2012中考数学压轴题精选精析(91-100例) 下载本文

(3)若抛物线的对称轴与OB交于点D,点P为线段DB上一点,过P作y轴的平行线,交抛物线于点M。问:是否存在这样的点P,使得四边形CDPM为等腰梯形?若存在,请求出此时点P的坐标;若不存在,请说明理由。(5分)

?b4ac?b2注:抛物线y?ax?bx?c(a≠0)的顶点坐标为???2a,4a?2???,对称轴公?式为x??b 2a

C y

答案:(1)过点C作CH⊥x轴,垂足为H

O y B CEMBQDPA x OHNA x ∵在Rt△OAB中,∠OAB=900,∠BOA=300,AB=2 ∴OB=4,OA=23

由折叠知,∠COB=300,OC=OA=23

∴∠COH=600,OH=3,CH=3 ∴C点坐标为(3,3)[来源:学+科+网]

(2)∵抛物线y?ax2?bx(a≠0)经过C(3,3)、A(23,0)两点

2??a??13?3a?3b? ∴? 解得:?

2b?23???0?23a?23b???? ∴此抛物线的解析式为:y??x2?23x

(3)存在。因为y??x2?23x的顶点坐标为(3,3)即为点C MP⊥x轴,设垂足为N,PN=t,因为∠BOA=300,所以ON=3t ∴P(3t,t)

作PQ⊥CD,垂足为Q,ME⊥CD,垂足为E

把x?3?t代入y??x2?23x得:y??3t2?6t

∴ M(3t,?3t2?6t),E(3,?3t2?6t) 同理:Q(3,t),D(3,1) 要使四边形CDPM为等腰梯形,只需CE=QD 即3??3t2?6t?t?1,解得:t1? ∴ P点坐标为(

??4

,t2?1(舍) 3

443,)

33 ∴ 存在满足条件的点P,使得四边形CDPM为等腰梯形,此时P点的坐为(

443,)

33

33、(安徽桐城白马中学模拟一).我们把一个半圆与抛物线的一部分合成的封闭图形称为“蛋圆”,如果一条直线与“蛋圆”只有一个交点,那么这条直线叫做“蛋圆”的切线.

如图1,点A、B、C、D分别是“蛋圆”与坐标轴的交点,已知点D的坐标为(0,-3),AB为半圆的直径,半圆圆心M的坐标为(1,0),半圆半径为2. (1) 请你求出“蛋圆”抛物线部分的解析式,并写出自变量的取值范围; (2)你能求出经过点C的“蛋圆”切线的解析式吗?试试看;

(3)开动脑筋想一想,相信你能求出经过点D的“蛋圆”切线的解析式.

A y C

O M B x D 答案:

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y?a(x?1)(x?3)(a≠0)

第10图 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y?ax2?bx?c(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

?a?b?c?0?a?1?? ∴?9a?3b?c?0,解之得:?b??2

?c??3?c??3??∴y=x2-2x-3 ········································································ 3分 自变量范围:-1≤x≤3······················································· 4分

解:(1)解方程x2?2x?3?0

得x1??3,x2?1 ······································································································· 1分

0)B(1,0) ················································· 2分 ∴抛物线与x轴的两个交点坐标为:C(?3,,设抛物线的解析式为

y?a(x?3)(x?1) ······································································································ 3分

∵A(3,6)在抛物线上

∴6?a(3?3)·(3?1) ∴a?1················································································· 4分 213········································································· 5分 ∴抛物线解析式为:y?x2?x?·

22131(2)由y?x2?x??(x?1)2?2 ········································································ 6分

222?2),对称轴方程为:x??1 ··································· 7分 ∴抛物线顶点P的坐标为:(?1,设直线AC的方程为:y?kx?b

∵A(3,,6)C(?3,0)在该直线上