Question

Find the equivalent polar form of the cartesian equation:

$x^2-y^2=2ax$.

Answer 1

Given,

$x^2-y^2=2ax;---\left(1\right)$

Let, $x=n\cos \theta ,y=n\sin \theta$ in $\left(1\right)$

or, $n^2({\cos }^2\theta -{\sin }^2\theta )=\left(2an\cos \theta \right)$

or, $n^2\left(\cos \theta \right)=\left(2an\cos \theta \right)$

This is the equivalent polars equation

equation $\left(t\right)$.