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 $(1 + 3sin(2theta))$ :

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

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

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