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!

  • 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
    ^^^^^^^^^^^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.

  • 0
    @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