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

<br /> x_0=-\frac{f(x)-\tan(2\arctan(\frac{df}{dx}))x}{\tan(2\arctan(\frac{df}{dx}))}<br />

basicly 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:

<br /> \frac{f(x)}{x-x_0}=\tan(2\arctan(\frac{df}{dx}))<br />

 

[wormhole]

ok, i took the formula
<br /> \tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}<br />

using this formula i can simplify the equation to become:

<br /> \frac{2\frac{df}{dx}}{1-{(\frac{df}{dx})}^2}=\frac{f(x)}{x-x_0}<br />

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.