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.