Parameterizing lines reflected in a parabola Points reflected by a parabolic mirror create images that appear to be at specific positions on the other side of the mirror. I am attempting to use geometry to find the map from each point in space to the apparent position of their image.
Given the parabolic mirror $y(x) = x^2/(4f)$, my calculations [*] show that a point $\langle x, y\rangle$ on one side of the mirror will be reflected to an image point apparently at
$$\hat x = 2f \left[\frac{y-f}{x} + \sqrt{\left(\frac{y-f}{x}\right)^2 + 1}\right]$$
$$\hat y = f + \left(\frac{x^2/4f - f}{x}\right) \hat{x}$$
For example, the reflection of a grid takes the following apparent shape:

I am attempting to figure out whether the bowed curves created by the horizontal lines form a particular named shape.  To do so, I am trying to reparametrize the curve in terms of some $\hat{x}(t)$ and $\hat{y}(t)$, but I'm unsure of how to proceed. For example, I previously suspected that they might be confocal parabolas (though probably not) and would like to rearrange terms so as to show what they are.

[*] My calculations:
I found this formula using the property that a ray of light pointed at the focus will be reflected parallel to the axis of the parabola, and vice versa. Given the point $\langle x,y\rangle$, we drop one vertical (paraxial) line from $\langle x,y\rangle$, which is reflected at the point of intersection with the parabola to a line $\ell_1$ pointing toward the focus. We drop a second line from $\langle x,y\rangle$ itself toward the focus, which is reflected at the point of intersection with the parabola into a vertical (paraxial) line $\ell_2$. The intersection of $\ell_1$ and $\ell_2$ yields the apparent position of the image point $\langle \hat{x}, \hat{y}\rangle$. This method breaks down when $\langle x, y\rangle$ is situated on the axis of the parabola, because then both $\ell_1$ and $\ell_2$ are coincident vertical lines — but I expect this to be a manageable point discontinuity.
The line $\ell_1$ lies between $\langle 0, f\rangle$ and $\langle  x, x^2/4f\rangle$. The line $\ell_2$ is a vertical line positioned at wherever the ray from $\langle 0, f\rangle$ toward $\langle x, y\rangle$ intersects the parabola (solvable by quadratic equation).

Note that as a sanity check, the map fixes points on the parabola: if $y = x^2/(4f)$, then $x^2 = 4fy$ and first equation becomes:
\begin{align*}
\hat{x} &= \frac{2f}{x}\left[(y-f) + \sqrt{(y-f)^2 + x^2}\right]\\
&= \frac{2f}{x}\left[(y-f) + \sqrt{(y-f)^2 + 4fy}\right]\\
&= \frac{2f}{x}\left[(y-f) + \sqrt{(y+f)^2}\right]\\
&= \frac{2f}{x}\left[2y\right]\\
&= \frac{4fy}{x}\\
&= \frac{x^2}{x}\\
&= x.
\end{align*}
Similarly, 
\begin{align*}
\hat y &= f + \left(\frac{x^2/4f - f}{x}\right) \hat{x}\\
&= f + \left(\frac{x^2/4f - f}{x}\right) x\\
&= f + x^2/4f - f\\
&= x^2/4f\\
&= y
\end{align*}

New insight — there's a convenient change of coordinates we can make. Instead of rectilinear coordinates $\langle x,y\rangle$, we can use $\langle x, \alpha\rangle$, where $\alpha$ is the angle formed between the point, the focus, and the x-axis.  Using one rectilinear coordinate and one focal angle makes sense, because parabolas turn vertical lines into lines angled through the focus and vice-versa, with slope dependent on horizontal position.
We can specify any point in 2D by giving its coordinates $\langle x, \alpha\rangle$. Then the coordinates of the reflected map are simply:
$$\begin{align*}\widehat{x} &= 2f\left[\frac{1+\sin{\alpha}}{\cos{\alpha}}\right]\\\widehat{\alpha}& = \arctan \frac{\frac{1}{4}x^2-f}{x}\end{align*}$$
Wonderfully, $\widehat{x}$ depends only on $\alpha$, and $\widehat{\alpha}$ depends only on $x$.
 A: Here's a derivation based using a left-opening parabola with the focus at the origin, with $f$ the distance from vertex to focus. That is, we take the parabola to have Cartesian equation
$$4 f ( x - f ) + y^2 = 0 \tag{1}$$
In polar form:
$$r = \frac{2f}{1+\cos\theta} \tag{2}$$


