Asymptotes are the lines which touch the curve at infinity.
Putting $u=\frac{1}{r}$, then
\begin{equation*}
\begin{split}
u &= \dfrac{\theta \cos \theta}{a \cos 2\theta} = F(\theta)\\
\end{split}
\end{equation*}
When $r \rightarrow \infty$, $u \rightarrow 0$, or
\begin{equation*}
\begin{split}
\dfrac{\theta \cos \theta}{a\cos 2 \theta} &= 0\\
\theta \cos \theta &= 0\\
\theta &= 0, \pm \dfrac{\pi}{2}
\end{split}
\end{equation*}
i.e. when $r \rightarrow \infty$, $\theta \rightarrow 0, \pm \dfrac{\pi}{2}$. So,
\begin{equation*}
\begin{split}
\theta_1 &= 0, \pm \dfrac{\pi}{2}
\end{split}
\end{equation*}
Differentiating $F(\theta)$ w.r.t $\theta$,
\begin{equation*}
\begin{split}
F'(\theta) &= \dfrac{a\cos 2 \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (\theta \cos \theta) – \theta \cos \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (a\cos 2\theta)}{a^2 \cos^2 2\theta}\\
F'(\theta) &= \dfrac{a\cos 2\theta (\cos \theta – \theta \sin \theta) + 2a\theta \cos \theta \sin 2 \theta}{a^2\cos^2 2 \theta}\\
\end{split}
\end{equation*}
The equation of the asymptote in case of polar curves is given by,
\begin{equation*}
\begin{split}
r \sin (\theta – \theta_1) &= \dfrac{1}{F'(\theta_1)}
\end{split}
\end{equation*}
So in our case, asymptotes are,
Plotting reveals that while first two are asymptotes, the third one does not seem to be.
My approach looks fine but why is this discrepancy between graph and my solution?
Best Answer
You miscalculated $F'\left(-\dfrac \pi 2\right)$. It is actually $\dfrac{-\pi}{2a}$. You will find it is the same line as for $\theta = \dfrac \pi 2$.These asymptotes are actually rays, not lines. $\dfrac \pi 2$ is the asymptote for the top side of the graph and $-\dfrac \pi 2$ is the asymptote for the bottom side of the graph.
However, there are more asymptotes that you missed. Each of those climbing curves has its own asymptote.