Question

What is the Cartesian form of `r^2+theta = -tan^2theta+4sec^2theta`?

Answer 1

We know that cartesian to polar coordinates is made by

`x=rcostheta`
`y=rsintheta`

From these , we have then `r^2=x^2+y^2` and `theta=arctan(y/x)` and `y/x=tantheta` and `sectheta=r/x=sqrt(x^2+y^2)/x`

`r^2+theta=-tan^2theta+4sec^2theta`

`x^2+y^2+arctan(y/x)=-y^2/x^2+4(x^2+y^2)/x^2`