Answer
The graphs of ${{r}^{2}}={{a}^{2}}\sin 2\theta $ and ${{r}^{2}}={{a}^{2}}\cos 2\theta $, $a\ne 0$, are shaped like propellers and are called lemniscates. The graph of ${{r}^{2}}={{a}^{2}}\sin 2\theta $ is symmetric with respect to the pole. The graph of ${{r}^{2}}={{a}^{2}}\cos 2\theta $ is symmetric with respect to the polar axis, the line $\theta =\frac{\pi }{2}$, and the pole.
Work Step by Step
After replacing $\theta $ with $-\theta $, if the resulting polar equation is the same as the given polar equation, then the polar curve of any polar equation is symmetric with respect to the polar axis.
It is given that
${{r}^{2}}={{a}^{2}}\sin 2\theta $ and ${{r}^{2}}={{a}^{2}}\cos 2\theta $
Replacing $\theta $ with $-\theta $, we get
$\begin{align}
& {{r}^{2}}={{a}^{2}}\sin 2\left( -\theta \right) \\
& {{r}^{2}}=-{{a}^{2}}\sin 2\theta
\end{align}$
and
$\begin{align}
& {{r}^{2}}={{a}^{2}}\cos 2\left( -\theta \right) \\
& {{r}^{2}}={{a}^{2}}\cos 2\theta
\end{align}$
Hence, only ${{r}^{2}}={{a}^{2}}\cos 2\theta $ is symmetric with respect to the polar axis.
After replacing $\left( r,\theta \right)$ with $\left( -r,-\theta \right)$, if the resulting polar equation is the same as the given polar equation, the polar curve of any polar equation is symmetric with respect to the line $\left( -r,-\theta \right)$ $\theta =\frac{\pi }{2}$
It is given that
${{r}^{2}}={{a}^{2}}\sin 2\theta $ and ${{r}^{2}}={{a}^{2}}\cos 2\theta $
Replace $\left( r,\theta \right)$ with $\left( -r,-\theta \right)$:
$\begin{align}
& {{\left( -r \right)}^{2}}={{a}^{2}}\sin 2\left( -\theta \right) \\
& {{r}^{2}}=-{{a}^{2}}\sin 2\theta
\end{align}$
and
$\begin{align}
& {{\left( -r \right)}^{2}}={{a}^{2}}\cos 2\left( -\theta \right) \\
& {{r}^{2}}={{a}^{2}}\cos 2\theta
\end{align}$
Hence, only ${{r}^{2}}={{a}^{2}}\cos 2\theta $ is symmetric with respect to the line $\theta =\frac{\pi }{2}$.
After replacing $r$ with $-r$, if the resulting polar equation is the same as the given polar equation, the polar curve of any polar equation is symmetric with respect to the pole.
It is given that
${{r}^{2}}={{a}^{2}}\sin 2\theta $ and ${{r}^{2}}={{a}^{2}}\cos 2\theta $
Replacing $r$ with $-r$ we get,
$\begin{align}
& {{\left( -r \right)}^{2}}={{a}^{2}}\sin 2\theta \\
& {{r}^{2}}={{a}^{2}}\sin 2\theta
\end{align}$
and
$\begin{align}
& {{\left( -r \right)}^{2}}={{a}^{2}}\cos 2\theta \\
& {{r}^{2}}={{a}^{2}}\cos 2\theta
\end{align}$
Hence, both ${{r}^{2}}={{a}^{2}}\sin 2\theta $ and ${{r}^{2}}={{a}^{2}}\cos 2\theta $ are symmetric with respect to the pole.
