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Math Help - Conic to Quadratic Form (How To?)

  1. #1
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    Conic to Quadratic Form (How To?)

    I am reviewing an old Linear Algebra textbook, and am going through a chapter about going from a Quadratic Form to a Conic Section.

    For example, starting with the following Quadratic Form:

    F(x,y) = a x^{2} + b x y + c y^{2}

    Let \vec{v} = \begin{bmatrix}x\\ y\end{bmatrix}

    Then find a symmetric Matrix, A, such that F(x,y) = A \vec{v}\cdot \vec{v}

    Then find an orthogonal matrix, Q, such that Q^{T}A Q = D, where D is a diagonal matrix.

    Also, let \vec{v'} = Q' \vec{v}

    At this point, we have enough information to determine if the original Quadratic Form represented a parabola, ellipse, or hyperbola.

    However, I would like to go in the reverse direction: from a Conic to a Quadratic Form.
    For example, say I have an ellipse of eccentricity, e.
    How can I work backwards to create the Q, D, and A matrices, compute their eigenvalues and eigenvectors, etc.

    Can anybody suggest an easy-to-understand reference for working through these steps?
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    Re: Conic to Quadratic Form (How To?)

    I worked through this problem, so I am posting my work (maybe somebody else would find this informative.)

    - - -

    I am interested in seeing a relationship for standard ellipses as the eccentricity varies, so there is no rotation and the major axis, a, is 1, and aligned along the x-axis.

    So, given the standard form of an ellipse

    \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and knowing that b = a(1 - \textit{e}^{2})

    the equation can be rearranged into the following form:

    (1 - \textit{e}^{2})^2 x^{2} + y^{2} = (1 - \textit{e}^{2})^2

    This means that the diagonal matrix, D, is

    D = \begin{bmatrix}(1 - \textit{e}^{2})^2 &0\\ 0 &1\end{bmatrix}

    Because there is no rotation (the a- and b-axes are aligned with the x- and y-axes), Q, which can be considered a rotation matrix, is

    Q = \begin{bmatrix}\cos(\theta) &-\sin(\theta) \\ \sin(\theta)  &\cos(\theta) \end{bmatrix} = \begin{bmatrix}1 &0 \\ \0  &1 \end{bmatrix}

    det(Q) is 1, so it does, indeed, serve as a rotation matrix.

    The A matrix can now be computed as follows:

    A = QDQ^{T}

    which works out to

    A = \begin{bmatrix}(1 - \textit{e}^{2})^2 &0\\ 0 &1\end{bmatrix}

    which is the same as the D matrix.

    det(A) = (1 - \textit{e}^{2})^2

    The equation in its original quadratic form is the same as the one found above:

    F(x,y) = (1 - \textit{e}^{2})^2 x^{2} + y^{2} = (1 - \textit{e}^{2})^2

    I had hoped to find some more information but, unfortunately, did not.
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  3. #3
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    Re: Conic to Quadratic Form (How To?)

    I get a different result from using the substitution b = a(1-e^2):

    \begin{align*}\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & = 1 \\ \dfrac{x^2}{a^2} + \dfrac{y^2}{a^2(1-e^2)^2} & = 1 \\ (1-e^2)^2x^2 + y^2 & = a^2(1-e^2)^2\end{align*}

    Edit: Also, from what I recall about eccentricity (which I haven't looked at in a long time, so forgive me if I am mistaken), e = \sqrt{1-\dfrac{b^2}{a^2}} so b=a\sqrt{1-e^2}, which would give:

    \begin{align*}\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & = 1 \\ \dfrac{x^2}{a^2} + \dfrac{y^2}{a^2(1-e^2)} & = 1 \\ (1-e^2)x^2 + y^2 & = a^2(1-e^2)\end{align*}
    Last edited by SlipEternal; October 19th 2013 at 08:56 PM.
    Thanks from DavidB
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    Re: Conic to Quadratic Form (How To?)

    Oops! You are correct! Thanks.

    p=a (1-e^2), another variable relevant to ellipses.

    b=a\sqrt{1-e^2}


    Goes to show why it is always good to have more than one pair of eyes look things over.
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  5. #5
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    Re: Conic to Quadratic Form (How To?)

    Quote Originally Posted by DavidB View Post
    Oops! You are correct! Thanks.

    p=a (1-e^2), another variable relevant to ellipses.

    b=a\sqrt{1-e^2}


    Goes to show why it is always good to have more than one pair of eyes look things over.
    Ah, ok. Then when using that substitution, I get (1-e^2)^2x^2 + y^2 = p^2, which is slightly different from what you got. Perhaps Q^\prime = aQ would work for your matrix. It is a rotation and stretch matrix. But, that would give you A^\prime = Q^\prime D (Q^\prime)^T with \det(A^\prime) = p^2.
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