# Match the following polar equation to the best description: r^2 = 39 / sin(2θ)

Now, I've guessed the answer, it's a hyperbola, and I know what a hyperbolic function looks like, but I'm having a hard time getting it there. Here's my work so far:

1. First off, $r^2 = x^2 + y^2$
2. Therefore, $x^2 + y^2 = \frac{39}{\sin(2θ)}$
3. Now, I'm thinking I should use the double angle formula: $\sin(2t) = 2\sin(t)\cos(t)$
4. That way, I get $x^2 + y^2 = \frac{39}{2\sin(θ)\cos(θ)}$
5. Now, $\sin(θ) = \frac{y}{r}$ and $\cos(θ) = \frac{x}{r}$
6. This turns the above equation into $x^2 + y^2 = \dfrac{39}{2\cdot\frac{y}{r}\cdot\frac{x}{r}}$
7. We can flip the r to the top... $x^2 + y^2 = \dfrac{39r}{x\cdot y}$
8. $r = \sqrt{x^2 + y^2}$
9. $x^2 + y^2 = \dfrac{39\sqrt{x^2 + y^2}}{xy}$
10. ... I feel like I'm doing this wrong... because when I multiply this thing out I get: $$2x^3y + 4x^3y^2 + 2xy^5 - 1521x^2 - 1521 y^2 = 0$$... What am I doing wrong? I have a tendency to make things harder than they need to be.

Any help is greatly appreciated! I check this site kind of obsessively, so I'll respond quickly!

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I've converted your equations into LaTeX. –  Zev Chonoles Jul 30 '11 at 16:18
Also, +1 for showing your work. Not only does it demonstrate that you've put effort into trying to solve it yourself, it also makes it easier to catch where the mistake was and explain it. –  Zev Chonoles Jul 30 '11 at 16:21
A quicker way: $r^2 \sin 2\theta = 39 \Leftrightarrow 2 (r \cos\theta) (r \sin\theta) = 39 \Leftrightarrow 2xy=39$. –  Hans Lundmark Jul 30 '11 at 18:25

You only flipped one $r$ up to the top, and you lost the 2 in the denominator. Note that $$2\cdot\frac{y}{r}\cdot\frac{x}{r}=\frac{2xy}{r^2}$$ so the equation in step 6, $$x^2+y^2=\frac{\quad39\quad}{2\cdot\frac{y}{r}\cdot\frac{x}{r}}$$ becomes $$x^2+y^2=\frac{39r^2}{2xy}.$$ Now you should be able to follow your original line of reasoning to derive the correct equation (the next step would be to express $r$ in terms of $x$ and $y$).

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Dang! Noob mistake! Thanks a lot :)! –  user13327 Jul 30 '11 at 16:25
No problem, glad to help :) –  Zev Chonoles Jul 30 '11 at 16:26
Alright, so two questions: 1. Can't you just cancel out the \r^2 on both sides, which leaves you with xy = 39/2? That doesn't really look like a hyperbola, does it? 2. If you change r^2 into x^2 + y^2, you get x^2 + y^2 = 39(x^2+y^2)/2xy ... and then I turned this into 2x^3y + 2xy^3 = 39x^2 + 39y^2. I still don't see how this turns out to be a hyperbola :( –  user13327 Jul 30 '11 at 16:42
$xy=\frac{39}{2}$ is a hyperbola - see here. Also, here is a plot of $xy=\frac{39}{2}$. –  Zev Chonoles Jul 30 '11 at 16:49
^^^^^^^^^^^win! –  user13327 Jul 30 '11 at 17:00

Start with $r^2=39/\sin(2\theta)$, expand $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and rearrange terms. Finally, remember that $x=r\cos(\theta)$ and $y=r\sin(\theta)$: \begin{align*} \frac{39}{2}&=r\;\cos(\theta)\;r\;\sin(\theta)\\\\ &= x\;y \end{align*} This is definitely a rectangular hyperbola.

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I now see that this equation for a hyperbola was pointed out earlier, but this derivation is a bit shorter, so I'll leave it for now. –  robjohn Jul 30 '11 at 22:10
A $\TeX$ tip: \sin and \cos make for nicer typesetting... –  Guess who it is. Jul 30 '11 at 22:15
@J.M.: Thanks! I caught it in the equation, but missed it in the inline. I thought  around \begin{align} was redundant; am I mistaken? – robjohn Jul 31 '11 at 0:29 @J.M.:Never mind. I now understand why I needed 4 backquotes at the end of each line in my align block; I was missing the  around it. Thanks again! –  robjohn Jul 31 '11 at 0:56