Translate $2xy=x^2-y^2$ to polar coordinates 
As per the title i need to translate $2xy=x^2-y^2$ into a polar coordinates function

Why it confuses me:
$$
x = r\cos\phi\\
y = r\sin\phi
$$
Then:
$$
2r^2\cos{\phi}\sin{\phi} = r^2\cos{\phi}-r^2\sin{\phi} = r^2(\cos^2{\phi} - \sin^2{\phi}) \\
r^2\sin{2\phi}=r^2\cos{2\phi} \\
sin{2\phi} = \cos{2\phi}
$$
Which gives:
$$
\phi = \frac{\pi}{8} + \frac{\pi n}{2}, \;\;\;\;
n \in \mathbb Z
$$
But it's just a set of $\phi$ values. How do i plot such a function? In polar coordinates $r$ is a function of some $\phi$, isn't that?
 A: After conversion to polar coordinates, $r$ disappears from the equations (but for $r=0$, which corresponds to the origin). This means that for a given $\phi$, $r$ is arbitrary and this describes an infinite straight line through the origin and with direction $\phi$.

By the way, you can obtain the same result by plugging the equation of a straight line through the origin, $y=mx$, which gives
$$2mx^2=x^2-m^2x^2$$ or
$$2m=1-m^2$$ or
$$m=\pm\sqrt2-1.$$
A: What to expect:
$x^2-2xy-y^2= 0;$
$(x-y)^2-2y^2 =0;$
$(x-y-√2y)(x-y+√2y)= 0;$
1)$x = y(1+√2)$,or $y= (1+√2)^{-1}x$;
2) $y=(1-√2)^{-1}x;$
1)and 2) are 2 lines passing through the origin.
Slopes: $m_1= -1+√2$ ; $m_2=-1-√2$.
Perhaps not so surprising any more your result in polar coordinates.
A: It is $$
\phi= \frac{\pi}{8} + \frac{\pi n}{2}, \;\;\;\;
n \in \mathbb Z
$$
and not $x$.
So $$x= r\cos (\frac{\pi}{8} + \frac{\pi n}{2})=$$
$$ = r(\cos \frac{\pi}{8} \cos \frac{\pi n}{2}-\sin \frac{\pi}{8} \sin \frac{\pi n}{2})$$
Now try to put $n=0,1,2,3,4...$. We see that: 
if we put $n=0,4,8...$ we get the same value.
  if we put $n=1,5,9...$ we get the same value.
...
So write $n= 4k+r$ where $r\in\{0,1,2,3\}$ and you will get 4 different families of solution.
A: The graph of $\phi =c$ where $c$ is a constant, is just a ray through the origin  which makes an angle of $\phi$ with the positive direction of $x$-axis.
A: Accoring to the equation $$\sin 2\phi=\cos 2\phi,$$ we can obtain $$\phi=\frac{4n+1}{8}\pi,n \in \mathbb {Z}.$$ But in fact, 


*

*$\phi=2k\pi+\dfrac{1}{8}\pi$,when $n=4k$;

*$\phi=2k\pi+\dfrac{5}{8}\pi$,when $n=4k+1$;

*$\phi=2k\pi+\dfrac{9}{8}\pi$,when $n=4k+2$;

*$\phi=2k\pi+\dfrac{13}{8}\pi$,when $n=4k+3$.


These are four rays，which are reverse pairwisely. In another word, they're two lines as follows  
 
Another method
You can directly solve the equation $$y^2+2xy-x^2=0.$$ Therefore, $$y=\frac{-2x \pm \sqrt{(2x)^2-4 \cdot 1 \cdot (-x^2)}}{2}=(-1 \pm \sqrt 2)x.$$ This also represents the two lines.
