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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

试题详情

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

试题详情

26. 解:方案一:由题意可得:

到甲村的最短距离为.······································································· (1分)

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,连接于点,则

.·········································································· (4分)

中,

两点重合.即点.············································· (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

于点,交于点,交于点

为点的最短距离,即

中,

两点重合,即点.

中,.············································· (10分)

在线段上任取一点,过于点,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)

试题详情

25. 解:(1)取中点,联结

的中点,.································· (1分)

.··········································································· (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.

试题详情

24. 解:(1)∵点上,

.

(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4a2b2也可)

 

试题详情

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

(Ⅲ)对于二次函数

由已知时,时,

,∴

于是.而,∴,即

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

∵关于的一元二次方程的判别式

,   

∴抛物线轴有两个公共点,顶点在轴下方.····························· 8分

又该抛物线的对称轴

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

试题详情

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.

试题详情

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)﹒

    

试题详情

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).

∴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=-3x=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

,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

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).

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

试题详情

19. 解:(1)在中,令

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

的解析式为···················································································· 2分

(2)由,得  ·························································· 4分

····································································································· 5分

······························································································· 6分

(3)过点于点

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

················································································································ 8分

由直线可得:

中,,则

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

·························································································· 10分

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

此抛物线开口向下,时,

当点运动2秒时,的面积达到最大,最大为.   

试题详情


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