Given a bivariate quadratic function: $$ y = g(x, w) = a x^2 + b x w + c w^2 + d x + e w + f $$ I think I know that if $b^2 - 4 a c < 0$, then it has a single point minimum (if $a > 0$) or maximum (if $a < 0$) at: $$ x = \frac{2 c d - b e}{b^2 - 4 a c} , w = \frac{2 a e - b d}{b^2 - 4 a c} $$ Please correct me if that is incorrect.
But if $b^2 - 4 a c = 0$ and $2 c d - b e = 0$ and/or(?) $2 a e - b d = 0$, then it has a minimum (if $a > 0$) or maximum (if $a < 0$) line rather than a single point. What I do not know how to do is to compute that line in terms of $a, b, c, d, e,$ and $f$, similar to how I can compute the single point above. Can anyone tell me how to compute that line? Thanks!
$$y = \begin{bmatrix} x & w\end{bmatrix}\begin{bmatrix} a & \frac{b}2 \\ \frac{b}2 & c\end{bmatrix}\begin{bmatrix} x \\ w\end{bmatrix} + \begin{bmatrix}d & e \end{bmatrix}\begin{bmatrix} x \\ w\end{bmatrix} + f$$
$$\begin{bmatrix} 2a & b \\ b & 2c\end{bmatrix}\begin{bmatrix} x \\ w\end{bmatrix} = -\begin{bmatrix} d \\ e\end{bmatrix}$$
If $b^2-4ac=0,$ $a>0$, then it is positive semidefinite. If $2ae-bd=0$, then the system is consistent. The line is
$$2ax + bw = -d$$
Similarly if $b^2-4ac=0,$ $a<0$, it is negative semidefinite and we get the same line.