# Hyperbolic Graph Problem

1. May 30, 2010

### trojsi

Hi,
I can't understand the last step of the answer.

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2. May 31, 2010

### HallsofIvy

Given that (k, -1) is on the graph of y= cosh(x)- 3sinh(x), show that
$$e^{2k}- e^k- 2= 0$$

First the part you say you understand, but I'll write it out so others can follow:

By definition
$$cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex] and [tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/itex] so [tex]cosh(x)- 3sinh(x)= \frac{e^x+ e^{-x}}{2}- \frac{3e^x- 3e^{-x}}{2}$$
$$= \frac{-2e^x+ 4e^{-x}}{2}$$

and the fact that (k, -1) is on the graph means that
$$cosh(k)- 3sinh(k)= \frac{-2e^k+ 4e^{-k}}{2}= -1$$

Multiplying through by 2 gives
$$-2e^k+ 4e^{-k}= -2$$

Now, for the step you say you don't understand: Multiply through by $e^k$ to get:
$$-2e^{2k}+ 4= -2e^{k}$$
amd divide by -2 to get
$$e^{2k}- 2= e^{k}[/itex] Finally, subtract $e^k$ from both sides: [tex]e^{2k}- e^k- 2= 0$$