Find the singular solution (envelope) of the one-parameter…

Find the singular solution (envelope) of the one-parameter family $(y-c)^2=(x-c)^3$ via the $C$-discriminant.

Step 1: Set up and solve the $C$-discriminant system with \texttt{sympy}

With $F(x,y,c)=(y-c)^2-(x-c)^3$, \[\frac{\partial F}{\partial c}=-2(y-c)+3(x-c)^2.\] Calling \texttt{sympy.solve([F, dF\_dc], [x, y])} (an independent elimination, not a check of the source's hand algebra) returns two branches: \[\{x=c,\ y=c\}\qquad\text{and}\qquad \left\{x=c+\tfrac49,\ \ y=c+\tfrac{8}{27}\right\}.\] The second branch, with $c=x-\tfrac49$, gives $y=x-\tfrac49+\tfrac{8}{27}=x-\tfrac{12}{27}+\tfrac{8}{27}=x-\tfrac{4}{27}$ — reproducing the source's claimed line. But \texttt{sympy} also turns up a first branch the source never mentions, so both must be checked before trusting either.

Step 2: Identify the extraneous branch

The branch $x=c,\ y=c$ traces the line $y=x$. Substituting $x=c,\,y=c$ into $F$ and $\partial F/\partial c$ both vanish trivially, because at $x=c$ the point $(c,c)$ is exactly the cusp of the semicubical-parabola member $(y-c)^2=(x-c)^3$ itself (both $y-c$ and $x-c$ vanish there). This is the classic $C$-discriminant "cusp locus," a known spurious companion to the true envelope — it must be tested against the ODE, not accepted on sight.

Step 3: Derive the family's ODE independently (resultant elimination)

Totally differentiating $F=0$ in $x$ (with $y'=y_p$) gives $\partial F/\partial x+\partial F/\partial y\cdot y_p=-3(x-c)^2+6c x-2c y_p-3x^2+2yy_p\big|_{\text{expanded}}$. Eliminating $c$ between $F=0$ and this derivative via \texttt{sympy.resultant} (treating both as polynomials in $c$) gives, after factoring, \[-(x-y)^3\big[\,27(x-y)+8y_p^3-12y_p^2\,\big]=0.\] The factor $(x-y)^3$ is exactly the extraneous cusp-locus branch from Step 2 (it carries no ODE information — it is a resultant artifact of the shared root). The genuine family ODE is the other factor: \[27(y-x)=8y_p^3-12y_p^2.\]

Step 4: Test each candidate against the true family ODE

For $y=x-\tfrac{4}{27}$ (slope $y_p=1$): $27(x-y)+8(1)-12(1)=27\!\left(\tfrac{4}{27}\right)-4=4-4=0$ — satisfies the genuine ODE factor. For $y=x$ (slope $y_p=1$): $27(0)+8-12=-4\ne0$ — fails it, confirming $y=x$ is only the spurious cusp locus, not a singular solution of the differential equation.

Step 5: Numerical tangency check

At $x_0=1$: $c_0=x_0-\tfrac49=\tfrac59$, and the envelope point is $\left(1,\ \tfrac{23}{27}\right)$. \texttt{sympy} confirms $F(1,\tfrac{23}{27},\tfrac59)=0$ exactly, and implicit differentiation of the family member $(y-c_0)^2=(x-c_0)^3$ at that point gives slope $y'=1$ — identical to the envelope line's own slope. This is genuine tangency, confirming $y=x-\tfrac{4}{27}$ is a true envelope, not merely an intersection locus.