How can I find a shape of a mirror which focuses all parallel beams in one point? I tried to do it in this way:
The mirror must be symmetric hence I assumed it has a center in the point $(0,0)$. The point which focuses all the beams is $ (0,a), \, a>0$.
Now
$$L_{1}:\frac{-1}{f^{'}(t)}x +C $$
is a line perpendicular to tangent to a mirror (a mirror given by a function $f: \mathbb{R} \rightarrow \mathbb{R}$ )at point $(t,f(t))$
$$L_2:t$$
is a beam
and
$$L_3:\frac{f(t)-a}{t}x +a $$
is a line going through both $(0,a)$ and $(t,f(t))$.
Now the requirement that angle between $L_1 $ and $L_2 $ and angle between $L_1 $ and $L_3 $ are equal should give a proper differential equation. Unfortunately, it is not. Am I doing any mistakes? Is there an easier way to do it?
Suppose $y=f(x)$ is the equation of the mirror. Let $\alpha$ be the angle between the tangent line at $(x_0,f(x_0))$ and incident ray. Clearly, $\tan\alpha=\frac1{f'(x_0)}$. Also, the angle between reflected ray and $x$-axis is $\frac{\pi}{2}+2\left(\frac\pi 2-\alpha\right)$. The equation of the reflected ray is thus $$\frac{y-f(x_0)}{x-x_0}=\tan\left(\frac{3\pi}{2}-2\alpha\right)=\frac{f'(x_0)^2-1}{2f'(x_0)}.$$ All of these rays (parameterized by $x_0$) should pass by focus $(0,a)$, hence the differential equation $$\frac{f(x_0)-a}{x_0}=\frac{f'(x_0)^2-1}{2f'(x_0)}.$$ This has as the general solution the $1$-parameter family of symmetric parabolas $$f(x_0)=\frac{x_0^2}{4C}+a-C,$$ corresponding to different focal distances.