# How do you convert x^2-y^2=1 to polar form?

Apr 28, 2016

${r}^{2} = \sec 2 \theta$.

#### Explanation:

$\left(x , y\right) = \left(r \cos \theta , r \sin \theta\right)$

So, ${x}^{2} - {y}^{2} = {r}^{2} \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) = {r}^{2} \cos 2 \theta = 1$.

The semi-asymptotes are given by opposites $\theta = \frac{\pi}{4} \mathmr{and} \theta = \frac{5 \pi}{4}$ and, likewise, opposites $\theta = \frac{3 \pi}{4} \mathmr{and} \theta = \frac{7 \pi}{4}$.

In cartesian form, these equations are bundled to the simple

form $x \pm y = 0$, but the rotation effect given by $\theta$ is missing.. , .