What the is the polar form of #y = y/x-(x+y)^2 #?

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Answer 1 · 2016-01-23T09:05:31

#r^2(1+sin 2theta)+r sin theta-tan theta=0#

To convert, use #y=r sin theta# and #x=r cos theta#

replace all #x# and #y# the equivalents

#y=y/x-(x+y)^2#
#r sin theta=(r sin theta)/(r cos theta)-( r cos theta+ r sin theta)^2#

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

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

#r sin theta= tan theta-r^2(1+sin 2theta)#

#r^2(1+sin 2theta)+r sin theta-tan theta=0#