Question

test for symmetry and graph the polar equation.
$r^{2}=64cos(2\theta)$
a. is the polar equation symmetrical with respect to the polar axis?
a. the polar equation failed the test for symmetry which means that the graph may or may
not be symmetric with respect to the polar axis.
b. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the polar axis.
c. yes.
b. is the polar equation symmetrical with respect to the line $\theta=\frac{pi}{2}$?
a. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the line $\theta=\frac{pi}{2}$.
b. yes.
c. the polar equation failed the test for symmetry which means that the graph may or may
not be symmetric with respect to the line $\theta=\frac{pi}{2}$.
c. is the polar equation symmetrical with respect to the pole?
a. yes.
b. the polar equation failed the test for symmetry which means that the graph is not
symmetric with respect to the pole.
c. the polar equation failed the test for symmetry which means that the graph may or may
not be symmetric with respect to the pole.

Explanation:

Step1: Test for symmetry about the polar axis ($\theta = 0$)

Replace $\theta$ with $-\theta$.
We know that $\cos(2(-\theta))=\cos(2\theta)$ (since $\cos(-x)=\cos(x)$).
So, $r^{2}=64\cos(2(-\theta)) = 64\cos(2\theta)$.
The equation remains the same, so it is symmetric about the polar axis.

Step2: Test for symmetry about the line $\theta=\frac{\pi}{2}$

Replace $(r,\theta)$ with $(-r,-\theta)$.
We get $(-r)^{2}=64\cos(2(-\theta))$, which simplifies to $r^{2}=64\cos(2\theta)$ (because $(-r)^{2}=r^{2}$ and $\cos(-x)=\cos(x)$).
The equation remains the same, so it is symmetric about the line $\theta = \frac{\pi}{2}$.

Step3: Test for symmetry about the pole ($r = 0$)

Replace $r$ with $-r$.
We get $(-r)^{2}=64\cos(2\theta)$, which simplifies to $r^{2}=64\cos(2\theta)$ (since $(-r)^{2}=r^{2}$).
The equation remains the same, so it is symmetric about the pole.

Answer:

a. C. Yes.
b. B. Yes.
c. A. Yes.