Locus and hyperbolic functions

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

working: http://gyazo.com/4c96af128d0293bce7f18029c2f54b0d
where have I gone wrong :(
 
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synkk said:
show that the locus of the point [itex]\left(\dfrac{a(cosh\theta + 1)}{2cosh\theta},\dfrac{b(cosh\theta - 1)}{2sinh\theta}\right)[/itex]
has equation x(4y^2 + b^2) = ab^2

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

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.
 
haruspex said:
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.

see below

Mark44 said:
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.

Apologies, I thought it was fine.

working:

[itex]x = \dfrac{a(cosh\theta + 1)}{2cosh\theta}[/itex][itex]y = \dfrac{b(cosh\theta - a)}{2sinh\theta}[/itex]

rearranging x = ... for cosh theta:

[itex]cosh\theta = \dfrac{a}{2x - a}[/itex]
subbing this into y = ...

[itex]y = \dfrac{b(\dfrac{2a-2x}{2x-a})}{2sinh\theta}[/itex]
rearranging for sinh:

[itex]sinh\theta = \dfrac{b(a-x)}{y(2x-a)}[/itex]

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

not sure where to go from here
 
haruspex said:
You're almost there. Just need to spot a factor.

spotted it, thank you.