Write the partial differential equation satisfied by all…
Write the partial differential equation satisfied by all conical surfaces with vertex at the point $(a,b,c)$, and solve it.
Explanation of Method
A surface is a cone with vertex $(a,b,c)$ exactly when every generator—the line joining $(a,b,c)$ to a surface point—lies entirely on the surface; equivalently, the normal is perpendicular to the generator direction $(x-a,y-b,z-c)$.
Step 1 · Derive the PDE
For $z=z(x,y)$, the normal is $(-p,-q,1)$. Perpendicularity to $(x-a,y-b,z-c)$ gives \[ -(x-a)p-(y-b)q+(z-c)=0 \;\Longrightarrow\; (x-a)\,p+(y-b)\,q=z-c. \]
Step 2 · Characteristic equations and first integrals
\[ \frac{dx}{x-a}=\frac{dy}{y-b}=\frac{dz}{z-c}. \] Directly: $\dfrac{y-b}{x-a}=c_1$ and $\dfrac{z-c}{x-a}=c_2$.
Step 3 · General solution
\[ z-c = (x-a)\,\Phi\!\left(\frac{y-b}{x-a}\right), \] where $\Phi$ is an arbitrary differentiable function.
Verification
With $P=x-a$, $Q=y-b$, $R=z-c$, both $c_1=(y-b)/(x-a)$ and $c_2=(z-c)/(x-a)$ directly satisfy $Pc_x+Qc_y+Rc_z=0$. For the PDE, with $u=(y-b)/(x-a)$: $p=\Phi(u)-\dfrac{(y-b)\Phi'(u)}{x-a}$, $q=\Phi'(u)$, so \[ (x-a)p+(y-b)q = (x-a)\Phi(u) = z-c. \checkmark \] Geometrically, scaling $(x,y,z)=(a,b,c)+\lambda(x_0-a,y_0-b,z_0-c)$ leaves the surface equation unchanged for all $\lambda$, confirming every generator line lies on the surface. $\checkmark$
Result
The PDE and its general solution are: