\[y=f(x)\]
is given by \[\kappa = \frac{\frac{d^2y}{dx^2}}{(1+ (\frac{dy}{dx})^2)^{\frac{3}{2}}}\]
.Sup[pose a hyperbola has equation
\[xy=K\]
then \[y=\frac{K}{x}\]
and \[\frac{dy}{dx}= - \frac{K}{x^2}\]
and \[\frac{d^2 y}{dx^2}= \frac{2K}{x^3}\]
.Then
\[\kappa = \frac{\frac{2K}{x^3}}{(1+(- \frac{K}{x^2})^2)^{\frac{3}{2}}}= \frac{Kx^3}{(x^4+K^2)^{\frac{3}{2}}}\]
.The degree of the denominator is 6 and the degree of the denominator is 4, so as
\[x \rightarrow \infty\]
the curvature tends to zero.