A quadric has the origin as centre if and only if its equation has no terms of degree $1$

Question

I'd like to know how to prove that if a(n affine) quadric of A^n (regarded as the affine space canonically associated to K^n, being a field) has the origin as centre, then its equation has no terms of degree $1$.

Answer 1

Let $\vec{q}\in V[x]$ be a vector in the two-dimensional vector space $V[x]$ over the field $\mathbb{R}$ defined by: $$\vec{q}(x)=(x,ax^2+bx+c) $$ For $\{a,b,c\}\subseteq\mathbb{R}$. It can be geometrically shown that the set $Q\subseteq V$ defined by: $$Q=\{\vec{q}(x)\in V[x]| x\in \mathbb{R}\}$$ has an axis of symmetry passing through $\vec{q}(0)=(0,c)$ iff $\vec{q}(x)=\vec{q}(-x)$. Assume this is the case, and observe that $\vec{q}$ can be rewritten as $\vec{q}=\vec{e}+\vec{o}$, where $\{\vec{e},\vec{o}\}\subseteq V[x]$ are defined as: $$\vec{e}(x)=\frac{1}{2}(\vec{q}(x)+\vec{q}(-x))$$ $$\vec{o}(x)=\frac{1}{2}(\vec{q}(x)-\vec{q}(-x))$$ We observe that $\vec{o}(-x)=-\vec{o}(x)$, so: $$\vec{q}(x)=\vec{q}(-x)\Leftrightarrow \vec{o}(x)\text{ is constant}$$ But the only way that $\vec{o}$ can be constant is if $b=0$, so: $$\vec{q}(x)=(x,ax^2+c)$$ To prove the general case for an arbitrary affine plane $A$ we recall that $V\cong A$ for every $A$, so there is an isomorphism $\phi:A\rightarrow V$ showing: $$\phi(Q')=Q$$ For a unique $Q'\subseteq A$, and a similar argument holds for every $\phi^{-1}(\vec{q})\in Q'$