Tangent to Hyperbolic function graph

[Karol]

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 ${\rm -csch(u)}$.
$$2x-2yy'=0~\rightarrow~\frac{x}{y}=y'=\frac{\cosh (u)}{\sinh (u)}=\frac{e^u+e^{-u}}{e^u-e^{-u}}$$
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 can't be 4 because it is a general expression, it should depend on u

 

[mfb]

Why do you expect the intersection with the y-axis to be at sinh(u)? That's the y-value where the tangent touches the hyperbola, it won't be identical with the y-axis intersection in general.

I don't understand the second part of your calculation at all.

 

[stevendaryl]

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 ${\rm -csch(u)}$.
$$2x-2yy'=0~\rightarrow~\frac{x}{y}=y'=\frac{\cosh (u)}{\sinh (u)}=\frac{e^u+e^{-u}}{e^u-e^{-u}}$$
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 can't be 4 because it is a general expression, it should depend on u

I don't see the point in going back and forth from exponentials to hyperbolic functions. If you stick to hyperbolic functions, then your equation for the tangent is:

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

If you plug $y=sinh(u), x = cosh(u)$ in, what do you get for $b$? Certainly not $4$.