Find the orthogonal trajectories of the family y^2=2p(x-a)…

Find the orthogonal trajectories of the family $y^2=2p(x-a)$, where $p$ is a fixed given constant (the shared semi-latus-rectum) and $a$ is the family's varying parameter.

Editor's note: resolving the ambiguity

The problem statement is genuinely ambiguous: $y^2=2p(x-a)$ has two symbols, $p$ and $a$, and it is not stated which one is the family's varying parameter — a real transcription ambiguity, not a source error to "correct." Compounding this, $p$ collides notationally with the standard convention $p=y'$ used for the parameter method throughout the rest of this problem set; here $p$ denotes a fixed constant (a semi-latus-rectum), not a derivative. This entry adopts the reading $p$ fixed, $a$ varying (the source's own section 3, after it abandons the confused, self-contradictory sections 1–2) and discards those earlier sections entirely, for the reason verified in Step 1 below.

Step 1: Why "both vary" is not well-posed here

If both $p$ and $a$ are treated as free (a genuine 2-parameter family), eliminating two constants generically requires two differentiations. \texttt{sympy}, differentiating $F=y^2-2p(x-a)$ twice in $x$ with $p,a$ held constant, gives \[F'=2yy'-2p,\qquad F''=2y(y'')+2(y')^2,\] so setting $F''=0$ (eliminating both $p$ and $a$ at once) gives exactly the second-order relation $yy''+(y')^2=0$ — reproducing the source's confused section-1 result. A second-order relation describes the whole 2-parameter family but is not the setting for the standard first-order orthogonal-trajectory method, which needs a single-parameter family and a first-order ODE. This confirms: only the "$p$ fixed, $a$ varies" reading gives a well-posed one-parameter problem, justifying the choice above.

Step 2: Family ODE (with $p$ fixed, $a$ varying)

With $p,a$ both constant for a single curve, \texttt{sympy} differentiates $F=y^2-2p(x-a)$ in $x$: \[\frac{dF}{dx}=2yy'-2p=0\ \Longrightarrow\ y'=\frac{p}{y}\] (the $-2pa$ contribution vanishes identically since $a$ is constant along a single curve). No elimination of $p$ is needed, or possible — $p$ is a given constant of the problem, not a family parameter.

Step 3: Orthogonal ODE and its solution

Negative reciprocal: \texttt{sympy.simplify}$(-1/(p/y))=-y/p$. This is linear/separable; \texttt{sympy.dsolve} on $y'=-y/p$ returns \[y=C\,e^{-x/p},\] matching the source. Direct substitution confirms $\dfrac{d}{dx}\!\left(Ce^{-x/p}\right)-\left(-\dfrac{Ce^{-x/p}}{p}\right)\equiv 0$.

Step 4: Slope-product sanity check

\texttt{sympy} confirms $\dfrac{p}{y}\cdot\left(-\dfrac{y}{p}\right)=-1$ identically, the required orthogonality condition.