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Let us take the simple parabola $x^2$.

A ray of light will bounce on it making equal angles to both sides.

The derivative of $x^2$ is $2x$ and its normal is $-1/2x$.

Given light rays coming straight from above we get, given a variable b that represents the x value of the vertical ray we get that the reflected ray in general is:

$\left(b^2-y\right)=\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-x\right)$

Graph: https://www.desmos.com/calculator/tqf9a8fnsr

In fact we see that this always pass through a point (focal point) for all b by watching the graph while sliding b.

I want to prove all points pass there analitically.

I substitute in $b=1/2$ and $b=(\sqrt3)/2$ as examples and I get the intersection (0,1/4) (also clearly seen in the graph).

Now I substitute (x=0,y=1/4) back in $\left(b^2-y\right)=\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-x\right)$ and I get an equation that is true for all b as you can check by sliding the slider. $\left(b^2-\frac{1}{4}\right)-\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-0\right)$ This means that all reflected rays of light pass by it.

How can I prove this last statement analytically?

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Apply the following identities

$ tan(x+ \frac \pi 2)=-cot(x)$

$cot(2x) = \frac{ \cot^2(x)-1 }{ 2\cot(x) }$

$ \cot( \tan^{-1}(x) ) =\frac 1x $

Get $$ \tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right) \\ = -\cot( 2 \tan^{-1}(2b) ) \\ =\frac { 1-\cot^2( \tan^{-1}(2b) ) }{ 2\cot( \tan^{-1}(2b) } \\ = \frac{ 1-\frac 1{4b^2} }{ \frac 1b } \\ = b-\frac 1{4b}$$

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  • $\begingroup$ Could you please add a small explanation or link for cot(tan^-1(x)) identity? $\endgroup$ Sep 24 '19 at 5:50
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    $\begingroup$ $$ \cot(\tan^{-1}(x) ) = \frac 1{\tan(\tan^{-1}(x) } =\frac 1x $$ $\endgroup$
    – WW1
    Sep 25 '19 at 0:58
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Another answer shows you how to reduce the trigonometric expression in your equation, but trigonometric functions can be avoided entirely.

As you’ve determined, a normal vector to the parabola at $x=b$ is $\mathbf n=(2b,-1)$. Using a well-known reflection formula, the reflection of an incident ray with direction $\mathbf v = (0,-1)$ has direction $$-\left(2{\mathbf n\cdot\mathbf v\over\mathbf n\cdot\mathbf n}\mathbf n-\mathbf v\right) = \left({-4b\over1+4b^2},{1-4b^2\over1+4b^2}\right).$$ We can discard the common nonzero denominator to obtain the parameterization $(b,b^2)+t(-4b,1-4b^2)$, $t\ge0$ of the reflected ray.

Observe now that the ray for $b=0$ is reflected onto itself, so if all of the reflected rays do pass through a common point, it must lie on the $y$-axis. Unless $b=0$, for the $x$-coordinate to be zero we must have $t=1/4$, which yields $y=1/4$. Therefore, all of the reflected rays pass through $(0,1/4)$.

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Apply direct trig formulas.

Use negative sign for tangent/cotangent in second quadrant

and tan expansion of double angle:

$$ \tan[2 \tan^{-1}2b + \pi/2]$$

$$ =\dfrac{-1}{ \tan [2\tan^{-1}2b] }$$

$$ =\dfrac{-1}{\big[ \dfrac{2 \cdot 2b}{1-(2b)^2}\big] }$$

$$\dfrac{1-4b^2}{4b}.$$

Including domain $-1<x< 1$ and one or more incident/reflected more rays for problem posing... is sufficient..

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