28. 解:(1)∵D(-8,0),∴B点的横坐标为-8,代入中,得y=-2.
∴B点坐标为(-8,-2).而A、B两点关于原点对称,∴A(8,2)
从而k=8×2=16
(2)∵N(0,-n),B是CD的中点,A,B,M,E四点均在双曲线上,
∴mn=k,B(-2m,-),C(-2m,-n),E(-m,-n)
=2mn=2k,
=
mn=
k,
=
mn=
k.
∴=
―
―
=k.∴k=4.
由直线及双曲线
,得A(4,1),B(-4,-1)
∴C(-4,-2),M(2,2)
设直线CM的解析式是,由C、M两点在这条直线上,得
,解得a=b=
∴直线CM的解析式是y=x+
.
(3)如图,分别作AA1⊥x轴,MM1⊥x轴,垂足分别为A1,M1
设A点的横坐标为a,则B点的横坐标为-a.于是,
同理
∴p-q=-
=-2
27. 解:(1)由题意:BP=tcm,AQ=2tcm,则CQ=(4-2t)cm,
∵∠C=90°,AC=4cm,BC=3cm,∴AB=5cm
∴AP=(5-t)cm,
∵PQ∥BC,∴△APQ∽△ABC,
∴AP∶AB=AQ∶AC,即(5-t)∶5=2t∶4,解得:t=
∴当t为秒时,PQ∥BC
………………2分
(2)过点Q作QD⊥AB于点D,则易证△AQD∽△ABC
∴AQ∶QD=AB∶BC
∴2t∶DQ=5∶3,∴DQ=
∴△APQ的面积:×AP×QD=
(5-t)×
∴y与t之间的函数关系式为:y=
………………5分
(3)由题意:
当面积被平分时有:=
×
×3×4,解得:t=
当周长被平分时:(5-t)+2t=t+(4-2t)+3,解得:t=1
∴不存在这样t的值
………………8分
(4)过点P作PE⊥BC于E
易证:△PAE∽△ABC,当PE=QC时,△PQC为等腰三角形,此时△QCP′为菱形
∵△PAE∽△ABC,∴PE∶PB=AC∶AB,∴PE∶t=4∶5,解得:PE=
∵QC=4-2t,∴2×=4-2t,解得:t=
∴当t=时,四边形PQP′C为菱形
此时,PE=,BE=
,∴CE=
………………10分
在Rt△CPE中,根据勾股定理可知:PC==
=
∴此菱形的边长为cm ………………12分
26. 解:方案一:由题意可得:,
点
到甲村的最短距离为
.······································································· (1分)
点
到乙村的最短距离为
.
将供水站建在点
处时,管道沿
铁路建设的长度之和最小.
即最小值为.········································································ (3分)
方案二:如图①,作点关于射线
的对称点
,则
,连接
交
于点
,则
.
,
.·········································································· (4分)
在中,
,
,
,
两点重合.即
过
点.············································· (6分)
在线段上任取一点
,连接
,则
.
,
把供水站建在乙村的
点处,管道沿
线路铺设的长度之和最小.
即最小值为
.··········· (7分)
方案三:作点关于射线
的对称点
,连接
,则
.
作于点
,交
于点
,交
于点
,
为点
到
的最短距离,即
.
在中,
,
,
.
.
,
两点重合,即
过
点.
在中,
,
.············································· (10分)
在线段上任取一点
,过
作
于点
,连接
.
显然.
把供水站建在甲村的
处,管道沿
线路铺设的长度之和最小.
即最小值为.································································ (11分)
综上,,
供水站建在
处,所需铺设的管道长度最短.········ (12分)
25. 解:(1)取中点
,联结
,
为
的中点,
,
.································· (1分)
又,
.··········································································· (1分)
,得
;······································ (2分)(1分)
(2)由已知得.··································································· (1分)
以线段
为直径的圆与以线段
为直径的圆外切,
,即
.·························· (2分)
解得,即线段
的长为
;······································································· (1分)
(3)由已知,以为顶点的三角形与
相似,
又易证得.··············································································· (1分)
由此可知,另一对对应角相等有两种情况:①;②
.
①当时,
,
.
.
,易得
.得
;······················································· (2分)
②当时,
,
.
.又
,
.
,即
,得
.
解得,
(舍去).即线段
的长为2.········································ (2分)
综上所述,所求线段的长为8或2.
24. 解:(1)∵点在
上,
∴,
∴,
∴.
