How do you convert #3=(x+2y)^2+3y# into polar form?

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Answer 1 · 2017-03-28T17:26:51

Please see the explanation.

Given: #3=(x+2y)^2+3y#

Here is the graph in Cartesian form:

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Substitute #rcos(theta)# for #x#:

#3 = (rcos(theta)+2y)^2 + 3y#

Substitute #rsin(theta)# for #y#:

#3 = (rcos(theta)+2rsin(theta))^2 + 3rsin(theta)#

Write in quadratic form:

#(cos(theta)+2sin(theta))^2r^2 + 3sin(theta)r - 3 = 0#

Use the positive root of the quadratic formula:

#r = (-3sin(theta)+sqrt(9sin^2(theta)+12(cos(theta)+2sin(theta))^2))/(2(cos(theta)+2sin(theta))^2)#

Here is the graph in polar form:

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