08年怀化市初中毕业学业考试试卷

如图13,在平面直角坐标系中,圆M经过原点O,且与x轴、y轴分别相交于A??8, 0?、B?0,?6?两点.(1)求出直线AB的函数解析式;

(2)若有一抛物线的对称轴平行于y轴且经过点M,顶点C在⊙M上,开口向下,且经过点B,求此抛物线的函数解析式;

(3)设(2)中的抛物线交x轴于D、E两点,在抛物线上是否存在点P,使得S?PDE?请求出点P的坐标;若不存在,请说明理由.

1S?ABC?若存在,10

参考答案

一、选择题(每小题2分,共20分) 二、填空题(每小题2分,共20分)

三、解答题

题号 答案

1 A

2 B

3 C

4 B

5 D

6 B

7 A

8 D

9 A

10 B

21.解

x?xx3??1?x?1?x?1??x?2?题号 答案

11

12

13 4

14

15

16 6

17

18

19

20

2?a?2b??a?2b??x?4 ?y?1?

120?0?<∠POC<110?

1 225?203670 16

···················································································································2分

x2?2x?3?x2?2x?x?2??x?1??x?2???x?1?x?1??x?2?

1 ······································································································· 5分 x?2213·························································7分 ?当x??时,原式的值为? ·

23??24311122. 解:摸到绿球的概率为:1??? ······················································1分

3261则袋中原有三种球共 3??18 (个) ·····························································3分

61?18?6 (个) ·所以袋中原有红球 ···························································5分 31?18?9 (个) ·袋中原有黄球 ·································································7分 2?y?x,?x1?1?x2??1?23.解:(1)解方程组?得, ············ 2分 ,?1?y??y1?1?y2??1?x?所以A、B两点的坐标分别为:A(1,1)、B(-1,-1) ····· 4分 (2)根据图象知,当?1?x?0或x?1时,正比例函数值大于反比24. 证明:(1)?四边形ABCD和四边形DEFG都是正方形

例函数值 ····························

?AD?CD,DE?DG,?ADC??EDG?90?,

??ADE??CDG,?△ADE≌△CDG, ·················· 3分 ?AE?CG ························································· 4分

(2)由(1)得 ?ADE??CDG,??DAE??DCG,又?ANM??CND,

?ANCN?MNDN,即AN?DN?CN?MN ············································7分

∴?AMN∽?CDN ················································································6分 25解:(1)由平移的性质得

AF//BC且AF?BC,△EFA≌△ABC,?四边形AFBC为平行四边形,

?S?EFA?S?BAF?S?ABC?3,?四边形EFBC的面积为9. ···········································································3分

(2)BE?AF.证明如下:由(1)知四边形AFBC为平行四边形

?BF//AC且BF?AC,又AE?CA,?BF//AE且BF?AE,?四边形EFBA为平行四边形又已知AB?AC,?AB?AE,

?平行四边形EFBA为菱形,?BE?AF ··························································5分

(3)作BD?AC于D,??BEC?15?,AE?AB,??EBA??BEC?15?,??BAC?2?BEC?30?,?在Rt?BAD中,AB?2BD.设BD?x,则AC?AB?2x,?S

?ABC?3,且S?ABC?12AC?BD?12?2x?x?x2,?x2?3,?x为正数,?x?3,?AC?23......................7分26.解:?1??i?BEAE?95,?设BE?9k,AE?5k?k为正数?,则在Rt?ABE中,?BEA?90?,AB?52106,AB2?BE2?AE2,.....................................2分2即??5?2106?????9k?2??5k?2,解得k?552,?BE?9?2?22.5?m?.故改造前坡顶与地面的距离BE的长为22.5米....................................................................4分27.解:?2?由?1?得AE?12.5,设BF?xm,作FH?AD于H,则FHAH?tan?FAH,由题意得22.5x?12.5?tan45?,即x?10.?坡顶B沿BC至少削进10m,才能确保安全.........................................................................7分因为租用甲种汽车为x辆,则租用乙种汽车?8?x?辆.

??4x?2?8?x?≥由题意,得?30,?3x?8?8?x?≥20. ·································································2分

?解之,得7?x?445. ················································································3分 即共有两种租车方案:

(1)

第一种是租用甲种汽车7辆,乙种汽车1辆;

第二种是全部租用甲种汽车8辆 ···································································5分 (2)第一种租车方案的费用为7?8000?1?6000?62000元 ···························6分 第二种租车方案的费用为8?8000?64000元 ··················································7分 所以第一种租车方案最省钱 ········································································· 8分 28.解:(1)设AB的函数表达式为y?kx?b.

3??0??8k?b,?k??,∵A??8,0?,B?0,?6?,∴?∴?4

?6?b.???b??6.∴直线AB的函数表达式为y??3x?6. ··························································3分 4(2)设抛物线的对称轴与⊙M相交于一点,依题意知这一点就是抛物线的顶点C。又设对称轴与x轴相交于点N,在直角三角形AOB中,AB?AO2?OB2?82?62?10.

?因为⊙M经过O、A、B三点,且?AOB?90,?AB为⊙M的直径,∴半径MA=5,∴N为AO的中点AN=NO=4,∴MN=3∴CN=MC-MN=5-3=2,∴C点的坐标为(-4,2). 设所求的抛物线为y?ax?bx?c

21?b????4,a??,?2a?2??则?2?16a?4b?c,??b??4, ??6?c.?c??6.????∴所求抛物线为y??(3)令?12x?4x?6 ··································································7分 212x?4x?6.?0,得D、E两点的坐标为D(-6,0)、E(-2,0),所以DE=4. 21又AC=25,BC?45,?直角三角形的面积S?ABC??25?45?20.

2111S?ABC,即?DE?y??20,?y??1. 假设抛物线上存在点p?x,y?使得S?PDE?10210当

y?1时,x??4?2;当y??1时,x??4?6.故满足条件的存在.它们是

P1?4?2,1,P2?4?2,1,P3?4?6,?1,P4?4?6,?1.

????????

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