How do you convert $8=(6x-4y)^2+2y-x$ into polar form?

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Answer 1 · 2018-01-01T15:54:35

$r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0$

The relation between polar coordinates $(r,theta)$ and Cartesian coordinates $(x,y)$ is $x=rcostheta$ , $y=rsintheta$ and $x^2+y^2=r^2$ .

Hence $8=(6x-4y)^2+2y-x$ can be written as

$8=36x^2+16y^2-48xy+2y-x$

or $8=36r^2cos^2theta+16r^2sin^2theta-48r^2sinthetacostheta+2rsintheta-rcostheta$

or $8=20r^2cos^2theta+16r^2-24r^2sin2theta+2rsintheta-rcostheta$

or $r^2(20cos^2theta-24sin2theta+16)+r(2sintheta-costheta)-8=0$

or $r^2(10(cos2theta-1)-24sin2theta+16)+r(2sintheta-costheta)-8=0$

or $r^2(10cos2theta-24sin2theta+6)+r(2sintheta-costheta)-8=0$

How do you convert 8=(6x-4y)^2+2y-x8=(6x-4y)^2+2y-x into polar form?