中考數(shù)學(xué)總復(fù)習(xí)《二次函數(shù)與平行四邊形綜合壓軸題》專項(xiàng)提升練習(xí)題附含答案

第第頁(yè)參考答案1．（1）解：把A?1,0B3,0代入0=a?b?20=9a+3b?2解得∴該拋物線的函數(shù)表達(dá)式為：y=2（2）當(dāng)x=0時(shí)y=?2∴C∵該拋物線的對(duì)稱軸為直線x=?∴D設(shè)點(diǎn)P∵PQ⊥CD∴Q∴PQ=?2?23∴PQ+QD=?∵?∴當(dāng)t=14時(shí)PQ+QD有最大值當(dāng)t=14時(shí)∴P1綜上：PQ+QD最大值為4924P（3）∵B∴原拋物線向左平移3個(gè)單位長(zhǎng)度后經(jīng)過原點(diǎn)∵原拋物線表達(dá)式為：y=23x2?∴新拋物線的表達(dá)式為：y=∵D∴E根據(jù)題意可得點(diǎn)F的橫坐標(biāo)為1①當(dāng)PE為平行四邊形的邊時(shí)：xE則?1+1=xG+∴點(diǎn)G的橫坐標(biāo)為1?把x=?14代入y=∴G或x則?1+xG=1+把x=94代入y=∴G②當(dāng)PE為平行四邊形的對(duì)角線時(shí)：x則?1+解得：x把x=?74代入y=∴G?綜上：G?14,?152．（1）解：分別把點(diǎn)A0,?3代入y=34x2所以拋物線的解析式為y=3（2）∵A∴OA=3OB=4AB=5直線AB的解析式為：y=設(shè)：Pm,3∴PE=?∴當(dāng)m=2時(shí)PE最大為3∵PE∥y軸PF⊥AB∴∠PEF=∠OAB∠PFE=∠AOB∴△PEF∽△BAO∴C∴C所以當(dāng)PE最大為3時(shí)△PEF的周長(zhǎng)最大為36此時(shí)P2,?（3）y=34則平移后的解析式為y=34設(shè)Mm,yN3當(dāng)MQ為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可得：m?1=4+32y?92=n可得m=13當(dāng)MB為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可得：m+4=?1+32y=n?92可得m=?72當(dāng)MN為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可得：m+32=?1+4y+n=?92可得m=3綜上符合條件的點(diǎn)M的坐標(biāo)為：M1323．解:（1）拋物線y=ax2+bx?3a≠0與x軸交于可得y=a(x+1)(x?3)=a∴?3a=?3解得a=1∴y=（2）設(shè)P(x,x2?2x?3)過P作PQ⊥AB于Q交由（1）可知y=令x=0,y=?3即C(0,?3)設(shè)BC解析式為：y=kx+b代入B3,00=3k+b?3=b解得∴BC解析式為：y=x?3∴M(x,x?3)∴PM=(x?3)?(∴S∴當(dāng)x=32時(shí)此時(shí)P(3（3）拋物線沿水平方向向左移動(dòng)2個(gè)單位可得y==∴頂點(diǎn)Q(?1,?4)∵P(①當(dāng)PQ是平行四邊形的一條邊時(shí)根據(jù)平移規(guī)律可得xN=當(dāng)x當(dāng)x∴N(52②當(dāng)PQ是平行四邊形的對(duì)角線時(shí)可知PQ中點(diǎn)(∵M(jìn)N中點(diǎn)也為(∴x∴y∴N(綜上所述：N(12,?74．