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