Conic Sections with Matrices I'm reading "Geometry" by Brannan, Esplen, and Gray. Any conic has an equation of the form
$$
Ax^{2} + Bxy + Cy^{2} + Fx + Gy + H = 0,
$$
where $A,\ B,\ C,\ F,\ G,\ H\ \in {\mathbb R}$  and not all of $A, B, C$ are zero. The matrix form of this equation is
$$x^{\sf T}{\cal A}x + J^{\sf T}x + H = 0,$$
where $x = \begin{bmatrix} x \\ y\end{bmatrix}$ is a vector in $\;\Bbb R^2\;$,
${\cal A} = \begin{bmatrix} A & B/2 \\ B/2 & C\end{bmatrix}$, and $J = \begin{bmatrix} F \\ G\end{bmatrix}$. Given the equation $$3x^2 - 10xy + 3y^2 + 14x -2y + 3 = 0,$$ we are asked to find what type of conic this is and its center (if it has one). 
You start by diagonalizing the matrix $A$ and constructing an orthogonal matrix, $P$, with the normalized eigenvectors of $A$. Now, the book says that it's important to make sure that columns of $P$ are arranged s.t. ${\rm det}\ (P) = 1$, so $P$ represents a rotation in the plane. My question is, why is it necessary that $P$ represents a rotation of $x$? Why can't ${\rm det}\ (P) = -1$, so $P$ represents a reflection followed by a rotation? I tried seeing where this change would cause a problem, but can't see it. Can anyone please clarify? 
 A: The curve is identified as a conic that has been rotated from the standard position. Having the determinant $+1$, the new axes are 'labeled' so that, visibly they are obtained by rotating the old ones by an angle (that can be worked out by looking at $P$). 
So it is all about the axes. Nothing much will go wrong if you don't bother checking the determinant of $P$ (often enough students are not even told about this, and they put whatever orientation on the axes).
A: The short answer is, since conic sections are symmetric, if a reflection followed by rotation exists to get a result, then just a rotation also exists to get the result.
The longer answer is similar to @Any's answer.  For whatever application you are using a determinant, the sign almost always reflects some choice of which direction you put positive and negative on your axises.  A conic section is a conic section regardless of which direction you put the signs of the axises, so we expect the sign of the determinant to be insignificant.  
Remark: consider doing yourself a huge favor and use $$Ax^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H = 0$$ as your canonical form for a conic section.  It saves you lots of trouble from having to carry lots of around unnecessary coefficients/fractions when working out problems.
A: $\newcommand{\+}{^{\dagger}}%
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It's better to write your equation as:
$$
x^{\sf T}{\cal A}x + \half\,\pars{J^{\sf T}x + x^{\sf T}J} + H = 0
$$
Then, set $x = v - \mu$ such that
\begin{align}
0&=\pars{v^{\sf T} - \mu^{\sf T}}{\cal A}\pars{v - \mu}
+
\half\,\bracks{J^{\sf T}\pars{v - \mu} + \pars{v^{\sf T} - \mu^{\sf T}}J} + H
\\[3mm]&=
v^{T}{\cal A}v + v^{\sf T}\pars{-{\cal A}\mu + \half\,J}
+
\pars{-\mu^{T}{\cal A} + \half\,J^{\sf T}}v + {\cal B}
\end{align}
$$\mbox{where}\quad
{\cal B} \equiv \mu^{\sf T}\mu - \half\,J^{\sf T}\mu - \half\,\mu^{\sf T}J + H
$$
Now, we choose $\mu$ such that $-{\cal A}\mu + \half\,J = 0$ which yields
$$
\mu = \half\,{\cal A}^{-1}J\quad\mbox{and}\quad
{\cal B} = {1 \over 4}J^{\sf T}{{\cal A}^{-1}}^{\sf T}{\cal A}^{-1}J
-
{1 \over 4}\,J^{\sf T}\pars{{\cal A}^{-1} + {{\cal A}^{-1}}^{\sf T}}J + H
$$
In terms of the vector $v$, we have "eliminated" the linear terms:
$$
v^{\sf T}{\cal A}v + {\cal B} = 0
$$
Can you take from here ?.
