Results 1 to 4 of 4

Math Help - Geometry...hard conic problem

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    296

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by zhupolongjoe View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by zhupolongjoe View Post
    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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Circle geometry - too hard for me :(
    Posted in the Geometry Forum
    Replies: 3
    Last Post: December 6th 2010, 12:28 AM
  2. Help with this hard geometry problem!
    Posted in the Geometry Forum
    Replies: 4
    Last Post: October 27th 2009, 02:21 AM
  3. Hard geometry problems! Help!
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 19th 2009, 01:27 AM
  4. Replies: 1
    Last Post: August 25th 2009, 01:49 AM
  5. algebra and geometry hard...
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 26th 2008, 01:23 PM

Search Tags


/mathhelpforum @mathhelpforum