Show that all tangent planes of the graph of g intersect in a common point

Question

For the equation $g(x,y) = x f(y/x)$, show that all tangent planes of the graph of $g$ intersect at a common point. I found the partial derivatives which are: \begin{align*} g_x &= (-y/x) f'(y/x) + f(y/x) \\ g_y &= f'(y/x) \end{align*} What do I do from here? Thanks so much!

Answer 1

In cylindrical / polar coords we have for a twisted surface e.g., catenoid

$$ z(x,y)= f(\theta) = g(y/x) $$

all planes passing through $r=0$ or $x=y=0. $

Answer 2

The tangent plane to a surface $z=g(x,y)$ in point $(x_0,y_0)$ is given by the following equation :

$$\tag{1}z=g(x_0,y_0)+(\tfrac{\partial g}{\partial x})_{|(x_0,y_0)}(x-x_0)+(\tfrac{\partial g}{\partial y})_{|(x_0,y_0)}(y-y_0)$$

As the surface you are interested in has equation $z=xf\left(\tfrac{y}{x}\right)$, (1) becomes :

$$z=x_0f(\tfrac{y_0}{x_0})+(x-x_0)\left(-\tfrac{y_0}{x_0}f'(\tfrac{y_0}{x_0})+f(\tfrac{y_0}{x_0})\right)+(y-y_0)f'(\tfrac{y_0}{x_0})$$

If you expand all, you will see that 4 terms can be cancelled, giving :

$$z=x\left(-\tfrac{y_0}{x_0}f'(\tfrac{y_0}{x_0})+f(\tfrac{y_0}{x_0})\right)+yf'(\tfrac{y_0}{x_0})$$

which is of the form $z=ax+by$, i.e., a plane passing through the origin, which is the common point to all these planes.

Here is a rather pretty illustration of a surface $z=xf(\tfrac{y}{x})$ with $f(x)=\arctan(x)$. The fact that all tangent planes are passing through the origin looks very plausible.

Fig. 1 : the "profile" of function $f$ (here given by $f(x)= $arctan $x$) can be seen as the intersection of the surface by vertical plane $x=1$.

Addendum:

1) It would be interesting to know if $z=xf(y/x)$ ($f$ being any differentiable function) is or isn't the general solution for the PDE (Partial Differential Equation) verified by all planes passing through the origin.

2) I just made a web search with ''$z=xf(y/x)$'' and/or ''planes passing through the origin''. I found a lot of duplicates such as these ones on Math. SE : (https://math.stackexchange.com/q/574594)(https://math.stackexchange.com/q/574572) and as ex. 4 p. 91 of the (excellent) book of M. Do Carmo (Differential Geometry of Curves and Surfaces).

3) I take the opportunity here to recall three different ways to obtain/to see equation (1) for the plane tangent in $M_0(x_0,y_0,z_0) \in (S)$ (surface with equation $z=g(x,y) \iff $ $\varphi(x,y,z)=z-g(x,y)=0$).

Let $M(x,y,z)$ be a generic point in $\mathbb{R^3}$. The tangent plane to $(S)$ in $M_0$ can be seen as the set of points $M$ such that:

• $\vec{M_0M}(x-x_0,y-y_0,z-z_0)$ is orthogonal to $\vec{grad} \varphi=(1,-\tfrac{\partial g}{\partial x},-\tfrac{\partial g}{\partial y})$, i.e., $\vec{M_0M}.\vec{grad}\varphi=0$, whence (1).

• the 3 following vectors (here with coordinates in columns) are coplanar:

$$\begin{pmatrix}1&0&(x-x_0)\\ 0&1&(y-y_0)\\ \tfrac{\partial g}{\partial x}&\tfrac{\partial g}{\partial y}&(z-z_0) \end{pmatrix}=0$$

(the last column represent the coordinates of $\vec{M_0M}$, the two others use the interpretation of partial derivatives as ''$x$-slopes'' and ''$y$-slopes'').

• we can ''reduce'' $g$ to its affine part, i.e. its terms of order 0 and 1 in the Taylor expansion of $g$ in the vicinity of $(x_0,y_0)$.