第第頁共23頁點(diǎn)P的縱坐標(biāo)為y=-fi+2t+3,p點(diǎn)〃的縱坐標(biāo)為y=3什9m55TOC\o"1-5"\h\z所以PM=y-y=-t2+2t+3+3t-9=-t2+13t+6.pm5555則s=s+sPFEPFMPEM=1PMFN+1PM-EH22=；PM.(FN+EH)=12+13+63+2255517132+289**101010010’所以當(dāng)上13時(shí)p/花的面積最大，最大值的立方根為3289x*【答案】由圖可知NP￡A，90。.①若ZPA￡=90。,作尸軸，11因?yàn)镺A=OE,^\^ZOAE=ZOEA=45°,所以NPAG=NAPG=45。,所以PG=AG,111所以上孑+2什3一3,即_/+七0,解得t=l或仁0（舍去）.②若NAP￡=90。,作PK±x^AQ±PK,222則PKE^AAQP,22所以2=,解得=〒或=^<-5（舍去）.綜上可知41或=T符合題意?8.(1)【答案】二?盧x-3過B,C兩點(diǎn),AB(3,0),C(0,-3),.?=%2+bx+c經(jīng)過B(3,0),C(0,-3),0=9+3+=-2A-3=，解得=3,A尸%2-2x-3.拋物線解析式為尸%2-2%-3，當(dāng)y=0時(shí),x2-2x-3=0,?.%=-1,%=3,AA(-1,0),??O,A=1,0B=OC=3,AZABC=45°,AC=10,AB=4,PE±%軸，AZEMB=ZEBM=45°,二?點(diǎn)P的橫坐標(biāo)為t,AEM=EB=3-t，連接AM,SAC=SAMC+SAMA1AB-OC=1AC-MN+2AB-EM,A1x4x3=1x10d+1x4(3-t),222Ad=2101.5(3)【答案】如圖2,圖2?？=x2-2%-3=(%-1)2-4,??對(duì)稱軸為x=1,?.?由拋物線對(duì)稱性可得D(2,-3),???CD=2.過點(diǎn)B作BK±CD交直線CD于點(diǎn)K,??四邊形OCKB為正方形，AZOBK=90°,CK=OB=BK=3,ADK=1,??BQ±CP,AZCQB=90°.過點(diǎn)O作OH±PC交PC延長(zhǎng)線于點(diǎn)H,OR±BQ交BQ于點(diǎn)I，交BK于點(diǎn)R,AZOHC=ZOIQ=ZOIB=90°,A四邊形OHQI為矩形，VZOCQ+ZOBQ=180°,AZOBQ=ZOCH,???△OHCSOIB,AOH=OI,A四邊形OHQI為正方形,AZQOI=45°,延長(zhǎng)KB至點(diǎn)G使BG=CS，連接OG,SR,??△OBG必OCS,AOG=OS,ZGOB=ZSOC,AZSOG=90°,AZGOR=45°,又OR=OR,???△OSRSOGR,ASR=GR,ASR=CS+BR,VZBOR+ZOBI=90°,ZOBI+ZTBK=90°,AZBOR=ZTBK,AtanZBOR=tanZTBK,A=,ABR=TK,VZCTQ=ZBTK,AZQCT=ZTBK,AtanZQCT=tanZTBK,設(shè)ST=TD=m,ASK=2m+1,CS=2-2m,TK=m+1=BR,SR=3-m,RK=2-m,在RtSKR中，VSK2+RK2=SR2,A(2m+1)2+(2-m)2=(3-m)2,

