Question

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

Answer 1

`x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)`

Explanation:

Relation between polar coordinates `(r,theta)` and rectangular coordinates is given by `x=rcostheta`, `y=rsintheta`, `r^2=x^2+y^2` and `tantheta=y/x`.

Hence, `r=-csc^2theta+4sec^2theta`

`hArrr=-1/sin^2theta+4/cos^2theta` or

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

`1=-r/y^2+(4r)/x^2=(r(-x^2+4y^2))/(x^2y^2)` or

`x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)`

graph{x^2y^2=(4y^2-x^2)sqrt(x^2+y^2) [-25.1, 25.08, -12.55, 12.55]}