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

Question

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!

Answer 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 $y$).

Answer 2

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.