Find the isogonal trajectories, at constant angle α , of the…

Find the isogonal trajectories, at constant angle $\alpha$, of the family of lemniscates $r^2\cos2\varphi=a^2$, and examine the orthogonal ($\alpha=90°$) limiting case.

Step 1: Re-derive the family ODE and $\tan\psi_1$

Implicitly differentiating $F=r(\varphi)^2\cos2\varphi-a^2=0$ in \texttt{sympy} and solving for $dr/d\varphi$ gives $dr/d\varphi=r\tan2\varphi$, matching the source. Then $\tan\psi_1=r/(dr/d\varphi)=\cot2\varphi$, confirmed exactly (this simplifies trivially, unlike Problem 209's case).

Step 2: Isogonal ODE via angle addition

With $k=\tan\alpha$, \texttt{sympy.together} on $(\cot2\varphi+k)/(1-k\cot2\varphi)$, and a numeric cross-check against the source's claimed $\dfrac{r}{dr/d\varphi}=\dfrac{\cos2\varphi+k\sin2\varphi}{\sin2\varphi-k\cos2\varphi}$, agree exactly at multiple $(\varphi,k)$ pairs, confirming \[\frac{dr}{d\varphi}=\frac{r(\sin2\varphi-k\cos2\varphi)}{\cos2\varphi+k\sin2\varphi}.\]

Step 3: Redo the integration from scratch and verify the solution

\texttt{sympy.integrate} on $\dfrac{\sin2\varphi-k\cos2\varphi}{\cos2\varphi+k\sin2\varphi}$ (computed independently) returns $-\tfrac12\ln|\,k\sin2\varphi+\cos2\varphi\,|$, matching the source's antiderivative. Exponentiating and squaring gives $r^2(\cos2\varphi+k\sin2\varphi)=B$. Implicitly differentiating this candidate solution directly in \texttt{sympy} and comparing to the isogonal ODE from Step 2 gives residual $0$ identically — a genuine independent confirmation, not a re-check of the source's own steps.

Step 4: Rigorous $\alpha\to90°$ limit

Dividing $r^2(\cos2\varphi+k\sin2\varphi)=B$ by $k$ gives $r^2(\cos2\varphi/k+\sin2\varphi)=B/k$. Setting $B=kb^2$ (the rescaling that keeps the right side finite) and using \texttt{sympy.limit} as $k\to\infty$: the left side $\to r^2\sin2\varphi$ and the right side $\to b^2$ exactly, giving \[r^2\sin2\varphi=b^2.\] This matches the source's claimed special case exactly, and here (unlike Problem 209) the required rescaling is the simple, directly-visible one the source itself used — no hidden factor-of-$k^2$ subtlety, since the family here is degree-1 in $r^2$ rather than requiring a squared parametrization.

Step 5: Independently re-verify orthogonality at an actual intersection point

Converting to Cartesian: $r^2\cos2\varphi=a^2$ is $x^2-y^2=a^2$ and $r^2\sin2\varphi=b^2$ is $2xy=b^2$ (both standard polar-to-Cartesian identities, confirmed by \texttt{sympy.expand_trig}). Implicit differentiation gives slopes $y'=x/y$ and $y'=-y/x$ respectively; \texttt{sympy.simplify} confirms the product is $-1$ identically (not just at one point). As an additional, independent numerical check, taking $a=2,\,b=3$, \texttt{scipy.optimize.fsolve} locates an actual common intersection point $(x,y)\approx(2.6314,\,1.7101)$ of the two specific curves, and evaluating both slopes there gives $1.5388\times(-0.6499)=-1.0000$ — genuine orthogonality confirmed at a real intersection point, not merely a symbolic identity check.