2013年龙岩市初中毕业、升学考试 下载本文

∴E坐标为(m, ∵E在反比例y=∴3m=23. m. ··········· 6分 3m)23图像上, x∴m1=2, m2=-2(舍去).

∴OE=22,EA=4?22,EG=2 ··················································· 7分 ∵4?22<2,

∴EA<EG.

∴以E为圆心,EA垂为半径的圆与y轴相离. ··································· 8分

(3) 存在.······························································································ 9分 方法一:假设存在点F,使AE⊥FE.过点F作FC⊥OB于点 C,过E点作EH⊥OB于点H.

设BF= x.

∵△AOB是等边三角形,

∴AB=OA=OB=4,∠AOB=∠ABO=∠A=60?.

1∴BC=FB·cos∠FBC=x

2FC=FB·sin∠FBC=3x 21∴AF=4-x,OC=OB -BC=4-x

2∵AE⊥FE

1∴AE=AF·cos∠A=2-x

21∴OE=O A-AE=x+2

21∴OH=OE·cos∠AOB=x?1,

4EH=OE·sin∠AOB=3x?3 43311x?3),F(4-x,x) ·∴E(x?1,··········································· 11分 4242

∵E、F都在双曲线y=

k的图象上, x3311x?3)=(4-x)x ∴(x?1)(42424解得 x1=4,x2=. ································································· 12分

5[来源学+科+网]

当BF=4时,AF=0,当BF=

BF不存在,舍去. AF416BF1时,AF=,······················································ 13分 ?. ·55AF4方法二:假设存在点F,使AE⊥FE.过E点作EH⊥OB于 H.

∵△AOB是等边三角形,设E(m,

3m),则OE=2m, AE=4-2m.

∴AB=OA=AB=4,∠AOB=∠ABO=∠A=60?. ∵Cos?A?AE1?, AF2∴AF=2AE=8-4m,FB=4m-4.

∴FC=FB·sin∠FBC=23m-23, BC=FB·cos∠FBC=2m-2. ∴OC=6-2m

∴F(6-2m, 23m-23). ······························································· 11分 ∵E、F都在双曲线y=

k上, x∴m·3m=(6-2m)(23m-23) 化简得:5m2-16m+12=0 解得: m1=2,m2=

6. ····································································· 12分 5当m=2时,AF=8-4m=0,BF=4,F与B重合,不合题意,舍去.

616164当m=时,AF=8-4m=BF=4-=.

55,55 ∴BF:FA?1:4. ·········································································· 13分

1125. (1)在菱形ABCD中, OA?AC?40,OD?BD?3022∵AC⊥BD

∴AD=302?402=50.

∴菱形ABCD的周长为200. ······················4分 (2) 过点M作MP⊥AD,垂足为点P. ①当0<t≤40 ∵Sin?OAD?3∴MP=t

51∴S??DN?MP

232t ··························································································· 6分 10②当40

MPOD3?? AMAD5 =

∵Sin?ADO=MPAO? MDAD4∴MP=(70?t)

5∴S?DMN?1DN?MP 222··································································· 8分 ??t2?28t??(t?35)2?490 ·

55?32t,0?t?40??10∴S??

2??(t?35)2?490,40?t?50??5当0<t≤40时,S随t的增大而增大,当t=40时,最大值为480. 当40<t≤50时,S随t的增大而减小,当t=40时,最大值为480. 综上所述,S的最大值为480. ······································································ 9分 (3)存在2个点P,使得∠DPO=∠DON.····················································· 10分 方法一:过点N作NF⊥OD于点F,

401203090??24DF=ND?Cos?ODA?30???18.则NF?ND?Sin?ODA?30?505505,

∴OF=12,∴tan?NOD?NF24??2 ·················································· 11分 OF12作?NOD的平分线交NF于点G,过点G作GH⊥ON于点H.

1111∴S?ONF?OF?NF?S?OGN?S?OFG?OF?FG?ON?GH?(OF?ON)?FG 2222OF?NF12?2424??∴FG=OF?ON12?1251?5

24GF1?52tan?GOF???∴ OF121?5设OD中垂线与OD的交点为K,由对称性可知:

11∴?DPK??DPO??DON??FOG ·················································· 12分

22DK152tan?DPK???∴PKPK1?5 15(5?1)∴PK= ················································································ 13分

2根据菱形的对称性可知,在线段OD的下方存在与点P关于OD轴对称的点P'.

15(5?1). ·········································· 14分 2方法二:如图,作ON的垂直平分线,交EF于点I,连结OI,IN.

∴存在两个点P到OD的距离都是过点N作NG⊥OD,NH⊥EF,垂足分别为G,H. 当t=30时,DN=OD=30,易知△DNG∽△DAO, ∴即

DNNGDG??. DAAOOD30NGDG??.504030[来源学科网ZXXK]

∴NG=24,DG=18. ············································································ 10分 ∵EF垂直平分OD,

∴OE= ED=15,EG=NH=3. ······························································ 11分 设OI=R,EI=x,则

在Rt△OEI中,有R2=152+x2 ① 在Rt△NIH中,有R2=32+(24-x)2 ② 15?x??2?由①、②可得:?

155?R???2∴PE=PI+IE=15?155. ·································································· 13分 2