Find the orthogonal trajectories of the lemniscate family…

Find the orthogonal trajectories of the lemniscate family $(x^2+y^2)^2=a^2(x^2-y^2)$.

Step 1: Re-derive the family ODE independently

With $F=(x^2+y^2)^2-a^2(x^2-y^2)$, \texttt{sympy} gives $dF/dx=-2a^2x+2a^2yy'+4x^3+4x^2yy'+4xy^2+4y^3y'$. Solving $F=0$ for $a^2=\dfrac{(x^2+y^2)^2}{x^2-y^2}$ and substituting, then solving for $y'$, \texttt{sympy} returns \[y'=\frac{x(3y^2-x^2)}{y(3x^2-y^2)},\] matching the source's family ODE exactly.

Step 2: Orthogonal ODE

Negative reciprocal: \texttt{sympy.simplify} confirms \[y'=\frac{y(3x^2-y^2)}{x(x^2-3y^2)}\left(=\frac{y(y^2-3x^2)}{x(3y^2-x^2)}\right),\] matching the source's stated form.

Step 3: Solve properly via $y=vx$ (redone from scratch)

Substituting $y=vx$ into the orthogonal ODE, \texttt{sympy} simplifies the right side to $\dfrac{v(v^2-3)}{3v^2-1}$, so \[x\frac{dv}{dx}=\frac{v(v^2-3)}{3v^2-1}-v=-\frac{2v(v^2+1)}{3v^2-1}\quad\text{(\texttt{sympy}-confirmed factorization)}.\] Hence $\dfrac{dx}{x}=-\dfrac{3v^2-1}{2v(v^2+1)}\,dv$. \texttt{sp.apart} on $\dfrac{3v^2-1}{v(v^2+1)}$ gives, exactly, \[\frac{3v^2-1}{v(v^2+1)}=\frac{4v}{v^2+1}-\frac{1}{v},\] and \texttt{sp.integrate} term-by-term gives $\displaystyle\int\frac{3v^2-1}{v(v^2+1)}\,dv=2\ln(v^2+1)-\ln v$. So \[\ln x=-\ln(v^2+1)+\tfrac12\ln v+\text{const}\ \Longrightarrow\ \frac{x^2(v^2+1)^2}{v}=K.\] Substituting back $v=y/x$ and simplifying with \texttt{sympy} collapses this to \[(x^2+y^2)^2=Kxy,\] which does land on the source's asserted final answer — the messy, incompletely-shown mid-calculation the source flagged with "this doesn't simplify nicely" was a real snag, but the destination it jumped to afterward turns out to be correct once redone rigorously.

Step 4: Confirm the solution symbolically

Implicit differentiation of $(x^2+y^2)^2=Cxy$ and elimination of $C$ (via the original relation) reproduces the orthogonal ODE $y'=\dfrac{y(3x^2-y^2)}{x(x^2-3y^2)}$ exactly — residual $0$, confirmed by \texttt{sympy}.

Step 5: End-to-end numerical cross-check

Pick the point $(x,y)=(2,1)$: from $(x^2+y^2)^2=a^2(x^2-y^2)$, this fixes $a^2=25/3$. Implicit differentiation of the original equation with $a^2=25/3$ fixed (not the derived ODE) gives family slope $y'=-2/11$ at $(2,1)$. The candidate orthogonal curve $(x^2+y^2)^2=Cxy$ through $(2,1)$ requires $C=25/2$; implicit differentiation gives orthogonal slope $y'=11/2$ there. Product: $\left(-\dfrac{2}{11}\right)\times\dfrac{11}{2}=-1$ — exact confirmation of perpendicularity.