# How to apply a transformation to a conic

I'm trying to read Undergraduate Algebraic Geometry by Miles Reid and I am having some difficulty understanding some of his statements. Specifically I'm trying to solve problem 1.3 which is to prove thay every conic in $\mathbb{R^2}$ can be put to one of the following forms by means of an affine transformation:

Ellipse; parabola; hyperbola; the empty set; a line ( $x=0$ ); a pair of lines ($xy=0$ ); parallel lines ( $x(x-1)=0$ ); 'double line' ( $x^2=0$)

Now, he gives a definition of an affine transformation $T$ in this space in terms of a 2x2 invertible matrix A, and a translation vector B, i.e. an affine transformation is one that has the form $T(x)=Ax+B$, where $x \in \mathbb{R^2}$

I thought maybe I could somehow apply this to the general equation for a conic, i.e. $ax^2+bxy+cy^2+dx+ey+f$, but I'm not sure how to do that or if it even makes any sense.

Any suggestions, hints or clarifications are appreciated.

Your general conic equation can be written as $$\mathbf x^TC\mathbf x = \begin{bmatrix}x&y&1\end{bmatrix} \begin{bmatrix} a&\frac b2&\frac d2 \\ \frac b2 & c & \frac e2 \\ \frac d2&\frac e2&f \end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix} = 0.$$ If you have an invertible point transformation $\mathbf x' = M\mathbf x$, then $$\mathbf x^TC\mathbf x = (M^{-1}\mathbf x')^TC(M^{-1}\mathbf x') = \mathbf x'^T(M^{-T}CM^{-1})\mathbf x',$$ so the conic’s matrix transforms as $C' = M^{-T}CM^{-1}$.
The invertible affine transformation $\tilde{\mathbf x} = A\tilde{\mathbf x}'+\tilde{\mathbf t}$ (note that I’m inverting your transformation for simplicity) can be represented by the $3\times 3$ matrix $$M^{-1} = \left[\begin{array}{c|c} A & \tilde{\mathbf t} \\ \hline \mathbf 0^T & 1\end{array}\right]$$ so that $\mathbf x = M^{-1}\mathbf x'$. Writing $C$ in block form as $$C = \left[\begin{array}{c|c} Q & \tilde{\mathbf b} \\ \hline \tilde{\mathbf b}^T & f \end{array}\right],$$ which corresponds to the form $\tilde{\mathbf x}^TQ\tilde{\mathbf x}+2\tilde{\mathbf b}^T\tilde{\mathbf x}+f = 0$ of the general conic equation, we then have $$M^{-T}CM^{-1} = \left[\begin{array}{c|c} A^T & \mathbf 0 \\ \hline \tilde{\mathbf t}^T & 1\end{array}\right] \left[\begin{array}{c|c} Q & \tilde{\mathbf b} \\ \hline \tilde{\mathbf b}^T & f \end{array}\right] \left[\begin{array}{c|c} A & \tilde{\mathbf t} \\ \hline \mathbf 0^T & 1\end{array}\right] = \left[\begin{array}{c|c} A^TQA & A^T(Q\tilde{\mathbf t}+\tilde{\mathbf b}) \\ \hline (Q\tilde{\mathbf t}+\tilde{\mathbf b})^T A & \mathbf t^TC\mathbf t \end{array}\right].$$ There are several things to note here: the quadratic part of the equation represented by $Q$ is only affected by the linear part of the affine transformation; the new constant term is the left-hand side of the untransformed equation evaluated at $\mathbf t$, the translation part of the transformation; and the signs of $\det C$ and $\det Q$ are preserved by affine transformations. If you examine the signs of these determinants for a canonical example of each the types of conics that you’ve listed, you’ll find that the combination of signs determines the type of conic—opposite signs for an ellipse, $\det Q=0$ and $\det C\ne0$ for a parabola, $\det C=0$ and $\det Q\lt0$ for a pair of intersecting lines, and so on. In fact, $\det Q$ is a multiple of the discriminant of the conic equation. It’s a bit tedious, but not very difficult, to work out appropriate transformations to bring an arbitrary conic into one of these canonical forms. (You might need to use the fact that a real symmetric matrix is orthogonally diagonalizable to prove existence of $A$ in all cases.) For example, $\det Q\ne0$ for any central conic, so the linear terms in the equation of a central conic can be eliminated by choosing $\tilde{\mathbf t} = -Q^{-1}\tilde{\mathbf b}$.
This is classic "analytic geometry". An affine transformation has the form \begin{align} x&=rx'+sy'+t\\ y&=ux'+vy'+w \end{align} where $rv-su\ne0$. So in the new coordinates, the conic becomes $$a(rx'+sy'+t)^2+b(rx'+sy'+t)(ux'+vy'+w )+c(ux'+vy'+w)^2+d(rx'+sy'+t)+e(ux'+vy'+w)+f=0$$ or $$a(rx'+sy')^2+b(rx'+sy')(ux'+vy')+c(ux'+vy)^2+ \text{linear and constant terms}=0.$$
Can you show that there are suitable $r$, $s$, $u$, $v$ such that $$a(rx'+sy')^2+b(rx'+sy')(ux'+vy')+c(ux'+vy)^2$$ is one of $x'^2+y'^2$, $x'y'$, $x'^2$ or zero? If so you can examine each case in turn. For instance $$x'^2+y'^2+\text{linear and constant terms}=0$$ will give either an ellipse (a circle even) or a point or the empty set (depending on the lower terms) etc.