Question

2. Let Σ C E3 be a regular surface, that meets a plane P in a single point p. Show that P has to coincide with the tangent plane of Σ at p. (Hint: use the Implicit Function Theorem.) 3 marksl

Answer 1

We may assume that S is given in the form

(x,y)↦((x,y,f(x,y))

where f is C1 in a neighborhood of (0,0), and furthermore that

f(0,0)=fx(0,0)=fy(0,0)=0 .

So 0=(0,0,0)∈S, and the tangent plane of S at 0 is the plane z=0

.

Assume now that we are given a plane

ax+by+cz=0,(a,b,c)≠(0,0,0),(1)

passing through 0, which is not the plane z=0. This entails that at least one of a, b is ≠0; so lets assume b≠0. The point 0 is a solution of the system

F(x,y,z):=   z−f(x,y)G(x,y,z):=ax+by+cz=0=0 .(2)

Now the matrix

[FyGyFzGz]0=[0b1c]

has determinant ≠0. The implicit function theorem then guarantees the existence of two C1-functions x↦ϕ(x) and x↦ψ(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

(x,ϕ(x),ψ(x))

satisfy the system (1). But this is saying that the plane (1) intersects S in a smooth curve passing through 0∈S.