$\newcommand\R{\mathbb R}\newcommand\c{\mathsf c}\newcommand\ep{\varepsilon}$The answer is no. Indeed, after clarifications given by the OP in comments and in the original post, the question can be stated as follows:
For $i=0,1,2$, let $$f_i(x,y)=a_{i,0}+a_{i,1}x+a_{i,2}y+a_{i,3}xy+a_{i,4}x^2+a_{i,5}y^2,$$ where the $a_{i,j}$ are real numbers. For a real $a>0$, let $M_a$ be the set of all matrices $(a_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $f(x,y):=f_1(x,y)^2-af_0(x,y)f_2(x,y)\ge0$ for some $(x,y)\in\R^2$. Is it true that the dimension (in whatever appropriate sense) of the complement $M_a^\c$ of $M_a$ to $\R^{3\times6}$ is strictly less that $0$$18$ (the dimension of $\R^{3\times6}$)?
Let $g(x,y):=1+x^2+y^2$. For real $\ep>0$, let $N_\ep$ denote the set of all matrices $(a_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $|a_{i,j}-b_j|<\ep$ for all $i=0,1,2,\,j=0,\dots,5$, where $(b_0,\dots,b_5)=(1,0,0,0,1,1)$. Then for $i=0,1,2$ and all real $x,y$ we have $$|f_i(x,y)-g(x,y)|<\ep(1+|x|+|y|+|xy|+x^2+y^2) \le2\ep g(x,y),$$ since $|x|\le(1+x^2)/2$, $|y|\le(1+y^2)/2$, and $|xy|\le(x^2+y^2)/2$.
So, taking now any $a\in(1,2)$ and any $\ep\in(0,\frac{\sqrt a-1}{2(\sqrt a+1)})$, we get ($\ep<1/2$ and) for all real $x,y$ $$f_1(x,y)^2-af_0(x,y)f_2(x,y)\le g(x,y)^2[(1+2\ep)^2-a(1-2\ep)^2]<0.$$ So, $N_\ep\subseteq M_a^\c$ and the dimension of $N_\ep$ is $18$. Thus, the dimension of $M_a^\c$ is $18\ne0$$18$, not $<18$.