Let $g(x)$ be the reflection of the continuous and differentiable curve $f(x)$ about the line $x\cos(\theta)+y\sin(\theta)=r$. Find $g(x)$. Also, examine the case in which a curve $f(x)$ is reflected about a curve $h(x)$ where $f(x)$ has a unique foot of the perpendicular on $h(x)$ for every point in its domain.
My progress:
I've learned the formula for reflection of a point $(x,y)$ in the line $ax+by+c=0$ in coordinate geometry as;
$\dfrac {X-x}{a}=\dfrac{Y-y}{b}=\dfrac{-2(ax+by+c)}{a^2+b^2}$
So, for a variable point $(t,f(t))$, the formula for the locus of the corresponding reflected point $(h,k)$ about the line will be;
$\dfrac{h-t}{\cos(\theta)}=\dfrac{k-f(t)}{\sin(\theta)}=-2(t\cos(\theta)+f(t)\sin(\theta)-r)$
$=>k-f(t)=(h-t)\tan(\theta) ...(i)$
And, $k=\cos(2\theta)f(t)-\sin(2\theta)t+2r\sin(\theta)...(ii)$
Similarly, $h=-[\sin(2\theta)f(t)+\cos(2\theta)t]+2r\cos(\theta)...(iii)$
Now, I'm stuck. How do I go from this $(h,k)$ notation to finding the function $g(x)$. I was also wondering if this would give us a new or interesting expression for the inverse of a function in terms of the function itself or if it will boil down to an equality of the kind $1=1$, by noting that the inverse is simply $f(x)$ reflected about the line $y=x$. Does anyone have any idea what I should do ahead of this? Also, I have no idea how to approach the general case, so I would really appreciate some help on that as well.
EDIT: I figured out how to do the case for a line; I just need to work backwards by taking the reflection of $(h,k)$ in the given line and substituting $t$ into $f(t)$ to get an implicit relation in $g(x)$.
$\dfrac{t-h}{\cos(\theta)}=\dfrac{f(t)-k}{\sin(\theta)}=-2(\cos(\theta)h+\sin(\theta)k-r)$
$=>f(t) = -\sin(2\theta)h+\cos(2\theta)k+2r\sin(\theta)$
And, $t=-\cos(2\theta)h-\sin(2\theta)k+2r\cos(\theta)$
Hence, $-\sin(2\theta)h+\cos(2\theta)k+2r\sin(\theta) = f(-\cos(2\theta)h-\sin(2\theta)k+2r\cos(\theta))$
$-\sin(2\theta)x+\cos(2\theta)g(x)+2r\sin(\theta) = f(-\cos(2\theta)x-\sin(2\theta)g(x)+2r\cos(\theta))$
There is a standard method of reflecting a point (x, y) about a line.
If the line equation is $\cos \theta \ x + \sin \theta \ y = r $, then the unit normal to this line is the vector $\hat{n} = (\cos \theta, \sin \theta) $ and a point on this line is $P = ( r \sec \theta , 0 ) $ if $ \cos \theta \ne 0 $ or $ P = ( 0, r \csc \theta ) $ if $ \cos \theta = 0 $.
The standard equation for reflecting $(x,y)$ about this line into the image $(x', y') $ is as follows
$ (x',y') = P + M ( (x, y) - P ) $
where
$ M = I_2 - 2 {\hat{n} \hat{n}}^T $
i.e.
$ M = \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} - 2 \begin{bmatrix} \cos^2 \theta && \cos \theta \sin \theta \\ \cos \theta \sin \theta && \sin^2 \theta \end{bmatrix} = \begin{bmatrix} - \cos(2 \theta) && - \sin(2 \theta) \\ - \sin(2\theta) && \cos(2 \theta) \end{bmatrix} $
Therefore, the reflection of $(x,y)$ is (assuming $\cos \theta \ne 0$ )
$$ \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} r \sec \theta \\ 0 \end{bmatrix} + \begin{bmatrix} - \cos(2 \theta) && - \sin(2 \theta) \\ - \sin(2\theta) && \cos(2 \theta) \end{bmatrix} \begin{bmatrix} x - r \sec \theta \\ y \end{bmatrix} $$
Noting that the inverse of $M$ is itself, then this last equation can also be written as
$$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} r \sec \theta \\ 0 \end{bmatrix} + \begin{bmatrix} - \cos(2 \theta) && - \sin(2 \theta) \\ - \sin(2\theta) && \cos(2 \theta) \end{bmatrix} \begin{bmatrix} x' - r \sec \theta \\ y' \end{bmatrix} $$
Now suppose that our function that we want to reflect is
$$ y = f(x) = x^2 + 1 $$
And that we want to reflect this function about the line $ \cos \theta \ x + \sin \theta \ y = 5 $, where $\theta = 120^\circ $. Then
$$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -10 \\ 0 \end{bmatrix} + \begin{bmatrix} \dfrac{1}{2} && \dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} && -\dfrac{1}{2} \end{bmatrix} \begin{bmatrix} x' + 10 \\ y' \end{bmatrix} $$
That is,
$ x = - 10 + \dfrac{1}{2}( (x' + 10 ) + \sqrt{3} y' ) $
$ y = \dfrac{1}{2} (\sqrt{3} (x' + 10 ) - y' ) $
Now, since $ y = x^2 + 1$ , then in terms of the image coordinates $(x',y')$ this means,
$ \dfrac{1}{2} (\sqrt{3} (x' + 10 ) - y' ) = 1 + \left( - 10 + \dfrac{1}{2}( (x' + 10 ) + \sqrt{3} y' ) \right)^2 $
To get rid of the $\dfrac{1}{2}$, multiply both sides by $4$, then
$ 2 (\sqrt{3} (x' + 10 ) - y' ) = 4 + \left( - 10 + x'+ \sqrt{3} y' ) \right)^2 $
Expanding,
$ 2 \sqrt{3} x' + 20 \sqrt{3} - 2 y' = 4 + 100 + x'^2 + 3 y'^2 - 20 x' - 20\sqrt{3} y' + 2 \sqrt{3} x' y' $
So that the final implicit equation of the reflected function is
$ x'^2 + 3 y'^2 + 2 \sqrt{3} x'y' - (20 + 2\sqrt{3} ) x' - ( 20 \sqrt{3} - 2 ) y' + 104 - 20 \sqrt{3} = 0 $
This equation related the $x'$ and $y'$ coordinates of the image (the reflection). To illustrate the correctness of this expression, I've plotted both the original function and the reflected implicit equation and also the line of reflection using Desmos on this desmos worksheet.