1. 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

2. 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.

3. can yu please explain me spectral theorem and corresponding eigen value

4. 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.

5. no can yu suggest me some other method

6. Originally Posted by prasum
no can yu suggest me some other method
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.

7. thanks