Conic sections, equations and properties

Question

I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know how to do it when the coeficient of $xy=0$, but I have troubles in the other case. Can someone recommend me a book, website o document where I can learn how to find the properties of a conic section given its equation please?

Answer 1

Given the equation of the conic in the form

$A x^2 + B xy + C y^2 + D x + E y + F = 0 $

The term containing $xy$ can be eliminated using a rotation. Define the change of variables as follows

$ x = x' \cos \theta - y' \sin \theta$

$ y = x' \sin \theta + y' \cos \theta $

A point (x, y) will have coordinates $(x',y')$ in a coordinate frame that is rotated by $\theta$ counter-clockwise with respect to the original frame.

Plug these in,

$\begin{equation} \begin{split} & A \left(x'^2 \cos^2 \theta + y'^2 \sin^2 \theta - x' y' \sin 2 \theta \right) \\ &+ B \left( \dfrac{1}{2} ( x'^2 - y'^2 ) \sin 2 \theta + x' y' \cos 2 \theta \right) \\ & + C \left(x'^2 \sin^2 \theta + y'^2 \cos^2 \theta + x' y' \sin 2 \theta \right) \\ & + D \left(x' \cos \theta - y' \sin \theta \right) + E \left(x' \sin \theta + y' \cos \theta \right) + F = 0 \end{split} \end{equation} $

Combining like terms,

$\begin{equation} \begin{split} & x'^2 \left( A \cos^2 \theta + C \sin^2 \theta + \dfrac{1}{2} B \sin 2 \theta \right) \\ & + y'^2 \left( A \sin^2 \theta + C \cos^2 \theta - \dfrac{1}{2} \sin 2 \theta \right) \\ & + x'y' \left( (C - A) \sin 2 \theta + B \cos 2 \theta \right) \\ & + D \left(x' \cos \theta - y' \sin \theta \right) + E \left(x' \sin \theta + y' \cos \theta \right) + F = 0 \end{split} \end{equation} $

Now we want to eliminate the $x' y' $ term, so we have to choose $\theta$ such that

$ (C - A) \sin 2 \theta + B \cos 2 \theta = 0 $

that is,

$ \tan 2 \theta = \dfrac{ B }{A - C} $

Now the equation is

$ A' x'^2 + C' y'^2 + D' x' + E' y' + F' = 0 $

where

$A' = A \cos^2 \theta + C \sin^2 \theta + \dfrac{1}{2} B \sin 2 \theta $

$C' = A \sin^2 \theta + C \cos^2 \theta - \dfrac{1}{2} B \sin 2 \theta $

$D' = D \cos \theta + E \sin \theta $

$E' = - D \sin \theta + E \cos \theta $

$F' = F $

Since rotation preserves lengths, you can find the axes, foci, vertices, and then rotate everything by the angle $\theta$ to obtain them in the original coordinate frame.