Find the surface orthogonal to the one-parameter family z^2…
Find the surface orthogonal to the one-parameter family \[ z^2 = axy, \] where $a$ is an arbitrary constant.
Explanation of Method
Eliminating $a$ gives $p=z/(2x)$, $q=z/(2y)$; the orthogonality PDE that results has both multiplier combinations $(x,-y,0)$ and $(x,y,z)$ make the denominator vanish, giving both first integrals directly.
Step 1 · PDE for the orthogonal surfaces
From $z^2=axy$: $p=z/(2x)$, $q=z/(2y)$, so $\mathbf n_1=(-z/(2x),-z/(2y),1)$. Orthogonality with $\mathbf n_2=(-p,-q,1)$, multiplied by $2xy$: \[ yz\,p+xz\,q=-2xy. \]
Step 2 · Characteristic equations and first integrals
\[ \frac{dx}{yz}=\frac{dy}{xz}=\frac{dz}{-2xy}. \tag{1} \]
Multipliers $(x,-y,0)$: denominator $0$, so $x^2-y^2=c_1$. \tag{I}
Multipliers $(x,y,z)$: denominator $xyz+xyz-2xyz=0$, so $x^2+y^2+z^2=c_2$. \tag{II}
Step 3 · General solution
\[ x^2+y^2+z^2 = \Phi(x^2-y^2), \] where $\Phi$ is an arbitrary differentiable function.
Verification
For $G=x^2+y^2+z^2-\Phi(x^2-y^2)$: $\nabla G=(2x(1-\Phi'),\,2y(1+\Phi'),\,2z)$; \[ \mathbf n_1\cdot\nabla G = -z(1-\Phi')-z(1+\Phi')+2z = 0. \checkmark \]
Result
The orthogonal family is: