y²= 4a(x+a)

so 2yy^' = 4a so a = yy^'/2

on putting a we get y²= 4yy^'/2(x+yy^'/2)      

so y² = yy^' (2x+yy^')  or y = 2xy^' + yy^'2    (1)

now on putting -1/y^' in the place of y^'

we get y² = -y/y^'[2x-y/y^']

so -yy^'2 = 2xy^' - y (2)

similarity of (1) and (2) shows that the given curve is self orthogonal