0
$\begingroup$

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 $x$, $y$, $y^\prime$."

  • 0
    Search for parabolic mirror – 2017-02-17
  • 1
    How hard is it to type the text of the photo into the question box so that it comes out consistent with the other text of this site and not sideways? – 2017-02-17
  • 0
    As well, photos of text aren’t searchable. – 2017-02-17
  • 1
    The problem is a little vague, is it like this imagin https://en.wikipedia.org/wiki/Parabolic_reflector#/media/File:Focus-balanced_parabolic_reflector.svg – 2017-02-18
  • 0
    Thank you for the link. It further led me to the geometric proof. But G N Berman asks for a calculus proof, using maxima minima concept. – 2017-02-18

1 Answers 1

2

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, $$ y=ax^2-\frac1{4a}\tag{5} $$