0  433111  433119  433125  433129  433135  433137  433141  433147  433149  433155  433161  433165  433167  433171  433177  433179  433185  433189  433191  433195  433197  433201  433203  433205  433206  433207  433209  433210  433211  433213  433215  433219  433221  433225  433227  433231  433237  433239  433245  433249  433251  433255  433261  433267  433269  433275  433279  433281  433287  433291  433297  433305  447090 

18. 解:(1)点轴上························································································· 1分

理由如下:

连接,如图所示,在中,

由题意可知:

轴上,轴上.············································································ 3分

(2)过点轴于点

中,

在第一象限,

的坐标为····························································································· 5分

由(1)知,点轴的正半轴上

的坐标为

的坐标为······························································································· 6分

抛物线经过点

由题意,将代入中得

  解得

所求抛物线表达式为:·························································· 9分

(3)存在符合条件的点,点.············································································ 10分

理由如下:矩形的面积

为顶点的平行四边形面积为

由题意可知为此平行四边形一边,

边上的高为2······································································································ 11分

依题意设点的坐标为

在抛物线

解得,

为顶点的四边形是平行四边形,

当点的坐标为时,

的坐标分别为

当点的坐标为时,

的坐标分别为.·················································· 14分

(以上答案仅供参考,如有其它做法,可参照给分)

试题详情

17. 解:(1)直线轴交于点,与轴交于点

······························································································· 1分

都在抛物线上,

 

抛物线的解析式为······························································ 3分

顶点····································································································· 4分

(2)存在····················································································································· 5分

··················································································································· 7分

·················································································································· 9分

(3)存在···················································································································· 10分

理由:

解法一:

延长到点,使,连接交直线于点,则点就是所求的点.

            ··························································································· 11分

过点于点

点在抛物线上,

中,

中,

····················································· 12分

设直线的解析式为

  解得

······································································································ 13分

  解得 

在直线上存在点,使得的周长最小,此时.········· 14分

解法二:

过点的垂线交轴于点,则点为点关于直线的对称点.连接于点,则点即为所求.···················································································· 11分

过点轴于点,则

同方法一可求得

中,,可求得

为线段的垂直平分线,可证得为等边三角形,

垂直平分

即点为点关于的对称点.··················································· 12分

设直线的解析式为,由题意得

  解得

······································································································ 13分

  解得 

在直线上存在点,使得的周长最小,此时.   1

试题详情

16.

解:(1)

(2)当时,过点作,交,如图1,

(3)①能与平行.

,如图2,则

,而

不能与垂直.

,延长,如图3,

,而

不存在.

试题详情

15. 解:(1)解法1:根据题意可得:A(-1,0),B(3,0);

则设抛物线的解析式为(a≠0)

又点D(0,-3)在抛物线上,∴a(0+1)(0-3)=-3,解之得:a=1

 ∴y=x2-2x-3···································································································· 3分

自变量范围:-1≤x≤3···················································································· 4分

      解法2:设抛物线的解析式为(a≠0)

          根据题意可知,A(-1,0),B(3,0),D(0,-3)三点都在抛物线上

          ∴,解之得:

y=x2-2x-3··············································································· 3分

自变量范围:-1≤x≤3······························································ 4分

      (2)设经过点C“蛋圆”的切线CEx轴于点E,连结CM

       在RtMOC中,∵OM=1,CM=2,∴∠CMO=60°,OC=

       在RtMCE中,∵OC=2,∠CMO=60°,∴ME=4

 ∴点CE的坐标分别为(0,),(-3,0) ·················································· 6分

∴切线CE的解析式为··························································· 8分

(3)设过点D(0,-3),“蛋圆”切线的解析式为:y=kx-3(k≠0) ·························· 9分

        由题意可知方程组只有一组解

  即有两个相等实根,∴k=-2············································· 11分

  ∴过点D“蛋圆”切线的解析式y=-2x-3····················································· 12分

试题详情

14. 解:(1)由题意可知,

解,得 m=3.     ………………………………3分

A(3,4),B(6,2);

k=4×3=12.    ……………………………4分

(2)存在两种情况,如图: 

①当M点在x轴的正半轴上,N点在y轴的正半轴

上时,设M1点坐标为(x1,0),N1点坐标为(0,y1).

