x^2=4y的焦点为(0,1),设过焦点(0,1)的直线为y=kx+1则令kx+1=x^2/4,即x^2-4kx-4=0由韦达定理得x1+x2=4k,x1x2=-4y1=k(x1)+1,y2=k(x2)+2所以y1+y2=k(x1+x2)+2=4k^2+2=6,所以k^2=1所以|AB|=|x1-x2|*根号(k^2+1)=根号{(k^...