Help finding ortho matrix for conic section eq

[Yoss]

I need to classify the conic given by the following equation:

$$x^2 + xy + 4y^2 + 3x - 9 = 0$$

I found $$A = \left(\begin{array}{cc}1 & \frac{1}{2}\\\frac{1}{2} & 4\end{array}\right)$$

I need to find an orthogonal matrix $$P$$ with $$detP = 1$$.

So I find the eigenvalues of $$A$$

I get $$(\lambda - 1)(\lambda - 4) - \frac{1}{4} = 0$$

or $$\lambda_1 = \frac{5 + \sqrt{10}}{2} , \lambda_2 = \frac{5 - \sqrt{10}}{2}$$

I can't seem to get a P, whose columns are the normalized solutions to

$$(\lambda - 1)x -.5y = 0$$
$$-.5x + (\lambda - 4)y = 0$$

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.

 

[Benny]

Well you know that for each of the eigenvalues you are only going to get one linearly independent eigenvector. It is not much of a problem even if you get a system with square roots all over the place because you can simply ignore one of the equations. So for your problem just pick which ever equation you want to solve and the solution will automatically satisfy the other one (assuming that you've arrived at the equations correctly).

 

[Yoss]

Well you know that for each of the eigenvalues you are only going to get one linearly independent eigenvector. It is not much of a problem even if you get a system with square roots all over the place because you can simply ignore one of the equations. So for your problem just pick which ever equation you want to solve and the solution will automatically satisfy the other one (assuming that you've arrived at the equations correctly).

I do understand that part. I know that for

$$\lambda_1 = \frac{5 + \sqrt{10}}{2}$$

will give

$$(3 + \sqrt{10})x - y = 0$$
$$-x + (-3 + \sqrt{10})y = 0$$

and that's just solving for one of the equations like you said. I get, for instance

$$\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}{\colv{1}{3 + \sqrt{10}}}$$.

For $$\lambda_2 = \frac{5 - \sqrt{10}}{2}$$

will give

$$(3 - \sqrt{10})x - y = 0$$
$$-x + (-3 - \sqrt{10})y = 0$$

$$\binom{-3 - \sqrt{10}}{1}$$

So $$P = \left(\begin{array}{cc}<br /> 1 & -3 - \sqrt{10}\\<br /> 3 + \sqrt{10} & 1<br /> \end{array}<br /> \right)$$

(I know it should be normalized, but that's a lot of tex. The length of each of the vectors is $$\sqrt{20}$$ so the determinant of this P should be 20, so the normalized form will equal 1.

$$detP = 1 - (-9 -6\sqrt{10} -10) = 20 + 6\sqrt{10}$$

Almost, but I can't seem to find a P where the determinant will equal exactly 20. That's what I'm finding confusing when messy numbers are involved, as opposed to nice integers with nice solutions.

I'm pretty sure I've gotten these calculations correct. Thanks for any advice.

 

[Yoss]

(I know it should be normalized, but that's a lot of tex. The length of each of the vectors is so the determinant of this P should be 20, so the normalized form will equal 1.

Ugh, I forgot to foil $$(3 + \sqrt{10})^2$$ when computing the distance. I guess this is the right answer after all.

 

[HallsofIvy]

If the problem is simply to classify the conic section, then just finding the eigenvalues is sufficient. One eigenvalue is positive, the other negative: that's a hyperbola.

 

[Yoss]

If the problem is simply to classify the conic section, then just finding the eigenvalues is sufficient. One eigenvalue is positive, the other negative: that's a hyperbola.

I am familiar with that, but we have to find the transformed center as well as the axis of rotation.