# Tangent to Hyperbolic functions graph

1. Oct 19, 2016

### Karol

1. The problem statement, all variables and given/known data
Show that the tangent to $x^2-y^2=1$ at points $x_1=\cosh (u)$ and $y_1=\sinh(u)$ cuts the x axis at ${\rm sech(u)}$ and the y axis at ${\rm -csch(u)}$.

2. Relevant equations
Hyperbolic sine: $\sinh (u)=\frac{1}{2}(e^u-e^{-u})$
Hyperbolic cosine: $\cosh (u)=\frac{1}{2}(e^u+e^{-u})$

3. The attempt at a solution
$$2x-2yy'=0~\rightarrow~\frac{x}{y}=y'=\frac{\cosh (u)}{\sinh (u)}=\frac{e^u+e^{-u}}{e^u-e^{-u}}$$
The equation of the tangent is $y=y'x+b$ and it's intersection with the y axis is at $y=\sinh (u)$:
$$y=y'x+b~~\rightarrow~~\sinh (u)=\frac{e^u+e^{-u}}{e^u-e^{-u}}(e^u+e^{-u})+b~~\rightarrow~~b=4$$
But b cant be 4 because it is a general expression, it should depend on u

2. Oct 19, 2016

### stevendaryl

Staff Emeritus
How are you getting 4 out of that? There is no reason to go back and forth between the hyperbolic trig functions and exponentials. In terms of the hyperbolic trig functions, you have:

$y = \frac{cosh(u)}{sinh(u)} x + b$

So plugging in $y=sinh(u), x = cosh(u)$ gives:

$sinh(u) = \frac{cosh(u)}{sinh(u)} cosh(u) + b$

So what do you get for $b$? (Remember the identity: $cosh^2(u) - sinh^2(u) = 1$)

3. Oct 19, 2016

### Karol

Thank you stevendaryl