# 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:

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

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?

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

• Could you please add a small explanation or link for cot(tan^-1(x)) identity? Sep 24 '19 at 5:50
• $$\cot(\tan^{-1}(x) ) = \frac 1{\tan(\tan^{-1}(x) } =\frac 1x$$
– WW1
Sep 25 '19 at 0:58

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 $$(0,1/4)$$.

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 and one or more incident/reflected more rays for problem posing... is sufficient..