<s>[INST] Parabolic mirror: why is $b^2-1/4 = \tan(2\tan^{-1}(2b)+{\pi}/{2})(b-0)$ for all $b$? 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?
 [/INST] 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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