Find the surface that passes through the circle z=h…

Find the surface that passes through the circle $z=h$, $x^2+y^2=a^2$ and is orthogonal to the family of surfaces \[ xy = \alpha z, \] where $\alpha$ is an arbitrary parameter.

Step 1 · PDE for the orthogonal surfaces

From $xy=\alpha z$: $y=\alpha p \Rightarrow p=y/\alpha=yz/(xy)=z/x$; similarly $q=z/y$. Orthogonality of $(-z/x,-z/y,1)$ with $(-p,-q,1)$, multiplied by $xy$: \[ yz\,p + xz\,q = -xy. \]

Step 2 · Characteristic equations and first integral

\[ \frac{dx}{yz} = \frac{dy}{xz} = \frac{dz}{-xy}. \tag{1} \] Multipliers $(x,-y,0)$: denominator vanishes, 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

\[ y^2+z^2 = \Phi(x^2-y^2). \tag{$*$} \]

Step 5 · Apply the initial condition

Parametrise $z=h$, $x^2+y^2=a^2$ by $t=x$: $c_1=2t^2-a^2$, $c_2=(a^2-t^2)+h^2$. With $t^2=(c_1+a^2)/2$: $c_2=(a^2+2h^2-c_1)/2$, so $\Phi(s)=(a^2+2h^2-s)/2$. Substituting $s=x^2-y^2$ into ($*$) and simplifying: \[ x^2+y^2+2z^2 = a^2+2h^2. \]

Verification

At $x^2+y^2=a^2$, $z=h$: $a^2+2h^2=a^2+2h^2$. $\checkmark$ For $G=x^2+y^2+2z^2-(a^2+2h^2)$: $\nabla G=(2x,2y,4z)$, and the family's normal is proportional to $(-yz,-xz,xy)$; their dot product is $-2xyz-2xyz+4xyz=0$. $\checkmark$