Find the equation of the surface whose arbitrary tangent…

Find the equation of the surface whose arbitrary tangent plane intersects the $Ox$ axis at a point whose abscissa is half the abscissa of the point of tangency.

Step 1 · Tangent plane and its intersection with $Ox$

The tangent plane at $(x_0,y_0,z_0)$ is $Z-z_0=p_0(X-x_0)+q_0(Y-y_0)$. Setting $Y=Z=0$: \[ X = x_0 + \frac{q_0y_0-z_0}{p_0}. \tag{1} \]

Step 2 · Impose the geometric condition

Requiring $X=x_0/2$ in (1): \[ \frac{q_0y_0-z_0}{p_0} = -\frac{x_0}{2} \;\Longrightarrow\; xp-2yq+2z=0, \] dropping subscripts (valid at every point), i.e. $xp-2yq=-2z$.

Step 3 · Characteristic equations and first integrals

\[ \frac{dx}{x}=\frac{dy}{-2y}=\frac{dz}{-2z}. \] From the last two ratios: $z/y=c_1$. From the first two: $x^2y=c_2$.

Step 4 · General solution

\[ z = y\,\Phi(x^2y), \] where $\Phi$ is an arbitrary differentiable function.

Verification

With $u=x^2y$: $p=2xy^2\Phi'(u)$, $q=\Phi(u)+x^2y\Phi'(u)$: \[ xp-2yq = 2x^2y^2\Phi'-2y\Phi-2x^2y^2\Phi' = -2y\Phi=-2z. \checkmark \] Geometrically, $q_0y_0-z_0=x_0^2y_0^2\Phi'$ and $p_0=2x_0y_0^2\Phi'$, so $X=x_0+\dfrac{x_0^2y_0^2\Phi'}{2x_0y_0^2\Phi'}=x_0+\dfrac{x_0}{2}-x_0=\dfrac{x_0}{2}$. $\checkmark$