Show that if all the normal lines to a surface pass through a fixed point, then the surface is (a portion of) a sphere.
I'm not sure how to parametrize my surface.
In the book there is an example that says we can define a regular parametrization
$x:U \to M$, where $U$ is an open set in $\mathbb{R^2}$ and $M$ is a surface
$x(u,v) = \alpha(u) + v\beta(u)$
From here if the parametrization is correct,
$x_u = \alpha'(u) + v\beta'(u)$
$x_v = \beta(u)$
<=>
$\alpha'(u) + v\beta'(u) – x_u = 0$
$\beta(u) – x_v = 0$
Then if we take the scalar product with $n$
$\alpha'(u) \cdot n + v\beta'(u)\cdot n – x_u\cdot n= 0$ <=> $\alpha'(u) \cdot n + v\beta'(u)\cdot n= 0$
$\beta(u) \cdot n – x_v \cdot n= 0$ <=> = $\beta(u) \cdot n = 0$ <=> $\beta(u)=0$ <=> $\beta'(u)=0$
Therefore
$\alpha'(u) \cdot n + v\beta'(u)\cdot n= 0$ <=> $\alpha'(u) \cdot n = 0$ <=> $\alpha'(u) = 0$
Since $\alpha'(u) = \beta'(u) =0$, then $x(u,v)$ is always a fixed distance from some point which means its the a part of the sphere.
Is this correct?
Best Answer
The hypothesis says that $x(u,v)= f(u,v) N (u,v)$, where $f$ is a scalar, and $N$ the unit normal vector (if we fix the origin to be this point). We have to prove that $f$ is constant.
${\partial x\over \partial u }= {\partial f\over \partial u }N+ f {\partial N\over \partial u }$.
As $N$ is a unit vector, ${\partial N\over \partial u }$ is a tangential vector (derive $<N,N>=1$, as well as ${\partial x\over \partial u }$. So computing the scalar product with $n$ we find ${\partial f\over \partial u }=0$, and similarly ${\partial f\over \partial v }=0$, whence the result.