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

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

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Answer 1 · 2018-07-26T01:25:04

$r = 0$ $r = 0$ and
$r = \frac{2 cos θ + 3 sin θ}{2 {cos}^{2} θ - {sin}^{2} θ}$ $r = ( 2 cos theta + 3 sin theta)/(2 cos^2 theta - sin^2 theta )$ .
This hyperbola passes through the pole $(O), r = 0$ $( O ), r = 0$ .

Explanation:

Conversion formula:

$(x . y) = r (cos θ, sin θ), r = \sqrt{x^{2} + y^{2}} \geq 0$ $( x. y ) = r ( cos theta, sin theta ), r = sqrt (x^2 + y^2) >= 0$

$2 x^{2} - y^{2} - 2 x - 3 y = 0$ $2x^2 - y^2- 2x - 3y = 0$ converts to

$r = 0$ $r = 0$ and

$r (2 {cos}^{2} θ - {sin}^{2} θ) - (2 cos θ + 3 sin θ)) = 0$ $r(2 cos^2 theta - sin^2 theta ) - ( 2 cos theta + 3 sin theta)) = 0$ .

This hyperbola passes through the pole $(O), r = 0$ $( O ), r = 0$ .

See graph.

graph{2x^2 - y^2- 2x - 3y = 0 [-20 20 -12 8] }

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

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Explanation:

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