2
$\begingroup$

For a general conic $Q(x,y)=ax^2+2hxy+by^2+2gx+2fy+c$ we define a matrix $A$ as follows:

$A=\left( \begin{matrix} a& h& g\\ h& b& f\\ g& f& c\end{matrix} \right)$.

Then we define $\tau=a+b$, $\delta=\left| \begin{matrix} a& h\\ h& b\end{matrix} \right| $ ve $\Delta=\left| \begin{matrix} a& h& g\\ h& b& f\\ g& f& c\end{matrix} \right|$ With these arguments, we say that $Q(x,y)$ gives us

an ellipse if $\Delta \neq 0$ and $\delta >0$ ,

a parabola if $\Delta \neq 0$ and $\delta =0$ ,

a hyperbola if $\Delta \neq 0$ and $\delta <0$ .

Now, my questions: Where does this $A$ matrix and all $\tau, \delta, \Delta$ definitions come from?Why do we need $\Delta \neq 0$?

$\endgroup$
2
  • $\begingroup$ That's a long story. However, I can answer your last question: If $\Delta=0$, you get a "degenerate" conic, which will be one or two lines, or a point, or the empty set. ... Check out the references in en.wikipedia.org/wiki/Matrix_representation_of_conic_sections $\endgroup$ Commented May 9, 2016 at 23:05
  • $\begingroup$ And for the rest, you need to know the conic representation in homogeneous coordinates, so that it comes clear that $\delta $ is related with the points at infinity, etc. $\endgroup$
    – G Cab
    Commented May 9, 2016 at 23:14

1 Answer 1

1
$\begingroup$

$Q(x,y)=ax^2+2hxy+by^2+2gx+2fy+c$ must be equal (with some corrections maybe) to $$(x,y,1)^TA(x,y,1)$$ For all $(x,y)$ vectors in the plane. If $\delta$ and $\Delta$ are simultaneously positive, this exactly means that $A$ is a positive definite matrix. (Notice that $A$ is symmetrical.) Then, it is obvious why the set of $(x,y)$ satisfying $$(x,y,1)^TA(x,y,1)=C, \ C >0 \ constant $$ is an ellipsis.

Other cases?... Play with them in similar ways.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .