1. Converting to Polar co-ordinates

Hi all,

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

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

My working so far is as follows,

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

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

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

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

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

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

$
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.

2. Originally Posted by Peleus
Hi all,

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

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

My working so far is as follows,

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

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

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

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

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

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

$
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.

Case 1: $r = 0$

Case 2: $1 = 3 \cos \theta \sin \theta = \frac{3}{2} \sin (2 \theta) \Rightarrow \sin (2 \theta) = \frac{2}{3}$

so you have four rays, all with terminus at the origin (alternatively, two lines passing through the origin). See http://www.wolframalpha.com/input/?i...%28y+%2B+3x%29

3. Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.

4. Originally Posted by Peleus
Any hints as to how it actually simplifies down to that? I'm not sure at what point I should be manipulating it to produce that result.
Delete the last line of your calcultation (which should have had a $\pm \sqrt{}$ in it by the way) and simplify the second last line.