∵ 四边形AN1M1B为平行四边形,

∴ 线段N1M1可看作由线段AB向左平移3个单位,

再向下平移2个单位得到的(也可看作向下平移2个单位,再向左平移3个单位得到的).

由(1)知A点坐标为(3,4),B点坐标为(6,2),

N1点坐标为(0,4-2),即N1(0,2);    ………………………………5分

M1点坐标为(6-3,0),即M1(3,0).    ………………………………6分

设直线M1N1的函数表达式为,把x=3,y=0代入,解得

∴ 直线M1N1的函数表达式为. ……………………………………8分

②当M点在x轴的负半轴上,N点在y轴的负半轴上时,设M2点坐标为(x2,0),N2点坐标为(0,y2). 

ABN1M1ABM2N2ABN1M1ABM2N2

N1M1M2N2N1M1M2N2.  

∴ 线段M2N2与线段N1M1关于原点O成中心对称.   

M2点坐标为(-3,0),N2点坐标为(0,-2).   ………………………9分

设直线M2N2的函数表达式为,把x=-3,y=0代入,解得

∴ 直线M2N2的函数表达式为.   

所以,直线MN的函数表达式为.  ………………11分

(3)选做题:(9,2),(4,5).  ………………………………………………2分

试题详情

13. 解:(1)分别过DC两点作DGAB于点GCHAB于点H. ……………1分

ABCD, 

DGCHDGCH. 

∴ 四边形DGHC为矩形,GHCD=1. 

DGCHADBC,∠AGD=∠BHC=90°,

∴ △AGD≌△BHC(HL). 

AGBH=3.  ………2分

∵ 在Rt△AGD中,AG=3,AD=5, 

DG=4.                

.    ………………………………………………3分

(2)∵ MNABMEABNFAB, 

MENFMENF. 

∴ 四边形MEFN为矩形. 

ABCDADBC,  

∴ ∠A=∠B. 

MENF,∠MEA=∠NFB=90°,  

∴ △MEA≌△NFB(AAS).

AEBF.     ……………………4分 

AEx,则EF=7-2x.  ……………5分 

∵ ∠A=∠A,∠MEA=∠DGA=90°,  

∴ △MEA∽△DGA

ME.     …………………………………………………………6分

.  ……………………8分

x时,ME<4,∴四边形MEFN面积的最大值为.……………9分

(3)能.   ……………………………………………………………………10分

由(2)可知,设AEx,则EF=7-2xME. 

若四边形MEFN为正方形,则MEEF. 

   即 7-2x.解,得 .  ……………………………………………11分

EF<4. 

∴ 四边形MEFN能为正方形,其面积为

试题详情

12. 解:(1).····················································································· 3分

(2)相等,比值为.················· 5分(无“相等”不扣分有“相等”,比值错给1分)

(3)设

在矩形中,

.···································································································· 6分

同理

.······························································································· 7分

,······························································································ 8分

解得

.······································································································ 9分

(4),·············································································································· 10分

.  12分

试题详情

11. 解:(1)设地经杭州湾跨海大桥到宁波港的路程为千米,

由题意得,································································································ 2分

解得

地经杭州湾跨海大桥到宁波港的路程为180千米.················································· 4分

(2)(元),

该车货物从地经杭州湾跨海大桥到宁波港的运输费用为380元.···························· 6分

(3)设这批货物有车,

由题意得,···························································· 8分

整理得

解得(不合题意,舍去),································································ 9分

这批货物有8车.···································································································· 10分

试题详情


同步练习册答案
关 闭