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

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Answer 1 · 2016-10-16T20:16:17

Please see the explanation for the process.

$r = 2(sin(theta) - cos(theta))/(sin^2(theta) - 4cos^2(theta))$

Write as:

$y^2 - 4x^2 - 2y - 2x = 0$

Substitute $rsin(theta)$ for every y and $rcos(theta)$ for every x:

$(rsin(theta))^2 - 4(rcos(theta))^2 - 2(rsin(theta)) - 2(rcos(theta)) = 0$

Put common factors outside of the ()s:

$(sin^2(theta) - 4cos^2(theta))r^2 - 2r(sin(theta) - cos(theta)) = 0$

Most the second term to the right:

$(sin^2(theta) - 4cos^2(theta))r^2 = 2r(sin(theta) - cos(theta))$

Divide both sides by $(sin^2(theta) - 4cos^2(theta))r$

$r = 2(sin(theta) - cos(theta))/(sin^2(theta) - 4cos^2(theta))$

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