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I understand how to reflect equations across the x and y axises as well as the line y=x. But how do you reflect an equation across more complex equations such as other slant lines and higher order equations?

Specifically I can easily reflect y=x^2 across y=x to produce the equation x=y^2. However how do you reflect y=x across the line y=x^2?

El Zapatilla
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    It depends on how you define reflection of a point. You might want to draw a straight line to the to the curve where it is perpendicular to a tangent and then continue the same distance the other side. – Henry Jun 16 '15 at 21:14
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    You cannot reflect an equation, but you can reflect the set of its solutions. The most obvious definition of reflection with respect to some curve is the one suggested by @Henry. Do note, though, that in general there will be some points for which the reflection isn't well defined: those are precisely the points where some of the perpendicular lines intersect. – A.P. Jun 16 '15 at 22:41
  • Sorry, yes my technical description wasn't very clear. I am considering reflection in the way that @Henry described. Is there a way of completing that kind of a transformation? – El Zapatilla Jun 17 '15 at 08:48
  • What do you mean by completing? In general, it may be hard to find the equation of the reflection of a curve with respect to another curve (as defined above). I'm not even sure that such an equation should always exist, i.e. if the reflection of an algebraic curve by another algebraic curve (when it is well defined) must again be an algebraic curve. – A.P. Jun 17 '15 at 11:04
  • I actually had a similar idea back in early 2015, I've already written a detailed answer here: https://math.stackexchange.com/questions/1837772/testing-for-symmetry-about-a-curve-line/4731735#4731735.

    Anyways, I can confirm that the curve provided by @Henry is indeed correct.

     – Edoardo Serra Jul 16 '23 at 18:42

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Take your example of trying to reflect the line $y=x$ in the parabola $y=x^2$.

Consider a point on the parabola $(t,t^2)$. The tangent there has gradient $2t$, so its normal can be expressed as $y=-\frac1{2t}x+t^2+\frac12$. This intersects the line $y=x$ at the point $\left(\dfrac{2t^3+t}{2t+1},\dfrac{2t^3+t}{2t+1}\right)$. So its reflected point has the parametric form $$\left(\dfrac{-2t^3+4t^2+t}{2t+1},\dfrac{2t^3+2t^2-t}{2t+1}\right)$$ which as far as I am concerned is the solution. I suspect trying to write this in the form $y=f(x)$ may involve solving at least a messy cubic equation.

The red curve below is what I think is the reflection of the green line in the blue parabola looks like. Much of it may seem counterintuitive, but at least the red curve passes through the points $(0,0)$ and $(1,1)$ as expected.

reflect line in parabola.png

Points on the green line above about $(3.0354,3.0354)$ get reflected three times in the parabola: in the right of the parabola onto the part of the higher red curve going to the top left, in the left of the parabola onto the lower red curve going to the top left, and in the part of the parabola roughly between $(-0.802, 0.644)$ and $(-0.5,0.25)$ onto the part of the red curve going to the bottom left. Points on the green line below about $(3.0354,3.0354)$ get reflected once near the centre of the parabola roughly between $(-0.5,0.25)$ and $(1.834, 3.363)$ onto the central loop and the part of the red curve going off to the top right.

Henry
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