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

1 Answer
Douglas K.
Nov 13, 2016

Please see the explanation for steps leading to the polar form:

Explanation:

Substitute #(rcos(theta))# for x and #r^2sin^2(theta)# for #y^2#:

#r^2sin^2(theta) + #2rcos(theta) - 5 = 0#

Use the positive root of the quadratic formula:

#r = {-2cos(theta) + sqrt(4cos^2(theta) + 20sin^2(theta))}/(2sin^2(theta))#

#r = {-cos(theta) + sqrt(cos^2(theta) + 5sin^2(theta))}/(sin^2(theta))#

#r = {-cos(theta) + sqrt(1 + 4sin^2(theta))}/(sin^2(theta))#

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