18. 解:(1)点在
轴上························································································· 1分
理由如下:
连接,如图所示,在
中,
,
,
,
由题意可知:
点
在
轴上,
点
在
轴上.············································································ 3分
(2)过点作
轴于点
,
在
中,
,
点
在第一象限,
点
的坐标为
····························································································· 5分
由(1)知,点
在
轴的正半轴上
点
的坐标为
点
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
以
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,
又
边上的高为2······································································································ 11分
依题意设点的坐标为
点
在抛物线
上
解得,,
,
以
为顶点的四边形是平行四边形,
,
,
当点
的坐标为
时,
点的坐标分别为
,
;
当点的坐标为
时,
点的坐标分别为
,
.·················································· 14分
(以上答案仅供参考,如有其它做法,可参照给分)
17. 解:(1)直线
与
轴交于点
,与
轴交于点
.
,
······························································································· 1分
点
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
作
于点
.
点在抛物线
上,
在中,
,
,
,
在中,
,
,
,
····················································· 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
作
的垂线交
轴于点
,则点
为点
关于直线
的对称点.连接
交
于点
,则点
即为所求.···················································································· 11分
过点作
轴于点
,则
,
.
,
同方法一可求得.
在中,
,
,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
16.
解:(1),
.
(2)当时,过
点作
,交
于
,如图1,
则,
,
,
.
(3)①能与
平行.
若,如图2,则
,
即,
,而
,
.
②不能与
垂直.
若,延长
交
于
,如图3,
则.
.
.
又,
,
,
,而
,
不存在.
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CE交x轴于点E,连结CM,
在Rt△MOC中,∵OM=1,CM=2,∴∠CMO=60°,OC=
在Rt△MCE中,∵OC=2,∠CMO=60°,∴ME=4
∴点C、E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
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).
∵ AB∥N1M1,AB∥M2N2,AB=N1M1,AB=M2N2,
∴ N1M1∥M2N2,N1M1=M2N2.
∴ 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的函数表达式为或
. ………………11分
(3)选做题:(9,2),(4,5). ………………………………………………2分
13. 解:(1)分别过D,C两点作DG⊥AB于点G,CH⊥AB于点H. ……………1分
∵ AB∥CD,
∴ DG=CH,DG∥CH.
∴ 四边形DGHC为矩形,GH=CD=1.
∵ DG=CH,AD=BC,∠AGD=∠BHC=90°,
∴ △AGD≌△BHC(HL).
∴ AG=BH==3. ………2分
∵ 在Rt△AGD中,AG=3,AD=5,
∴ DG=4.
∴ . ………………………………………………3分
(2)∵ MN∥AB,ME⊥AB,NF⊥AB,
∴ ME=NF,ME∥NF.
∴ 四边形MEFN为矩形.
∵ AB∥CD,AD=BC,
∴ ∠A=∠B.
∵ ME=NF,∠MEA=∠NFB=90°,
∴ △MEA≌△NFB(AAS).
∴ AE=BF. ……………………4分
设AE=x,则EF=7-2x. ……………5分
∵ ∠A=∠A,∠MEA=∠DGA=90°,
∴ △MEA∽△DGA.
∴ .
∴ ME=.
…………………………………………………………6分
∴ . ……………………8分
当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AE=x,则EF=7-2x,ME=.
若四边形MEFN为正方形,则ME=EF.
即 7-2x.解,得
. ……………………………………………11分
∴ EF=<4.
∴ 四边形MEFN能为正方形,其面积为.
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.
即.······································································································ 9分
(4),·············································································································· 10分
. 12分
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
10.
9.
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