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Math Help - How to reduce a conic to it's canonical form

  1. #1
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    How to reduce a conic to it's canonical form

    I'm having trouble with this :
    5x^2 + 4xy + 8y^2 -32 x - 56y + 80 = 0
    What i know :
    square form is : 5x^2 + 4xy + 8y^2 so the matix is A = 5 2
    2 8
    from the characteristic eq P(λ) = 0 ; |5-λ 2|
    . | 2 8-λ|
    so i reach λ^2 - 13λ +36 = 0
    delta = 25
    λ1,2 = 9 , 4
    using the formula sgn(λ1-λ2)=sgn(a12) to get detR=+1 (if a12 is pos then sgn(a12) = +1 if neg = -1)
    so λ1 = 9 and λ2=4
    for λ1 x2=2x1 so v1(1,2)
    for λ2 x1=-2x2 so v2(-2,1)
    e1 = v1 / || v1 || = (1/sqrt5 , 2/sqrt5)
    e2 = v2 / || v2 || = (-2/sqrt5, 1/sqrt5)
    so R = (1/sqrt5 -2/sqrt5)
    (2/sqrt5 1/sqrt5)

    now the rotation (x) = R(x`) => x = 1/sqrt5 ( x`-2y` ) and y = 1/sqrt5(2x`+y`)
    . (y) (y`)
    My problem starts here i don't know how to replace x and y in the eq of the conic. I have this problem solved the next part would be to reach this :

    9x`^2 - 144/sqrt5 *x` +4y`^2 +8/sqrt5 * y` +80 = 0 , i just can't get to that idk what i'm doing wrong can someone explain to me this next step in detail.
    Last edited by BoneSinger; July 6th 2012 at 04:21 AM.
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  2. #2
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    Re: How to reduce a conic to it's canonical form

    You've done all of the hard work. You have x=\frac{1}{\sqrt{5}}(x'- 2y') and [tex]y= \frac{1}{\sqrt{5}}(2x'+ 5y'). Now put those into " x^2", " xy", and y^2. Of course we will have, in each \frac{1}{\sqrt{5}}\frac{1}{\sqrt{5}}= \frac{1}{5} so we really just need to do 5x^2=5(\frac{1}{5}) (x'- 2y')^2= x'^2- 4x'y'+  4 y'^2, xy=4(\frac{1}{5} (x'- 2y')(2x'+ y')= \frac{4}{5}(2x'^2- 3x'y'- 2y'^2), and 8y^2= \frac{8}{5}(2x'+ y')^2= \frac{8}{5}(4x'^2+ 4x'y'+ y'^2). Put each of those into the equation, and "combine like terms".

    Also, -32x= -\frac{32}{\sqrt{5}}(x'+ 2y'), -56y= -\frac{56}{\sqrt{5}}(2x'- y').
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  3. #3
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    Re: How to reduce a conic to it's canonical form

    so funny for me 1/sqrt 5 * 1/sqrt 5 = 1/ 2sqrt 5...

    thnx for the help
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    Re: How to reduce a conic to it's canonical form

    I have one more but this one i just can't find it's vectors v1 or v2
    7x^2 - 8xy - y^2 - 2x - 4y -1 = 0
    so i reached at λ1 = 3-4*sqrt2 and λ2 = 3+4*sqrt2; i dunno if this is right or not please check my math out.

    and for this one idk what to do cose delta = 0 so λ1=λ2 and when i make the systems idk how to continue
    x^2 - 4xy + y^2 + 6x - 12y +8 = 0
    P(
    λ) = 0 => λ^2 - 2λ + 1 = 0
    Last edited by BoneSinger; July 7th 2012 at 06:43 AM.
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  5. #5
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    Re: How to reduce a conic to it's canonical form

    First, it's not a good idea to post a new question in the same thread. People who have already responded may not look at it again.

    Yes, we can write the quadratic portion as \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}7 & -4 \\ -4 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}. That has eigenvalue equation \left|\begin{array}{cc}7-\lambda & -4 \\ -4 & -1-\lambda\end{array}\right| = (7- \lamba)(-1- \lambda)- 16= \lambda^2- 6\lambda- 23= 0 and that has roots 3+ 4\sqrt{2} and 3- 4\sqrt{2}. An eigenvector corresponding to eigenvalue 3- 4\sqrt{2} must satisfy 7x- 4y= (3+ 4\sqrt{2})x which is the same as 4y= (4- 4\sqrt{2})x so is of the form \left<x, (1-\sqrt{2})x\right>. An eigenvector corresponding to eigenvalue 3+ 4\sqrt{2} must satisfy 7x- 4y= (3- 4\sqrt{2})x which is the same as 4y= (4+ 4\sqrt{2})x so is of the form \left<x, (1+\sqrt{2})x\right>.

    Let x'= y- (1-\sqrt{2})x and y'= y- (1+\sqrt{2})x so that the axes are rotated to be parallel to those eigenvectors. It really isn't necessary to determine the angle but if you want to know it, look at \theta= tan^{-1}(1-\sqrt{2}) and tan^{-1}(1+ \sqrt{2}). Those angles are complemetary so you can rotate through either angle reversing x' and y'.

    For x^2- 4xy+ y^2+ 6x- 12y+ 8= 0, the coefficient matrix is \begin{bmatrix}1 & -2 \\ -2 & 1\end{bmatrix} which has characteristic equation \left|\begin{array}{cc}1-\lambda & -2 \\ -2 & 1-\lambda\end{array}\right|= (1-\lambda)^2-4= \lambda^2- 2\lambda- 3= (\lambda- 3)(\lambda+ 1)= 0 so that \lambda= 3 and \lambda= -1. They are NOT the same. You appear to have forgotten the "4xy" part.

    (Of course, it is possible that the two eigenvalues are the same. But then you would still have two independent eigenvectors corresponding to that eigenvalue. The conic would be a circle, if the common eigenvalue is positive, or an equilateral hyperbola, if negative.)
    Last edited by HallsofIvy; July 7th 2012 at 12:16 PM.
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  6. #6
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    Re: How to reduce a conic to it's canonical form

    Can you please continue the first ex to the end i'm kinda lost it ; It would help me a lot to see it to the end ; for the second yep ... another silly mistake.
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