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Homework Help: Calc Optimization - Point on an ellipse closest to origin.

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data
    x^2 - 2xy + 6y^2 = 10
    Find the point on the ellipse closest to the origin (0,0).


    2. Relevant equations



    3. The attempt at a solution

    Absolutely no one in my class can solve this. We've been to the math lab and none of the helpers there know how to solve it. I think the only person who knows how to solve it here is my professor, and he essentially dodges our requests for an example. I know how to solve this if the ellipse is, for example, x^3+4y^2 = 7. The problem is, this ellipse has 2xy in it and I don't know how to solve for y to plug in the distance formula.


    For x^3 + 4y^2 = 10:

    y = [(10-x^3)/4]^.5

    D = sq rt {x^2 + [(10-x^3)^.5]^2}

    When D is a min, you have the closest point to the origin.

    So, I essentially just need to know how to solve x^2 - 2xy + 6y^2 = 10 for y, or an alternate means to solve this problem.

    Thanks!
     
  2. jcsd
  3. Nov 15, 2007 #2
    Complete the square with respect to y
     
  4. Nov 15, 2007 #3
    Thanks

    How do I do that?
     
  5. Nov 15, 2007 #4
    It is a rather involved process. I assume you know how to complete the square for something like [itex] ax^2+bx+c [/itex]. In this case treat x as a constant and complete the square for y
     
  6. Nov 15, 2007 #5
    I understand how I could take the derivative of that with respect to y... but I have no idea how to go about completing the square like that...
     
  7. Nov 15, 2007 #6
    [tex] 6y^2-2xy+x^2 = 6 \left( y^2-\frac{1}{3} xy \right) + x^2 [/tex]
    [tex] 6 \left( y^2-\frac{1}{3} xy + \frac{1}{36} x^2 - \frac{1}{36} x^2 \right) + x^2 [/tex]
    [tex] 6 \left( y^2-\frac{1}{3} xy + \frac{1}{36} x^2 \right) - \frac{1}{6} x^2 + x^2 [/tex]
    [tex] 6 \left( y- \frac{1}{6} x \right)^2 +\frac{5}{6} x^2[/tex]

    Edit: Sorry I destroyed the Typesetting so bad, trying to do math in two variables and type set at the same time threw me off
     
    Last edited: Nov 15, 2007
  8. Nov 15, 2007 #7
    Thank you so much! I think I can figure this one out now.
     
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