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

1 Answer
Nov 4, 2016

(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