$y^{2} = 4ax$ is a parabola with focus at $(a,0)$ and vertex at origin.
$-y^{2} = 4a(x-2a)$ is a parabola with focus at $(a,0)$ and vertex at $(2a,0)$.
$y^{2} = 4\frac{a}{2}(x-\frac{a}{2})$ is a parabola with focus at $(a,0)$ and vertex at $(\frac{a}{2},0)$.
I know that $y^{2} = 4a(x-b)$, where $a$ and $b$ are arbitrary constants that can take any real value, represents a family of parabolas along the $X$ axis (a coaxial family of parabolas). Similarly, is there any way to represent (an equation or something) to represent the family of all parabolas with focus at $(a,0)$ (a confocal family of parabolas)? Also, how to represent the family of all parabolas with focus at $(a,0)$ and having the vertex on $X$ axis (a confocal and coaxial family of parabolas)?
Let the focus be $F = (a, 0)$, and let the vertex be $V = (b , 0)$ where $b \ne a $, then
Define $p = a - b $
Then the equation of the parabola with focus $F$ and vertex $V$ is
$ 4 p (x - b) = y^2 $
Substitute $p$,
$ 4 (x - b) (a - b) = y^2 \tag{1}$
And this equation represents all parabolas having a fixed focus $(a, 0)$ and whose axis is along the $X$ axis.
Now if you rotate this parabola by an angle $\theta$ about the focus $F$, then the image $(u,v)$ of the point $(x, y)$ which is on the original parabola is given by
$ (u , v) = (a, 0) + R(\theta) ( x - a, y ) \tag{2}$
where
$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} \tag{3}$
Solving for $(x,y)$ in terms of $(u,v)$ we get
$ (x, y) = (a, 0) + R^{-1}(\theta) ( u - a , v ) \tag{4} $
where
$ R^{-1}(\theta) = $ R = \begin{bmatrix} \cos \theta && \sin \theta \\ -\sin \theta && \cos \theta \end{bmatrix} \tag{5}$
That means that
$ x = a + \cos \theta ( u - a) + \sin \theta v \tag{6}$
$ y = - \sin \theta (u - a) + \cos \theta v \tag{7}$
Since the point $(x,y)$ is on the original parabola, it satisfies its equation, so by plugging in the above two equations into the equation of the parabola that we derived earlier (equation $(1)$), we get
$4 (a - b) ( \cos \theta ( u - a) + \sin \theta v + a - b) = (- \sin \theta (u - a) + \cos \theta v )^2 \tag{8}$
At the final stage, replace the variable name $u$ with $x$ and the variable name $v$ with $y$ , to obtain,
$ 4 (a - b) ( \cos \theta ( x - a) + \sin \theta y + a - b) = (- \sin \theta (x - a) + \cos \theta y )^2 \tag{9}$
For example, suppose $a = 1, b = 0.5 $ and $\theta = \dfrac{\pi}{4}$, then by feeding this equation to Desmos it generates the following plot. You can change the rotation parameters $a$ and $b$ and the angle $s$ to view all possible parabolas that are all confocal at $(a, 0)$.