\(a\ne0\)
Từ đề bài ta có:
\(\left\{{}\begin{matrix}a+b+3=0\\\frac{12a-b^2}{4a}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-b-3\\b^2-16a=0\end{matrix}\right.\)
\(\Rightarrow b^2-16\left(-b-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=-4\Rightarrow a=1\\b=-12\Rightarrow a=9\end{matrix}\right.\)
Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=x^2-4x+3\\y=9x^2-12x+3\end{matrix}\right.\)