TITLE: If a plane intersects a regular surface at exactly one point, then it is the tangent plane
QUESTION [8 upvotes]: Question
Let a regular surface, $S$, intersect a plane, $P$, at only one point, $p_0 = (x_0, y_0, z_0)$ in $\mathbb{R}^3$. Show that the plane coincides with the tangent plane to the surface at $p_0.$
Remarks
The problem seems so easy and unassuming, however, I have been stuck on it for several days. I have come up with proofs but I am unconvinced by any so far. Please provide as elementary a proof as possible. Nothing greater than multivariable calculus and some linear algebra (if necessary). I have a feeling that it may be best solved by contradiction but my attempt does not follow this strategy. 
Attempted Proof
Let $S$ be a regular surface and $P$ be a plane which passes through $p_0$. Locally, any regular surfaces may be represented by a function $z=f(x,y)$. 
Hence the tangent plane, $L(x,y)$, to $S$ at $p_0$ is 
\begin{align*}
L(x,y) &= f(x_0, y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0).
\end{align*}
Generally, a plane, $P$, is given by 
$$
A(x-x_0) + B(y-y_0) + C(z-z_0) = 0.
$$ 
$$
\hat{n}\cdot(\bar{p}-\bar{p_0})
$$
where $\hat{n} = (A,B,C)$ is the normal vector to the plane and $p=(x,y,z)$ is any point on the plane.
Note that since $p$ is on the plane it cannot be on the surface by hypothesis.
Here is where I start to get stuck. Help would be appreciated.
I'm sure that there is a simple way of thinking about, and proving it. Thanks

REPLY [5 votes]: This is a purely local theorem. We may assume that $S$ is given in the form
$$(x,y)\mapsto\bigl((x,y,f(x,y)\bigr)$$
where $f$ is $C^1$ in a neighborhood of $(0,0)$, and furthermore that
$$f(0,0)=f_x(0,0)=f_y(0,0)=0\ .$$
So ${\bf 0}=(0,0,0)\in S$, and the tangent plane of $S$ at ${\bf 0}$ is  the plane $z=0$. 
Assume now that we are given a plane
$$ax+by+cz=0,\qquad(a,b,c)\ne(0,0,0),\tag{1}$$
passing through ${\bf 0}$, which is not the plane $z=0$. This entails that at least one of $a$, $b$ is $\ne0$; so lets assume $b\ne0$. The point ${\bf 0}$ is a solution of the system
$$\eqalign{F(x,y,z):=\ \ \ \quad z-f(x,y)&=0\cr
G(x,y,z):=\quad ax+by+cz&=0\ .\cr}\tag{2}$$
Now the matrix
$$\left[\matrix{F_y&F_z\cr G_y&G_z\cr}\right]_{\bf 0}=\left[\matrix{0&1\cr b&c\cr}\right]$$
has determinant $\ne0$. The implicit function theorem then guarantees the existence of two $C^1$-functions $x\mapsto\phi(x)$ and $x\mapsto\psi(x)$, defined in a neighborhood of $x=0$ and taking the value $0$ at $0$, such that all points $(x,y,z)$ of the form
$$\bigl(x,\phi(x),\psi(x)\bigr)$$
satisfy the system $(1)$. But this is saying that the plane $(1)$ intersects $S$ in a smooth curve passing through ${\bf 0}\in S$.