How do you convert # x^2 + 6xy + y^2=5/2# to polar form?

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Answer 1 · 2016-11-10T19:41:27

Please see the explanation for steps leading to the answer:
#r = sqrt(5/(2(1 + 3sin(2theta))))#

Substitute #r^2# for #x^2 + y^2#

#r^2 + 6xy = 5/2#

Substitute #rcos(theta)# for x and #rsin(theta)# for y:

#r^2 + 6r^2cos(theta)sin(theta) = 5/2#

Factor out #r^2#

#r^2(1 + 6cos(theta)sin(theta)) = 5/2#

Substitute #sin(2theta)# for #2cos(theta)sin(theta)#:

#r^2(1 + 3sin(2theta)) = 5/2#

Divide both sides by #(1 + 3sin(2theta))#:

#r^2 = 5/(2(1 + 3sin(2theta)))#

#r = sqrt(5/(2(1 + 3sin(2theta))))#