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Complex Numbers ~ Locus

  1. Mar 18, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    The curve on the Argand diagram for which
    [tex]|z-2|+|z-4|=10[/tex]
    is an ellipse. Find the co-ordinates of its centre and the lengths of its major and minor axes.


    2. Relevant equations
    [tex]z=x+iy[/tex]


    3. The attempt at a solution
    I could only find the centre by using common sense in a way: ([itex]\frac{2+4}{2}[/itex],0) = (3,0)
    No idea how to answer the second part though.
     
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  3. Mar 18, 2009 #2

    tiny-tim

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    Hi Mentallic! :smile:

    Hint: the major axis is obviously along the real axis,

    so the ends of the major axis are also, and the ends of the minor axis are on a line parallel to the imaginary axis :wink:
     
  4. Mar 18, 2009 #3

    HallsofIvy

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    You can interpret |z- a| as the distance from z to a. An ellipse has the property that the total distance from a point on the ellipse to the two foci is a constant. In other words |z- 2|+ |z- 4|= 10 tells you that the foci of the ellipse are 2 and 4.
     
  5. Mar 18, 2009 #4

    Mentallic

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    Yes I realized that too :smile: but the distances to these end-points...?

    If I were to take a stab at it, I'd go about it like this:

    [tex]\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+y^2}=10[/tex]

    where y=0 since end-points are real.

    Therefore, [tex]x-2+x-4=10[/tex]
    x=8

    So, distance 8 from centres of each modulus.

    [tex](2-8,0) = (-6,0)[/tex] and
    [tex](4+8,0) = (12,0)[/tex]

    Is this a legitimate approach?

    Similarly for the minor axes:

    x=3 since they lie on the perpendicular bisector of (2,0) and (4,0)

    Hence, [tex]\sqrt{1+y^2}+\sqrt{1+y^2}=10[/tex]

    [tex]y=\pm 2\sqrt{6}[/tex]

    So I'm guessing the minor end-points are [tex](3,2\sqrt{6}) (3,-2\sqrt{6})[/tex] ?

    Once again, this is a total guess so please correct me where I'm wrong.
     
  6. Mar 18, 2009 #5

    Mentallic

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    So the centre of the ellipse is (3,0) and has focii at (2,0) and (4,0).

    So for the general cartesian form of an ellipse: [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
    [tex]ae=1[/tex] where e=eccentricity.

    I'm lost from here...
     
  7. Mar 18, 2009 #6

    tiny-tim

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    Hi Mentallic! :smile:

    (have a square-root: √ :wink:)
    Yes, and now just say "so 8 is one end of the major axis"

    … and then find the other end by using the "opposite value" of the ||s.

    (why is that a guess? :confused:)

    Yes, that's fine, except you mean 3 ±2(√6)i :smile:
     
  8. Mar 18, 2009 #7

    HallsofIvy

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    For an ellipse, [itex]c^2= a^2- b^2[/itex] where a and b are the distances from the center to the vertices and c is the distance from the center to the focus. So you know that [itex]a^2- b^2= 1[/itex].
     
  9. Mar 18, 2009 #8

    Mentallic

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    Thanks for the correction. I also took on a more geometric approach and found half the minor axis to be [tex]\sqrt{5^2-1}=2\sqrt{6}[/tex] through an isosceles triangle and the pythagorean theorem which supported my algebraic result.

    Sorry, but I haven't studied ellipses much at all. I remember a few general results but don't know their significance and I'm once again unsure what to do with this relationship of [itex]a^2-b^2=1[/itex]
     
  10. Mar 19, 2009 #9

    tiny-tim

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    Hi Mentallic! :smile:

    Start by looking at the PF Library pages on conic and eccentricity :wink:
     
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