# Thread: length of large and small axis of the ellipse

1. ## length of large and small axis of the ellipse

I need to calculate this elipse:

$c^2=4x_1^2+3x_2^2-2\sqrt2{x_1x_2}$

1. c2=1
2. c2=4

I need to calculate direction and the length of large and small axis of the ellipse.

(hint: own vector and Eigenvalues)

Thanks for your help.

2. ## Re: length of large and small axis of the ellipse

Hey mikeno80.

This looks like a problem involving quadratic forms:

Principal axis theorem - Wikipedia, the free encyclopedia

3. ## Re: length of large and small axis of the ellipse

One method: if we rotate and xy-coordinate system through angle $\theta$, we have new coordinatex x' and y' such that:
$x= x'cos(\theta)+ y'sin(\theta)$
$y= -x'sin(\theta)+ y' cos(\theta)$

Replace x and y in your equation by those, combine like powers of x' and y' and choose $\theta$ so that the coefficient of x'y' is 0.

Another method: write the equation in matrix form, $\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}4 & -\sqrt{2} \\ -\sqrt{2} & 3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= c$.

Since that is a symmetric matrix, it can be "diagonalized". Find its eigenvalues and eigenvectors. Taking P to be the matrix having the eigenvectors as columns, we can write $P^{-1}AP= D$ where D is the diagonal matrix having the eigenvalues of A on its diagonal. Rewrite the equation $x^TAx= c$ as [tex](x^TP)P{-1}AP(P^{-1}x)= yDy= c[tex]. Since D is diagonal, that equation will have no "xy" term.

I notice that your "hint" refers to eigenvalues and eigenvectors so apparently you already know the second method. So where is your difficulty? Have you found the eigenvalues? Have your found the eigenvectors?

4. ## Re: length of large and small axis of the ellipse

I did find $\lambda_1=5$ and $\lamda_2=2$.

$v_1=(-\sqrt{2}, 1)$ and $v_2=(\frac{1}{\sqrt{2}}, 1)$.

Is this correnct and whati is the length of the axis?