Results 1 to 5 of 5

Math Help - Quadratic Forms???

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    7

    Quadratic Forms???

    im not really sure if this is to do with quadratic forms or not but tbh i have no idea of where to start with this question, i think it has something to do with diagonilising a matrix somewhere along the line but apart from that im clueless

    by using suitable transformations express
    x^2+x-8+5xy-6y+2y^2 = 0
    in the form
    AX^2+BY^2=1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    General Equation of a Conic

    Hello rebirthflame
    Quote Originally Posted by rebirthflame View Post
    im not really sure if this is to do with quadratic forms or not but tbh i have no idea of where to start with this question, i think it has something to do with diagonilising a matrix somewhere along the line but apart from that im clueless

    by using suitable transformations express
    x^2+x-8+5xy-6y+2y^2 = 0
    in the form
    AX^2+BY^2=1
    What you are being asked to do here is to transform a general equation of the second degree into a standard form for a conic. The usual method, starting with

    ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0

    is to let

    x = X\cos\theta - Y\sin\theta

    y = X\sin\theta + Y\cos\theta

    which corresponds to a rotation of the axes through an angle \theta, in order to eliminate the term in xy. When you do this, it leads to the value of \theta given by

    \tan 2\theta = \frac{2h}{a-b}

    You then complete the squares for X and Y to get the equation in the required form.

    In the example you've been given:

    \tan 2\theta = -5

    But this doesn't seem to give straightforward (i.e. easy) values for \sin\theta and \cos\theta ( \sin\theta = \sqrt{\frac{1 + \sqrt{26}}{2\sqrt{26}}}, for instance).

    Are you sure you have the right numbers here?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2008
    Posts
    7
    the numbers are defnetly right, ive had a word with my lecturer and he says like you that i need to transform the curve so that we get the equation in a quadratic form then he said something about completing squares and then using eigenvalues or diagonolising matrices, your method looks like it would work however i also know there is more than one way to do this, you are looking at eliminating the values of xy which wont be a quadratic form, i know what to do when i get to the quadratic form i think. its just getting there and choosing the suitable trandformation that is the problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    your quadratic form can be written as a matrix multiplication as
    <br />
x^2+5xy+2y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & \frac{5}{2} \\ \frac{5}{2} & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}<br />

    We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is \left|\begin{array}{cc}1-\lambda & \frac{5}{2} \\ \frac{5}{2} & 2-\lambda\end{array}\right|= \lambda^2- 3\lambda+ 2- \frac{25}{4} = \lambda^2- 3\lambda- \frac{17}{4}= 0

    You can find the eigenvalues by solving that equation using the quadratic formula. They are not, as already pointed out, "easy" numbers: something like, I think \frac{3}{2}\pm \frac{\sqrt{37}}{4}. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. P^{-1}AP will be the diagonal matrix having the eigenvallues as entries.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2008
    Posts
    7
    Quote Originally Posted by HallsofIvy View Post
    your quadratic form can be written as a matrix multiplication as
    <br />
x^2+5xy+2y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & \frac{5}{2} \\ \frac{5}{2} & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}<br />

    We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is \left|\begin{array}{cc}1-\lambda & \frac{5}{2} \\ \frac{5}{2} & 2-\lambda\end{array}\right|= \lambda^2- 3\lambda+ 2- \frac{25}{4} = \lambda^2- 3\lambda- \frac{17}{4}= 0

    You can find the eigenvalues by solving that equation using the quadratic formula. They are not, as already pointed out, "easy" numbers: something like, I think \frac{3}{2}\pm \frac{\sqrt{37}}{4}. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. P^{-1}AP will be the diagonal matrix having the eigenvallues as entries.

    this would be correct if my equation was in the quadratic for, however i cant just discount the other values that are in the equation can i? i think i have a slightly better understanding of what it is am i supposed to do now.

    i need to transform the curve to get rid of all the variables apart from the ones of the form ax^2+bxy+cy^2 this will centre my curve about the origin as far as i understand. then i can diagonalize the matrix as hallsofivy has done this rotates the curve so that it is of the form AX^2+BY^2=1

    i could be completely wrong but i feel this is what im supposed to do. i'm ok with everything after getting it into a quadratic form its just getting to that point which is troubling me so much. i've looked on the interenet but i cant find any way of transforming a polnomial of the form i have at the start
    x^2+x-8+5xy-6y+2y^2 = 0 into a quadratic form.
    Last edited by mr fantastic; March 3rd 2009 at 04:11 PM. Reason: Fixed some latex
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Forms
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 16th 2010, 03:30 AM
  2. quadratic forms
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: May 4th 2010, 12:48 PM
  3. Binary Quadratic Forms
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: March 23rd 2010, 05:11 PM
  4. quadratic forms
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: April 11th 2009, 04:17 PM
  5. Quadratic forms
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 18th 2007, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum