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如图,过点M作MG⊥x轴于点G. ∵Q(-2k,0)、R(5k,0)、G(?∴QO?2k,QR?7k,OG??S?QNR?12?QR?ON?1232?349??322?k,0?、N(0,-10ak)、M?k,?ak?,
4?2??2?72k,ON?10ak,MG?32k,QG?2494ak.
2?7k?10ak?35ak.
??121212?QO?ON?212(ON?GM)?OG?1?(10ak??7?3212?QG?GM32k?12?72k?494ak
2?2k?10ak?(29?15?3?21434943 ?2498ak)?2498)ak.
∴S?QNM:S?QNR?(ak):(35ak)?3:20. ??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S?QNM:S?QNR?3:20. ??1分 综上所知,S?QNM:S?QNR的值为3:20. ??1分 21.解:
(1)m=-5,n=-3 (2)y=
43x+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?h2?CN?h2?MN?H2
(CM+CN)h=MN﹒H
CM?CNH?MNh
又 H=
CM?CNMN
MNCM?CN?1h化简可得 (CM+CN)﹒故
1CM?1CN?1h
y
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22. 解:( 1)由已知得:?c=3,b=2
?c???1?b?c?0解得
∴抛物线的线的解析式为y??x2?2x?3 (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=S?ABO?S梯形BOFD?S?DFE ==
1212AO?BO??1?3?1212(BO?DF)?OF?12?2?4
12EF?DF
(3?4)?1?=9 (3)相似 如图,BD=BE=DE=
2BG?DG222?21?1?222 BO?OEDF?EF2??3?3?32 2?4?25 2222所以BD2?BE2?20, DE2?20即: BD2?BE2?DE2,所以?BDE是直角三角形
AOBDBOBE22所以?AOB??DBE?90?,且所以?AOB??DBE.
??,
23. 解(Ⅰ)当a?b?1,c??1时,抛物线为y?3x2?2x?1, 方程3x2?2x?1?0的两个根为x1??1,x2??113.
??0?和?,∴该抛物线与x轴公共点的坐标是??1,··············································· 2分 0?. ·
?3(Ⅱ)当a?b?1时,抛物线为y?3x2?2x?c,且与x轴有公共点.
对于方程3x2?2x?c?0,判别式??4?12c≥0,有c≤. ······································· 3分
31①当c?13时,由方程3x2?2x?113?0,解得x1?x2???13.
?此时抛物线为y?3x2?2x?与x轴只有一个公共点??,································ 4分 0?. ·
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13②当c?时,
, .
13x1??1时,y1?3?2?c?1?cx2?1时,y2?3?2?c?5?c由已知?1?x?1时,该抛物线与x轴有且只有一个公共点,考虑其对称轴为x???y1≤0,?1?c≤0,应有? 即?
?5?c?0.?y2?0.,
解得?5?c≤?1. 综上,c?13或?5?c≤?1. ··············································································· 6分
(Ⅲ)对于二次函数y?3ax2?2bx?c,
由已知x1?0时,y1?c?0;x2?1时,y2?3a?2b?c?0, 又a?b?c?0,∴3a?2b?c?(a?b?c)?2a?b?2a?b. 于是2a?b?0.而b??a?c,∴2a?a?c?0,即a?c?0.
∴a?c?0. ············································································································· 7分 ∵关于x的一元二次方程3ax2?2bx?c?0的判别式
??4b2?12ac?4(a?c)2?12ac?4[(a?c)2?ac]?0,
∴抛物线y?3ax2?2bx?c与x轴有两个公共点,顶点在x轴下方.······························ 8分 又该抛物线的对称轴x??b3a,
y 由a?b?c?0,c?0,2a?b?0, 得?2a?b??a, ∴??31b3a?23O 1 x .
又由已知x1?0时,y1?0;x2?1时,y2?0,观察图象,
可知在0?x?1范围内,该抛物线与x轴有两个公共点. ············································10分
24. 解:(1)∵点F在AD上, ∴AF?2a,
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∴DF?b?2a, ∴S△DBF?12DFAB?12×(b?2a)×b?12b?222ab.
(2)连结AF, 由题意易知AF∥BD, ∴S△DBF?S△ABD?12b.
2(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△BFD的边BD?得最大、最小值.
如图②所示CF2?BD时,
S△BFD的最大值=S△BF?1212?2b?????2b????2b22b2??b2?2ab2a??, ?2??b2?2ab2a??, ?2?2b,故当F点到BD的距离取得最大、最小值时,S△BFD取
2DS△BFD的最小值=S△BF2D??第二种情况:当b=2a时,存在最大值,不存在最小值;
S△BFD的最大值=
b?2ab22.(如果答案为4a2或b2也可)
F D O C E F1 B G A F2 25. 解:(1)取AB中点H,联结MH,
?M为DE的中点,?MH∥BE,MH?12(BE?AD). ································ (1分)
又?AB?BE,?MH?AB. ··········································································· (1分)
?S△ABM?12AB?MH,得y?212x?2(x?0); ······································ (2分)(1分)
2(2)由已知得DE?(x?4)?2. ·································································· (1分)
?以线段AB为直径的圆与以线段DE为直径的圆外切,
111122?MH?AB?DE,即(x?4)??2?(4?x)?2?. ·························· (2分)
?2222?44解得x?,即线段BE的长为; ······································································ (1分)
33(3)由已知,以A,N,D为顶点的三角形与△BME相似,
又易证得?DAM??EBM. ··············································································· (1分)
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由此可知,另一对对应角相等有两种情况:①?ADN??BEM;②?ADB??BME. ①当?ADN??BEM时,?AD∥BE,??ADN??DBE.??DBE??BEM.
····················································· (2分) ?DB?DE,易得BE?2AD.得BE?8; ·
②当?ADB??BME时,?AD∥BE,??ADB??DBE. ??DBE??BME.又?BED??MEB,?△BED∽△MEB.
?DEBE?BEEM,即BE2?EM?DE,得x2?122?(x?4)?2?(x?4).
2222解得x1?2,x2??10(舍去).即线段BE的长为2. ······································· (2分) 综上所述,所求线段BE的长为8或2.
26. 解:方案一:由题意可得:MB?OB,
······································································ (1分) ?点M到甲村的最短距离为MB. ·
?点M到乙村的最短距离为MD.
?将供水站建在点M处时,管道沿MD,MB铁路建设的长度之和最小.
即最小值为MB?MD?3?23. ······································································· (3分) 方案二:如图①,作点M关于射线OE的对称点M?,则MM??2ME,连接AM?交OE∥于点P,则PE 12?AM?2BM?6,?PE?3.·········································································· (4分)
AM.
在Rt△DME中,
?DE?DM?sin60?23??32?3,ME?12DM?12?23?3,
··········································· (6分) ?PE?DE,?P,D两点重合.即AM?过D点. ·
在线段CD上任取一点P?,连接P?A,P?M,P?M?,则P?M?P?M?. ?AP??P?M??AM?,
?把供水站建在乙村的D点处,管道沿DA,DM线路铺设的长度之和最小.
即最小值为AD?DM?AM?? O
30 ?AM22?MM??6?(23)?43. ·········· (7分)
M? 22北 东 B 甲村 A P C P? D M E
M?
F
G?G A O 30 ?B F N H M E C N? D (第25题答案图①) (第25题答案图②)
方案三:作点M关于射线OF的对称点M?,连接GM,则GM??GM.
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2011中考数学专题复习 - 压轴题(含答案)
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