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Using polar coodinates, find all the points $(x,y)\ne 0$ such that $\frac{2xy}{x^2+y^2}=\frac{1}{2}$.

I thought I could somehow substitute $x=r\cos(\theta), y=r\sin(\theta)$ but I'm not sure how to proceed. What strategy should I use to tackle these types of problems? Thanks!

EDIT: The numerator should be $2xy$

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    $\begingroup$ Why didn't you try ? $\endgroup$ – Yves Daoust Feb 17 at 8:59
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$x = r\cos\theta, y = r\sin\theta$

$\frac{2xy}{x^2+y^2} = \frac{2r^2\sin\theta\cos\theta}{r^2} = \sin 2\theta$

Solve $\sin 2\theta = \frac 12$ in the first four quadrants (only two quadrants are applicable).

Then, to get the solution in terms of $x$ and $y$, think about the equation(s) of line(s) subtending that angle with the horizontal axis. What is the gradient of those line(s)?

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  • $\begingroup$ I have $(r\cos(5\pi/12),rsin(5\pi/12))$ and $(r\cos(\pi/12),rsin(\pi/12)) \forall x\in \mathbb{R}$. Is this correct? $\endgroup$ – Sam Kirkiles Feb 17 at 8:57
  • $\begingroup$ I would probably express it as $(x, x\tan\frac{\pi}{12}) \cup (x,x\tan\frac{5\pi}{12}) \ \forall x \in \mathbb{R} \backslash {0}$ to eliminate the dummy variable $r$. If you want to make your life a bit harder, you can explicitly work out the value of the trigonometric ratios in terms of irrational numbers, but I don't think that's necessary. $\endgroup$ – Deepak Feb 17 at 9:40
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The substitution is exactly what you should use, except that you don't have $x=r\cos (x)$, but rather $x=r\cos(\phi)$.

Then, using the fact that $\sin^2\phi + \cos^2\phi = 1$, the denominator on your right side should simplify quite a bit.

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As $(x,y)\neq0$, you have $x=r\cos(\theta)$ and $y=r\sin(\theta)$ for $r>0$ and $\theta \in [0,2\pi]$.

Your equation become

$$\frac{2xy}{x^2+y^2}=\frac{2r^2 \cos(\theta)\sin(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))} = 2\cos(\theta)\sin(\theta)=\sin(2\theta)$$

So you get to solve $$\sin(2\theta) = \frac{1}{2}$$

Which is now pretty simple with basic trigonometry knowledge.

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By your work $$r^2=4r^2\cos\theta,$$ which gives $$r=0,$$ which is impossible, or $$1=4\cos\theta\sin\theta,$$ which is $$\sin2\theta=\frac{1}{2}.$$

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  • $\begingroup$ Thanks! Could you check the edit? I had an error in the equation. $\endgroup$ – Sam Kirkiles Feb 17 at 8:43
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    $\begingroup$ Michael.Changed numerator: xy now. $\endgroup$ – Peter Szilas Feb 17 at 8:54
  • $\begingroup$ Thank you, Peter $\endgroup$ – Michael Rozenberg Feb 17 at 8:58

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