How do you convert $3 y = 2 x^{2} - 2 x - 7 y^{2}$ $3y= 2x^2-2x-7y^2$ into a polar equation?

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How do you convert $3 y = 2 x^{2} - 2 x - 7 y^{2}$ $3y= 2x^2-2x-7y^2$ into a polar equation?

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Answer 1 · 2017-10-08T18:32:25

$7 r {sin}^{2} θ - 2 r {cos}^{2} θ + 2 cos θ + 3 sin θ = 0$ $7rsin^2theta-2rcos^2theta+2costheta+3sintheta=0$

Explanation:

$y = r sin (θ)$ $y=rsin (theta)$ & $x = r cos (θ)$ $x=rcos (theta)$
$3 r sin (θ) = 2 r^{2} {cos}^{2} (θ) - 2 r cos (θ) - 7 r^{2} {sin}^{2} (θ)$ $3r sin(theta)=2r^2cos^2(theta)-2rcos(theta)-7r^2sin^2(theta)$
$7 r^{2} {sin}^{2} θ - 2 r^{2} {cos}^{2} θ + 2 r cos θ + 3 r sin θ = 0$ $7r^2sin^2theta-2r^2cos^2theta+2rcostheta+3rsintheta=0$
$7 r {sin}^{2} θ - 2 r {cos}^{2} θ + 2 cos θ + 3 sin θ = 0$ $7rsin^2theta-2rcos^2theta+2costheta+3sintheta=0$

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How do you convert $3 y = 2 x^{2} - 2 x - 7 y^{2}$ $3y= 2x^2-2x-7y^2$ into a polar equation?

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How do you convert 3y= 2x^2-2x-7y^2 3y=2x2−2x−7y23y= 2x^2-2x-7y^2 into a polar equation?

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How do you convert $3 y = 2 x^{2} - 2 x - 7 y^{2}$ $3y= 2x^2-2x-7y^2$ into a polar equation?