$$\kappa(t)=\frac{2}{\left(4 t^2+1\right)^{3/2}}.$$
For the normal vector of $\alpha$, we first calculate the tangent vector
$$T_\alpha(t)=\frac{\alpha'(t)}{|\alpha'(t)|}\Rightarrow N_\alpha(t)=J T_\alpha(t)=\left(-\frac{2 t}{\sqrt{4 t^2+1}},\frac{1}{\sqrt{4 t^2+1}}\right).$$ Finally we have $$\beta(t)=\alpha(t)+N(t)/\kappa(t)=\left(-4 t^3,3 t^2+\frac{1}{2}\right).$$ The picture of both curves is:
where the blue color corresponds with the parabola and the green one is its evolute. We check the property proved in classroom that the tangent line of $\alpha$ is the normal line of $\beta$ at the corresponding point. We do it at $t=1$, with $\alpha(1)=(1,1)$ and $\beta(1)=(-4,7/2)$.
Question: Let do with the curve $y=\sin(x)$ at the point $(0,0)$.
