Parabola Evolute

$\begin{displaymath} y=x^2, \end{displaymath}$

(1)

the parametric equation and its derivatives are

$\begin{displaymath} \matrix{x=t\cr y=t^2\cr}\qquad \matrix{x'=t\cr x''=0\cr}\qquad \matrix{y'=2t\cr y''=2.\cr} \end{displaymath}$

(2)

$\begin{displaymath} R={(x'^2+y'^2)^{3/2}\over x'y''-x''y'} = {(1+4t^2)^{3/2}\over 2}. \end{displaymath}$

(3)

$\begin{displaymath} \hat {\bf T} = {1\over\sqrt{1+4t^2}} \left[{\matrix{1\cr 2t\cr}}\right], \end{displaymath}$

(4)

so the parametric equations of the evolute are

$\displaystyle \xi$	$\textstyle =$	$\displaystyle -4t^3$	(5)
$\displaystyle \eta$	$\textstyle =$	$\displaystyle {\textstyle{1\over 2}}+3t^2,$	(6)

and

$\displaystyle -{\textstyle{1\over 4}}\xi$	$\textstyle =$	$\displaystyle t^3$	(7)
$\displaystyle {\textstyle{1\over 3}} (\eta-{\textstyle{1\over 2}})$	$\textstyle =$	$\displaystyle t^2$	(8)

$\begin{displaymath} {\textstyle{1\over 3}}(\eta-{\textstyle{1\over 2}})=(-{\textstyle{1\over 4}}\xi)^{2/3} \end{displaymath}$

(9)

$\begin{displaymath} {\textstyle{1\over 3}}(\eta-h) = \left({-{2\xi\over 8}}\right)^{2/3}= {\textstyle{1\over 4}}(2\xi)^{2/3}. \end{displaymath}$

(10)

The Evolute is therefore

$\begin{displaymath} \eta={\textstyle{3\over 4}} (2\xi)^{2/3}+{\textstyle{1\over 2}}. \end{displaymath}$

(11)

This is known as Neile's Parabola and is a Semicubical Parabola. From a point above the evolute three normals can be drawn to the Parabola, while only one normal can be drawn to the Parabola from a point below the Evolute.