Question

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

Answer 1

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

Explanation:

for polar conversion

`r^2=x^2+y^2`

`x=rcostheta`

`y=rsintheta`

for

`y=x-2y+x^2-3y^2`

substituting from above.

`rsintheta=rcostheta-2rsintheta+r^2cos^2theta-3r^2sin^2theta`

cancelling and rearranging for `""r`

`cancel(r)sintheta=cancel(r)costheta- 2cancel(r)sintheta+cancel(r^2)^rcos^2theta-3cancel(r^2)^rsin^2theta`

`sintheta=costheta-2sintheta+rcos^2theta-3rsin^2theta`

`sintheta-costheta+2sintheta=rcos^2theta-3rsin^2theta`

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

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

Answer 2

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

Explanation:

The conversion formula is `(x, y)=r(cos theta, sin theta)`

Making substitutions and reorganizing for the explicit form

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

This represents a hyperbola, with center at (-1/2, -1/2). See graph.

graph{x^2-3y^2+x-3y=0 [-2.5, 2.5, -1.25, 1.25]}