How do you convert $2 y = 3 y^{2} - 4 x^{2} - 2 x$ $2y=3y^2-4x^2 -2x$ into a polar equation?

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How do you convert $2 y = 3 y^{2} - 4 x^{2} - 2 x$ $2y=3y^2-4x^2 -2x$ into a polar equation?

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Answer 1 · 2017-03-27T15:13:50

$r = \frac{2 (sin θ + cos θ)}{3 {sin}^{2} θ - 4 {cos}^{2} θ}$ $r=(2(sintheta+costheta))/(3sin^2theta-4cos^2theta)$

Explanation:

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and Cartesian coordinates $(x, y)$ $(x,y)$ is given by $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Hence, $2 y = 3 y^{2} - 4 x^{2} - 2 x$ $2y=3y^2-4x^2-2x$

$\Leftrightarrow 2 r sin θ = 3 r^{2} {sin}^{2} θ - 4 r^{2} {cos}^{2} θ - 2 r cos θ$ $hArr2rsintheta=3r^2sin^2theta-4r^2cos^2theta-2rcostheta$

i.e. $r^{2} (3 {sin}^{2} θ - 4 {cos}^{2} θ) = 2 r (sin θ + cos θ)$ $r^2(3sin^2theta-4cos^2theta)=2r(sintheta+costheta)$

or $r = \frac{2 (sin θ + cos θ)}{3 {sin}^{2} θ - 4 {cos}^{2} θ}$ $r=(2(sintheta+costheta))/(3sin^2theta-4cos^2theta)$

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How do you convert $2 y = 3 y^{2} - 4 x^{2} - 2 x$ $2y=3y^2-4x^2 -2x$ into a polar equation?

1 Answer

Explanation:

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How do you convert 2y=3y^2-4x^2 -2x 2y=3y2−4x2−2x2y=3y^2-4x^2 -2x into a polar equation?

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How do you convert $2 y = 3 y^{2} - 4 x^{2} - 2 x$ $2y=3y^2-4x^2 -2x$ into a polar equation?