How do you convert $x^{2} + 6 x y + y^{2} = \frac{5}{2}$ $x^2 + 6xy + y^2=5/2$ to polar form?

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Answer 1 · 2016-11-10T19:41:27

Please see the explanation for steps leading to the answer:
$r = \sqrt{\frac{5}{2 (1 + 3 sin (2 θ))}}$ $r = sqrt(5/(2(1 + 3sin(2theta))))$

Substitute $r^{2}$ $r^2$ for $x^{2} + y^{2}$ $x^2 + y^2$

$r^{2} + 6 x y = \frac{5}{2}$ $r^2 + 6xy = 5/2$

Substitute $r cos (θ)$ $rcos(theta)$ for x and $r sin (θ)$ $rsin(theta)$ for y:

$r^{2} + 6 r^{2} cos (θ) sin (θ) = \frac{5}{2}$ $r^2 + 6r^2cos(theta)sin(theta) = 5/2$

Factor out $r^{2}$ $r^2$

$r^{2} (1 + 6 cos (θ) sin (θ)) = \frac{5}{2}$ $r^2(1 + 6cos(theta)sin(theta)) = 5/2$

Substitute $sin (2 θ)$ $sin(2theta)$ for $2 cos (θ) sin (θ)$ $2cos(theta)sin(theta)$ :

$r^{2} (1 + 3 sin (2 θ)) = \frac{5}{2}$ $r^2(1 + 3sin(2theta)) = 5/2$

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

$r^{2} = \frac{5}{2 (1 + 3 sin (2 θ))}$ $r^2 = 5/(2(1 + 3sin(2theta)))$

$r = \sqrt{\frac{5}{2 (1 + 3 sin (2 θ))}}$ $r = sqrt(5/(2(1 + 3sin(2theta))))$

How do you convert x2+6xy+y2=52 x^2 + 6xy + y^2=5/2 to polar form?