发布时间 : 星期一 文章2012中考数学压轴题精选精析（91-100例）更新完毕开始阅读2b6db9b3856a561252d36fa1
（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