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Thread: Quadratic Forms???

  1. #1
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    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
    $\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$
    in the form
    $\displaystyle AX^2+BY^2=1$
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  2. #2
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    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
    $\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$
    in the form
    $\displaystyle AX^2+BY^2=1$
    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

    $\displaystyle ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

    is to let

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

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

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

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

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

    In the example you've been given:

    $\displaystyle \tan 2\theta = -5$

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

    Are you sure you have the right numbers here?

    Grandad
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    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.
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  4. #4
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    your quadratic form can be written as a matrix multiplication as
    $\displaystyle
    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}
    $

    We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is $\displaystyle \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}$$\displaystyle = \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 $\displaystyle \frac{3}{2}\pm \frac{\sqrt{37}}{4}$. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. $\displaystyle P^{-1}AP$ will be the diagonal matrix having the eigenvallues as entries.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    your quadratic form can be written as a matrix multiplication as
    $\displaystyle
    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}
    $

    We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is $\displaystyle \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}$$\displaystyle = \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 $\displaystyle \frac{3}{2}\pm \frac{\sqrt{37}}{4}$. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. $\displaystyle P^{-1}AP$ 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 $\displaystyle ax^2+bxy+cy^2$ 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 $\displaystyle AX^2+BY^2=1$

    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
    $\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$ into a quadratic form.
    Last edited by mr fantastic; Mar 3rd 2009 at 04:11 PM. Reason: Fixed some latex
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