3
$\begingroup$

I have $x^2+2xy-2y^2+x-4y=0$ and I have to find its canonical form, but I'm a little confused.. I'd like to understand very well what I have to do.. Can you help me, please? Thanks!

  • 1
    First of all, decide whether it is an ellipse, a hyperbola, or a parabola. This is easy.2012-09-14
  • 0
    May have a look into : http://math.stackexchange.com/questions/194535/ellipse-question2012-09-14
  • 0
    @Siminore, yes it is easy! :) But then.. what do I have to do? I'd like to understand the steps needed to get the canonical form. Could you help me?2012-09-14
  • 0
    @labbhattacharjee I'm sorry, thank you, but my Prof doesn't like this proceeding...2012-09-14
  • 0
    @sunrise What are you allowed to use, then?2012-09-15
  • 0
    @Siminore I can only use translations + rotations.. I can't use the derivation of rotation formula. I have to find the centre of the conic and then, using the equations set $\begin {cases}x=X+ \alpha \\ y=Y+\beta \end{cases}$ $(\alpha, \beta$ are center coords), I have to substitute these x and y in the equation of the conic. Then, I know that I have to diagonalize a matrix but.. which matrix? and then? What do I have to do?2012-09-15

1 Answers 1

3

You want to eliminate the term involving $xy$. The simplest way for this example is to notice that $x^2+2xy=(x+y)^2-y^2$. So we use new variables $X=x+y, Y=y$ or $x=X-Y, y=Y$. Making this substitution gives $X^2-3Y^2+X-5Y=0$, and so it is a hyperbola.