# ellipse area

• Nov 16th 2010, 11:39 AM
prasum
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
• Nov 16th 2010, 12:13 PM
FernandoRevilla
The ellipse can be written:

$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) $(x,y)^t=P(x',y')^t$ 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,

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

Regards.
• Nov 17th 2010, 12:06 AM
prasum
can yu please explain me spectral theorem and corresponding eigen value
• Nov 17th 2010, 12:35 AM
FernandoRevilla
Quote:

Originally Posted by prasum
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.

Regards.
• Nov 17th 2010, 12:48 AM
prasum
no can yu suggest me some other method
• Nov 17th 2010, 05:29 AM
FernandoRevilla
Quote:

Originally Posted by prasum
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:

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

Regards.
• Nov 17th 2010, 06:59 AM
prasum
thanks