For a curve $y=f(x)$, the curvature is $$ k=\frac{|y''|}{(1+y'^2)^{3/2}} $$ If the curvature is a constant, there are 3 cases \begin{align} \frac{y''}{(1+y'^2)^{3/2}}=0 \tag{1}\\ \frac{y''}{(1+y'^2)^{3/2}}=\frac{1}{R} \tag{2} \\ \frac{y''}{(1+y'^2)^{3/2}}=-\frac{1}{R} \tag{3} \end{align} where $R>0$. Obviously, the solution of (1) is line. I solution (2) and (3), the solution of (2) and (3) always be $$ R^2=(x-C_2R)^2+(y-C_2R)^2 \\ R^2=(x-C_2R)^2-(y-C_2R)^2 $$ they are circle and hyperbola. But no matter how I solute it , I can't get the hyperbola like $$ R^2=-(x-C_2R)^2+(y-C_2R)^2 $$ Besides, there is not more general hyperbola which focus is not at x-axis and y-axis. But obviously, the curvature should be a intrinsic concept, so the general hyperbola should be contained in the solution of (1),(2),(3). How to explain it ?
In fact, I think it is because I assume $y=f(x)$, but I can't think it out.