# 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).

• I see. Thank you for the explanation. – Adam Nov 29 '13 at 5:27

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.

• Thank you for the explanation and tip :) – Adam Nov 29 '13 at 5:27

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 ?.