（1）解：∵二次函數(shù)的最小值為?1點(diǎn)M(1，∴二次函數(shù)頂點(diǎn)為(1∴可設(shè)二次函數(shù)解析式為y=a將點(diǎn)O(0，0)代入得解得：a=1∴二次函數(shù)解析式為y=(x?1)（2）如圖連接OP對(duì)于y=x2?2x當(dāng)y=0時(shí)解得：x1=0∴A(2，∵點(diǎn)P在拋物線y=x∴點(diǎn)P的縱坐標(biāo)為t2∵點(diǎn)P在第四象限∴0<t<2∴S====?=?∴當(dāng)t=34時(shí)S取最大值最大值為y∴此時(shí)P3（3）設(shè)N(n分類討論：①當(dāng)AB為對(duì)角線時(shí)∵A(2，0)B(0∴由中點(diǎn)坐標(biāo)公式得x∴2+0=1+n解得：n=1∴y∴N(1，②當(dāng)AM為對(duì)角線時(shí)同理可得2+1=n+0解得：n=3∴N(3，③當(dāng)AN為對(duì)角線時(shí)同理可得2+n=0+1解得：n=?1∴N(?1，綜上可知點(diǎn)N的坐標(biāo)為：(1，?1)或(3，5．（1）解：當(dāng)x=0時(shí)y=3即C(0,3)當(dāng)y=0時(shí)?x2+2x+3=0解得x=?1x=3即A(?1,0)S△ABC（2）若∠BMN=90°如圖1BM=(3?t)BN=2t△BMN∽△BOCBMBO=BNBC∴2t=2(3?t)若∠BNM=90°時(shí)如圖2BM=(3?t)BN=2t∵△BMN∽△BCOBNOB=BM3?t=2×2t綜上所述：t=1或t=3（3）如圖3若CB為對(duì)角線即CP1∥QBP1若CB為邊即CB∥PQ設(shè)Pa,bD1,bPQ=CB即a?1化簡(jiǎn)得a2解得a=?2或a=4．當(dāng)a=?2時(shí)b=?即P2當(dāng)a=4時(shí)b=?即P3綜上所述：P12,3P26．（1）解：在y=?12x+3中令x=0則y=3；令y=0∴B把A?1，0Ba?b+c=0∴a=?∴拋物線解析式為y=?1（2）解：如圖所示過點(diǎn)D作DT⊥x軸于T交直線BC于G延長(zhǎng)FE交DT于H設(shè)Dm，1∴DG=?∵DT⊥x軸OC⊥x軸∴DT∥OC∴∠OCB=∠DGE∵B∴OC=3∴BC=∴cos∠OCB=cos∠DGE=OC在Rt△DEG中∴EG=DH?在Rt△EGH中∴EF=FH?EH=∵∠HEG+∠HED=90°=∠HGE+∠HEG∴∠HED=∠HGE∴cos∴在Rt△DEH中∴EF?5==∵2∴當(dāng)m=74時(shí)EF?5（3）解：∵平移前的拋物線解析式為y=?∴平移后的拋物線解析式為y=?1設(shè)直線AC的解析式為y=kx+∴?k+∴k=3∴直線AC的解析式為y=3x+3；設(shè)點(diǎn)G的坐標(biāo)為s當(dāng)QA為對(duì)角線時(shí)則QG和AC為平行四邊形的邊∴QG∥AC∴可設(shè)直線QG的解析式為y=同理可求出直線QG的解析式為y=3x?1聯(lián)立y=3x?1y=?1解得x=?6+93∴點(diǎn)G的橫坐標(biāo)為?6+932或當(dāng)QG為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可知此時(shí)QG的中點(diǎn)一定在直線AC上∴?∴?∴4解得s=?6+29∴點(diǎn)G的橫坐標(biāo)為?6+292或綜上所述點(diǎn)G的橫坐標(biāo)為?6+932或?6?932或7．（1）解：將ABC點(diǎn)的坐標(biāo)代入解析式得9a?3b+c=0解得a=?1∴拋物線的解析式為y=?x（2）解：存在點(diǎn)P的坐標(biāo)為?1+132,如圖設(shè)M為線段AC的中點(diǎn)∵OA=OC=3∴直線OM為線段AC的垂直平分線直線OM與拋物線必有兩個(gè)交點(diǎn)且都是滿足條件的點(diǎn)P．