What conditions would make a system of two quadratic equations have one real solution?

The equation we use is in the form, $ax^2+bx+c= 0$. We also use the vertex form $a(x-p)^2+q$. In the textbook, it says the two parabolas need to have the same vertex, but different a values. However, I also graphed examples where the vertex is different but the a value is the same. So they only intersect once and the two graphs are parallel.

If we have have $y = f(x)$ and $y=g(x)$ [1], a solution will satisfy $f(x)-g(x)=0$, hence we need $f(x)-g(x)$ to have exactly one solution. Quadratic equations always have exactly two solutions, so either $f(x)-g(x)$ is linear, or it has a multiple root. If it's linear, it has the form $y = mx+B$ for some $m,B$ (I'm using a capital $B$ to distinguish it from the $b$ in $y = ax^+bx+x$), which means that the $a$ for $f$ and $g$ are the same, but their $b$ are different (if their $b$ are the same, $f(x)-g(x)$ would be constant, and so would have infinite solutions). If it has a multiple root, then it's of the form $y = A(x-r)^2$ for some $A,r$, which means that the $q$ and $p$ for $f(x)$ and $g(x)$ are the same.

[1] $x = y^2$ can also be considered a quadratic equation, but I'm assuming you're dealing with equations of the form $y = f(x)$.