Find all surfaces z=varphi(x,y) for which the projection of…

Find all surfaces $z=\varphi(x,y)$ for which the projection of $\overrightarrow{OM}$ onto the (upward) unit normal equals $\dfrac{z-f(p,q)}{\sqrt{1+p^2+q^2}}$, with $f$ homogeneous of degree $n\neq1$; and for $f$ an entire homogeneous quadratic, find the surface through the parabola $x=0$, $2z=y^2$.

Step 1 · Derive the Clairaut PDE

The unit normal (acute angle with $Oz$) is $(-p,-q,1)/\sqrt{1+p^2+q^2}$, so the stated condition $\mathbf n\cdot\overrightarrow{OM}=(z-f)/\sqrt{1+p^2+q^2}$ becomes, after cancelling denominators, \[ z = px+qy+f(p,q). \]

Step 2 · Complete and singular integrals in general

Complete integral: $z=ax+by+f(a,b)$. For the singular integral, $\partial_a,\partial_b=0$ give $x=-f_a$, $y=-f_b$; by Euler's theorem $af_a+bf_b=nf$, so $-ax-by=nf$, and combined with $ax+by=z-f$ from the complete integral, \[ f(a,b) = \frac{z}{1-n}, \] which together with $x=-f_a(a,b)$, $y=-f_b(a,b)$ determines the singular integral implicitly.

Step 3 · Determine $f$ for the quadratic case

Let $f(p,q)=Ap^2+Bpq+Cq^2$. On $x=0$, $z=y_0^2/2$: the strip condition gives $b=q_0=y_0$; substituting into $z=ax+by+f(a,b)$ at $x=0$ forces (matching coefficients in $y_0$ for all $a$) \[ A=0,\quad B=0,\quad C=-\tfrac12, \qquad\text{i.e.}\qquad f(p,q)=-\frac{q^2}{2}. \]

Step 4 · Singular integral through the parabola

With $z=ax+by-b^2/2$: $\partial_a=0$ gives $x=0$; $\partial_b=0$ gives $b=y$. Substituting: \[ z = 0+y\cdot y-\frac{y^2}{2} = \frac{y^2}{2}. \]

Verification

$z=y^2/2$: $p=0$, $q=y$, and $px+qy+f(p,q)=y^2-\tfrac{y^2}{2}=\tfrac{y^2}{2}=z$. \checkmark At $x=0$: $2z=y^2$. \checkmark