Find the surface orthogonal to the one-parameter family z =…
Find the surface orthogonal to the one-parameter family \[ z = axy, \] where $a$ is an arbitrary constant.
Explanation of Method
Eliminating $a$ from the family gives $p=z/x$, $q=z/y$, so its normal is $\propto(-z/x,-z/y,1)$. Requiring the normal of an unknown surface to be perpendicular to this at every point gives the PDE of the orthogonal family, which is then solved by Lagrange's method.
Step 1 · PDE for the orthogonal surfaces
From $z=axy$: $p=ay=z/x$, $q=ax=z/y$, so the family's normal is $\mathbf n_1=(-z/x,-z/y,1)$. An orthogonal surface has normal $\mathbf n_2=(-p,-q,1)$ with $\mathbf n_1\cdot\mathbf n_2=0$: \[ \frac{zp}{x}+\frac{zq}{y}+1=0 \;\Longrightarrow\; yz\,p+xz\,q=-xy. \]
Step 2 · Characteristic equations and first integral
\[ \frac{dx}{yz}=\frac{dy}{xz}=\frac{dz}{-xy}. \tag{1} \] Multipliers $(x,-y,0)$ make the denominator vanish identically: $x\,dx-y\,dy=0 \Rightarrow x^2-y^2=c_1$. \tag{I}
Step 3 · Second integral
Multipliers $(x,y,z)$: denominator $xyz+xyz-xyz=xyz$; equating to the first ratio and simplifying 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. Every surface in this family is orthogonal to every surface $z=axy$.
Verification
For $G=y^2+z^2-\Phi(x^2-y^2)$: $\nabla G=(-2x\Phi',\,2y(1+\Phi'),\,2z)$, and with $\mathbf n_1=(-z/x,-z/y,1)$: \[ \mathbf n_1\cdot\nabla G = 2z\Phi'-2z(1+\Phi')+2z = 0. \checkmark \]
Result
The orthogonal family is: