2
$\begingroup$

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!

  • 2
    I've converted your equations into LaTeX.2011-07-30
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    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.2011-07-30
  • 1
    A quicker way: $r^2 \sin 2\theta = 39 \Leftrightarrow 2 (r \cos\theta) (r \sin\theta) = 39 \Leftrightarrow 2xy=39$.2011-07-30

2 Answers 2

1

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 $y$).

  • 0
    Dang! Noob mistake! Thanks a lot :)!2011-07-30
  • 0
    No problem, glad to help :)2011-07-30
  • 1
    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 :(2011-07-30
  • 1
    $xy=\frac{39}{2}$ is a hyperbola - see [here](http://en.wikipedia.org/wiki/Hyperbola#Rectangular_hyperbola_with_horizontal.2Fvertical_asymptotes_.28Cartesian_coordinates.29). Also, [here is a plot of $xy=\frac{39}{2}$](http://www.wolframalpha.com/input/?i=Plot%5B39%2F%282x%29%2C{x%2C-5%2C5}%5D).2011-07-30
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    ^^^^^^^^^^^win!2011-07-30
1

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.2011-07-30
  • 1
    A $\TeX$ tip: `\sin` and `\cos` make for nicer typesetting...2011-07-30
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    @J.M.: Thanks! I caught it in the equation, but missed it in the inline. I thought `$$` around `\begin{align}` was redundant; am I mistaken?2011-07-31
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    @J.M.:Never mind. I now understand why I needed 4 backquotes at the end of each line in my `align` block; I was missing the `$$` around it. Thanks again!2011-07-31