Locus and hyperbolic functions

[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 :(

 

[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.

 

[Mark44]

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 :(

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.

 

[synkk]

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

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:

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

 

[haruspex]

-b^2a^2 + 2ab^2x - b^2x^2 = 4y^2x^2 - 4ay^2x

not sure where to go from here

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

 

[synkk]

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

spotted it, thank you.