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Center of a Central Conic

  1. Jun 12, 2005 #1
    How do I find the center of a central conic with the equation [tex]ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0[/tex] Is there some easy forumlua for both the ellipse and hyperbola? Thanks!
     
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  3. Jun 12, 2005 #2

    arildno

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    Do you know how to reduce a conic to standard form through rotation&translation?
     
  4. Jun 12, 2005 #3
    I don't remember, but I can use it, and it is in the book I am looking at. Thanks! I will see if I can get it from here, thanks!
     
  5. Jun 12, 2005 #4
    Just a quick question, about whether I am doing this correct. Because this thing looks nasty.


    I used

    [tex]cot(2\theta) = \frac{a-b}{2h} [/tex]

    Then got

    [tex]sin(\theta) = \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)} [/tex]

    and

    [tex]cos(\theta) = \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)} [/tex]

    Then

    [tex] x = x' [ \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)} ] - y' [ \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)} ] [/tex]



    [tex] y = x' [ \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)} ] + y' [ \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)} ] [/tex]


    Does that look correct so far? And then, ughh, I have to substitue that x and y into the original equation, and then see if I can figure the centers from that? Seems like a pain :cry:




    Also, what is translation?

    Thanks
     
    Last edited: Jun 12, 2005
  6. Jun 12, 2005 #5

    OlderDan

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    If you succeed in rotating, you will not have any terms that involve xy products. If you had something of the form

    ax^2 + bx + cy^2 + dy + e = 0

    you could complete the squares of the x terms and the y terms separately and wind up with something of the form

    (x-h)^2/A^2 +or- (y-k)^2/B^2 = 1

    which is a ellipse or hyperbola, depending on the sign, centered at (h,k). That's the translation part.
     
  7. Jun 12, 2005 #6
    Thanks for the clarification.

    I am beginning to think the approach I was trying is not a good one.

    Here is what I am trying to prove.

    [tex]\phi (x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0[/tex]
    for [tex] ab - h^2 \neg 0[/tex] That is: ab - h^2 not equal to 0

    Show that the center of the central conic [tex]\phi (x,y) [/tex] is the interesection of the lines [tex] \frac{\partial \phi}{\partial x} = 0 [/tex] and [tex] \frac{\partial \phi}{\partial y} = 0 [/tex]

    Do you think the approach I am trying, by first finding the center of the central conic is a good route? Thanks
     
    Last edited: Jun 12, 2005
  8. Jun 13, 2005 #7

    OlderDan

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    I think there is a better way. It is easier to visualize for the case of an ellipse. Think about what it means for dy/dx = 0 and dy/dx approaches infinity for an ellipse, and how you might use the location of a pair of points that satisfy either one of those conditions to locate the center of the ellipse. How could you find the points (x,y) that satisfy those conditions? Then relate those points to the zeros of the partial derivatives.

    For the hyperbola, consider the same interpretations of dy/dx = 0 and dy/dx approaches infinity. Only one of the two can be satisfied for points on the hyperbola, but all you need is one pair of points that satisfies one of those two conditions to locate the center. You can still show that the intersection of the lines obtained from the partial derivatives is the center of the hyperbola.
     
  9. Jun 13, 2005 #8
    That sounds like a great idea, and I tried it, but I did not come up with exactly what I wanted.

    I got dy/dx, using implicit differentiation, to be: [tex]\frac{-2ax - 2hy - 2g}{2hx + 2by + 2f}[/tex] So, when dy/dx = 0, then -2ax - 2hy - 2g = 0. You could also say from that, that when dy/dx = 0, then 2ax + 2hy + 2g = 0. Also, when dy/dx = [tex] \infty [/tex] then 2hx + 2by + 2f = 0.

    Now, when I did the partial differentiation, I got the same answers, which is why I did not like it.

    I got

    [tex] \frac{\partial \phi}{\partial x} = 2ax + 2hy + 2g [/tex]

    and

    [tex] \frac {\partial \phi}{\partial y} = 2hx + 2by + 2f [/tex]

    From here I am stuck, I am unsure of really what to do. It seems like it is close, but just not there.

    I guess you could say that for the first part, dy/dx = 0, that [tex]y = \frac{-ax - g}{h}[/tex] Which is a line, but from there I am uncertain, other than finding the other line, which is [tex] y = \frac {-hx - f}{b}[/tex]
     
  10. Jun 13, 2005 #9

    OlderDan

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    The correspondence between the dy/dx results and the partial derivatives is exactly what you want. Think of any ellipse you can imagine, centered anywhere, with any aspect ration, with axes at any angle of rotation relative to the x-y axes. The solution to [tex]dy/dx = 0[/tex] is a line that will intercept that ellipse at two special points. The solution to [tex]dy/dx = \infty [/tex] is a line that intercepts the ellipse at two other special points. Think about what these points mean, and where these lines must intersect based on the symmetry of an ellipse.

    As I said before, you really only need one of these lines and the intercepts with the ellipse to locate its center. Use that fact to help you interpret the same equations for the hyperbola.
     
  11. Jun 13, 2005 #10
    I must not being seeing it.

    Here is a diagram of the ellipse.


    [​IMG]

    There is the line where the slope is 0, and the line where the slope is [tex] \infty[/tex] Now, I can see how you can use it to find the center. But the problem says that the intersection of these two will be the center. [tex] \frac{\partial \phi}{\partial x} and \frac{\partial \phi}{\partial y}[/tex]

    Maybe I am misinterpreting something here, but those two are equal to the 0 and infinity lines, which to me intersect up at the top. Now, if you used the infinty line at either point of either of the zero lines, and the zero lines at either point of the infity lines, I can see how that would be the center, but to me that is not what the problem said, which I why I think I am not interpreting something correctly.
     
  12. Jun 13, 2005 #11

    OlderDan

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    Take a look at these example cases created using the graphing feature at

    http://www.quickmath.com/

    See where the blue lines (partial derivatives) intercept the curves. The horizontal and vertical lines are just to guide your eye to see what is special about these intercepts
     

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  13. Jun 13, 2005 #12
    Ahh I see, very cool, thanks!
     
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