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? 

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).
