How do you convert $3 x^{2} - y^{2} = \frac{x^{2}}{2 x - 3 y}$ $3x^2-y^2=x^2/(2x-3y)$ to polar form?

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Answer 1 · 2016-07-10T03:45:28

$r = \frac{{cos}^{2} θ}{(2 cos θ - 3 sin θ) (4 {cos}^{2} θ - 1)}$ $r=(cos^2 theta)/((2cos theta-3 sin theta)(4cos^2 theta -1))$

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How do you convert 3x^2-y^2=x^2/(2x-3y)3x2−y2=x22x−3y3x^2-y^2=x^2/(2x-3y) to polar form?