y'=x/2

Let: the coordinates of a (t, t 2/4)

Then, the slope of the tangent passing through point A is t/2, and the point is inclined:

Y-t 2/4 = t/2 (x-t), finishing:

Y = t * x/2-t 2/4, bringing the coordinates of p into the equation:

Finishing: t 2-3t-4 = 0

Solve t 1=4, t2=- 1.

So a (4,4), B(- 1, 1/4)

PA:y=2x-4

PB:y=- 1/2*x- 1/4