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#