Find the normal form of a conic

In summary, the student struggled to find the eigenvalues and eigenvectors of a symmetric matrix and eventually found that the eigenvalues are 5/2 and -1/2.
  • #1
sara_87
763
0

Homework Statement



(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 don't know what to do about the 1 on the right hand side.
Thank you.
 
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  • #2
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
[tex]\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[/tex]

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
[tex]\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[/tex]
What kind of conic section that is will, of course, depend upon whether [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are positive, negative, or 0.
 
  • #3
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
the e-values are: sqrt13/2 and -sqrt13/2
 
  • #5
i'm struuggling to find the eigenvectors
 
  • #6
sara_87 said:
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?
No. A matrix giving a quadratic form is always symmetric.
[tex]\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[/tex]
[tex]= \left(\begin{array}{cc}x- 3/2y \\-3/2 x+ y\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)[/tex]
[tex]= x^2- 3xy+ y^2[/tex]
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
[tex]\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!
 
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  • #7
so eigenvalues are (sqrt5)/2 and -(sqrt5)/2
?
 
  • #8
What in the world are you doing?

What are the roots of [itex](\lambda- 1)^2= 9/4[/itex]?
 
  • #9
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
i am struggling to find the eigen vectors.
 
  • #11
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)[/tex]
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
oh ok. i worked on it and got the same result for -1/2.
so now what will i do with the =1?
 
  • #13
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
oh sorry i made a mistake, the eigenvector is not the same, it is <1, 1>
 
  • #15
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
so x=(v-u)/2
and y=(v+u)/2
 
  • #17
Yes, now substitute those in your equation: x2- 3xy+ y2= 1.
 
  • #18
this gives:
(5u^2)-(v^2)=4
 

What is the normal form of a conic?

The normal form of a conic is a simplified equation that represents a conic section such as a circle, ellipse, parabola, or hyperbola. It is written in the form Ax^2 + By^2 + Cx + Dy + E = 0, where A and B are coefficients and x and y are variables.

How do you find the normal form of a conic?

To find the normal form of a conic, you need to complete the square for both the x and y terms. This involves factoring out the coefficients of the x^2 and y^2 terms, and then adding and subtracting a constant to the other side of the equation. Once the equation is in the correct form, the coefficients can be identified.

What is the significance of the normal form of a conic?

The normal form of a conic allows for easier identification of the type of conic (circle, ellipse, parabola, or hyperbola) and its major features, such as the center, vertices, and foci. It also makes it easier to graph the conic and solve problems involving it.

Can any conic be put into normal form?

Yes, any conic can be put into normal form. However, it may require different methods depending on the type of conic. For example, a circle can be easily put into normal form by completing the square, while an ellipse may require other techniques such as rotation or translation.

Are there any limitations to using the normal form of a conic?

The normal form of a conic is only applicable to conic sections that can be represented by an equation in the form Ax^2 + By^2 + Cx + Dy + E = 0. This excludes degenerate conics, such as a single point or a pair of intersecting lines, which cannot be represented by this form. Additionally, the normal form may not be useful for solving certain problems involving conics, such as those that require a parametric form of the equation.

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