Level Curves of xy/(x^2+y^2)

  • Thread starter notnottrue
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  • #1
notnottrue
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I am trying to find the level curves for the function g(x,y)= k = xy/(x^2+y^2).
I get, x^2+y^2-xy/k=0.
I know this is an ellipse, but I do not know how to factor, and find values of k for which the level curves exist.
 

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  • #2
Sourabh N
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I am trying to find the level curves for the function g(x,y)= k = xy/(x^2+y^2).
I get, x^2+y^2-xy/k=0.
I know this is an ellipse, but I do not know how to factor, and find values of k for which the level curves exist.

If the level curves have to be ellipses, you can use the constraint on a general conic section (any quadratic equation in two variables) for it to be an ellipse. Applying the constraint to your quadratic equation will give you the values of k.
 
  • #3
HallsofIvy
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There are various ways to change "rotated" conic sections (that have an "xy" term) to "standard form" (without the xy). One is to rotate the coordinate system by writing [itex]x= x'cos(\theta)+ y'sin(\theta)[/itex], [itex]y= -x'sin(\theta)+ y'cos(\theta)[/itex]. Put those into the equation and combine all x'y' terms. Choose [itex]\theta[/itex] to make its coefficient 0.

A method that is more 'advanced' but simple to compute is to write the second order terms as a matrix multiplication: write [itex]ax^2+ bxy+ cy^2[/itex] as
[tex]\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}a & b/2 \\ b/2 & c\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}[/tex]
and find the eigenvalues and eigenvectors of that matrix. The eigenvalues will be the coefficients of x2 and y2 and the eigenvectors will give the directions for the rotated coordinate system.

(I "distributed" the b as the b/2 off diagonal terms to make this a symmetric matrix which guarentees that it has real eigenvalues and independent eigenvectors.)
 

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