Two equal parabolas with foci at S and S' touch each other at point P, such that PS'=PS. If the parabola with focus S is fixed, find the locus of S' 
Two equal parabolas with foci at S and S' touch each other at point P,
such that PS'=PS. If the parabola with focus S is fixed, find the
locus of S'

Hint given: Let the common tangent be L. PS and PS' are equally inclined to L which implies that P,S,S' are collinear or SS' is perpendicular to L. Thus the locus will be a parabola or directrix of a fixed parabola
My attempt:

I can't figure out how to prove that PS and PS' are equally inclined to L. I've assumed the fixed parabola to be a standard one, with vertex at the origin. If we join S and S' and construct a perpendicular from P to SS', we can prove the two triangles are congruent. Thus S,S',P have to be collinear, or SS' is perpendicular to L.
Case 1:S,S',P are collinear

Assume P to be $(at^2,2at)$ and S to be (h,k)
Then, $(h-a)^2+k^2=2a(t+1)^2$, where t is the parameter. This looks like it will give a parabola upon plotting the locus, but I can't think of a proof.
Case 2: SS' is perpendicular to L
Then SS' will be the normal to the parabola, and the mirror image of the focus of the parabola about the tangent lies on the directrix. So the locus o S' in this case will be a straight line.
 A: Let's take parabola $y^2 = 4 ax$ as the fixed parabola. The point you need to note is that both parabola are equal. In other words, the other parabola is $y^2 = 4 ax$ but rotated and / or translated.
On your question as to why the angles are same while you can show using algebra it is not really necessary. Here is some insight that may help. There are two ways to get the common tangent -
First - you fix point $P$ on the original parabola $y^2 = 4 ax$ and then rotate it around point $P$ till the tangents at $P$ align. The new parabola will be opening up in the negative x-direction.
Second - we take $P'$ on the original parabola such that it is mirror image of point $P$ about x-axis. We now shift the parabola such that $P'$ comes to point $P$ and then rotate the parabola counter-clockwise till tangents align and the new parabola also opens towards positive x-direction.

As mentioned above, there will be two equal parabola at a given point sharing the common tangent  - one that opens towards positive x direction and the other towards negative x direction. $S, P$ and $S'$ will be collinear for the parabola that opens towards negative x-direction and $S S'$ will be perpendicular to the tangent line for the parabola that opens towards positive x-direction.
Case $a)$ - $S, P$ and $S'$ are collinear
As $P (a t^2, 2 a t)$ is the midpoint of segment $SS'$,
$at^2 = \frac{a + x_0}{2}, 2 at = \frac{y_0}{2}$
So coordinates of $S'$ is $(2at^2 -a, 4 a t)$ and the locus can be written as,
$x = 2 a \cdot \left(\frac{y}{4a}\right)^2 - a \implies y^2 = 8 a (x + a) $
For case $b)$, you can geometrically show it or find the intersection of the tangent line at $P(at^2, 2at)$ and the perpendicular line through $S$. As the intersection is the midpoint, you can find coordinates of $S'$ which comes to $(-a, 2 a t)$ and that is directrix of the fixed parabola.
