Find the surface orthogonal to the one-parameter family xy =…
Find the surface orthogonal to the one-parameter family \[ xy = az^2, \] where $a$ is an arbitrary constant.
Explanation of Method
Since $xy=az^2$ is just $z^2=(1/a)xy$ with a relabelled parameter, it describes the same geometric family as Problem 65, so the orthogonal family necessarily coincides.
Step 1 · PDE for the orthogonal surfaces
From $xy=az^2$: $p=z/(2x)$, $q=z/(2y)$ (identical to Problem 65), giving the same orthogonality PDE $yz\,p+xz\,q=-2xy$.
Step 2 · Characteristic equations and first integrals
\[ \frac{dx}{yz}=\frac{dy}{xz}=\frac{dz}{-2xy}. \] As before, multipliers $(x,-y,0)$ and $(x,y,z)$ both make the denominator vanish: \[ x^2-y^2=c_1, \qquad x^2+y^2+z^2=c_2. \]
Step 3 · General solution
\[ x^2+y^2+z^2 = \Phi(x^2-y^2), \] where $\Phi$ is an arbitrary differentiable function — the same orthogonal family found in Problem 65, as expected since $z^2=axy$ and $xy=az^2$ describe the same surfaces.
Verification
Identical to Problem 65's verification, since the PDE and first integrals coincide. $\checkmark$
Result
The orthogonal family is: