Question

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

Answer 1

`(x^2+y^2)(y^2-4*root2(x^2+y^2))+y^3=0`

Explanation:

Let's rewrite the given function in polar coordinates exclusively in term of `sin(theta)` so that it turns into `R=-sin(theta)+4/sin^2(theta)`
By recalling that `x=R*cos(theta)` and `y=R*sin(theta)`, it follows that `sin(theta)=y/R` and `R=root2(x^2+y^2)`.

By replacing both of them it into the given function we get
`root2(x^2+y^2)=-y/root2(x^2+y^2)+4(x^2+y^2)/y^2`

After some simple algebraic manipulations we get
the implicit form of the curve on the x-y plane

`(x^2+y^2)y^2+y^3-4(x^2+y^2)root2(x^2+y^2)=0`

in other terms

`(x^2+y^2)(y^2-4*root2(x^2+y^2))+y^3=0`