(2)连结, 由题意易知
,
∴.
(3)正方形AEFG在绕A点旋转的过程中,F点的轨迹是以点A为圆心,AF为半径的圆.
第一种情况:当b>2a时,存在最大值及最小值;
因为的边
,故当F点到BD的距离取得最大、最小值时,
取得最大、最小值.
如图②所示时,
的最大值=
的最小值=
第二种情况:当b=2a时,存在最大值,不存在最小值;
的最大值=
.(如果答案为4a2或b2也可)
23. 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
22. 解:( 1)由已知得:解得
c=3,b=2
∴抛物线的线的解析式为
(2)由顶点坐标公式得顶点坐标为(1,4)
所以对称轴为x=1,A,E关于x=1对称,所以E(3,0)
设对称轴与x轴的交点为F
所以四边形ABDE的面积=
=
=
=9
(3)相似
如图,BD=
BE=
DE=
所以,
即:
,所以
是直角三角形
所以,且
,
所以.
21.解:
(1)m=-5,n=-3
(2)y=x+2
(3)是定值.
因为点D为∠ACB的平分线,所以可设点D到边AC,BC的距离均为h,
设△ABC AB边上的高为H,
则利用面积法可得:
(CM+CN)h=MN﹒H
又 H=
化简可得 (CM+CN)﹒
故
20. 解:(1)如图,过点B作BD⊥OA于点D.
在Rt△ABD中,
∵∣AB∣=,sin∠OAB=
,
∴∣BD∣=∣AB∣·sin∠OAB
=×
=3.
又由勾股定理,得
∴∣OD∣=∣OA∣-∣AD∣=10-6=4.
∵点B在第一象限,∴点B的坐标为(4,3). ……3分
设经过O(0,0)、C(4,-3)、A(10,0)三点的抛物线的函数表达式为
y=ax2+bx(a≠0).
由
∴经过O、C、A三点的抛物线的函数表达式为
……2分
(2)假设在(1)中的抛物线上存在点P,使以P、O、C、A为顶点的四边形为梯形
①∵点C(4,-3)不是抛物线的顶点,
∴过点C做直线OA的平行线与抛物线交于点P1 .
则直线CP1的函数表达式为y=-3.
对于,令y=-3
x=4或x=6.
∴
而点C(4,-3),∴P1(6,-3).
在四边形P1AOC中,CP1∥OA,显然∣CP1∣≠∣OA∣.
∴点P1(6,-3)是符合要求的点. ……1分
②若AP2∥CO.设直线CO的函数表达式为
将点C(4,-3)代入,得
∴直线CO的函数表达式为
于是可设直线AP2的函数表达式为
将点A(10,0)代入,得
∴直线AP2的函数表达式为
由,即(x-10)(x+6)=0.
∴
而点A(10,0),∴P2(-6,12).
过点P2作P2E⊥x轴于点E,则∣P2E∣=12.
在Rt△AP2E中,由勾股定理,得
而∣CO∣=∣OB∣=5.
∴在四边形P2OCA中,AP2∥CO,但∣AP2∣≠∣CO∣.
∴点P2(-6,12)是符合要求的点. ……1分
③若OP3∥CA,设直线CA的函数表达式为y=k2x+b2
将点A(10,0)、C(4,-3)代入,得
∴直线CA的函数表达式为
∴直线OP3的函数表达式为
由即x(x-14)=0.
∴
而点O(0,0),∴P3(14,7).
过点P3作P3E⊥x轴于点E,则∣P3E∣=7.
在Rt△OP3E中,由勾股定理,得
而∣CA∣=∣AB∣=.
∴在四边形P3OCA中,OP3∥CA,但∣OP3∣≠∣CA∣.
∴点P3(14,7)是符合要求的点. ……1分
综上可知,在(1)中的抛物线上存在点P1(6,-3)、P2(-6,12)、P3(14,7),
使以P、O、C、A为顶点的四边形为梯形. ……1分
(3)由题知,抛物线的开口可能向上,也可能向下.
①当抛物线开口向上时,则此抛物线与y轴的副半轴交与点N.
可设抛物线的函数表达式为
(a>0).
即
如图,过点M作MG⊥x轴于点G.
∵Q(-2k,0)、R(5k,0)、G(、N(0,-10ak2)、M
∴
∴
……2分
②当抛物线开口向下时,则此抛物线与y轴的正半轴交于点N,
同理,可得
……1分
综上所知,的值为3:20.
……1分
19. 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
.
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