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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- Nov 16th 2010, 10:39 AMprasumellipse 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

- Nov 16th 2010, 11:13 AMFernandoRevilla
The ellipse can be written:

$\displaystyle E\equiv (x,y)\begin{pmatrix}{3}&{2}\\{2}&{3}\end{pmatrix}\ begin{pmatrix}{x}\\{y}\end{pmatrix}=1

$

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

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

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

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

Then,

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

Regards. - Nov 16th 2010, 11:06 PMprasum
can yu please explain me spectral theorem and corresponding eigen value

- Nov 16th 2010, 11:35 PMFernandoRevilla
- Nov 16th 2010, 11:48 PMprasum
no can yu suggest me some other method

- Nov 17th 2010, 04:29 AMFernandoRevilla
The equations of any rotation are:

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

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

$\displaystyle

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

$

Regards. - Nov 17th 2010, 05:59 AMprasum
thanks