阳光家教网 www.ygjj.com 中考（初三复习）数学资料 5?k?3?0?3k?b???9 解得? 5?5b??3?b???33??3??y?553?3 ··············································································································· 13分 933?55x???3x?3?3?1037??y??93 解得? ?M?， ?????77???y??103?y??3x?3??7??3103???在直线AC上存在点M，使得△MBF的周长最小，此时M???7，?． 1 7??18. 解：（1）点E在y轴上 ·································································································· 1分 理由如下：
连接AO，如图所示，在Rt△ABO中，?AB?1，BO?3，?AO?2
?sin?AOB?1?，??AOB?30 2?由题意可知：?AOE?60
??BOE??AOB??AOE?30??60??90?
··················································································· 3分 ?点B在x轴上，?点E在y轴上． （2）过点D作DM?x轴于点M
?OD?1，?DOM?30?
?在Rt△DOM中，DM??点D在第一象限，
13，OM? 22?31?D···································································································· 5分 ?点的坐标为???2，? ·2??由（1）知EO?AO?2，点E在y轴的正半轴上
2) ?点E的坐标为(0，······································································································ 6分 ?点A的坐标为(?31)， ·
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 ?抛物线y?ax2?bx?c经过点E，
?c?2
由题意，将A(?31)，，D??31?，代入y?ax2?bx?2中得 ??22???8??3a?3b?2?1a???9?? 解得 ?3?31b?2??a??b??53?422?9?853······························································· 9分 x?2 ·?所求抛物线表达式为：y??x2?99（3）存在符合条件的点P，点Q． ··················································································· 10分 理由如下：?矩形ABOC的面积?AB?BO?3 ?以O，B，P，Q为顶点的平行四边形面积为23．
由题意可知OB为此平行四边形一边， 又?OB?3
?OB边上的高为2··············································································································· 11分
2) 依题意设点P的坐标为(m，853x?2上 ?点P在抛物线y??x2?99853??m2?m?2?2
99解得，m1?0，m2??53 8?53?2??P2)，P2?1(0，??8，?
???以O，B，P，Q为顶点的四边形是平行四边形，
?PQ∥OB，PQ?OB?3， 2)时， ?当点P1的坐标为(0，A
y E F C D O M x B 阳光家教网 www.ygjj.com 中考（初三复习）数学资料 点Q的坐标分别为Q1(?3，2)，Q2(3，2)；
?53?2?当点P2的坐标为???8，?时，
??点Q的坐标分别为Q3???133??33?，2Q，2?，． ······················································· 14分 ?4?????8???8?32x?3中，令y?0 4（以上答案仅供参考，如有其它做法，可参照给分） 19. 解：（1）在y??3??x2?3?0
4?x1?2，x2??2
?A(?2，0)，B(2，0) ·························································· 1分 又?点B在y??C E y 3x?b上 4A N 3?0???b
23b?
2M D O P B x 33?BC的解析式为y??x? ··························································································· 2分
4232?y??x?3?x1??1???4（2）由?，得?9
33y1??y??x???4??42?x2?2 ································································ 4分 ?y?0?29??0) ?C??1，?，B(2，4??9 ·············································································································· 5分 4199?S△ABC??4?? ······································································································· 6分
242（3）过点N作NP?MB于点P ?EO?MB ?NP∥EO
?△BNP∽△BEO ·············································································································· 7分 BNNP?? ·························································································································· 8分 BEEO?AB?4，CD?
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 由直线y??33?3?x?可得：E?0，? 42?2?35，则BE? 22?在△BEO中，BO?2，EO??62tNP，?NP?t ····································································································· 9分 ?5352216?S??t?(4?t)
25312S??t2?t(0?t?4) ·································································································· 10分
55312S??(t?2)2? ············································································································ 11分
5512?此抛物线开口向下，?当t?2时，S最大?
512?当点M运动2秒时，△MNB的面积达到最大，最大为．
520. 解：（1）如图，过点B作BD⊥OA于点D. 在Rt△ABD中， ∵∣AB∣=35,sin∠OAB=
5, 5 ∴∣BD∣=∣AB∣·sin∠OAB =35×
又由勾股定理，得 AD? ?5=3. 52AB?BD (35)2?32?6
2∴∣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
2
y=ax+bx(a≠0).
1?a?,??16a?4b??3?8??由?
?100a?10b?0?b??5.??4∴经过O、C、A三点的抛物线的函数表达式为y?125x?x. ??2分 84（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
阳光家教网 www.ygjj.com 中考（初三复习）数学资料 ①∵点C（4，-3）不是抛物线y?1258x?4x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y?18x2?54x，令y=-3?x=4或x=6. ∴??x1?4,?x2?6,?y1??3;??y2??3. 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. ②若AP2∥CO.设直线CO的函数表达式为y?k1x. 将点C（4，-3）代入，得4k31??3.?k1??4. ∴直线CO的函数表达式为y??34x. 于是可设直线AP32的函数表达式为y??4x?b1. 将点A（10，0）代入，得?34x?152. ∴直线AP3152的函数表达式为y??4x?2.
?由??y??3x?15.?42?x2?4x?60?0，即（x-10）（x+6）=0. ??y?18x2?5?4x∴??x1?10,??y?x2??6 1?0;?y2?12;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
AP222?P2E?AE?122?162?20.
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. 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
1分 1分 ????
2010中考数学专题复习 - 压轴题（含答案）