∵A?3,0∴點(diǎn)M的坐標(biāo)為?設(shè)OM的解析式為y=kx將點(diǎn)M?32即直線OM的解析式為y=?x聯(lián)立y=?x得?解得x1=∴點(diǎn)P的坐標(biāo)為?1+132,（3）能點(diǎn)E的坐標(biāo)為?2,1或?3+172,將y=?x2?2x+3配方∴頂點(diǎn)D的坐標(biāo)為?1,4由x=?b2a∵A?3,0∴直線AC的方程為y=x+3聯(lián)立x=?1y=x+3得點(diǎn)E在直線AC上設(shè)E①當(dāng)點(diǎn)E在線段AC上時(shí)點(diǎn)F在點(diǎn)E上方則F∵EF=DN∴?解得x=?2或x=?1（舍）∴點(diǎn)E的坐標(biāo)為?2,1；②當(dāng)點(diǎn)E在線段AC（或CA）延長(zhǎng)線上時(shí)點(diǎn)F在點(diǎn)E下方則F∵EF=DN∴x+3?解得x=?3+17∴點(diǎn)E的坐標(biāo)為?3+172綜上所述：滿足條件的點(diǎn)E坐標(biāo)為?2,1或?3+172,8．解：發(fā)現(xiàn)：令y=0時(shí)則1解得：x令x=0時(shí)則有y=?4∴A?2,0B8,0設(shè)直線BC的解析式為y=kx+b則有：8k+b=0解得：k=∴直線BC的解析式為y=1拓展：過點(diǎn)P作PH∥y軸交BC于點(diǎn)H如圖所示：設(shè)點(diǎn)Pm,14∴PH=?∴S∵0<m<8且函數(shù)S△PBC∴當(dāng)m=4時(shí)△PBC的面積最大此時(shí)點(diǎn)P4,?6探究：由拋物線解析式y(tǒng)=14∴D∴DE=∵四邊形PMED為平行四邊形∴DE∥PM,DE=PM由題意知Mm,12∴PM=?∴?解得：m1=5,m∴當(dāng)m=5時(shí)四邊形PMED為平行四邊形．9．（1）解：令y=0則1解得x=4或x∴A?1,0令x=0則y∴C設(shè)直線AD的函數(shù)表達(dá)式為y=kx+b將A?1,0D5,3的坐標(biāo)代入得解得：k=∴y=1（2）解：存在理由如下：設(shè)Pa,0則Ea,∵P線段AB上的一個(gè)動(dòng)點(diǎn)∴E在x軸下方∴EF=∵?∴當(dāng)a=2時(shí)EF有最大值最大值為92（3）解：存在點(diǎn)M的坐標(biāo)為0,?22+14,4+14設(shè)Mm,1∵B4,0①當(dāng)平行四邊形對(duì)角線為BN和DM時(shí)則4+n解得：m=0n=1或m=4n=5（當(dāng)m=4時(shí)M4,0與B點(diǎn)重合∴點(diǎn)M的坐標(biāo)為0,?2；②當(dāng)平行四邊形對(duì)角線為BM和DN時(shí)則4+m解得：m=2+14n=1+∴點(diǎn)M的坐標(biāo)為2+14,4+綜上所述點(diǎn)M的坐標(biāo)為0,?22+14,4+1410．（1）解：將點(diǎn)A?1,0Ba?b?1=0解得：a=∴拋物線的表達(dá)式為：y=1（2）存在理由如下：∵y=當(dāng)x=0時(shí)y=∴點(diǎn)C設(shè)點(diǎn)P當(dāng)AB為對(duì)角線時(shí)另一條對(duì)角線為CP∴x+02解得：x=2y=1∴此時(shí)點(diǎn)P2,1當(dāng)AC為對(duì)角線時(shí)另一條對(duì)角線為BP∴?1+02=3+x∴此時(shí)點(diǎn)P(?4,?1)；當(dāng)AP為對(duì)角線時(shí)另一條對(duì)角線為BC∴?1+x2=3+0∴此時(shí)點(diǎn)P4,?1綜上所述點(diǎn)P的