The general parabola $y = f(x)$ with a given root $p$ (i.e., given $x$-coordinate of the endpoint) has the form
$$y = f(x) = - A (x - p) (q - x),$$
where $A$ and $q$ are arbitrary constants. The condition that the parabola passes through $(a, b)$ is that $f(a) = b$, so that
$$\boxed{A(a - p)(a - q) = b} .$$
By symmetry the vertex of the parabola has $x$-coordinate $\frac{p + q}{2}$, so the requirement that the parabola has (for $A > 0$) maximum height (i.e., $y$-coordinate) $h$ there is $f\left(\frac{p + q}{2}\right) = h$, or
$$\boxed{A (p - q)^2 = -4 h} .$$
Multiplying the equations and canceling $A$ gives $$-4h (a - p) (a - q) = b (p - q)^2 .$$ This is a quadratic equation in $q$, and expanding, collecting like terms in $q$, and applying the quadratic formula gives an expression for $q$ in terms of $a, b, h, p$. Substituting this expression in either of the above boxed equations (the first is easier to use, since it is linear, not quadratic, in $q$) and solving gives an expression for $A$. The result is: $$\boxed{A = \frac{z}{(a - p)^2}, \qquad q = \frac{a z + (p - a) b}{z}, \qquad \textrm{where} \qquad z := b - 2 h + 2 \sqrt{h (h - b)}}.$$
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.
Best Answer
Given a point $(a,b)$ and a horizontal line $y=k$, in $\mathbb{R^2}$.
Let the locus of the points which are equally far away from the point and the line be denoted by $(x,y)$.
Then distance between $(x,y)$ and the line is just $|y-k|$
The distance between the point $(x,y)$ and $(a,b)$ is $\sqrt{(x-a)^2 + (y-b)^2}$
Equating the two:
$|y-k|=\sqrt{(x-a)^2 + (y-b)^2}$
$(y-k)^2=(x-a)^2+(y-b)^2$
Expanding and rearranging: $y=\frac{x^2-2ax+(a^2+b^2-k^2)}{2b-2k}$
Given any quadratic function, you can find the unique value for $a,b,k$ (thus the diretrix and focus).