We know that 5 points in general position determine a conic section uniquely. I wonder if there is a generalization. Consider the equation 
$F(x\_1,\dots, x\_n) =0$ where $F$ is a polynomial of degree $d$ (over $R$). Is there a 
(sharp) bound on the number of points $(x\_1,...,x\_n)$ (in general position) which uniquely determines $F$ ? 



How many points determine an algebraic surface ?
The answer has essentially been given by J. C. Ottem in a comment. I just put it here (with a couple details) so that the question is "answered".


The space of degree $d$ polynomials in n+1 variables has dimension $\binom{n+d}{d}$ (coefficient count), so hypersurfaces of degree d in $\mathbb{P}^n$ are parameterized by a projective space of dimension $N:=\binom{n+d}{d}-1$. Asking that the hypersurface goes through a given point is a linear equation on the coefficients of polynomials.


If the base field (or domain) is infinite, then the conditions imposed by general points are independent. You prove this by induction on the number of points: assume that hypersurfaces of degree d in $\mathbb{P}^n$ through $k-1 < N-1$ general points are parameterized by $\mathbb{P}^{N-k+1}$. **EDIT**: Choose one of these hypersurfaces $X$; it does *not* contain all of the points in $\mathbb{P}^n$, as the base field is infinite. So if the $k$-th point is out of $X$ (which we can assume as the points are general) the parameter space of hypersurfaces containing all $k$ points is strictly conained in $\mathbb{P}^{N-k+1}$, which means the last condition is linearly independent and defines a hyperplane $\mathbb{P}^{N-k}$.


So $N$ is the number of general points that uniquely determine a hypersurface of degree $d$.


