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

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Answer 1 · 2016-12-08T14:11:21

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

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 · 2016-12-08T14:22:30

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

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]}