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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