Parameterizing lines reflected in a parabola

Question

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.

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[*] 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).

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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*}

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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 $x$.

Answer 1

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: $$r = \frac{2f}{1+\cos\theta} \tag{2}$$

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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 $q\to p$.)

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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.

Answer 2

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.

Answer 3

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".