Question

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

Answer 1

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

Explanation:

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.