Exploring a Conic Section: 2X² - Y² - 4xy - 4X - 8Y + 14 = 0

[Darkbalmunk]

Homework Statement

describe the conic section.
a) 2X² - Y² - 4xy - 4X - 8Y + 14 = 0

convert to standard form of a conic section and identify its graph.
b) X² + Y² - 2XY + 16X + 16Y = 0

Homework Equations

a) 2X² - Y² - 4xy = 4X + 8Y - 14

b) X² + Y² - 2XY = -16X - 16Y

aX² + bY² + cZ² + 2dXY + 2eXZ + 2fYZ = 0

|a d e|
|d b f| = symetric matrix A
|e f c|

The Attempt at a Solution

im actually trying to get the symetric matrix from the quadratic forms but i only know how to use the formula i have in my relevant equations. and looking at my textbook it always assume that if it has kX+kY+kZ in the formula that they are zero.

 

[HallsofIvy]

$$Ax^2+ Bxy+ Cy^2= \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}A & \frac{B}{2} \\ \frac{B}{2} & C\end{bmatrix}\begin{bkmatrix}x \\ y\end{bmatrix}$$
LaTeX doesn't seem to be working. That is
[A B/2]
[B/2 A ]

 

[Darkbalmunk]

the problem is when i try to get the eigenvalues of det(A-lamdaI) = 4X + 8Y - 14
So what do i do with the 4X + 8Y - 14

 

[HallsofIvy]

What? Since A itself contains only numbers, not "X" or "Y", how could the determinant be equal to 4X+ 8y- 14?

The characteristic equation is given by
$$\left|\begin{array}{cc}2-\lambda & -2 \\ -2 & -1-\lambda\end{array}\right|= \lambda^2- \lambda- 4= 0$$.

Now find the corresponding eigenvectors and construct matrix P having those eigenvectors as columns. Then P^{-1}AP= D where D is the diagonal matrix having -1 and 4 on the diagonal. Your original equation is [X, Y]A[X, Y]^T+ B[X, Y]+ 14= 0 where B is [4, 8]. [X,Y]P(P^{-1}AP)+ B[X, Y]P+ 14P= 0. If you let [X', Y']= [X, Y]P, you will have a quadratic in X' and Y' that does not contain a X'Y' term.

 

[Darkbalmunk]

2X² - Y² - 4xy - 4X - 8Y + 14 = 0

aX² + bY² + 2cXY = 0

|a c|
|c b|

The problem I am having is i can make the equation into
2X² - Y² - 4xy - 4X - 8Y = -14

but in class nor in the book explains how i can find a symetric matrix when i have -4X and -8Y because to get the conics the teacher explained its [XY] [matrix A] |X|
|Y|
and when the problems included -4x-8y i became completely lost.

 

[vela]

If all you want to do is identify the type of conic section it is, you can ignore the linear and constant terms Dx+Ey+F. They don't change the basic shape of the conic section. As HallsofIvy has said, you want to construct the 2x2 matrix from the coefficients A, B, and C, and then find its eigenvalues and eigenvectors. From the eigenvalues, you can identify the type of curve you have.

From the eigenvectors, you can figure what the new variables x' and y' are to get rid of the cross term. When you rewrite the original equation in terms of x' and y', you'll have

$$A'x'^2 + C'y'^2 + D'x'+E'y'+F=0$$

Then it's essentially just a matter of completing the square twice and possibly dividing by a constant to get the equation into standard form.