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

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

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Answer 1 · 2016-02-25T05:37:26

$3 r = 3 sec θ - 2 tan θ sec θ$ $3r=3sectheta-2tanthetasectheta$

Explanation:

To convert the equation $2 y = - 3 x^{2} + 3 x$ $2y=−3x^2+3x$ into a polar equation, we should use $y = r cos θ$ $y=rcostheta$ and $x = r sin θ$ $x=rsintheta$ .

Putting these, the equation becomes

$2 r sin θ = - 3 r^{2} {cos}^{2} θ + 3 r cos θ$ $2rsintheta=-3r^2cos^2theta+3rcostheta$ dividing by $r$ $r$ and transposing

or $3 r {cos}^{2} θ = 3 cos θ - 2 sin θ$ $3rcos^2theta=3costheta-2sintheta$

or $3 r = 3 sec θ - 2 tan θ sec θ$ $3r=3sectheta-2tanthetasectheta$

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

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

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