y=(t^2+ t+ 1)x^2-2(a+ t)^2x+ t^2+ 3at+ b对任何实数t,都经过定点P(1,0)
即有t^2+t+1-2(a^2+2at+t^2)+t^2+3at+b=0
(1-a)t+1+b-2a^2=0
故有1-a=0,1+b-2a^2=0
即有a=1,b=1
y=(t^2+t+1)x^2-2(1+t)^2x+t^2+3t+1
令y=0得到x1+x2=2(1+t)^2/(t^2+t+1),x1x2=(t^2+3t+1)/(t^2+t+1)
|x1-x2|^2=4(1+2t+t^2)^2/(t^2+t+1)^2-4(t^2+3t+1)/(t^2+t+1)
=[4(t^2+1)^2+16t(t^2+1)+16t^2-4(t^2+1)^2-16t(t^2+1)-12t^2]/(t^2+t+1)^2
故有|x1-x2|=2t/(t^2+t+1)=2/(t+1/t+1)