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

Question

1507 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-28T17:26:51

Please see the explanation.

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

Here is the graph in Cartesian form:

![Desmos.com]( useruploads.socratic.org )

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:

![Desmos.com]( useruploads.socratic.org )

How do you convert 3=(x+2y)^2+3y3=(x+2y)^2+3y into polar form?