坐標(biāo)為2,1或?4,?1或4,?1；（3）y=當(dāng)x=2時(shí)y=?1∴D(2,?1)設(shè)H(p,q)過點(diǎn)A作AH⊥DM于點(diǎn)H過點(diǎn)H作TK∥x軸過點(diǎn)A作AT⊥TK于點(diǎn)T過點(diǎn)D作DK⊥TK當(dāng)點(diǎn)M在AD上方時(shí)如圖所示：∵∠ADM=45°∴△ADH是等腰直角三角形∴AH=DH,∵∠∴△AHT≌△HDK(AAS∴AT=HK,TH=DK∵AT=q,HK=2?p,TH=p+1,DK=q+1∴q=2?p解得：p=1∴H(設(shè)直線DH的解析式為y=cx+d將點(diǎn)H(1,1)1=c+d解得：c=?2∴直線DH解析式為y=?2x+3令x=0得y=3∴M(0,3當(dāng)點(diǎn)M在AD下方時(shí)如圖所示：同理可得AT=HK,TH=DK∵AT=?q,HK=2?p,TH=p+1,DK=?q?1∴?q=2?p解得：p=0∴H(0,?2)此時(shí)點(diǎn)M與點(diǎn)H重合即點(diǎn)M(0,?2)綜上可得：M的坐標(biāo)為(0,?2)或0,3．11．（1）解：根據(jù)題意拋物線頂點(diǎn)為(?1,設(shè)拋物線為y=a(x+1)拋物線過點(diǎn)C(0,5)∴a=?拋物線解析式為y=?1（2）令?解得：x∴A(?5,0)B(3,0)．如圖作FD⊥AC于D∵OA=5OC=5∴∠CAO=45°．設(shè)AF=m則MF=10在△MEF中F∴(解得m1=2，m∴AF=2∴點(diǎn)Q的橫坐標(biāo)為?3．又點(diǎn)Q在拋物線y=?1∴Q(?3,4)（3）設(shè)直線AC的解析式y(tǒng)=kx+n由題意得?5k+n=0解得：k=1∴直線AC的解析式y(tǒng)=x+5．由已知點(diǎn)QNF及點(diǎn)PME橫坐標(biāo)分別相同．設(shè)F(t,0)E(t+1,0)N(t,t+5)M(t+1,t+6)Q(t,?1在矩形平移過程中以PQNM為頂點(diǎn)的平行四邊形有兩種情況：①點(diǎn)QP在直線AC同側(cè)時(shí)QN=PM．∴(?解得：t=?3．∴M(?2,3)．②點(diǎn)QP在直線AC異側(cè)時(shí)QN=MP．∴(?解得t∴M?2+6,3+∴符合條件的點(diǎn)M是(?23)?2+6，3+612．解:（1）∵平行四邊形OABC中A(6,0)C(4,3)∴BC=OA=6BC∥x軸∴xB=xC設(shè)拋物線y=ax2+bx+c經(jīng)過點(diǎn)B∴100a+10b+c=316a+4b+c=3a+b+c=0∴拋物線解析式為y=?1（2）如圖1作點(diǎn)E關(guān)于x軸的對(duì)稱點(diǎn)E′連接E′F交∵C(4,3)∴OC=∵BC∥OA∴∠OEC=∠AOE∵OE平分∠AOC∴∠AOE=∠COE∴∠OEC=∠COE∴CE=OC=5∴xE=∴直線OE解析式為y=∵直線OE交拋物線對(duì)稱軸于點(diǎn)F對(duì)稱軸為直線：x=?∴F(7,∵點(diǎn)E與點(diǎn)E′關(guān)于x軸對(duì)稱點(diǎn)P在x∴E′∴當(dāng)點(diǎn)FPE′在同一直線上時(shí)PE+PF=P設(shè)直線E′F∴9k+?=?37k+?=73∴直線E當(dāng)?83x+21=0時(shí)∴當(dāng)PE+PF的值最小時(shí)點(diǎn)P坐標(biāo)為638（3）存在滿足條件的點(diǎn)MN使得以點(diǎn)MNHE為頂點(diǎn)的四邊形為平行四邊形．