Answer
symmetric with respect to the pole, the polar axis, and the line $\theta=\frac{\pi}{2}$.
Work Step by Step
1. Given $r^2cos\theta=5$, to test symmetry with respect to the pole, replace $r$ with $-r$, we have $(-r)^2cos\theta=5$ which does not change the original equation, thus it is symmetric with respect to the pole.
2. To test symmetry with respect to the polar axis, replace $\theta$ with $-\theta$, we have $(r)^2cos(-\theta)=5$ which does not change the original equation, thus it is symmetric with respect to the polar axis.
3. To test symmetry with respect to the line $\theta=\frac{\pi}{2}$ (y-axis), rewrite the equation as $(\pm\sqrt {x^2+y^2})x=5$, we can see that it is symmetric with respect to the y-axis ($\theta=\frac{\pi}{2}$) because $\pm x$ gives the same value.
