高考数学压轴专题新备战高考《平面解析几何》经典测试题附答案解析

【高中数学】数学《平面解析几何》复习资料一、选择题rrrrrrrrrrrrr,b,cabab2,acbc1bc1a）75735-331．．．．2A222rrrra与ra0，b，cxy,=3br1rrracbc1y2x3y0x222y02x3y100x222.rrr，brrrrbr0，abab2=3a与arcx,y，r1rrracbc1y2x3y0x22，2r又br2cx2y2，10x2y2x3y2021305y2x3y0，x222221x2y2x3y02223575.212222x2,FC,B,CCC:y12F21242121AFBFC）122326．2．3．．2D|||2121|2||2221121232632＝F|)＝＝.21223ABCDAB4，BCCDAD2）33．．．3．2342B2(0)p.pypx2mBm，2(0)Cx2pyp123332p3p则242p(m2.x2y224C:1(a0,b0)Fya2bC，BO||（）5a||345233413．．．．AABABA,B||||5a3Cabc,,.bayxABABabayxcaabacabb20,B,yxc令x0C由由bbyxaccayxcabc5a2cbA,a得a2bab3byxa2222aa2c2b1ab4ab022222或abab9a22222ba1bab2C,B22ba故a2ab|FB|ybb24a242c.B|FC|yaccaba4a522222Cb.xy22F,Fb52b212yxP(3,)yPFPF)012．．204C，∴PFPF(23,1)(23,1)3410．12xy2216F、FM1246MFMF0MFNMFN）2112．122．242C6n2，MFmMFnMFMFm设设，1212NFtNF4tt6.21MFmMFn，，12xy22∵F、F11246na4，FF2c210.12∴mMF0，∵MF12∴MFMF，12∴mn4c40，222∴m即2mn401624，∴mn12，6n2nmn2，222m，，NFtNF2at4t，设21在RtNMFt262，4t221t6，∴∴MN628，11MFN86.12S12．x2y27C:1，OF为CFCPOQPQ（）4P,Q2468CQ90.OPQP，POFOPQQ90POF4OP23nPOQPOQ∴PQ3OP6.60，OPQ90OP23，C，QVPOQP.8CF,FCPFFP1212C）2354A3x,4x,5x.23xc4xc5x，和c3x,4x,5x.2c若c3x若c4x若c5x2a5x4xxe3；2a2c2a5x3x2xe2；2a2c2a4x3xxe5.2aA.x22y229－y2(，2ab(()25234345Apyx，22；1y；x，2则c55；．yPd021ddd）212655．．2．55Ay24xPFd,P1dMFdF1,034y90xd1223912ddMFd5122242minmin.F,F3xy3a(a0)222y28axP12）123x2x1．．x4．x．Cx2y2a2a2a2ca.1c2342a2a2.12PFPFPF6a.12PFPF2a212a3x8ax(xa.223a.Px,yx2y240bxayaC:ab00a2b21xx2yy2C002,4,．．．．Bbxay2a0bxay0d，xxyy1d1C2200xyb22C:1(a0,b0)yxab2a2bxay0，Px,ybxay4a0∵00c，bxay4a0bxay0da2b2xxyy1d1，22C004ac1e41∴eca，e故．C的右支没有公共点得出d1x22y2abFF，，AB12ab2FAB4FF23）ABFF于，，21212x2xy2xy2xy2222y1132119．．．．2396Cb2AB4FF23c3，4a，b，12ax22y21ab）F，FB12ab2b24，aFAB4FF23c3，ABFF于，，21212aba3，6，bc222xy221．96x2y21a0,b0,,M若FFa2b2121cosFMF,2()412123y2x3xyx．．y．yx．3AMFMF2aFM,,在1212.xy1ab022Q,,MFF2ab2122a12,:MF4a,MF22a2112F在:12222MFFFF12212121214(2c)(4a)(2a)24a2a:222:ca:bca4aaa222222:b3abxQ:ya:y3x.xy22FF110abP2a2b2PFxPFF3023与）212y2xy3x．．y3x．y2x．Bb2c1P2aba2c23，c3，Fc,0Pc,y∵∴，2b2ax22y21y，ab2b2∴，2aPFF30∵，12b2∴2，12ab22a，12aba则2ab2．22y2x．．Ba,b,c.．C:y4x为COF2PPF4VPOF3．3．．．22By24xP(x,y)由x(1)4即x3P1S|y|.24xx1,由y2:Px1,M=4,x3将x3yx423,设P(x,y)x(1)4则2y11|y|2313POF.=.22P点P.A，，CAC20BABABxCAPx2722yP16436A到APBPB）P2,,．．．77773217,45,162．3Bxy22到APB1xaa2b2a15到APB272x2y2x2y1x.1P3664xy22到APB1xaa2b2a15c=17b8到APBx2y2到APB1x2xy21x135322P,77x2y21xB.x2y21，9点BAAA.CABAA．.ABCDDDaABCDEF11111()面//面ABEC1BF1FC1111a2a2a．a．．．22D设，I、CDFHICC1H11HI.CD，，I、、GCDH111则ABEG又//，A1QF/