parabola

(a) y = mx + b ........(1)
y^2 = 4ax ...........(2)
Sub (1) into (2)
(mx + b)^2 = 4ax
m^2x^2 + 2bmx + b^2 - 4ax = 0
m^2x^2 + (2bm - 4a)x + b^2 = 0
For the line to be a tangent, delta = 0
(2bm - 4a)^2 - 4m^2b^2 = 0
4m^2b^2 + 16a^2 - 16abm - 4m^2b^2 = 0
16a(a - bm) = 0
so (a - bm) = 0, b = a/m.
(b)
Let the other tangent be y = nx + c.......(3)
Sub (3) into (2) and put delta = 0, we will come to the same result as above, but this time is a - cn = 0, c = a/n.
Now n = - 1/m since the 2 tangents are perpendicular to each other, so c = a/(-1/m) = - am
so equation of the other tangent is : y = - x/m - am.
(c)
Since the intersecting point is on both lines :
y = mx + a/m............(4)
y = - x/m - am ..........(5)
From (4) my = m^2x + a ..........(6)
From (5) my = - x - am^2 ...........(7)
(6) - (7)
0 = m^2x + a + x + am^2
0 = m^2 ( x + a) + (x + a)
0 = (m^2 + 1)(x + a)
so the locus is x + a = 0 or x = - a.