠, ∴APE BPK ??≌, ∴45K AEP ∠=∠=?, ∴AEP PEB ∠=∠, ∴PE 平分AEB ∠;

(3)连接AO 并延长交圆O 于点M ,连接OB 、OH 、OP 、OD 、DM

∵45ADE ∠=?,90AED ∠=?, ∴DE AE =, ∵OA 、OD 为半径, ∴OA OD =, ∵OE OE =, ∴DEO AEO ??≌, ∴1

452

AEO OED AED ∠=∠=∠=?, ∴90OEP ∠=?, ∵AM 为圆O 的直径, ∴90ADM ∠=?, ∵弧AD =弧AD , ∴ABD AMD ∠=∠,

在Rt ADM ?中,8AD =,4

sin 5

AMD ∠=,则10AM =, ∴5OA OB ==,

由题易证四边形OAPB 为正方形, ∴对角线AB 垂直平分OP ,AB OP =, ∵H 在AB 上, ∴OH PH =, 在Rt OAP ?中,252OP OA ==

延长EO 交AD 于K ,

∵DE AE =,可证OK AD ⊥,DOK ABD ∠=∠, ∴4DK KE ==,3OK =,1OE = ∴在Rt OEP ?中,227PE OP OE =-= 在Rt OEH ?中,222OH OE EH =+ ∵OH PH =,7EH PE HP PH =-=- ∴()2

2217PH PH =+-

∴257

PH =

. 【点睛】

本题考查了圆的综合题,圆的性质,等腰三角形的性质,相交弦定理,正弦定理,勾股定理,灵活运用这些性质定理解决问题是本题的关键.

9.如图①②,在平面直角坐标系中,边长为2的等边CDE ?恰好与坐标系中的OAB ?重合,现将CDE ?绕边AB 的中点(G G 点也是DE 的中点),按顺时针方向旋转180?到△1C DE 的位置. (1)求1C 点的坐标;

(2)求经过三点O 、A 、1C 的抛物线的解析式; (3)如图③,

G 是以AB 为直径的圆,过B 点作G 的切线与x 轴相交于点F ,求切

线BF 的解析式;

(4)抛物线上是否存在一点M ,使得:16:3AMF OAB S S ??=.若存在,请求出点M 的坐标;若不存在,请说明理由.

【答案】(1)13)C ;(2)23333y x x =

-;(3)323

33

y x =+

;(4)1283834,,2,33M M ???- ? ? ????

. 【解析】 【分析】

(1)利用中心对称图形的性质和等边三角形的性质,可以求出. (2)运用待定系数法,代入二次函数解析式,即可求出.

(3)借助切线的性质定理,直角三角形的性质,求出F ,B 的坐标即可求出解析式. (4)当M 在x 轴上方或下方,分两种情况讨论. 【详解】

解:(1)将等边CDE ?绕边AB 的中点G 按顺时针方向旋转180?到△1C DE , 则有,四边形'OAC B 是菱形,所以1C 的横坐标为3, 根据等边CDE ?的边长是2, 利用等边三角形的性质可得13)C ; (2)

抛物线过原点(0,0)O ,设抛物线解析式为2y ax bx =+,

把(2,0)A

,C '

代入,得420

93a b a b +=???+=??

解得3

a =

,b =

抛物线解析式为2y x x =

-;

(3)90ABF ∠=?,60BAF ∠=?,

30AFB ∴∠=?, 又2AB =, 4AF ∴=, 2OF ∴=,

(2,0)F ∴-,

设直线BF 的解析式为y kx b =+,

把B ,(2,0)F -

代入,得20

k b k b ?+=??

-+=??,

解得k =

b = ∴直线BF

的解析式为33

y x =

+

(4)①当M 在x

轴上方时,存在2()M x ,

211

:[4)]:[216:322

AMF OAB S S ??=???=,

得2280x x --=,解得14x =,22x =-, 当14x =

时,244y , 当12x =-

时,2(2)(2)y =--=

1M ∴

,2(M -; ②当M 在x

轴下方时,不存在,设点2()M x x ,

211

:[4)]:[216:322

AMF OAB S S ??=-???=,

得2280x x -+=,240b ac -<无解,

上宝中学2016学年初三第一学期英语期中复习试卷(三)

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