焦点为(1,0),所以p=2,抛物线方程为y^2=4x
a=1时,点斜式(y-0)/(x-1)=2 解得y=2x-2
代入得(2x-2)^2=4x 化简得x^2-3x+1=0
设A(x1,2x1-2) B(x2,2x2-2)
韦达定理得x1+x2=3 x1x2=1
AB的距离为sqrt((x1-x2)^2+(2x1-2-2x2+2)^2)=sqrt((x1+x2)^2-4x1x2+(2x1-2x2)^2)
=sqrt((x1+x2)^2-4x1x2+4((x1+x2)^2-4x1x2)=sqrt(5((x1+x2)^2-4x1x2))
=sqrt(5*(3^2-4*1))=sqrt(25)=5
AB的长为5