Find the surface that passes through the line y=x , z=h and…

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

Step 1 · PDE for the orthogonal surfaces

Eliminating $\alpha=(x^2+y^2+z^2)/x$ from $\nabla(x^2+y^2+z^2-\alpha x)$ gives a normal proportional to $(x^2-y^2-z^2,\,2xy,\,2xz)$. Orthogonality with $(-p,-q,1)$ gives, after multiplying by $x$: \[ (x^2-y^2-z^2)\,p + 2xy\,q = 2xz. \]

Step 2 · Characteristic equations and first integral

\[ \frac{dx}{x^2-y^2-z^2} = \frac{dy}{2xy} = \frac{dz}{2xz}. \tag{1} \] From the last two ratios: $dy/y=dz/z \Rightarrow z/y=c_1$. \tag{I}

Step 3 · Second integral

Multipliers $(2x,2y,2z)$ give denominator $2x(x^2+y^2+z^2)$; equating to the second ratio and simplifying: \[ \frac{d(x^2+y^2+z^2)}{x^2+y^2+z^2} = \frac{dy}{y} \;\Longrightarrow\; \frac{x^2+y^2+z^2}{y} = c_2. \tag{II} \]

Step 4 · General solution

\[ \frac{x^2+y^2+z^2}{y} = \Phi\!\left(\frac{z}{y}\right). \tag{$*$} \]

Step 5 · Apply the initial condition

Parametrise $y=x$, $z=h$ by $t=x$: $c_1=h/t$, $c_2=(2t^2+h^2)/t=2t+h^2/t$. With $t=h/c_1$: $c_2=2h/c_1+hc_1$, so $\Phi(s)=2h/s+hs$. Substituting $s=z/y$ into ($*$) and clearing denominators: \[ z(x^2+y^2+z^2) = 2hy^2+hz^2 \;\Longleftrightarrow\; z(x^2+y^2+z^2-hz) = 2hy^2. \]

Verification

At $y=x$, $z=h$: $h(2x^2+h^2-h^2)=2hx^2=2hy^2$. $\checkmark$ For $G=z(x^2+y^2+z^2-hz)-2hy^2$, computing $\nabla G$ and dotting with the family's normal $(x^2-y^2-z^2,\,2xy,\,2xz)$ collapses (after grouping) to $4xG=0$ on the surface. $\checkmark$