Let M=(x_0,y_0,z_0) be a point on a surface, P=(x_0,y_0,0)…

Let $M=(x_0,y_0,z_0)$ be a point on a surface, $P=(x_0,y_0,0)$ its orthogonal projection onto $z=0$, and $N$ the intersection of the normal at $M$ with the plane $z=0$. Let $O$ be the origin.
1° Find the general form of surfaces such that $\angle NOP = \pi/4$.
2° Find the surface passing through the $Ox$ axis.
3° Which quadric surfaces have this property?

Step 1 · Locate $N$ and set up the angle condition

The normal at $M$ meets $z=0$ at $N=(x_0+pz_0,\,y_0+qz_0,\,0)$. Writing $\tan(\angle NOP)$ via the tangent-subtraction formula for the azimuths of $\overrightarrow{OP}$ (angle $\arctan(y/x)$) and $\overrightarrow{ON}$ (angle $\arctan\bigl(\tfrac{y+qz}{x+pz}\bigr)$), and setting it equal to $\tan(\pi/4)=1$: \[ \frac{z(xq-yp)}{x^2+y^2+z(xp+yq)} = 1 \;\Longrightarrow\; z\bigl[(x-y)q-(x+y)p\bigr] = x^2+y^2. \tag{PDE} \]

Step 2 · Rewrite in cylindrical coordinates

With $x=r\cos\phi$, $y=r\sin\phi$, one finds $xq-yp=z_\phi$ and $xp+yq=rz_r$, so the PDE becomes \[ z(z_\phi - rz_r) = r^2. \] Setting $w=z^2/2$ (so $w_\phi=zz_\phi$, $w_r=zz_r$) linearises this: \[ w_\phi - r\,w_r = r^2. \]

Step 3 · Solve the linear PDE for $w$

Characteristics: $\dfrac{dr}{-r}=\dfrac{d\phi}{1}=\dfrac{dw}{r^2}$. First integral: $re^\phi=c_1$. Second: $dw=-r\,dr \Rightarrow w+r^2/2=c_2$. Substituting $w=z^2/2$: $z^2+r^2=2c_2$, so the general solution is \[ z^2+r^2 = \Phi(re^\phi), \] where $\Phi$ is an arbitrary differentiable function; equivalently in Cartesian coordinates, $\Phi\!\left(\sqrt{x^2+y^2}\,e^{\arctan(y/x)}\right)$.

Step 4 · Part 2: surface through the $Ox$ axis

On $Ox$: $y=0$, $z=0$, $r=x$, $\phi=0$, so $x^2=\Phi(x)$, giving $\Phi(t)=t^2$. Hence $z^2+r^2=(re^\phi)^2=r^2e^{2\phi}$, i.e. $z^2=r^2(e^{2\phi}-1)$, or in Cartesian coordinates: \[ z^2 = (x^2+y^2)\!\left(e^{2\arctan(y/x)}-1\right). \]

Step 5 · Part 3: quadric surfaces

For $z^2+r^2=\Phi(re^\phi)$ to define a quadric (degree-2 polynomial), the simplest and essentially only possibility is $\Phi\equiv\lambda^2$ (constant), giving \[ x^2+y^2+z^2 = \lambda^2, \] a sphere centred at the origin.

Verification

On the sphere $x^2+y^2+z^2=\lambda^2$: $p=-x/z$, $q=-y/z$, so $xq-yp=0$ and $xp+yq=-(x^2+y^2)/z$, giving \[ z\bigl[(x-y)q-(x+y)p\bigr] = z\!\left(0+\frac{x^2+y^2}{z}\right) = x^2+y^2. \checkmark \]