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Find the normal form of a conic

  • Thread starter sara_87
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1. 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 dont know what to do about the 1 on the right hand side.
Thank you.
 

Answers and Replies

HallsofIvy
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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.
 
763
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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?
 
763
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the e-values are: sqrt13/2 and -sqrt13/2
 
763
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i'm struuggling to find the eigenvectors
 
HallsofIvy
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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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763
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so eigenvalues are (sqrt5)/2 and -(sqrt5)/2
???
 
HallsofIvy
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What in the world are you doing?

What are the roots of [itex](\lambda- 1)^2= 9/4[/itex]?
 
763
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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!
 
763
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i am struggling to find the eigen vectors.
 
HallsofIvy
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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?
 
763
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oh ok. i worked on it and got the same result for -1/2.
so now what will i do with the =1?
 
HallsofIvy
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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".
 
763
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oh sorry i made a mistake, the eigenvector is not the same, it is <1, 1>
 
HallsofIvy
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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.
 
763
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so x=(v-u)/2
and y=(v+u)/2
 
HallsofIvy
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Yes, now substitute those in your equation: x2- 3xy+ y2= 1.
 
763
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this gives:
(5u^2)-(v^2)=4
 

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