发布时间 : 星期六 文章2009年全国中考数学压轴题精选精析(福建)更新完毕开始阅读42e7b1b43c1ec5da51e270d5
?CF2?CE2?EF2?12?22?5
······································································································ 6分 ?CF?5 ·在Rt△DEF中,DE?4,EF?2
?DF2?DE2?EF2?42?22?20
?DF?25
由(1)得C??1,?1?,D?4,?1?
?CD?5
?CD2?52?25
?CF2?DF2?CD2 ····················································································· 7分 ??CFD?90°
······························································································· 8分 ?CF⊥DF ·
55?3?方法二:由 (1)知AF?1????,AC?
44?4??AF?AC ································································································· 6分
同理:BF?BD[来源:Z|xx|k.Com]
2??ACF??AFC ?AC∥EF
??ACF??CFO
??AFC??CFO ························································································ 7分
同理:?BFD??OFD
??CFD??OFC??OFD?90°
即CF⊥DF ································································································ 8分 (3)存在.
解:如图3,作PM⊥x轴,垂足为点M ·········· 9分 又?PQ⊥OP[来源:学科网]
y P ?Rt△OPM∽Rt△OQP
F O C E M l D 图3
Q x PMOM?? PQOP?PQPM? ············································ 10分 OPOM2设P?x,x??x?0?,则PM???14??12x,OM?x 4①当Rt△QPO∽Rt△CFD时,
PQCF51················································································· 11分 ??? ·
OPDF252
12xPM41??? OMx2解得x?2
·································································································· 12分 ?P,? ·1?21②当Rt△OPQ∽Rt△CFD时,
PQDF25················································································· 13分 ???2 ·
OPCF512xPM4???2 OMx解得x?8
?P2?816,?
综上,存在点P···························· 14分 ,,?使得△OPQ与△CDF相似. ·?、P2?8161?21
4.(09年福建泉州)28.(13分)在直角坐标系中,点A(5,0)关于原点O的对称点为点C.
(1)请直接写出点C的坐标;
(2)若点B在第一象限内,∠OAB=∠OBA,并且点B关于原点O的对称点为点D. ①试判断四边形ABCD的形状,并说明理由;
②现有一动点P从B点出发,沿路线BA—AD以每秒1个单位长的速度向终点D运动,另一动点Q从A点同时出发,沿AC方向以每秒0.4个单位长的速度向终点C运动,当其中一个动点到达终点时,另一个动点也随之停止运动.已知AB=6,设点P、Q的运动时间为t秒,在运动过程中,当动点Q在以PA为直径的圆上时,试求t的值. (09年福建泉州28题解析)28.(本小题13分) 解:(1)C(-5,0)………………………(3分) (2)①四边形ABCD为矩形,理由如下:
如图,由已知可得:A、O、C在同一直线上,且 且
OB=OD,∴四边形
ABCD
是平行四边
OA=OC;B、O、D在同一直线上,
形.…………………………………(5分)
∵∠OAB=∠OBA∴OA=OB,即AC=2OA=2OB=BD ∴四边形ABCD是矩形.…………………(7分) ②如图,由①得四边形ABCD是矩形 ∴∠CBA=∠ADC=90°……………(8分) 又AB=CD=6,AC=10 ∴由勾股定理,得BC=AD= =∵
AC2?AB2?102?62=8…………………………………(9分)
106?8?25,?14,∴0≤t≤14.……………………(10分) 0.41当0≤t≤6时,P点在AB上,连结PQ.
∵AP是直径,∴∠PQA=90°…………………………………(11分)[来源:Z,xx,k.Com] 又∠PAQ=∠CAB,∴△PAQ∽△CAB ∴
PAAQ6?t0.4t??,即,解得t=3.6…………………………(12分) CAAB106
当6<t≤14时,P点在AD上,连结PQ, 同理得∠PQA=90°,△PAQ∽△CAD ∴
PAAQt?60.4t??,即t-6,解得t=12. CAAD108综上所述,当动点Q在以PA为直径的圆上时,t的值为3.6或12.……(13分)
5.(09年福建厦门)26.(11分)已知二次函数y=x2-x+c.
(1)若点A(-1,a)、B(2,2n-1)在二次函数y=x2-x+c的图象上,求此二次函数的最小值;
(2)若点D(x1,y1)、E(x2,y2)、P(m,n)(m>n)在二次函数y=x2-x+c的图象上,且D、E两点关于坐标原点成中心对 3
称,连接OP.当22≤OP≤2+2时,试判断直线DE与抛物线y=x2-x+c+的交点个数,并说明理由.
(09年福建厦门26题解析) (1)解:法1:由题意得
?n=2+c,
??2n-1=2+c. ……………………………………1分
??n=1,
解得?c=-1.
……………………………………2分
法2:∵ 抛物线y=x2-x+c的对称轴是x=1
2
,
且 12-(-1) =2-1
2,∴ A、B两点关于对称轴对称. ∴ n=2n-1 ……1分 ∴ n=1,c=-1. ……2分 ∴ 有 y=x2-x-1 ……3分 =(x-12)2-5
4.
∴ 二次函数y=x2-x-1的最小值是-5
4. ……4分
(2)解:∵ 点P(m,m)(m>0), ∴ PO=2m.
∴ 22≤2m ≤2+2.
∴ 2≤m≤1+2. ……5分 法1: ∵ 点P(m,m)(m>0)在二次函数y=x2-x+c的图象上, ∴ m=m2-m+c,即c=-m2+2m. ∵ 开口向下,且对称轴m=1, ∴ 当2≤m≤1+2 时,
有 -1≤c≤0. ……6分 法2:∵ 2≤m≤1+2, ∴ 1≤m-1≤2. ∴ 1≤(m-1)2≤2.
∵ 点P(m,m)(m>0)在二次函数y=x2-x+c的图象上, ∴ m=m2-m+c,即1-c=(m-1)2. ∴ 1≤1-c≤2.
∴ -1≤c≤0. ……6分
8
∵ 点D、E关于原点成中心对称, 法1: ∴ x2=-x1,y2=-y1.
?y1=x1-x1+c,
∴ ? 2
?-y1=x1+x1+c.
∴ 2y1=-2x1, y1=-x1. 设直线DE:y=kx. 有 -x1=kx1.
由题意,存在x1≠x2.
∴ 存在x1,使x1≠0. ……7分 ∴ k=-1.
∴ 直线DE: y=-x. ……8分 法2:设直线DE:y=kx.
则根据题意有 kx=x2-x+c,即x2-(k+1) x+c=0. ∵ -1≤c≤0, ∴ (k+1)2-4c≥0.
2
∴ 方程x2-(k+1) x+c=0有实数根. ……7分 ∵ x1+x2=0, ∴ k+1=0. ∴ k=-1.
∴ 直线DE: y=-x. ……8分 y=-x,??3322
若 ?则有 x+c+=0.即 x=-c-. 3288
??y=x-x+c+8.
333
① 当 -c-=0时,即c=-时,方程x2=-c-有相同的实数根,
888
3
即直线y=-x与抛物线y=x2-x+c+有唯一交点. ……9分
8333
② 当 -c->0时,即c<-时,即-1≤c<-时,
8883
方程x2=-c-有两个不同实数根,
8
3
即直线y=-x与抛物线y=x2-x+c+有两个不同的交点. ……10分
8333
③ 当 -c-<0时,即c>-时,即-<c≤0时,
8883
方程x2=-c-没有实数根,
8
3
即直线y=-x与抛物线y=x2-x+c+没有交点. ……11分
8