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))#