Parabola - three tangents and the focus?

I could not solve it on my own, but found the solution from a book on co-ordinate geometry by S. L. Loney which is reproduced as under.

The equations of tangents to the parabola y^2 = 4ax at the points

P(at^2, 2at), Q(at'^2, 2at') and R(at"^2, 2at") are

y = (1/t) (x + at^2) ... ( 1 )

y = (1/t') (x + at'^2) ... ( 2 ) and

y = (1/t") (x + at"^2) ... ( 3 ).

Their points of intersection are

A [att', a(t + t')], B [at't", a(t' + t")] and C [at"t, a(t" + t)]

If they lie on the circle

x^2 + y^2 + 2gx + 2fy + c = 0, then

a^2t^2t'^2 + a^2 (t + t')^2 + 2gatt' + 2fa(t + t') + c = 0 ... (4)

a^2t'^2t"^2 + a^2 (t' + t")^2 + 2gat't" + 2fa(t' + t") + c = 0 ... (5)

a^2t"^2t^2 + a^2 (t" + t)^2 + 2gat"t + 2fa(t" + t) + c = 0 ... (6)

Subtracting (5) from (4) and dividing by a (t - t"), we have

a [t'^2 (t + t") + t + t" + 2t'] + 2gt' + 2f = 0

Similarly, from ( 5 ) and ( 6 ),

a [t"^2 (t' - t) + t + t' + 2t"] + 2gt" + 2f = 0

Solving,

2g = - a (1 + tt' + t't" + t"t) and 2f = - a (t + t' + t" - tt't")

Substituting in ( 4 ),

c = a^2 (tt' + t't" + t"t)

=> the equation of the circumcircle is

x^2 + y^2 - ax (1 + tt' + t't" + t"t) - ay (t + t' + t" - tt't") + a^2 (tt' + t't" + t"t) = 0

which clearly passes through the focus (a, 0).

What are you like, mathematician testing the peeps sitting out in yahoo answers?

Cant draw in here.. thought i'd mention main things,

To proove lamberts theorem

Assuming, the tangents at the extremities of a chord passing through focus F, meet on the directrix. This means that the directrix is the polar of the focus, while the focus is the pole of the directrix with respect to the parabola.

Let the tangent at C intersect tangents AX and BX in points Y and Z, respectively. Theorem of similar triangles, applied twice, gives ∠FXY= ∠FBX = ∠FVY,

which tells us that the quadrilateral XYFX is cyclic.