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?
 A: 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 $$   \tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)   
\\ = -\cot( 2 \tan^{-1}(2b)  )
\\ =\frac { 1-\cot^2(  \tan^{-1}(2b)  )  }{  2\cot(  \tan^{-1}(2b)   } 
\\ = \frac{ 1-\frac 1{4b^2}  }{ \frac 1b  }
\\ = b-\frac 1{4b}$$ 
A: 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 $(0,1/4)$.
A: 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..
