- #1
Karol
- 1,380
- 22
Homework Statement
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)}##.
Homework Equations
Hyperbolic sine: ##\sinh (u)=\frac{1}{2}(e^u-e^{-u})##
Hyperbolic cosine: ##\cosh (u)=\frac{1}{2}(e^u+e^{-u})##
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 can't be 4 because it is a general expression, it should depend on u