# How do you convert 9=(-2x+y)^2-5y+3x into polar form?

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1s2s2p Share
Mar 14, 2018

9=4r^2cos^2(theta)-4r^2sinthetacostheta+r^2sin^2(theta)-5rsintheta+3rcostheta=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))

#### Explanation:

$x = r \cos \theta$
$y = r \sin \theta$

$9 = {\left(- 2 \left(r \cos \theta\right) + r \sin \theta\right)}^{2} - 5 r \sin \theta + 3 r \cos \theta$

$9 = 4 {r}^{2} {\cos}^{2} \left(\theta\right) - 4 {r}^{2} \sin \theta \cos \theta + {r}^{2} {\sin}^{2} \left(\theta\right) - 5 r \sin \theta + 3 r \cos \theta$

9=r(sintheta(r(sintheta−4costheta)−5)+costheta(4rcostheta+3))

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