The locus of the middle points of the parallel chords is a line parallel to the axis of the parabola

Eliminating $$t,\,x = {\left( {\frac{{y - 1}}{2}} \right)^2} + 1{\text{ or }}{\left( {y - 1} \right)^2} = 4\left( {x - 1} \right)$$
Putting $$y - 1 = Y,\,x - 1 = X$$     the equation becomes $${Y^2} = 4X$$
So, the equation of the directrix is $$X + 1 = 0\,\,\,\,\, \Rightarrow x = 0.$$

Parabola $${x^2} = 16y,$$   axis is the $$y$$-axis, focus $$ = \left( {0,\,4} \right)$$
It is clear if point $$P\left( {a,\,2a} \right)$$   lies interior region then ordinate $$ \in \left( {0,\,4} \right) \Rightarrow 0 < 2a < 4 \Rightarrow 0 < a < 2$$

The sum of the ordinates of the feet of three normals $$ = {y_1} + {y_2} + {y_3} = 0.$$

$$\eqalign{ & {\text{Here, }}C = \left( {\frac{{2at_1^2 + at_2^2}}{3},\,\frac{{4a{t_1} + 2a{t_2}}}{3}} \right) \cr & {\text{It line on }}y = 0 \cr & \therefore \,\,\frac{{4a{t_1} + 2a{t_2}}}{3} = 0 \cr} $$

An end of the latus rectum $$ = \left( {a,\,2a} \right).$$   The equation of the tangent at $$\left( {a,\,2a} \right)$$  is $$y.2a = 2a\left( {x + a} \right),$$    i.e., $$y = x + a.$$   The normal at $$\left( {a,\,2a} \right)$$  is $$y + x = 2a + a,$$   i.e., $$y + x = 3a.$$   Solving $$y = 0$$  and $$y = x + a,$$   we get $$x = - a,\,y = 0.$$    Solving $$y = 0,\,y + x = 3a,$$    we get $$x = 3a,\,y = 0.$$   The area of the triangle with vertices $$\left( {a,\,2a} \right),\,\left( { - a,\,0} \right),\,\left( {3a,\,0} \right)$$     is $$4{a^2}.$$

The equation of the normal at $$\left( { - 2 + {t^2},\,2t} \right)$$   is
$$\eqalign{ & y - 2t = \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( { - 2 + {t^2},\,2t} \right)}}}}.\left( {x + 2 - {t^2}} \right) \cr & {\text{or }}y - 2t = - \frac{{2t}}{2}.\left( {x + 2 - {t^2}} \right) \cr & {\text{or }}tx + y = 2t - 2t + {t^3} \cr & {\text{or }}tx + y = {t^3} \cr} $$
It passes through $$\left( {k,\,0} \right)$$  if $$tk = {t^3}{\text{ or }}t\left( {{t^2} - k} \right) = 0$$
It has three real values of $$t$$ if $$k > 0.$$  So, the set of points $$\left( {k,\,0} \right)$$  is such that $$k > 0.$$

If $$a$$ and $$b > 0,$$  then graphic representation would be as follows :

$${S_2}$$ will contain $${S_1},$$
if latus rectum of $${S_2} > $$  latus rectum of $${S_1}$$
$$\eqalign{ & \Rightarrow 4b > 4a \cr & \therefore \,b > a > 0 \cr} $$

Let $$R\left( { - 4a,\,k} \right)$$   be any point on the line $$x = - 4a.$$   The equation of chord of contact $$PQ$$  w.r.t. $$P\left( { - 4a,\,k} \right)$$   is

$$y.k = 2a\left( {x - 4a} \right)......\left( 1 \right)$$
Making equation of parabola $${y^2} = 4ax$$   homogeneous using $$\left( 1 \right),$$ we get
$$\eqalign{ & {y^2} = 4ax\left( {\frac{{2ax - yk}}{{8{a^2}}}} \right) \cr & \Rightarrow 8{a^2}{x^2} - 8{a^2}{y^2} - 4akxy = 0 \cr} $$
This represents the pair of straight lines $$AP$$  and $$AQ.$$  Since coefficient of $${x^2} + $$  coefficient of $${y^2} = 0$$
$$\therefore \,\angle PAQ = {90^ \circ }$$    i.e. chord of contact $$PQ$$  subtends a right angle at the vertex.

Let the equation of circle be
$${x^2} + {\left( {y - k} \right)^2} = {r^2}$$
It touches $$x-y=0$$

$$\eqalign{ & \Rightarrow \left| {\frac{{0 - k}}{{\sqrt 2 }}} \right| = r \cr & \Rightarrow k = r\sqrt 2 \cr} $$
$$\therefore $$ Equation of circle becomes
$${x^2} + {\left( {y - k} \right)^2} = \frac{{{k^2}}}{2}.....({\text{i}})$$
It touches $$y = 4 - {x^2}$$   as well
$$\therefore $$ Solving the two equations
$$\eqalign{ & \Rightarrow 4 - y + {\left( {y - k} \right)^2} = \frac{{{k^2}}}{2} \cr & \Rightarrow {y^2} - y\left( {2k + 1} \right) + \frac{{{k^2}}}{2} + 4 = 0 \cr} $$
It will give equal roots $$\therefore $$ $$D=0$$
$$\eqalign{ & \Rightarrow {\left( {2k + 1} \right)^2} = 4\left( {\frac{{{k^2}}}{2} + 4} \right) \cr & \Rightarrow 2{k^2} + 4k - 15 = 0 \cr & \Rightarrow k = \frac{{ - 2 + \sqrt {34} }}{2} \cr & \therefore r = \frac{k}{{\sqrt 2 }} = \frac{{ - 2 + \sqrt {34} }}{{2\sqrt 2 }} \cr} $$