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

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Answer 1 · 2017-02-06T09:15:37

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

This equation represents an ellipse. See graph.

graph{x^2+2y^2+3x-y=0 [-4, 1, -1, 1.5]}

Using the conversion formula #(x, y) = r(costheta, sintheta)#,

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

This includes pole r = 0, at two approach ( slope ) angles

#theta=psi=tan^(-1)3 and pi+tan^(-1)(3)#

In respect of this ellipse, it is easy to find center as

#C(-3/2, 1/4), a =sqrt(19/8), b=sqrt(19/16), e =1/sqrt2 # and a

focus at #S( -3/2+sqrt19/4, 1/4)#. The major axis axis is y = 1/4.

Now, the polar equation referred to S as pole and major axis as

#theta=0# is as simple as

#((sqrt152)/16)/r=1+1/sqrt2costheta#.

Interested readers can work out this simplification in polar using

befitting transformation in polar.