How do you convert $9=(2x+y)^2-3y-x$ into polar form?

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Answer 1 · 2018-04-27T07:39:25

$r=9/(2(cos^2theta+1)+2sin(2theta)-3sintheta-costheta)$

We will use:
$x=rcostheta$
$y=rsintheta$

$9=(2rcostheta+rsintheta)^2-3rsintheta-rcostheta$

$9=r((2costheta+sintheta)^2-3sintheta-costheta)$

$r=9/((2costheta+sintheta)^2-3sintheta-costheta)$

$r=9/(4cos^2theta+4costhetasintheta+2sin^2theta-3sintheta-costheta)$

$r=9/(2(2cos^2theta+sin^2theta)+2sin(2theta)-3sintheta-costheta)$

$r=9/(2(cos^2theta+1)+2sin(2theta)-3sintheta-costheta)$

How do you convert 9=(2x+y)^2-3y-x9=(2x+y)^2-3y-x into polar form?