解：用点差法＋共线。A(x1,y1),B(x2,y2),AB中点M(xo,yo),焦点F(2,0).则有x1+x2=2xo,y1+y2=2yo,又A,B在曲线上有y1^2=8x1,y2^2=8x2,两式相减得AB斜率k=(y1-y2)/(x1-x2)=8/(y1+y2)=4/yo=tana,得yotana=4.又AB,MP共线得k(AB)=k(MP),即k=yo/(xo-2)=4/yo,得yo^2=4xo-8.易得AB中垂线方程y=-(1/tana)(x-xo)+yo,令y=0,得P点横坐标xP=xo+yotana=xo+4.于是得|FP|=xP-xF=xo+2.由于1-cos2a=1-(cos^2a-sin^2a)=1-(cos^2a-sin^2a)...
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