# Locus and hyperbolic functions!

1. May 1, 2013

### synkk

show that the locus of the point $\left(\dfrac{a(cosh\theta + 1)}{2cosh\theta},\dfrac{b(cosh\theta - 1)}{2sinh\theta}\right)$
has equation x(4y^2 + b^2) = ab^2

working: http://gyazo.com/4c96af128d0293bce7f18029c2f54b0d
where have I gone wrong :(

2. May 1, 2013

### haruspex

Your 5th and 6th lines each have the form <some expression> = 1. The second should have something other than 1 on the right hand side.

3. May 1, 2013

### Staff: Mentor

I have to say that the image of your work is very poor quality. IMO, you are lucky that haruspex took the time to decipher your hard-to-read work. I'm sure that many others wouldn't bother.

4. May 2, 2013

### synkk

see below

Apologies, I thought it was fine.

working:

$x = \dfrac{a(cosh\theta + 1)}{2cosh\theta}$

$y = \dfrac{b(cosh\theta - a)}{2sinh\theta}$

rearranging x = ... for cosh theta:

$cosh\theta = \dfrac{a}{2x - a}$
subbing this into y = ...

$y = \dfrac{b(\dfrac{2a-2x}{2x-a})}{2sinh\theta}$
rearranging for sinh:

$sinh\theta = \dfrac{b(a-x)}{y(2x-a)}$

now using $cosh^2\theta - sinh^2\theta = 1$
$\dfrac{a^2}{(2x-a)^2} - \dfrac{b^2(a-x)^2}{y^2(2x-a)^2} = 1$
$y^2a^2 - b^2(a-x)^2 = y^2(2x-a)^2$
$y^2a^2 - b^2(a^2-2ax + x^2) = y^2(4x^2 - 4ax + a^2)$
$-b^2a^2 + 2ab^2x - b^2x^2 = 4y^2x^2 - 4ay^2x$

not sure where to go from here

5. May 2, 2013

### haruspex

You're almost there. Just need to spot a factor.

6. May 3, 2013

### synkk

spotted it, thank you.