How do you write polar equations of hyperbolas, from the polar equations of their asymptotes?

Jul 6, 2018

See the explanation. Details about easy to remember polar equations of asymptotes and hyperbola are included.

Explanation:

The polar equation of a straight line is

$r \cos \left(\theta - \alpha\right) = p \ge 0 ,$

$\theta \in \left(\alpha - \frac{\pi}{2} , \alpha + \frac{\pi}{2}\right)$, with $\left(p , \alpha\right)$ as the

foot of the perpendicular from the pole r = 0, on the straight line.

So, the polar equation of hyperbolas having

$r \cos \left(\theta - \alpha\right) = a$ and

$r \cos \left(\theta - \beta\right) = b$

as asymptotes is

$\left(r \cos \left(\theta - \alpha\right) - a\right) \left(r \cos \left(\theta - \beta\right) - b\right) = C$

Note that $\alpha , \beta , a , b \mathmr{and} C$ are parameters.

For example, consider the asymptotes $x \pm y = 1$.

Their polar equations are $r \left(\cos \theta \pm \sin \theta\right) = 1$.

So, the polar equations of the related hyperbolas are

$\left(r \left(\cos \theta + \sin \theta\right) - 1\right) \left(r \left(\cos \theta - \sin \theta\right) - 1\right) = C$.

Upon setting C = 1, this gives the hyperbola

$r \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) - 2 \cos \theta = 0$

that has the Cartesian form

${x}^{2} - {y}^{2} - 2 x = 0$,

and this in the standard form is

${\left(x - 1\right)}^{2} - {y}^{2} = 1$

Graph for the illustrative hyperbola

$\left(r \left(\cos \theta + \sin \theta\right) - 1\right) \left(r \left(\cos \theta - \sin \theta\right) - 1\right) = 1$.
graph{((x-1)^2-y^2-1)((x-1)^2-y^2)=0[-2 6 -2 2]}