設(shè)AH與OE相交于點(diǎn)G(t,13∵AH⊥OE于點(diǎn)GA(6,0)∴∠AGO=90°∴A∴∴解得：t1=0（舍去）∴G設(shè)直線AG解析式為y=dx+e∴6d+e=0275d+e=∴直線AG:y=?3x+18當(dāng)y=3時(shí)?3x+18=3解得：x=5∴H(5,3)∴HE=9?5=4點(diǎn)HE關(guān)于直線x=7對(duì)稱①當(dāng)HE為以點(diǎn)MNHE為頂點(diǎn)的平行四邊形的邊時(shí)如圖2則HE∥MNMN=HE=4∵點(diǎn)N在拋物線對(duì)稱軸：直線x=7上∴xM=7+4或7?4當(dāng)x=3時(shí)y當(dāng)x=11時(shí)y∴M(3,209②當(dāng)HE為以點(diǎn)MNHE為頂點(diǎn)的平行四邊形的對(duì)角線時(shí)如圖3則HEMN互相平分∵直線x=7平分HE點(diǎn)F在直線x=7上∴點(diǎn)M在直線x=7上即M為拋物線頂點(diǎn)∴∴M(7,4)綜上所述點(diǎn)M坐標(biāo)為(3,209)(11,13．（1）解：令y=0解得x1=?1∴A?1,0將C點(diǎn)的橫坐標(biāo)x=2代入y=?x2∴C(2,3)設(shè)直線AC的解析式為y=kx+b則3=2k+b0=?k+b解得：∴直線AC的函數(shù)解析式是y=x+1；（2）設(shè)P點(diǎn)的橫坐標(biāo)為x(?1≤x≤2)則PE的坐標(biāo)分別為：P(x,x+1)E(x,?∵P點(diǎn)在E點(diǎn)的下方PE=(?∴當(dāng)x=12時(shí)PE的最大值（3）存在這樣的點(diǎn)F①如圖連接點(diǎn)C與拋物線和y軸的交點(diǎn)那么CG∥x軸∴F點(diǎn)的坐標(biāo)是(?3,0)；②如圖則AF=CG=2∵A點(diǎn)的坐標(biāo)為(?1,0)∴F點(diǎn)的坐標(biāo)為(1,0)；③如圖此時(shí)CG兩點(diǎn)的縱坐標(biāo)互為相反數(shù)∴G點(diǎn)的縱坐標(biāo)為?3代入拋物線中即可得：G點(diǎn)的坐標(biāo)為(7+1設(shè)直線GF的解析式為y=x+?將G點(diǎn)代入后可得出直線GF的解析式為y=x?4?令y=0則x=4+因此直線GF與x軸的交點(diǎn)F的坐標(biāo)為(4+70)④如圖同③可求出F的坐標(biāo)為(4?70)綜上：存在這樣的點(diǎn)F坐標(biāo)為(?3,0)(1,0)(4+70)(4?714．（1）解：∵拋物線y=ax2+3∴16a+6+c=0c=6解得∴該拋物線的表達(dá)式為y=?3（2）解：∵y=?∴拋物線的對(duì)稱軸為直線x=1∵點(diǎn)A和點(diǎn)B(4,0)關(guān)于直線x=1對(duì)稱∴A(?2,0)∴AB=4?(?2)=6∵C(0,6)∴OC=6∴S如圖2過點(diǎn)P作PE∥y軸交BC于點(diǎn)E設(shè)BC所在直線的解析式為：y=kx+6過點(diǎn)B(4,0)∴k=?32即BC設(shè)Pt,?34∴PE=?∴S∴1∴S∵?∴當(dāng)t=2時(shí)S四邊形ABPC有最大值∴點(diǎn)P的坐標(biāo)為(2,6)．（3）解：拋物線的表達(dá)式為y=?34x2+∴y=?34x2+3①如圖所示四邊形BPNM為平行四邊形∴PN∥BM且PN=BM∴點(diǎn)N的縱坐標(biāo)為154y=?34x2+∴點(diǎn)N的坐標(biāo)為?1,∴PN=3?(?1)=4設(shè)點(diǎn)M(∵PN=BM∴4?m1=4則m1②如圖所示四邊形BPMN是平行四邊形過點(diǎn)P作PE⊥x軸于E過點(diǎn)N作NF⊥x軸于F∴PB∥MNPM∥BNPB=MN,PM=PN∴△BPM≌△MNB∴PE=NF=154且BE=FM=1,設(shè)M(∴??34n2+32當(dāng)n=1?14時(shí)n?m2=1即1?14?m2=1則m2=?