Find the surface orthogonal to the one-parameter family xyz…
Find the surface orthogonal to the one-parameter family \[ xyz = a, \] where $a$ is an arbitrary constant.
Explanation of Method
The family's normal is $\nabla(xyz-a)=(yz,xz,xy)$ directly (no elimination needed); orthogonality then gives a Lagrange PDE solved by the same two multiplier tricks as Problem 63.
Step 1 · PDE for the orthogonal surfaces
Normal to $xyz=a$: $\mathbf n_1=(yz,xz,xy)$. Orthogonality with $\mathbf n_2=(-p,-q,1)$: \[ -yzp-xzq+xy=0 \;\Longrightarrow\; yz\,p+xz\,q=xy. \]
Step 2 · Characteristic equations and first integral
\[ \frac{dx}{yz}=\frac{dy}{xz}=\frac{dz}{xy}. \tag{1} \] Multipliers $(x,-y,0)$: denominator $xyz-xyz=0$, so $x^2-y^2=c_1$. \tag{I}
Step 3 · Second integral
Multipliers $(x,y,-z)$: denominator $xyz+xyz-xyz=xyz$; equating to the first ratio gives $y\,dy-z\,dz=0$: \[ y^2-z^2 = c_2. \tag{II} \]
Step 4 · General solution
\[ y^2-z^2 = \Phi(x^2-y^2), \] where $\Phi$ is an arbitrary differentiable function.
Verification
For $G=y^2-z^2-\Phi(x^2-y^2)$: $\nabla G=(-2x\Phi',\,2y(1+\Phi'),\,-2z)$; \[ \mathbf n_1\cdot\nabla G = -2xyz\Phi'+2xyz(1+\Phi')-2xyz = 0. \checkmark \]
Result
The orthogonal family is: