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Math Help - ellipse area

  1. #1
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    ellipse area

    if A represents the area of the ellipse 3(x^2)+4xy+3(y^2)=1 then the value of (3sqrt(5)/(pi))*A
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    MHF Contributor FernandoRevilla's Avatar
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    The ellipse can be written:

    E\equiv (x,y)\begin{pmatrix}{3}&{2}\\{2}&{3}\end{pmatrix}\  begin{pmatrix}{x}\\{y}\end{pmatrix}=1<br />

    By the Spectral Theorem, there is an orthogonal transformation (i.e. preserving distances) (x,y)^t=P(x',y')^t such that the ellipse's equation is:

    \lambda_1x'^2+\lambda_2y'^2=1

    where \lambda_1,\lambda_2 are the corresponding eigenvalues. In this case \lambda_1=5,\lambda_2=1. So,

    E\equiv \dfrac{x'^2}{(1/\;\sqrt[]{5})^2}+\dfrac{y'^2}{1^2}=1

    Then,

    A=\pi ab=\pi/\;\sqrt[]{5}.

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    can yu please explain me spectral theorem and corresponding eigen value
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by prasum View Post
    can yu please explain me spectral theorem and corresponding eigen value
    First of all: Did you cover it?. There are another methods for this problem.

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    no can yu suggest me some other method
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by prasum View Post
    no can yu suggest me some other method
    The equations of any rotation are:

    G_{\alpha } \equiv\begin{Bmatrix} x=x'\cos \alpha-y'\sin \alpha\\y=x'\sin \alpha+y'\cos \alpha\end{matrix}

    Substitute in the ellipse's original equation and make \textrm{coef}(x'\cdot y')=0, you will obtain \alpha=\pi/4 and you can fiind the canonical equation:

     <br />
E\equiv \dfrac{x'^2}{(1/\;\sqrt[]{5})^2}+\dfrac{y'^2}{1^2}=1<br />

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    thanks
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