How do you convert #xy=x-2y+y^2 # into a polar equation?

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Answer 1 · 2016-02-20T06:54:38

#r=(cos theta-2*sin theta)/(sin theta* cos theta-sin^2 theta)#

just use the transformation equivalent

#x=r cos theta# and #y=r sin theta# in the equation

#xy=x-2y+y^2#

so that #xy=x-2y+y^2# becomes

#r cos theta* r sin theta=r cos theta-2* r sin theta+ (r sin theta)^2#

canceling some of the r

#r* cos theta* sin theta= cos theta-2* sin theta+ r sin^2 theta#

simplify to obtain

#r* (cos theta* sin theta- sin^2 theta)= cos theta-2* sin theta#

and

#r=(cos theta-2* sin theta)/(cos theta* sin theta- sin^2 theta)#

have a nice day!