How do you convert # r= - sin theta costheta +cos^2theta# to Cartesian form?

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Answer 1 · 2016-08-03T05:49:48

If the polar coordinate of a point having cartesian coordinate #(x,y) be (r,theta)# then they are related as follows.

#x=rcostheta ,y=rsintheta and r=sqrt(x^2+y^2)#

Now given polar equation is

#r=-sinthetacostheta+cos^2theta#

#=>r^3=-rsintheta*rcostheta+r^2cos^2theta#

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

#=>(x^2+y^2)^3=(x^2-xy)^2#

#=>x^6+y^6+3x^4y^2+3x^2y^4-x^4+2x^3y-x^2y^2=0#