一道例题的变式与探究

张忠旺尹玉玲(201600)(061001):Mk(k0),M.56()70.,.,,,,.,,:AOBM,Mk(k0),M.1O,x,y,1,Mxy=k(x0).:1?1NAOBM,Mk(k0),M.kNAOBM,2NAOBM,MPLOAP,MQLOBQ,OPMQk,M.kNAOBM,3NAOBM,MOA,OBME,MF,OPMQk,M.NAOB=2A,O,NAOBx,2,2(1)M(x,y),OAy=xtanA,OBy=-xtanA,MMPLOAP,MQLOBQ,|MP|=xtanA-y1+tan2A=xsinA-ycosA,|MQ|=xsinA+ycosA.|MP||MQ|=kMx2sin2A-y2cos2A=kx\ksinA.3(2)M(x0,y0),MPy-y0=-cotA(x-x0),xcotA+y-x0cotA-y0=0;MQxcotA-y-x0cotA+y0=0.|MP|=x0sinA-y0cosA,|MQ|=x0sinA+y0cosA,|OP|=x0cotA+y01+cot2A=x0cosA+y0sinA,|OQ|=x0cosA-y0sinA.OPMQS=SvOPQ+SvMPQ=12(|OP|#|OQ|+|MP|#|MQ|)sin2A=12(x20-y20)sin2A=k.x20-y20=2ksin2A,Mx2-y2=2ksin2Ax\2ksin2A.(3)M(x,y),MMPLOAP,MQL48数学通报2007年第46卷第1期OBQ,|MP|=xsinA-ycosA,|MQ|=xsinA+ycosA,SOEMF=|MF||ME|sin2A=|MP|sin2A#|MQ|sin2A#sin2A=1sin2A(x2sin2A-y2cos2A)=kMx2sin2A-y2cos2A=ksin2A(x\2kcotA).21?(1)M?(2)M,P,Q,MPMQ?(3)MMEMF,MEMF?(1)a2b2a2+b2;(2),x2-y2=a2,12a2sin2A;(3)12ab;(3).x2a2-y2b2=1,bx-ay=0,bx+ay=0,M(x0,y0),P,Q.OPA,tanA=ba,|MP|=bx0-ay0a2+b2,|MQ|=bx0+ay0a2+b2,sin2A=2aba2+b2,SOEMF=|MF||ME|sin2A=|MP|sin2A#|MQ|sin2A#sin2A=b2x20-a2y20a2+b2#a2+b22ab=12ab.31OM,OEF,OEF?EF?,SvOEF=2k,MEF,M(x0,y0),E(2x0,0),F(0,2y0)EFx0y+y0x=2x0y0,xy=kx0y+y0x=2x0y0x=x0,y=y0.4EF,EF.54,3?MEF,,MEF.x2a2-y2b2=1,bx-ay=0,bx+ay=0,M(x0,y0),Mbx2x0x-a2y0y=a2b2.A(x1,y1),B(x2,y2),bx-ay=0b2x0x-a2y0y=a2b2,x1=a2bbx0-ay0y1=ab2bx0-ay0x2=a2bbx0+ay0y2=-ab2bx0+ay0ABM.bx-ay=0A,tanA=ba,sin2A=2aba2+b2,SvOAB=12|OA||OB|sin2A=12aba2+b2|bx0-ay0|#aba2+b2|bx0+ay0|#2aba2+b2=a3b3b2x20-a2y20=ab.54,,?lP,Q,E,F,EP=QF.PQEF,.492007年第46卷第1期数学通报64?NAOBE,F,OEFS,EFM?6NAOB=2A,O,NAOBx,,E(x1,x1tanA),F(x2,-x2tanA),M(x,y),|OE|=x1cosA,|OF|=x2cosA,SvOEF=12|OE||OF|sin2A=12#x1x2cos2Asin2A=x1x2tanA.x1x2tanA=S(1)MEF,x1+x2=2x(x1-x2)tanA=2yx1+x2=2x(2)x1-x2=2ycotA(3)(2)(3)x1x2=x2-y2cot2A(4)(1)(4)x1x2x2sin2A-y2cos2A=S2sin2A(x\ScotA).776NAOBE,F,OEFS(1)vOEFG?(2)vOEFH?(3)vOEFI?(1)x2sin2A-y2cos2A=2Ssin2A9x\23ScotA,().(2)NAO=2A,,,E(x1,x1tanA),F(x2,-x2tanA),M(x,y),OFy-x1tanA=1tanA(x-x1)(1)OEy+x2tanA=-1tanA(x-x2)(2)(1)(2)x=(1-tan2A)(x1+x2)2y=(1-tan2A)(x2-x1)2tanA(3)x1x2tanA=S(4)(3)(4)x1x2x2-k2y2=2k(1-ktanA)2tanA,x2cos2A-y2sin2A=2Scos2Acot2A0bA45b,Hx2cos2A-y2sin2A=2Scos2Acot2A(x\(1-tan2A)ScotA);A=45b,H(0,0);45bA90b,Hx2cos2A-y2sin2A=2Scos2Acot2A(x[(1-tan2A)ScotA);(3)Ix2cos2A-y2sin2A=S2sin2Ax\(1+tan2A)2ScotA().8NAOBE,F,OEFS.(1)EF?,?(2)EFM?(3)vOEFGHI?(1)E(x1,x1tanA),F(x2,x2tanA),x1x2=ScotA,|EF|=(x1-x2)2+[(x1+x2)tanA]2\4x1x2tan2A=4ScotA#tan2A\2StanAx1=x2=ScotA,EF.(2)EFMScotA;(3)vOEFGHI2ScotA3,|1-tan2A|ScotA,1+tan2A2ScotA.50数学通报2007年第46卷第1期