So F(0, p/2) means F(0, 1/4a), and the directrix l:y=-p/2 means y=- 1/4a.

(1) The slope of the straight line L does not exist.

It's easy to have only one intersection, it doesn't matter.

(2) Set a straight line L:y=kx+ 1/4a (over-focus).

x 1 =(k+sqrt(k2+ 1))/2a，x2 = (k-sqrt (k 2+ 1))

Then y1= (2k2+2ksqrt (k2+1)+1)/4a, y2 = (2k2-2ksqrt (k2+1)+1)/.

Let A(x 1, y 1) and B(x2, y2).

By "the distance from the point on the parabola to the focus is equal to the distance to the directrix"

af=- 1/4a-y 1=-(k^2+ksqrt(k^2+ 1)+ 1)/2a

bf=- 1/4a-y2=-(k^2-ksqrt(k^2+ 1)+ 1)/2a

So1/AF+1/BF =-(2a/(k2+1+ksqrt (k2+1))+2a/(k2+1-ksqrt (k

=-2 * 2a * (k2+1)/(k2+1) (equally divided)

=-4a

To sum up, 1/AF+ 1/BF=-4a.