Originally Posted by

**Peleus** Hi all,

I've got to convert this equation to polar co-ordinates.

$\displaystyle

x^2 + 2y^2 = y(y + 3x)

$

My working so far is as follows,

$\displaystyle

x^2 + 2y^2 = y^2 + 3xy

$

$\displaystyle

x^2 - 3xy + y^2 = 0

$

Let $\displaystyle y = (r \cos\theta)^2$ and $\displaystyle x = (r \sin\theta)^2$

$\displaystyle

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

$

$\displaystyle

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

$

$\displaystyle

r^2 (\cos ^2\theta + \sin ^2\theta) = 3(r \cos\theta)(r \sin\theta)

$

$\displaystyle

r = \sqrt{3(r \cos\theta)(r \sin\theta)}

$

I can't see this being right with r on the RHS though, anyone able to help? Cheers.