的直线y?kx?1交抛物线于点C?2,3?． （1）求直线AC及抛物线的解析式； （2）若直线y?kx?1与抛物线的对称轴交于 点E，以点E为中心将直线y?kx?1顺时针 旋转90?得到直线l，设直线l与y轴的交点
为P，求?APE的面积；
（3）若G为抛物线上一点，是否存在x轴上的
点F，使以B、E、F、G为顶点的四边形为平行四边形，若存在，直接写出点F的坐标；若不存在，请说明理由．
九、解答题（本题满分8分）
25.已知?AOB?90?，OM是?AOB的平分线．将一个直角RPS的直角顶点P在射线OM上移动，点P不与点O重合.
（1）如图，当直角RPS的两边分别与射线OA、OB交于点C、D时，请判断PC与PD的数量关系，并证明你的结论；
GD3的值； PD，求OD2（3）若直角RPS的一边与射线OB交于点D，另一边与直线OA、直线OB分别交于点C、E，且以P、D、E为顶点的三角形与?OCD相似，请画出示意图；当OD?1时，直接写出OP的长.
（2）如图，在（1）的条件下，设CD与OP的交点为点G，且PG?APRCGODMBS
昌平区2008—2009学年第二学期初三年级第一次统一练习
数学试卷答案及评分参考
一、选择题（共8道小题，每小题4分，共32分）
题号 答案
1 D
2 D
3 B
4 A
5 C
6 C
7 B
8 C
二、填空题（共4道小题，每小题4分，共16分）
题号
9
10
11
12
答案
x?1 2
15°
x13?6y,
(?1)n?1x2n?1 ny三、解答题（共5道小题，每小题5分，共25分） 13．解：27?(1??)0?2sin60???2
?33?1?2?3··················································································· 4分 ?2 ·
2································································································ 5分 ?43?1． ·
14．解：x(x?x)?x(3x?1)?4
=x?x?3x?x?4 ··················································································· 2分 =4x?4． ··································································································· 3分 当x?1?0时，x=1． ················································································ 4分
333323222
原式?4?1?4?8． ······················································································ 5分 15．解：分母因式分解，得
x6············································ 1分 ??1 ·
x?1?x?1??x?1?方程两边同乘?x?1??x?1?，得x?x?1??6??x?1??x?1? ·································· 3分 解得 x??5． ······························································································ 4分 经检验，x??5是原分式方程的解． ································································· 5分 16．证明：∵四边形ABCD是矩形， ??A??D?90?，
EFDAAB?DC．???????????????2分
在△AEB和△DFC中，
?AB?DC，? ??A??D，?AE?DF,?BC?△AEB≌△DFC．···················································································· 4分 ?BE?CF． ······························································································· 5分
17．解：解方程组??2x?y?4,?x?3,得? ????????????????2分
?x?y?5?y??2.············································································· 3分 ?点A的坐标为?3,?2?． ·∵点A(3,?2)在双曲线y?k上， x??2?k. 3解得k??6． ······························································································· 4分
6····································································· 5分 ?该双曲线的解析式为y??． ·x四、解答题（共2道小题，每小题5分，共10分）
18．解：如图，分别过点D、E作DF?BC于点F，EH?BC于点H． ?EH∥DF，
?DFB??DFC??EHB??EHC?90?．
又?A?90?，AD∥BC， ??ABC?90? .
?四边形ABFD是矩形． ∵AB?2AD?4,
ADE?AD?2．
?BF?AD?2，DF?AB?4． ·················· 1分
在Rt△DFC中，?C?45，
?BFHC?FC?DF?4． ························································································· 2分 又∵E是CD的中点，
1?EH?DF?2． ······················································································ 3分
2?HC?EH?2． ?FH?2．
?BH?4． ································································································· 4分
在Rt△EBH中，
······················································ 5分 ?BE?BH2?EH2?42?22?25． ·
19．（1）证明：如图，连结OA． ??AFB?30?, 点F在⊙O上， ??AOB?60?． ??CBD?60?, ??CBD??AOB．
?OA∥BC ．???????1分 又?BC?AD, ?OA?AD ． ∵点A在?O上，
··················································································· 2分 ?AD是?O的切线．·
（2）解：∵?AOB?60?，OA?OB，
??OAB是等边三角形． ∵AB?6，
?OA?AB?6．
在Rt△OAD中，?OAD?90，
??tan?AOD?AD， OA?AD?6?tan60??63．
1?S?OAD??63?6?183． ······································································ 3分
2?S扇形AOB60???62??6?， ········································································ 4分
360··················································································· 5分 ?S阴影?183?6?． ·
五、解答题（本题满分6分）
20．解：（1）图1中，丙得票所占的百分比为35%． ············································ 1分 补全图2见下图． ························································································ 2分 三名候选人得票情况统计图
80706050403020100得票数甲乙丙
（2）∵x甲?50?30%?75?40%?93?30%?72.9，
30%?40%?30v?30%?80?40%?70?30%x乙??75.8，
30%?40%?30p?30%?90?40%?68?30%x丙??77.4． ·················································· 5分
30%?40%?30%∴丙被录用． ································································································ 6分