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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?

Gaurang Tandon
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  • Compute $u$ and $v$, then replace $x$ and $y$, respectively, in the original equation with $X+u$ and $Y+v$ and see what you get. – amd Jan 25 '18 at 03:35
  • @amd I understand, I would get back the same result as I have described. But I still wonder why $c'$ magically has such a beautiful value of $\Delta/\delta$ :/ For a comparison, the values of $u$ and $v$ have weird terms like $ag-bf$ and $bf-ch$ :( – Gaurang Tandon Jan 25 '18 at 03:58
  • Not that weird, if you ask me. The homogeneous coordinates of the center are the last row/column of the inverse matrix, so you’re going to see various minors of it. – amd Jan 25 '18 at 04:25

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