I am on my way of converting the ellipse $ Q(x,y) = ax^2 + by^2 + 2hxy + 2fy + 2gx+c$ to its more friendly form $S(X^\prime, Y^\prime) = \dfrac{X{{^\prime}^2} }{\alpha'^2} + \dfrac{Y{{^\prime}^2} }{\beta'^2} - 1 = 0$.
I know how to eliminate the term $xy$, by rotating the axes by $\tan(2\theta)=\frac{2h}{a-b}$.
But before that rotation, we need to translate the axes so as to eliminate the $x$ and $y$ terms first. I came across this method:
Compute $c'=\frac{\Delta}{\delta}$, where $\delta =\begin{vmatrix} a & h \\ h & b\end{vmatrix}$ and $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$.
So, the new equation is ${R(X,Y) = Q(X + u, Y + v)} = aX^2 + b Y^2 + 2hXY + c^\prime$, where $(u, v)$ is the center of the conic, and is given by $\begin{cases}\dfrac{\partial Q}{\partial x} = 0 \\ \dfrac{\partial Q}{\partial y} = 0 \end{cases}$.
My question: What if the proof/reasoning behind $c'=\frac{\Delta}{\delta}$? How does it so easily replace the $c$, and then we can drop the $x$ and $y$ terms. Why does it work?