Light rays from a point source, reflected from a searchlight mirror, are directed as a parallel beam. What surface of revolution is the mirror? The question comes from GN Berman, and hints given are as follows: "Paraboloid of revolution. Let the plane $Oxy$ be a meridian plane of the mirror surface, the required line lies in this plane. The needed differential equation is derived by equating the tangents of the angles of incidence and reflection expressed in terms of $x$, $y$, $y^\prime$."

![enter image description here](

From the diagram, $$ \tan(\theta)=\frac xy\tag{1} $$ and $$ \tan(\theta/2)=-\frac{\mathrm{d}y}{\mathrm{d}x}\tag{2} $$ Using the formula for $\tan(2\theta)$, we get $$ \frac xy=\frac{-2\frac{\mathrm{d}y}{\mathrm{d}x}}{1-\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\tag{3} $$ Equation $(2)$ is satisfied by $y=ax^2-b$, where $$ \begin{align} \frac{x}{ax^2-b} &=\frac{-4ax}{1-4a^2x^2}\\\ &=\frac{x}{ax^2-\frac1{4a}}\tag{4} \end{align} $$ Therefore, $b=\frac1{4a}$. That is, $$ y=ax^2-\frac1{4a}\tag{5} $$