# Geometry...hard conic problem

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• September 22nd 2010, 07:23 AM
zhupolongjoe
Geometry...hard conic problem
Identify the conic with equation x^2+3xy+4y^2-7=0, and find its center/vertex and axis.

My A matrix is (1 3/2; 3/2 4), which gives eigenvalues 1/2(5+3sqrt(2)) and 1/2(5-3sqrt(2))....these don't seem right as in every other problem, it comes out much nicer. Am I doing something wrong?
• September 22nd 2010, 12:39 PM
Opalg
Quote:

Originally Posted by zhupolongjoe
Identify the conic with equation x^2+3xy+4y^2-7=0, and find its center/vertex and axis.

My A matrix is (1 3/2; 3/2 4), which gives eigenvalues 1/2(5+3sqrt(2)) and 1/2(5-3sqrt(2))....these don't seem right as in every other problem, it comes out much nicer. Am I doing something wrong?

You have the correct eigenvalues, and you could continue to find the eigenvectors that way. But there is another method which seems to work better in this case.

When you rotate the axes through an angle $\theta$, x becomes $x\cos\theta - y\sin\theta$ and y becomes $\x\sin\theta + y\cos\theta$. Make those substitutions in the equation $x^2+3xy+4y^2-7=0$, to get $(x\cos\theta - y\sin\theta)^2 + 3(x\cos\theta - y\sin\theta)(\x\sin\theta + y\cos\theta) + 4(\x\sin\theta + y\cos\theta)^2-7=0$. If you choose $\theta$ so that the xy-term is zero then you will have found the directions of the principal axes. The coefficient of xy is $6\sin\theta\cos\theta +3(\cos^2\theta-\sin^2\theta) = 3\sin2\theta + 3\cos2\theta.$ This will be zero when $\tan2\theta=-1.$ So $2\theta = -\pi/4$, or $\theta=-\pi/8$. That gives you the orientation of the conic, and you should then be able to complete the question.

The numbers that come in the eigenvalue/eigenvector equations are related to the trig functions of $\pi/8$, which is what makes them messy.
• September 22nd 2010, 03:23 PM
zhupolongjoe
But the professor wants us to use the eigenvalue transformation technique to practice. So I've got it down to ((5+3sqrt(2))*x'^2+(5-3sqrt(2))*y'^2)/14=1.

So I believe this to be an ellipse with center (0,0). Is that correct? Also would the major axis be x=0 and the minor be y=0?
• September 22nd 2010, 11:53 PM
Opalg
Quote:

Originally Posted by zhupolongjoe
But the professor wants us to use the eigenvalue transformation technique to practice. So I've got it down to ((5+3sqrt(2))*x'^2+(5-3sqrt(2))*y'^2)/14=1.

So I believe this to be an ellipse with center (0,0). Is that correct? Also would the major axis be x=0 and the minor be y=0?

Ellipse – yes. Centre (0,0) – yes. But the major axis is x'=0, not x=0, and similarly for the minor axis. You need to give the direction of the line x'=0 in terms of the original coordinates x and y (which according to my method of solution should give a line making an angle $\pi/8$ with the original x-axis).
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