Hyperbolic functions

How can I show that the hyperbola $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ can be represented in the form x=acoshu , y=bsinhu ?

How can I show that the hyperbola $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ can be represented in the form x=acoshu , y=bsinhu ?

That's an ellipse.

You need $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$.

Probably just a typo but if not that could be your problem.

Substitute the x,y and show that the Cartesian eq. is satisfied for all $u$. Do note the comment above, though.

I dont understand what you mean by substitute the x,y ?

You have $x = a \cosh u$ and $y = b \sinh u$ ... so sub them in, and give a reason why the resulting equation holds for all $u$.

Oh, ops, I thought I had to prove the x values would be in the form $x = a \cosh u$ and $y = b \sinh u$ from the equation. I suppose thats how im supposed to do it.

Thanks again RDK

You can do that too, but (a) there are infinitely many ways in which you can parameterise this curve (*), and (b) it's much easier to just take the simple proposition like this and show that it holds in these cases.

(*) I.e. note that the parameterisation $x = a \cosh (2u), \quad y = b \sinh (2u)$ is valid as well. If $u \in (-\infty, \infty)$ for your parameterisation, then it doesn't matter what you put in the argument in this case (as long as its surjective on $(-\infty, \infty)$) and we don't need to adjust the domain. If we had, say, $u \in [-2,2]$ for your parameterisation, then this parameterisation would hold if we define $u \in [-1,1]$ instead and produce the exact same thing.