- #1
Yoss
- 27
- 0
I need to classify the conic given by the following equation:
[tex] x^2 + xy + 4y^2 + 3x - 9 = 0[/tex]
I found [tex] A = \left(\begin{array}{cc}1 & \frac{1}{2}\\\frac{1}{2} & 4\end{array}\right)[/tex]
I need to find an orthogonal matrix [tex]P[/tex] with [tex]detP = 1[/tex].
So I find the eigenvalues of [tex]A[/tex]
I get [tex](\lambda - 1)(\lambda - 4) - \frac{1}{4} = 0[/tex]
or [tex]\lambda_1 = \frac{5 + \sqrt{10}}{2} , \lambda_2 = \frac{5 - \sqrt{10}}{2}[/tex]
I can't seem to get a P, whose columns are the normalized solutions to
[tex] (\lambda - 1)x -.5y = 0 [/tex]
[tex] -.5x + (\lambda - 4)y = 0 [/tex]
for each eigenvalue. Is there a certain way you deal with these when your eigenvalues are not 'nice'? Should I just deal with them in decimal notation? Thanks for any advice.
[tex] x^2 + xy + 4y^2 + 3x - 9 = 0[/tex]
I found [tex] A = \left(\begin{array}{cc}1 & \frac{1}{2}\\\frac{1}{2} & 4\end{array}\right)[/tex]
I need to find an orthogonal matrix [tex]P[/tex] with [tex]detP = 1[/tex].
So I find the eigenvalues of [tex]A[/tex]
I get [tex](\lambda - 1)(\lambda - 4) - \frac{1}{4} = 0[/tex]
or [tex]\lambda_1 = \frac{5 + \sqrt{10}}{2} , \lambda_2 = \frac{5 - \sqrt{10}}{2}[/tex]
I can't seem to get a P, whose columns are the normalized solutions to
[tex] (\lambda - 1)x -.5y = 0 [/tex]
[tex] -.5x + (\lambda - 4)y = 0 [/tex]
for each eigenvalue. Is there a certain way you deal with these when your eigenvalues are not 'nice'? Should I just deal with them in decimal notation? Thanks for any advice.