How do you convert $2=(-x+2y)^2-y-x$ into polar form?

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Answer 1 · 2018-02-11T08:23:32

$r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)-2=0$

Given:
$2=(-x+2y)^2-y-x$
$x=rcostheta$
$y=rsintheta$

Now,
$2=(-rcostheta+2rsintheta)^2-rsintheta-rcostheta$

$2=r^2(-costheta+2sintheta)^2-r(sintheta+costheta)$

$2=r^2((costheta)^2-2costheta(2sintheta)+(2sintheta)^2)-rsintheta-rcostheta$

$2=r^2cos^2theta-4r^2costhetasintheta+4r^2sin^2theta-rsintheta-rcostheta$

$2=r^2(1+cos2theta)/2-4r^2(sin2theta)/2+4r^2(1-cos2theta)/2-r(sintheta+costheta)$

Simplifying further,
$2=r^2(0.5+0.5cos2theta-2sin2theta+2-2cos2theta)-r(sintheta+costheta)$

$2=r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)$
Or

$r^2(2.5-1.5cos2theta-2sin2theta)-r(sintheta+costheta)-2=0$

How do you convert 2=(-x+2y)^2-y-x2=(-x+2y)^2-y-x into polar form?