Rotation of axes to Eliminate the xy-term in a quadratic equation

[Randomers]

Homework Statement

Rotation of axes to Eliminate the xy-term in a quadratic equation

3x^2 - 10xy + 3y^2 + 16 √2x - 32 = 0

2. The attempt at a solution

I do not know the method that should be taken to solve this sort of question. That is what I am looking for. If anyone can point me in the right direction or help me through this type of problem I would very much appreciate it.

 

[M Quack]

You are looking for the principal axes of an ellipse. And because of the 16\sqrt{2}x term, I guess that it is not even centered around (0,0)

 

[Randomers]

The thing is, I don't know what I'm meant to do with this. How do I go about rotating axes of an equation?

 

[HallsofIvy]

If your function is something like f(x,y)= Ax^2+ Bxy+ Cy^2+ linear terms, there are two methods of doing this, one involving more calculation, the other "deeper" and more sophisticated.

The one involving more calculation is this: let x= x'cos(\theta)+ y'sin(\theta) and y= -x'sin(\theta)+ y'cos(\theta). Put those into the formula so that you have x'^2, x'y', and y'^2, each with coefficients depending on known number A, B, C, and \theta. Set the coefficient of x'y' equal to 0 and solve for \theta.

More sophisticated: write f(x,y)= Ax^2+ Bxy+ Cy^2 in matrix form:
\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}A & B/2 \\ B/2 & C\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}
and solve for the eigenvalues and eigenvectors for that 2 by 2 matrix. Since it is a symmetric matrix, it has real eigenvalues and 2 independent eigenvectors. If you take the new x', y' axes in the direction of the eigenvectors, the matrix becomes diagonal, having the eigenvalues on the main diagonal so that your equation becomes
\begin{bmatrix}x' & y'\end{bmatrix}\begin{bmatrix}\lambda_0 & 0 \\ 0 & \lambda_1\end{bmatrix}\begin{bmatrix}x' \\ y'\end{bmatrix}= \lambda_0x'^2+ \lambda_1y'^2