How do you convert #y=-y^2-3x^2-xy # into a polar equation?

1 Answer
May 26, 2018

#r=-(sintheta)/(sin^2theta+3cos^2theta+costhetasintheta)#

Explanation:

Rewrite as:
#y^2+3x^2+xy=-y#

Substitute in:
#x=rcostheta#
#y=rsintheta#

#(rsintheta)^2+3(rcostheta)^2+(rcostheta)(rsintheta)=-rsintheta#

#r^2(sintheta)^2+3r^2(costheta)^2+r^2(costhetasintheta)=-rsintheta#

Divide both sides by #r#

#r(sintheta)^2+3r(costheta)^2+r(costhetasintheta)=-sintheta#

Factorise out #r#:
#r(sin^2theta+3cos^2theta+costhetasintheta)=-sintheta#

Make #r# the subject:
#r=-(sintheta)/(sin^2theta+3cos^2theta+costhetasintheta)#