Evolute of Parabola

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Theorem

The evolute of the parabola $y = x^2$ is the curve:

$27 X^2 = 16 \paren {Y - \dfrac 1 2}^3$


Proof

From Parametric Equations for Evolute: Formulation 1:

$\begin {cases}

X = x - \dfrac {y' \paren {1 + y'^2} } {y} \\ Y = y + \dfrac {1 + y'^2} {y} \end{cases}$

where:

$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
$y'$ and $y$ denote the derivative and second derivative respectively of $y$ with respect to $x$.


Thus we have:

\(\ds y'\) \(=\) \(\ds 2 x\)
\(\ds y\) \(=\) \(\ds 2\)


and so:

\(\ds X\) \(=\) \(\ds x - \dfrac {2 x \paren {1 + \paren {2 x}^2} } 2\)
\(\ds \) \(=\) \(\ds x - x - 4 x^3\)
\(\ds \) \(=\) \(\ds -4 x^3\)


and:

\(\ds Y\) \(=\) \(\ds x^2 + \dfrac {1 + \paren {2 x}^2} 2\)
\(\ds \) \(=\) \(\ds \frac {2 x^2 + 1 + 4 x^2} 2\)
\(\ds \) \(=\) \(\ds \frac {6 x^2 + 1} 2\)


Then:

\(\ds Y\) \(=\) \(\ds \frac {6 \paren {-\frac X 4}^{2/3} + 1} 2\) substituting for $X$
\(\ds Y\) \(=\) \(\ds 3 \paren {-\frac X 4}^{2/3} + \frac 1 2\) substituting for $X$
\(\ds \leadsto \ \ \) \(\ds \frac 1 3 \paren {Y - \frac 1 2}\) \(=\) \(\ds \paren {\frac {X^2} {16} }^{1/3}\)
\(\ds \leadsto \ \ \) \(\ds 27 X^2\) \(=\) \(\ds 16 \paren {Y - \dfrac 1 2}^3\)


The parabola (blue) and its evolute (red) are illustrated below:


ParabolaAndEvolute.png

$\blacksquare$


Sources

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