How do you convert $2=(2x+6y)^2-y-3x$ into polar form?

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Answer 1 · 2016-10-07T12:29:54

Pola fom is $4r^2(1+8sin^2theta+6sinthetacostheta)-r(sintheta+3costheta)-2=0$

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

Therefore $2=(2x+6y)^2-y-3x$ can be written as

$2=(2rcostheta+6rsintheta)^2-rsintheta-3rcostheta$

or $2=r^2(4cos^2theta+36sin^2theta+24sinthetacostheta)-r(sintheta+3costheta)$

or $2=r^2(4+32sin^2theta+24sinthetacostheta)-r(sintheta+3costheta)$

or $4r^2(1+8sin^2theta+6sinthetacostheta)-r(sintheta+3costheta)-2=0$

How do you convert 2=(2x+6y)^2-y-3x2=(2x+6y)^2-y-3x into polar form?