How to make mirrored parabola

Question

I try to figure it out how to make mirror of the parabola to some line.

For example like that:

In that example my original parabola is: $f_p(x) = x^2\ \ $ - (red line)

My mirror center line is: $f_c(x) = x\ \ \ $ - (green line)

And my mirrored parabola is: $f_m(x) = \sqrt { x }\ \ $ - (blue line - of course it's just half of parabola but it's enough for me)

But what in case if my mirror center line is for example: $f_c(x) = 0.3x$

And my parabola that I want to mirror is more complicated like: $f_p(x) = ax^2+bx+c$

How to make mirrored parabola to such line (just for remind: half of parabola would be great enough for me).

For any help thanks in advance.

Answer 1

For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = \frac{x}{a}$.

The intuition here is to swap the variables to achieve the mirroring: \begin{equation*} f(x) = y = ax^2 \iff \frac{y}{a} = x^2 \overset{\textrm{mirror}}{\implies} y^2 = \frac{x}{a}. \end{equation*} The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!). This covered all parabolas.

Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.

Answer 2

Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by: $$ x'={2my+(1-m^2)x\over1+m^2},\quad y'={2mx-(1-m^2)y\over1+m^2}. $$ To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get: $$ 4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0. $$