Problem source

\begin{questionparts}
		\item 
Show that in polar coordinates, the gradient of any curve at the point
$(r,\theta)$ is 
\[
\left.\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\tan\theta+r\right)\right/\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}-r\tan\theta\right).
\]
\begin{center}
\begin{tikzpicture}
    % Horizontal line L
    \draw (0,0) -- (6.54,0);
    \node at (-0.47,0.07) {$L$};
    
    % Point O
    \node [right] at (4.13,-0.22) {$O$};
    
    % Parabola (rotated 270 degrees in the original)
    \draw plot[domain=-3:3, samples=200] ({5.75-\x*\x/6}, {\x});
    
    % Horizontal line segment
    \draw (2,1.5) -- (5.42,1.5);
    
    % Line from bottom to top point
    \draw (3.73,-0.74) -- (5.42,1.5);
    
    % Arrow on horizontal segment
    \draw[-{Stealth[length=3pt]}] (3,1.5) -- (4,1.5);
    
    % Arrow on downward segment
    \draw[-{Stealth[length=3pt]}] (5.42,1.5) -- (4.99,0.93);
    
    % Final line segment
    \draw (3.84,0.78) -- (6.62,2.05);
\end{tikzpicture}
\end{center}
\item A mirror is designed so that any ray of light which hits one side of the mirror and which is parallel to a certain fixed line $L$ is reflected through a fixed point $O$ on $L$. For any ray hitting the mirror, the normal to the mirror at the point of reflection bisects the angle between the incident ray and the reflected ray, as shown in the figure. Prove that the mirror intersects any plane containing $L$ in a parabola.
\end{questionparts}

Solution source

\begin{questionparts}
\item Suppose our curve is $r(\theta)$, then $y = r \sin \theta, x = r \cos \theta$ and

\begin{align*}
&& \frac{\d y}{\d \theta} &= \frac{\d r}{\d \theta} \sin \theta + r \cos \theta \\
&& \frac{\d x}{\d \theta} &= \frac{\d r}{\d \theta} \cos \theta - r \sin \theta \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{\d y}{\d \theta} \Bigg / \frac{\d x}{\d \theta} \\
&&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r \cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r \sin \theta} \\
&&&= \frac{\frac{\d r}{\d \theta} \tan\theta + r }{\frac{\d r}{\d \theta}  - r \tan\theta}
\end{align*}
as required.
\item Set up a system of polar coordinates such that the origin is at $O$ and all points in the plane containing $L$ are represented by $(r, \theta)$.

The constraint we have is that the angle of the normal, is $\frac12 \theta$. Let $\tan \tfrac12 \theta = t$, then $\tan \theta = \frac{2t}{1-t^2}$

\begin{align*}
&& \tan \frac12 \theta &= -\frac{\frac{\d r}{\d \theta}  - r \tan\theta}{\frac{\d r}{\d \theta} \tan\theta + r } \\
\Rightarrow && t &= -\frac{r'-r\frac{2t}{1-t^2}}{r' \frac{2t}{1-t^2}+r} \\
&&&= \frac{2tr-(1-t^2)r'}{2tr'+(1-t^2)r} \\
\Rightarrow && (2t^2+1-t^2)r' &= (2t-t+t^3)r \\
&& (1+t^2)r' &= t(t^2+1) r \\
\Rightarrow && r' &= t r \\
\Rightarrow && \frac{\d r}{\d \theta} &= \tan \tfrac12 \theta r \\
\Rightarrow && \int \frac1r \d r &= \int \tan \frac12 \theta \d \theta \\
&& \ln r &= -2\ln \cos \tfrac12 \theta+C \\
\Rightarrow && r\cos^2 \frac12 \theta &= C \\
\Rightarrow && r + r\cos \theta &= D \\
\Rightarrow && r &= D-x \\ 
\Rightarrow && x^2 + y^2 &= D^2 - 2Dx + x^2 \\
\Rightarrow && y^2 &= D^2-2Dx
\end{align*}

Therefore it is a parabola
\end{questionparts}