- #1

notnottrue

- 10

- 0

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter notnottrue
- Start date

- #1

notnottrue

- 10

- 0

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.

- #2

Sourabh N

- 631

- 0

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

Science Advisor

Homework Helper

- 43,021

- 970

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 x

(I "distributed" the b as the b/2 off diagonal terms to make this a

Share:

- Replies
- 21

- Views
- 763

- Last Post

- Replies
- 2

- Views
- 362

- Last Post

- Replies
- 1

- Views
- 590

- Last Post

- Replies
- 3

- Views
- 634

- Replies
- 11

- Views
- 412

- Last Post

- Replies
- 8

- Views
- 780

- Last Post

- Replies
- 2

- Views
- 282

- Last Post

- Replies
- 4

- Views
- 535

- Last Post

- Replies
- 2

- Views
- 735

- Last Post

- Replies
- 10

- Views
- 894