I recently studied quadrics at university, in linear algebra, but professor has skipped parts of proofs which I think are important.
We have defined a function $\phi$: $\mathbb{R}^n\to\mathbb{R}$ by
$$\phi(x)=\sigma(x,x)+2f(x)+\alpha$$ being $\sigma$ a non-zero bilinear form, $f$ a linear form and $\alpha\in\mathbb{R}$. We have been told a quadric is the set$C=\{x\in\mathbb{R}^n|\phi(x)=0\}$ for some $\phi$.
We have then defined the centre of a non-empty quadric as a point $z\in\mathbb{R}^n$ such that $\forall x\in C: 2z-x\in C$. In a following theorem we have seen $$\phi(2z-x)=\phi(x)+4(\sigma(z,z-x)+f(z-x))$$ for all x ( it is trivial). It is a direct consequence that z is a centre of C if and only if $\sigma(z,z-x)+f(z-x)=0$ $\forall x\in C$. However, to calculate these centres, the professor solves the equation system given by $\sigma(z,y)+f(y)=0$ $\forall y\in\mathbb{R}^n$ ( I don't know how to write this system here, but you know what I mean)
It's obvious that every solution of this system has to be a centre of C, but how can we prove that every center is a solution of the system? If we aren't able to prove this, then we can't assure, for example, that a parabola has no centers.
EDIT: How would you prove that a non-empty quadric which is not the null space of $\sigma$ isn't contained in any hyperplane? I think with this proof I could make up the rest. Thank you