Testing for Symmetry Test the polar equation for symmetry to the polar axis, the pole, and the line $\theta =\pi /2$

$r^2=4\cos 2\theta$ .

The given polar equation $r^2=4\cos 2\theta$.

If the polar equation $r^2=4\cos 2\theta ,(r,\theta )$ can be replaced

by $(r,-\theta )$ or $(-r,\pi -\theta )$ , then the graph is symmetric to the pole.

If the polar equation $r^2=4\cos 2\theta ,(r,\theta )$ can be replaced by $(-r,\theta )$ or $(r,\pi +\theta )$ , then the graph is symmetric to the pole.

If the polar equation $r^2=4\cos 2\theta ,(r,\theta )$ can be replaced by $(r,\pi -\theta )$ or $(-r,-\theta )$ , then the graph is symmetric to the line $\theta =\frac{\pi }{2}$.

The given polar equation is $r^2=4\cos 2\theta$

Testing for the symmetry of the line $\theta =\frac{\pi }{2}$ .

Replacing $(r,\theta )=(-r,-\theta )$ and simplifying the equation

If the resulting equation $r^2=4\cos 2\theta$ is equal to the original equation then it is symmetry about the line$\theta =\frac{\pi }{2}$ .

$\begin{array}{l}\text{Invalid <m:msup> element}=4\cos (-2\theta )\\ r^2=4\cos 2\theta \end{array}$

The polar equation $r^2=4\cos 2\theta$ is not symmetric to the line $\theta =\frac{\pi }{2}$ .

Replacing $(r,\theta )=(r,-\theta )$ and simplifying the equation $r^2=4\cos 2\theta$ then it is symmetry about the polar axis.

$\begin{array}{c}{r}^2=4\cos (-2\theta )\\ r^2=4\cos 2\theta \end{array}$

The polar equation $r^2=4\cos 2\theta$ is symmetric to the polar axis.

Replacing $(r,\theta )=(-r,\theta )$ and simplifying the equation

If the resulting equation is equal to the original equation $r^2=4\cos 2\theta$ is symmetric to the pole.

$\begin{array}{c}{(-r)}^2=4\cos 2\theta \\ r^2=4\cos 2\theta \end{array}$

The polar equation $r^2=4\cos 2\theta$ is symmetric to the pole.