x= 2at and y=at^2 are the parametric equations of the parabola x^2=4ay
At t= 1, x=2a and y=a ------------> Let this be the point P (2a,a)
At t= -1, x= -2a and y=a -----------> Let this be the point Q (-2a,a)
Thats just the basic working first
Gradient of PQ= a-a/2a - (-2a)
= 0
Therefore the gradient of the line PQ is horizontal, i.e. parallel to the X-axis. (This is the answer to ii))
Now using the point gradient formula, to find the equation of PQ
y-a = 0(x-2a)
y-a=0
y=a --------------> (*)
Therfore the equation of PQ is y=a.
Now it is known that the focal point of of the parabola x^2=4ay is (0,a), and this focal point clearly passes through the line y=a (*). Therefore the line PQ is a focal chord, as it passes through the focal point (0,a)
