What the is the polar form of #x^2-y^2 = 6x-x^2y-y #?

1 Answer
Shwetank Mauria
Nov 30, 2017

#r^2cos^2thetasintheta+rcos2theta+sintheta-6costheta=0#

Explanation:

The relation between polar coordinates #(r,theta)# and Cartesian or rectangular coordinates #(x,y)# is given by

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2#

Hence we can write #x^2-y^2=6x-x^2y-y# as

#r^2cos^2theta-r^2sin^2theta=6rcostheta-r^3cos^2thetasintheta-rsintheta#

or #r^3cos^2thetasintheta+r^2cos2theta+r(sintheta-6costheta)=0#

or #r^2cos^2thetasintheta+rcos2theta+sintheta-6costheta=0#

Related questions
See all questions in Converting Between Systems
Impact of this question
1730 views around the world
You can reuse this answer
Creative Commons License