Find the normal form of a conic

1. Mar 31, 2008

sara_87

1. The problem statement, all variables and given/known data

(a) find the normal form of the following quadratic form
(b) describe the conic section

(x^2)-3xy+(y^2)=1

i know we must put it as a matrix first but i dont know what to do about the 1 on the right hand side.
Thank you.

2. Mar 31, 2008

HallsofIvy

Staff Emeritus
Ignore it to start with. There are several ways to "normalize" a quadratic but since you specifically mention "matrices", I suspect what you are doing is writing the quadratic as
$$\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= 1$$

Since the middle matrix is symmetric, it can be diagonalized. Find its eigenvalues and eigenvectors to diagonalize it and then in the x', y' coordinates having the eigenvectors as axes, the matrix will be diagonal and
$$\left(\begin{array}{cc} x' & y'\end{array}\right)\left(\begin{array}{cc}\lambda_1 & 0 \\ 0 & \lambda_2\end{array}\right)\left(\begin{array}{c}x' \\ y'\end{array}\right)= 1$$
What kind of conic section that is will, of course, depend upon whether $\lambda_1$ and $\lambda_2$ are positive, negative, or 0.

3. Mar 31, 2008

sara_87

this helps alot. thanx.
but in your matrix, u wrote -3/2 twice....but surely one of them must be positive.
so if e-values are -ve, it's hyperbola, if +ve, ellipse and if zero, circle
is that right?

4. Mar 31, 2008

sara_87

the e-values are: sqrt13/2 and -sqrt13/2

5. Mar 31, 2008

sara_87

i'm struuggling to find the eigenvectors

6. Mar 31, 2008

HallsofIvy

Staff Emeritus
No. A matrix giving a quadratic form is always symmetric.
$$\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= 1$$
$$= \left(\begin{array}{cc}x- 3/2y \\-3/2 x+ y\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)$$
$$= x^2- 3xy+ y^2$$
Because it is always symmetric, a quadratic form always has real eigenvalues.

If both eigenvalues have the same sign, then you have an ellipse (or its "special forms", circle, point or empty set). If the eigenvalues have different signs then a hyperbola, if one is 0, then parabola.

The eigenvalue equation is
$$\left|\begin{array}{cc}1-\lambda & -3/2 \\ -3/2 & 1-\lambda\end{array}\right|= (1- \lambda)^2- 9/4= 0[/itex] so the eigenvalues are much simpler than you got! Last edited: Mar 31, 2008 7. Mar 31, 2008 sara_87 so eigenvalues are (sqrt5)/2 and -(sqrt5)/2 ??? 8. Mar 31, 2008 HallsofIvy Staff Emeritus What in the world are you doing? What are the roots of $(\lambda- 1)^2= 9/4$? 9. Mar 31, 2008 sara_87 Oh my god hahahahahahahaha wt was i doing? i'm so ashamed of myself!! i expanded wrong. so the eigenvalues are 5/2 and -1/2 so now i cud find the eigen vectors! 10. Apr 2, 2008 sara_87 i am struggling to find the eigen vectors. 11. Apr 2, 2008 HallsofIvy Staff Emeritus The fact that 5/2 is an eigenvalue means that, for any eigenvector <x, y> [tex]\left(\begin{array}{cc}1 & -3/2 \\ -3/2 & 1\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= \left(\begin{array}{c}(5/2)x \\ (5/2)y \end{array}\right)$$
which, equating the two rows gives two equations: x- (3/2)y= (5/2)x and -(3/2)x+ y= (5/2)y. Because 5/2 is an eigenvector those two equations are NOT independent: they both reduce to y= -x. Any x,y satisfying that gives an eigenvector corresponding to eigenvalue 5/2: in particular, taking x= 1, y= -1, <1, -1> is an eigenvector corresponding to eigenvalue 5/2.

Can you do the same thing for eigenvalue -1/2?

12. Apr 2, 2008

sara_87

oh ok. i worked on it and got the same result for -1/2.
so now what will i do with the =1?

13. Apr 2, 2008

HallsofIvy

Staff Emeritus
No, two different eigenvalues cannot have the same eigenvectors. In fact, eigenvectors, corresponding to different eigenvalues, of a symmetric matrix must be perpendicular.

Please show me exactly how you "got the same result for -1/2".

14. Apr 2, 2008

sara_87

oh sorry i made a mistake, the eigenvector is not the same, it is <1, 1>

15. Apr 2, 2008

HallsofIvy

Staff Emeritus
And, of course, the vector <1, 1> points along the line y= x or y- x= 0 while the vector <1, -1> points along the line y= -x or y+ x= 0. Let u= y- x, v= y+ x (so that x= ? , y= ?) and write the equation in terms of the variables u and v by replacing x and y with their expressions in terms of u and v.

16. Apr 2, 2008

sara_87

so x=(v-u)/2
and y=(v+u)/2

17. Apr 3, 2008

HallsofIvy

Staff Emeritus
Yes, now substitute those in your equation: x2- 3xy+ y2= 1.

18. Apr 3, 2008

sara_87

this gives:
(5u^2)-(v^2)=4