m=-2（舍去）,m=2,TOC\o"1-5"\h\z；?ST=TD=1,TK=3,22tanNTBK=—=—^-3=—,AtanNPCD—222過點(diǎn)P作PE±%軸于E交CD于F；，：CF=OE=t,APF'=11,APE=11+3,AP,-1-3,222A-11-3=t2-21-3.At=0（舍去）,t=3,2122AMN=d=29.⑴【答案】8,30.令-3%令-3%2123%+833=0,解得解得%=-12,%=8,AA(8,0),B(0,83),NOBA=30°,OA的長(zhǎng)為8.(2)【答案】①證明:：DE〃AB,A=又：OM=AM,AOH=BH,又?「BN=AN,AHN//AM,A四邊形AMHN是平行四邊形.②點(diǎn)D在該拋物線的對(duì)稱軸上.理由:如圖，過點(diǎn)D作DR±y軸于F,VRtCDE0RtABO,ANHDG=NOBA=30°,???DE/AB,???BND=ZD=30°,NDHG=/OBA=30°,又：HN//OA,AZBNH=NBAO=60°,AZDNH=30°,ZD=ZHND,ADH=HN=10A=4,2.??在RtDHE中,DE=DHsin30°=1x4=2,2A點(diǎn)D的橫坐標(biāo)是-2,-3又???拋物線的對(duì)稱軸是直線％=--=一『=-2,22X--312A點(diǎn)D在該拋物線的對(duì)稱軸上.(3)【答案】連接MN,易知MN±%軸,M(4,0),N(4,43),ON=NA=OA,ZONA=ZBAO=ZNOA=60°,MN=1OB=43,ZONM=30°,2?「RtCDE0RtABO,AZONA=ZC=60°,ZCDE=ZOBA=30°,AAB/CE,AAB±DE,AZEMA=ZOMK=30°,ZPKD=60°,AZOMK=ZCDE=ZONM=30°.AOD=OM=4,DM=MN=43,ADN=OD+ON=4+8=12,設(shè)D(%,y)，則％=-(DMcos30°-OM)=-43X^3-4=-2,y=-(DMssin30°)=-43x1=-23,AD(-2,-23).設(shè)DQ與％軸交于點(diǎn)H,則一二一,即瓷==3。尺=103,AQ(-2,103).=——一設(shè)DM的方程為尸k%+仇將點(diǎn)D,M的坐標(biāo)代入得-24++==j3,解得3,Ayq%-1.-3設(shè)K，工-心，過點(diǎn)。作％的平行線交K/于點(diǎn)J,則J(%,-23).33K解得％=-63-2,—x-^~^=-6-23,K3K3：.K(-63-2,-6-23).過點(diǎn)P作KD的垂線，垂足為T,則PT=KTtanNQKD=3KT,ZPDK=45°,PT=TD=小KT,：.KT+3KT=KD=12,KT=6(3-1),：,PK=12(3-l),.*.P(-63-2,103-18),：.PQ=-2-(6￥2)2+10￥(10字18)2=(63)2+182=123.10.(1)【答案】???拋物線尸42+陵+c經(jīng)過點(diǎn)5(6,0),C(0,6),...-|><62+6+=0,解得=2,=6,一d拋物線的解析式為尸吳+2%+6.j=-1x2+2x+6=-^(x-2)2+8,，頂點(diǎn)。的坐標(biāo)為(2,8).(2)【答案】如圖，過點(diǎn)尸作bG,%軸，垂足為點(diǎn)G.'D??D(2,8),B(6,0),??DE=8,BE=6-2=4.VZFBA=ZBDE,??tanZFBA=tanZBDE,.\=即-12+2+66-化簡(jiǎn),得a2-5a-6=0或a2-3a-18=0.解得a=-1,a=6(舍去),a=-3,a=6(舍去)?.點(diǎn)F-1,7或點(diǎn)F-3,-9.1222(3)【答案】①如圖，當(dāng)點(diǎn)M在％軸的上方時(shí)，設(shè)點(diǎn)M(x,y).V四邊形MPNQ是正方形,???MS=SP,即2-%=-卜2+2x+6.解得x=3-17,x;3+17.Vy>0,即-1x2+2x+6>0,/.-2<x<6./.x=3-17.???PQ=2MS=2(2-x)=-2+217,；.Q(2,-2+217).②如圖，當(dāng)點(diǎn)M在x軸的下方時(shí)，設(shè)點(diǎn)M(x,y).V四邊形MPNQ是正方形,???MT=PT，即2-x=--；2+2+6,解得x=1-17,x2=1+17.一一一._一.一、一一/一.