∴點(diǎn)M的坐標(biāo)為(?14,0)或③如圖所示四邊形BNPM為平行四邊形∴PN∥MB,PN=MB=4B(4,0)∴設(shè)M(m3,0)∴m3=8即點(diǎn)M的坐標(biāo)為綜上所示點(diǎn)M的坐標(biāo)為(0,0)或(?14,0)或(1415．解：（1）將A(?1,0),B(3,0)代入y=?∴?1?b+c=0解得b=2∴y=?x（2）拋物線的對(duì)稱軸上存在點(diǎn)P使得△ACP的周長(zhǎng)最小理由如下：∵y=?∴拋物線的對(duì)稱軸為直線x=1∵AB點(diǎn)關(guān)于直線x=1對(duì)稱∴PA=PB∴△ACP的周長(zhǎng)=AC+AP+CP=AC+PB+CP≥AC+BC∴當(dāng)BCP三點(diǎn)共線時(shí)△ACP的周長(zhǎng)有最小值當(dāng)x=0時(shí)y=3∴C(0,3)∴AC=∴△ACP的周長(zhǎng)的最小值為10+3設(shè)直線BC的解析式為y=kx+m∴m=3解得k=?1∴y=?x+3∴P(1,2)（3）設(shè)M(x,?當(dāng)AC為平行四邊形的對(duì)角線時(shí)∴?1=x+n解得x=0n=?1（舍）或∴M(2,3)；當(dāng)AM為平行四邊形的對(duì)角線時(shí)∴?1+x=n解得x=0n=?1（舍）或∴M(2,3)；當(dāng)AN為平行四邊形的對(duì)角線時(shí)∴?1+n=x解得x=1+7n=2+∴M(1+7,?3)或綜上所述：M點(diǎn)橫坐標(biāo)為2或1+7或1?16．解：（1）（1）∵點(diǎn)A(?5,0)在拋物線y=?x∴0=?∴c=5即拋物線解析式為：y=?當(dāng)x=0時(shí)有y=5∴點(diǎn)C的坐標(biāo)為(0,5)；（2）過P作PE⊥AC于點(diǎn)E過點(diǎn)P作PF⊥x軸交AC于點(diǎn)H垂足為F如圖：∵A(?5,0)∴OA=OCAC=∴S當(dāng)PE取最大值時(shí)三角形ACP面積為最大值．∵OA=OC∴△AOC是等腰直角三角形∴∠CAO=45°∵PF⊥x軸∴∠AHF=45°=∠PHE∴△PHE是等腰直角三角形∴PE=∴當(dāng)PH最大時(shí)PE最大設(shè)直線AC解析式為y=kx+b將A(?5,0)C(0,5)代入得：?5k+b=0∴k=1∴直線AC解析式為y=x+5設(shè)Pm,?m2?4m+5∴PH=∵a=?1<0∴當(dāng)m=?52時(shí)PH∴此時(shí)PE最大為PE=∴△ACP面積的最大值：S即面積最大值為：1258（3）存在．∵y=?∴拋物線的對(duì)稱軸為直線x=?2設(shè)點(diǎn)N的坐標(biāo)為?2,m點(diǎn)M的坐標(biāo)為x,?分三種情況：①當(dāng)AM為平行四邊形ANMC的對(duì)角線時(shí)如圖∵A(?5,0)C(0,5)∴xC?x解得x=3.∴?∴點(diǎn)M的坐標(biāo)為3,?16②當(dāng)AN為平行四邊形ANMC的對(duì)角線時(shí)如圖方法同①可得x=?7∴?∴點(diǎn)M的坐標(biāo)為(?7,?16)；③當(dāng)AC為平行四邊形ANMC的對(duì)角線時(shí)如圖∵A(?5,0)C(0,5)∴線段AC的中點(diǎn)H的坐標(biāo)為?5+02,0+52∴x+(?2)2=?52∴?∴點(diǎn)M的坐標(biāo)為?3,8綜上點(diǎn)M的坐標(biāo)為：?3,8或3,?16或(?7,?16)．