How do you convert $2 y = - 2 x^{2} + 3 x y$ $2y= -2x^2+3xy$ into a polar equation?

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Answer 1 · 2016-09-25T19:18:49

Substitute $r sin (θ)$ $rsin(theta)$ for y and $r cos (θ)$ $rcos(theta)$ for x and then do some algebra.

$2 r sin (θ) = - 2 {(r cos (θ))}^{2} + 3 (r cos (θ)) (r sin (θ))$ $2rsin(theta) = -2(rcos(theta))^2 + 3(rcos(theta))(rsin(theta))$

There is a common factor of $r^{2}$ $r^2$ on the right:

$2 r sin (θ) = r^{2} {- 2 {cos}^{2} (θ) + 3 cos (θ) sin (θ))}$ $2rsin(theta) = r^2{-2cos^2(theta) + 3cos(theta)sin(theta))}$

Flip things a bit:

$r^{2} {3 cos (θ) sin (θ)) - 2 {cos}^{2} (θ)} = 2 r sin (θ)$ $r^2{3cos(theta)sin(theta))-2cos^2(theta)} = 2rsin(theta)$

Remove a common r:

$r {3 cos (θ) sin (θ)) - 2 {cos}^{2} (θ)} = 2 sin (θ)$ $r{3cos(theta)sin(theta))-2cos^2(theta)} = 2sin(theta)$

Divide by everything in the {}s:

$r = \frac{2 sin (θ)}{3 cos (θ) sin (θ) - 2 {cos}^{2} (θ)}$ $r = (2sin(theta))/(3cos(theta)sin(theta)-2cos^2(theta))$

There you have $r (θ)$ $r(theta)$

How do you convert 2y= -2x^2+3xy 2y=−2x2+3xy2y= -2x^2+3xy into a polar equation?