(1) B(1,0) y=2x^2-2,不解释。(2)CO=2BO,BD=m-1,所以DP=2(m-1)或(m-1)/2所以P(m,正负2(m-1))或(m,正负(m-1)/2).(3)P在第一象限,PQ⊥PB,kPQ*kPB=-1当P(m,2(m-1))时,kBP=2设Q(t,2t^2-2),则kPQ=(2t^2-2m)/(t-m)=-1/2;整理得:4t^2+t-5m=0因为t>0,所以 t=[(-1)+√(1+80m)]/8; 当P(m,(m-1)/2)时,kBP=1/2设Q(t,2t^2-2),则kPQ=(2t^2-2-(m-1)/2)/(t-m)=-2整理得4t^2+4t-5m-3=0 ... |