Find the orthogonal trajectories of the family of parabolas…

Find the orthogonal trajectories of the family of parabolas $y=ax^2$.

Step 1: Differentiate and eliminate $a$

\texttt{sympy} gives $y'=2ax$ for $y=ax^2$. Solving the original equation for $a=y/x^2$ and substituting: \[y'=2x\cdot\frac{y}{x^2}=\frac{2y}{x},\] matching the source's family ODE exactly.

Step 2: Form the orthogonal ODE

The orthogonal-trajectory slope is the negative reciprocal: \texttt{sympy.simplify}$\left(-1\big/\frac{2y}{x}\right)=-\dfrac{x}{2y}$, matching the source.

Step 3: Separate and integrate

$-\dfrac{x}{2y}=\dfrac{dy}{dx}\Rightarrow 2y\,dy=-x\,dx$. \texttt{sympy.integrate} gives $\int 2y\,dy=y^2$ and $\int -x\,dx=-\dfrac{x^2}{2}$, so \[y^2=-\frac{x^2}{2}+C_1 \iff \frac{x^2}{2}+y^2=C_1 \iff x^2+2y^2=C,\] matching the source's family of ellipses exactly.

Step 4: Verify the slope-product identity symbolically

Implicit differentiation of $x^2+2y^2=C$ in \texttt{sympy} gives $y'=-\dfrac{x}{2y}$, identical to Step 2. Multiplying the two family slopes, \[\frac{2y}{x}\cdot\left(-\frac{x}{2y}\right)=-1\] identically (confirmed symbolically, not just at a sample point), the required orthogonality condition.