1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the normal form of a conic

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data

    (a) find the normal form of the following quadratic form
    (b) describe the conic section

    (x^2)-3xy+(y^2)=1

    i know we must put it as a matrix first but i dont know what to do about the 1 on the right hand side.
    Thank you.
     
  2. jcsd
  3. Mar 31, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Ignore it to start with. There are several ways to "normalize" a quadratic but since you specifically mention "matrices", I suspect what you are doing is writing the quadratic as
    [tex]\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= 1[/tex]

    Since the middle matrix is symmetric, it can be diagonalized. Find its eigenvalues and eigenvectors to diagonalize it and then in the x', y' coordinates having the eigenvectors as axes, the matrix will be diagonal and
    [tex]\left(\begin{array}{cc} x' & y'\end{array}\right)\left(\begin{array}{cc}\lambda_1 & 0 \\ 0 & \lambda_2\end{array}\right)\left(\begin{array}{c}x' \\ y'\end{array}\right)= 1[/tex]
    What kind of conic section that is will, of course, depend upon whether [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are positive, negative, or 0.
     
  4. Mar 31, 2008 #3
    this helps alot. thanx.
    but in your matrix, u wrote -3/2 twice....but surely one of them must be positive.
    so if e-values are -ve, it's hyperbola, if +ve, ellipse and if zero, circle
    is that right?
     
  5. Mar 31, 2008 #4
    the e-values are: sqrt13/2 and -sqrt13/2
     
  6. Mar 31, 2008 #5
    i'm struuggling to find the eigenvectors
     
  7. Mar 31, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No. A matrix giving a quadratic form is always symmetric.
    [tex]\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= 1[/tex]
    [tex]= \left(\begin{array}{cc}x- 3/2y \\-3/2 x+ y\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)[/tex]
    [tex]= x^2- 3xy+ y^2[/tex]
    Because it is always symmetric, a quadratic form always has real eigenvalues.

    If both eigenvalues have the same sign, then you have an ellipse (or its "special forms", circle, point or empty set). If the eigenvalues have different signs then a hyperbola, if one is 0, then parabola.

    The eigenvalue equation is
    [tex]\left|\begin{array}{cc}1-\lambda & -3/2 \\ -3/2 & 1-\lambda\end{array}\right|= (1- \lambda)^2- 9/4= 0[/itex]
    so the eigenvalues are much simpler than you got!
     
    Last edited: Mar 31, 2008
  8. Mar 31, 2008 #7
    so eigenvalues are (sqrt5)/2 and -(sqrt5)/2
    ???
     
  9. Mar 31, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What in the world are you doing?

    What are the roots of [itex](\lambda- 1)^2= 9/4[/itex]?
     
  10. Mar 31, 2008 #9
    Oh my god
    hahahahahahahaha
    wt was i doing? i'm so ashamed of myself!! i expanded wrong.
    so the eigenvalues are 5/2 and -1/2
    so now i cud find the eigen vectors!
     
  11. Apr 2, 2008 #10
    i am struggling to find the eigen vectors.
     
  12. Apr 2, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The fact that 5/2 is an eigenvalue means that, for any eigenvector <x, y>
    [tex]\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= \left(\begin{array}{c}(5/2)x \\ (5/2)y \end{array}\right)[/tex]
    which, equating the two rows gives two equations: x- (3/2)y= (5/2)x and -(3/2)x+ y= (5/2)y. Because 5/2 is an eigenvector those two equations are NOT independent: they both reduce to y= -x. Any x,y satisfying that gives an eigenvector corresponding to eigenvalue 5/2: in particular, taking x= 1, y= -1, <1, -1> is an eigenvector corresponding to eigenvalue 5/2.

    Can you do the same thing for eigenvalue -1/2?
     
  13. Apr 2, 2008 #12
    oh ok. i worked on it and got the same result for -1/2.
    so now what will i do with the =1?
     
  14. Apr 2, 2008 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, two different eigenvalues cannot have the same eigenvectors. In fact, eigenvectors, corresponding to different eigenvalues, of a symmetric matrix must be perpendicular.

    Please show me exactly how you "got the same result for -1/2".
     
  15. Apr 2, 2008 #14
    oh sorry i made a mistake, the eigenvector is not the same, it is <1, 1>
     
  16. Apr 2, 2008 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And, of course, the vector <1, 1> points along the line y= x or y- x= 0 while the vector <1, -1> points along the line y= -x or y+ x= 0. Let u= y- x, v= y+ x (so that x= ? , y= ?) and write the equation in terms of the variables u and v by replacing x and y with their expressions in terms of u and v.
     
  17. Apr 2, 2008 #16
    so x=(v-u)/2
    and y=(v+u)/2
     
  18. Apr 3, 2008 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, now substitute those in your equation: x2- 3xy+ y2= 1.
     
  19. Apr 3, 2008 #18
    this gives:
    (5u^2)-(v^2)=4
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find the normal form of a conic
  1. Jordan normal form (Replies: 3)

Loading...