17．（1）解：當(dāng)x=0時(shí)y=∴點(diǎn)C的坐標(biāo)為0，令y=解得x=?4或x=2∴點(diǎn)A的坐標(biāo)為?4，0點(diǎn)B的坐標(biāo)為設(shè)直線AC的解析式為y=kx+b∴?4k+b=0∴k=?1∴直線AC的解析式為y=?x?4；（2）解：如圖所示延長(zhǎng)PD交x軸于F∵C0，∴OC=OA=4∴∠OAC=∠OCA=45°∵PD∥y軸即PD⊥x軸∴∠DFA=90°∴∠PDE=∠ADF=45°∵PE⊥AC即∠PED=90°∴∠DPE=45°=∠PDE∴DE=PE∴PD=∴S∴要使S△PDE最大則要使PD設(shè)Pm，1∴PD=?m?4?∵?∴當(dāng)m=?2時(shí)PD最大即S△PDE∴點(diǎn)P的坐標(biāo)為?2，（3）解：設(shè)M的坐標(biāo)為s，tN當(dāng)BC為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可知：2+0解得t=?4∴由對(duì)稱性可知點(diǎn)M與點(diǎn)C關(guān)于對(duì)稱軸對(duì)稱∴點(diǎn)M的坐標(biāo)為?2，當(dāng)BM為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可知：2+s解得t=?4∴由對(duì)稱性可知點(diǎn)M與點(diǎn)C關(guān)于對(duì)稱軸對(duì)稱∴點(diǎn)M的坐標(biāo)為?2，當(dāng)BN為對(duì)角線時(shí)由平行四邊形對(duì)角線中點(diǎn)坐標(biāo)相同可知：2+n解得t=4當(dāng)y=1解得x=?1?17或∴點(diǎn)M的坐標(biāo)為?1?17，4綜上所述點(diǎn)M的坐標(biāo)為?2，?4或?1?1718．（1）解：①把y=0代入y=x0=解得：x=±1∵A在B的左邊∴點(diǎn)A的坐標(biāo)為?1,0；把x=32代入y=∴點(diǎn)E的坐標(biāo)為：3∵四邊形ACDE為平行四邊形點(diǎn)C的坐標(biāo)是0∴設(shè)點(diǎn)D的坐標(biāo)為xD,y解得：x∴點(diǎn)D的坐標(biāo)為52②連接CE過點(diǎn)E作x軸的垂線垂足為M過點(diǎn)C作CN⊥EM垂足為N如圖所示：設(shè)點(diǎn)C的坐標(biāo)0,n點(diǎn)E的坐標(biāo)m,∵四邊形ACDE為平行四邊形∴將AC沿AE平移可與ED重合則點(diǎn)D的坐標(biāo)為：m+1,∵點(diǎn)D在拋物線上∴m解得：n=2m+1∴點(diǎn)C的坐標(biāo)為：∵?ACDE的面積是12∴S∵S△ACE=S∴6=整理得：m解得：m1=2∴點(diǎn)E的坐標(biāo)為：2,3．（2）解：∵拋物線的頂點(diǎn)坐標(biāo)為0,?1F是原點(diǎn)O關(guān)于拋物線頂點(diǎn)的對(duì)稱點(diǎn)∴點(diǎn)F的坐標(biāo)為：0,?2∵拋物線的對(duì)稱軸為y軸∴AB關(guān)于y軸對(duì)稱∴B設(shè)直線BF的解析式為y=kx+b則k+b=0解得：k=2∴直線BF的解析式為y=2x?2同理可得：直線AF的解析式為y=?2x?2設(shè)直線l的解析式為：y=tx+n聯(lián)立y=tx+n消去y得：x∵直線l與拋物線只有一個(gè)公共點(diǎn)∴△=解得：n=?聯(lián)立y=2x?2y=tx?t解得：x聯(lián)立y=?2x?2y=tx?t解得：x∵AB關(guān)