Question

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

Answer 1

`( x^2 + y^2 )^1.5( x^2 - 4 sqrt( x^2 + y^2 ))+ x^4 = 0`

Explanation:

Using the conversion formula

`( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 + y^2 ) >= 0`,

`r = - sin^2theta + 4 csc^2theta in [0, oo ), theta ne kpi`,

`k = 0, +-1, +-2, +-3, ...` becomes

`x^2( x^2 + y^2 )^1.5 = - x^4 + 4 ( x^2 + y^2 )^2` giving

`( x^2 + y^2 )^1.5( x^2 - 4 sqrt( x^2 + y^2 ))+ x^4 = 0`

Graph of `r = - sin^2theta + csc^2theta`:
graph{(y-(x^4 + ( x^2 + y^2 )^1.5( x^2 -4 sqrt( x^2 + y^2 ))))=0[-40 40 -20 20]}

Observe the inner loop bracing the pole, from blow.
graph{(y-(x^4 + ( x^2 + y^2 )^1.5( x^2 -4 sqrt( x^2 + y^2 ))))=0[-2 2 -1.5 0.5]}