Find the evolute of the parabola y^2=4px
Find the evolute of the parabola $y^2=4px$.
Explanation of Method
The source's first attempt — finding the envelope of the tangent-line family by $t$-elimination — reproduces the original parabola and is explicitly abandoned by the source as "incorrect." This is expected: the envelope of a curve's own tangent lines is always the curve itself, a general fact, not a special failure of this curve. Editorial correction: "evolute = orthogonal trajectory of the tangent lines" is not the standard characterization; the orthogonal trajectories of a curve's tangent-line family are its involutes, not its evolute. The evolute is properly the envelope of the normal lines, equivalently the locus of centers of curvature $\left(x-y'\frac{1+y'^2}{y''},\ y+\frac{1+y'^2}{y''}\right)$ — the method the source correctly switches to. That curvature-center computation is redone fully symbolically here with \texttt{sympy} (not re-checking the source's hand algebra), and the final closed form is confirmed by two independent methods: direct substitution and polynomial resultant elimination.
Step 1: Why the tangent-envelope attempt was doomed from the start
The envelope of the family of tangent lines to any smooth curve is, by definition, that same curve — each tangent line touches the curve at exactly the point being envelope-eliminated, so eliminating the parameter returns the original relation. The source's finding "$Y^2=4pX$, which is incorrect" is therefore not a computational slip but an instance of this general fact; it is evidence that "orthogonal trajectory of the tangents" and "envelope of the tangents" are two different constructions, not that the algebra was wrong. (The orthogonal trajectories of the tangent family are the parabola's involutes — a different, legitimate curve family, not pursued further here since the problem asks for the evolute.)
Step 2: Compute $y',y''$ from $y^2=4px$ with \texttt{sympy}
Parametrizing by $y$ (so $x=y^2/(4p)$), \texttt{sympy} confirms $y'=2p/y$ (from implicit differentiation of $y^2=4px$) and, differentiating again via the chain rule $y''=\frac{d y'}{dy}\cdot y'$, \[y''=-\frac{4p^2}{y^3},\] matching the source.
Step 3: Build $X_c(y),Y_c(y)$ fully symbolically
Substituting into $X_c=x-y'\dfrac{1+y'^2}{y''}$ and $Y_c=y+\dfrac{1+y'^2}{y''}$ and calling \texttt{sympy.simplify} directly (not re-deriving the source's by-hand fraction chain) gives \[X_c=2p+\frac{3y^2}{4p}=\frac{3y^2+8p^2}{4p},\qquad Y_c=-\frac{y^3}{4p^2},\] matching the source's claimed center-of-curvature formulas exactly (residual $0$ against both).
Step 4: Confirm the final closed form by direct substitution
Substituting $X_c(y),Y_c(y)$ from Step 3 into $(X_c-2p)^3-\tfrac{27p}{4}Y_c^2$ and calling \texttt{sympy.simplify}, the expression reduces identically to $0$ for all $y$ — the claimed evolute equation holds along the entire parametrized curve, not just at isolated points.
Step 5: Independent cross-check via resultant elimination
As a second, algebraically independent method, eliminating $y$ between $X-X_c(y)=0$ and $Y-Y_c(y)=0$ via \texttt{sympy.resultant} (clearing denominators first) gives, after factoring, \[-16p^3\left(-4X^3+24pX^2-48p^2X+27pY^2+32p^3\right)=0.\] Expanding $(X-2p)^3-\tfrac{27p}{4}Y^2=0$ into $-4X^3+24pX^2-48p^2X+32p^3+27pY^2=0$ shows this is exactly the same polynomial — the resultant elimination independently reproduces the same relation obtained by direct substitution in Step 4.
Result
Final answer:
The source's tangent-envelope attempt was correctly abandoned (it necessarily reproduces the original parabola, a general fact about envelopes of tangent lines) but its framing "evolute = orthogonal trajectory of the tangent lines" is not standard — that construction actually yields the involutes; the evolute is properly the envelope of the normals / locus of curvature centers, the method the source switches to. That computation is confirmed correct end-to-end by two independent \texttt{sympy} methods (direct substitution and resultant elimination), both giving exact residual $0$: \[\boxed{\,(x-2p)^3=\frac{27p}{4}\,y^2\,}\] a semicubical parabola (Neil's parabola), the genuine evolute of $y^2=4px$.