Now, let $P = (p,q) = r (\cos\theta,\sin\theta)$ be our reflecting point. The horizontal line through $P$ meets the parabola at $A$; line $\overleftrightarrow{PF}$ meets the parabola at $B$. Then, the horizontal line through $B$ meets the $\overleftrightarrow{AF}$ at the reflected point, $P^\prime$. The reader can verify these calculations:
$$A =\left(\frac{4f^2-q^2}{4f},q\right)=\left(\frac{4 f^2 - r^2 \sin^2\theta}{4 f}, r \sin\theta\right) \qquad B = \frac{2f}{1+\cos\theta}\left(\cos\theta,\sin\theta\right) \tag{3}$$
so that
$$P^\prime = \left(\frac{4 f^2 - r^2 \sin^2\theta}{2r (1 + \cos\theta)}, \frac{2 f \sin\theta}{1 + \cos\theta}\right) = B - \frac{\left(\;r(1+ \cos\theta)-2f\;\right) \left(\;r(1-\cos\theta)+2f\;\right)}{2 r(1 + \cos\theta)}\;(1, 0) \tag{4}$$
Since points on the parabola satisfy $(2)$, we see immediately that, for such points, $P^\prime$ reduces to $B$, which coincides with $P$. $\square$
Note that the Cartesian form of $P^\prime$ is
$$P^\prime = \frac{1}{2\left(p+\sqrt{p^2+q^2}\right)}\;\left(\;4 f^2 - q^2, 4fq\;\right) = -\frac{p-\sqrt{p^2+q^2}}{2q^2}\;\left(\;4 f^2 - q^2, 4fq\;\right) \tag{5}$$
This can be put into agreement with OP's formulas by appropriate coordinate transformations. (Move the origin to the vertex via $p\to p+f$; then rotate $90^\circ$ via $p \to -q$ and $q\to p$.)

As for the reflections of lines perpendicular to the parabola's axis ... Such a line has polar equation $r = k\sec\theta$ for some $k$, so that the corresponding reflected curve comes from eliminating $\theta$ from the system
$$(x,y) \;=\; \left(\;\frac{4 f^2 \cos^2\theta - k^2\sin^2\theta}{2 \cos\theta (1 + \cos\theta)}, \frac{2 f \sin\theta}{1 + \cos\theta}\;\right) \tag{6}$$
The identity $\sin^2\theta+\cos^2\theta = 1$ lets us write our system in terms of $\cos\theta$. Eliminating that (with the help of Mathematica's Resultant[] command) yields
$$\left(2kx + y^2-4f^2\right)^2 = 4 k^2 \left(x^2+y^2\right) \tag{7}$$
Here are the associated curves for $k=f$, $2f$, and $3f$.

Whether such a curve has a name, I do not know. It is definitely not a parabola. 
Note that each curve is doubly-asymptotic with horizontal lines through the ends of the parabola's latus rectum (since the $y$ coordinate of $(6)$ approaches $2f$ as $\theta$ approaches $\pi/2$). Also, when $y=0$, equation $(7)$ reduces to $x = f^2/k$; as $k$ grows without bound, the "vertex" of the curve approaches the origin, i.e., the parabola's focus.
A: Proceeding according to your calculations [*] I got the transformation formulas a bit different with yours.
I got for the reflected point $(x_r,y_r)$ and the real point $(x,y)$
$$
x_r(x,y,f) = -\frac{2 f \left(\sqrt{(f-y)^2+x^2}+f-y\right)}{x}\\
y_r(x,y,f) = \frac{\left(\sqrt{(f-y)^2+x^2}+f-y\right) \left(f \left(\sqrt{(f-y)^2+x^2}+f+y\right)-\frac{x^2}{2}\right)}{x^2}
$$
Follows a plot for $f = 0.5$. The parabola is in red. The parabola focus is a black dot, In green is the real grid and in light blue is the reflected grid. It is shown also a green point (real) and a light blue point (reflected)
I hope this helps.

A: Actually, according to the sketch you made, thus with a negative $f$, the equation of the curves are:
$$
\left\{ \matrix{
  x = 2f\left( {{{v - f} \over u} \mp \sqrt {\left( {{{v - f} \over u}} \right)^{\,2}  + 1} } \right) \hfill \cr 
  y = f + {1 \over u}\left( {{{u^{\,2} } \over {4f}} - f} \right)x \hfill \cr}  \right.
$$
and for positive $u$ we shall take the $-$ sign and v.v..
It is then understood that for $u=0$ we shall take the limit of the above expressions.
Let's rewrite them putting $f =- h$, so not to get confusion with the signs
$$
\left\{ \matrix{
  h =  - f \hfill \cr 
  x = 2h\left( {\sqrt {\left( {{{v + h} \over u}} \right)^{\,2}  + 1}  - {{v + h} \over u}} \right) \hfill \cr 
  y =  - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x \hfill \cr}  \right.\quad \left| {\;0 < x,u,h} \right.
$$
Then you already have the parametric equations:   


*

*at constant $u$, you have the line
$$
y =  - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x
$$
passing through the focus;

*at constant $v$, you get
$$
\eqalign{
  & x = 2h\left( {\sqrt {\left( {{{v + h} \over u}} \right)^{\,2}  + 1}  - {{v + h} \over u}} \right) =   \cr 
  &  = {h \over {h + v}}u - {h \over {4\left( {h + v} \right)^{\,3} }}u^{\,3}  + O\left( {u^{\,5} } \right)  \cr 
  & y =  - h + {1 \over u}\left( {h - {{u^{\,2} } \over {4h}}} \right)x =   \cr 
  &  =  - {h \over {h + v}} - {{h^{\,2}  + h\left( {h + v} \right)^{\,2} } \over {4\left( {h + v} \right)^{\,3} }}u^{\,2}  + O\left( {u^{\,4} } \right) \cr} 
$$
and thus the curves are "simil-parabolas".
