Finding the Mirror Shape for Focusing Light: What's the Next Step?

[wormhole]

i'm trying to find a mirror shape which focuses a light at some specific point $x_0$
the initial equation i derived for determining the shape of the mirror is:
(assuming that light rays fall parallel to x-axis - light source is very far from the mirror)
f(x) is the shape I'm trying to determine

$$x_0=-\frac{f(x)-\tan(2\arctan(\frac{df}{dx}))x}{\tan(2\arctan(\frac{df}{dx}))}$$

basically this is an expression for a line passing through point $x_0$ and point on
f(x) where light reflected.
so $\tan(2\arctan(\frac{df}{dx}))$ is a incline of this line


from the initial equation i got to this point and I'm not sure what to do next:

$$\frac{f(x)}{x-x_0}=\tan(2\arctan(\frac{df}{dx}))$$

 

[wormhole]

ok, i took the formula
$$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$$

using this formula i can simplify the equation to become:

$$\frac{2\frac{df}{dx}}{1-{(\frac{df}{dx})}^2}=\frac{f(x)}{x-x_0}$$

is it possible to solve it?

 

[qbert]

i get $f = \pm \sqrt{ (x+c)^2 - (x-x_0)^2 }$ as
a solution. (where c is an arbitrary constant, hopefully
not the same as -x₀.)

 

[HallsofIvy]

i get $f = \pm \sqrt{ (x+c)^2 - (x-x_0)^2 }$ as
a solution. (where c is an arbitrary constant, hopefully
not the same as -x₀.)

Well, I'm impressed with what you have done so far!
I assume you mean the light is focussed at the point (x₀,0).

You might as well multiply out the terms in the square root:
$$f(x)= \sqrt{x^2+ 2cx+ c^2- x^2+2x_0x- x_0^2}$$

You can't determine c from the information given: any parabola with focus at (x₀,0) will focus light as required.
And don't write $f(x)= \pm \sqrt{...}$. A function is single valued. You can, of course, write x as a function of y.