How do you convert #r = 12/(4 + 8 sinx)# into cartesian form?

1 Answer
Shwetank Mauria
Nov 5, 2017

#x^2-3y^2+12y-9=0#

Explanation:

As #x# is used in Cartesian or rectangular corrdinate, I would use #theta# for angle in place of #x#. The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is given by

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2# and #theta=tan^(-1)(y/x)#

Hence #r=12/(4+8sintheta)#

or #4r+8rsintheta=12#

or #4sqrt(x^2+y^2)=12-8y#

or #sqrt(x^2+y^2)=3-2y#

or #x^2+y^2=9+4y^2-12y#

or #x^2-3y^2+12y-9=0#

which isthe equation of hyperbola.
graph{x^2-3y^2+12y-9=0 [-9.75, 10.25, -3, 7]}

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