解：可设A(a,a^2),B(b,b^2).则所求的距离d=(a^2+b^2)/2.由|AB|=4===>(a-b)^2+(a^2-b^2)^2=16.===>（2d+1/2)^2-16=(2ab+1/2)^2≥0.===>2d+1/2≥4.====>d≥7/4.===>dmin=7/4.
