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Ellipse: geometric equivalence of two definitions

  1. Jul 29, 2008 #1
    I've been stuck with this problem:

    An ellipse can be defined as
    1) locus of points for which is constant the sum of the distances from two fixed points (foci)
    2) locus of points for which is constant the ratio between the distances from a fixed point (focus) and a fixed line (directrix). The ratio value is the eccentricity of the ellipse.

    I would like to find a pure geometric demonstration that property 1) implies 2) and/or vice-versa.

    It's easy to demonstrate the equivalence between the two definitions if one translates the properties into analytical relationships between the coordinates of the locus points (cartesian or polar). I'm not interested in that.

    I would like to find a (possibly elegant) proof of the equivalence between definition 1) and 2) via pure geometric arguments (a la Euclide): with no coordinates but just geometric relationships, based on ruler and compass.

    I've searched the net and various textbook but haven't find nothing similar (even if I have found out several interesting geometrical constructions based on properties 1) or 2) that could be the first half of the bridge to proving the equivalence, i.e. the Boscovich circle seems interesting).

    Any help from a true geometry lover will be appreciated.
     
  2. jcsd
  3. Jul 29, 2008 #2

    tiny-tim

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    Hi lemma28! :smile:
    ooh! me! I luurve geometry! :!!)

    Use Dandelin spheres … see the PF Library entry "conic". :smile:
     
  4. Jul 29, 2008 #3

    HallsofIvy

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    You can't prove the equivalence of those- they are not equivalent. (2) is true for any conic section. To get an ellipse you have to add that that ratio is larger than 0 and less than 1.

     
  5. Jul 29, 2008 #4
    You're right. Naturally I intended
    ...
    2) locus of points for which the ratio between the distances from a fixed point (focus) and a fixed line (directrix) has a constant value that is larger than 0 and less than 1.
    ...
     
  6. Aug 5, 2008 #5
    Thanks tiny-tim for the suggestion about the Dandelin spheres. They do the work in the simplest way.
    But I was not completely satisfied since I looked for demonstrations using "plane" geometry.
    So I’ve worked a little bit more on it and I think that these are the “geometric” proofs in 2 dimensions).
    I post them in case someone else is interested and in case someone can show me a better and simpler way to do it, or some mistakes.
    Especially the point 1 of part 2 (just sketched here) seems quite awkward to me, it should be simpler than that, but I haven’t found a better way to do it.
    The attached file contains the figure.

    First part
    Given the locus of points for which is constant the sum of the distances from two fixed points F1 and F2 then for any point belonging to the locus it is
    PF1=e*PH1,
    where PH1 is the distance form an appropriate fixed line (directrix) perpendicular to the line joining F1 and F2.


    Demonstration
    Let it be
    F1F2=2c;
    PF1+PF2=2a;
    PF1=f1;
    PF2=f2;
    PH1=d1;
    PH2=d2;

    First consider the triangle F1F2P, and construct the circumscribed circle around it.
    The perpendicular bisector of the segment F1F2 meets the circle in V (and the quadrilateral VF1F2P is a cyclic quadrilateral). Let’s call N the intersection of this line with the line joining F1 and F2, and we’ll call
    NF1=m1;
    NF2=m2;
    Since F1V=F2V, it follows, for the chord properties, that the segment PV bisects the angle F1PF2, so that F1PN=NPF2=F1F2V=,F2F1V=[tex]\alpha[/tex].
    Applying the angle bisector theorem to the triangle F1PF2 we have
    f1:m1=f2:m2
    and also
    (f1+f2):f1=(m1+m2):m1 –> 2a:f1=2c:m1 –> m1/f1=c/a and
    (f1+f2):f2=(m1+m2):m2 –> 2a:f2=2c:m2 –> m2/f2=c/a

    The triangles H1PF1 and F1PN are similar (they have the same angle [tex]\alpha[/tex] and two alternate interior angles). So are the triangles H2PF2 and F2PN.
    So it is
    m1:f1=f1:d1=c/a and m2:f2=f2:d2=c/a
    whence f1+f2=c/a*(d1+d2) (that proves the statement, since d1+d2 is the length of the fixed segment H1H2 and c/a=e is constant).
    Combining the proportions it is
    m1:d1=m2:d2=(c/a)2 and the same for the sums.
    So the sum of the distances PF1+PF2 is mean proportional between the lengths H1H2 and F1F2


    Second part
    Given the locus of points for which the ratio between the distances from a fixed point F1 (focus) and a fixed line (directrix) has a constant value that is larger than 0 and less than 1, then
    1) There exists a second focus F2 and a second directrix for which the same relation applies
    2) For any point P belonging to the locus the sum of the distances of P to the foci F1 and F2 has a constant value.


    Demonstration of 1)
    Basically, starting from the focus F1, the directrix, a point P belonging to the locus, and calling H1 the perpendicular projection of P on the directrix it’s possible to build the triangle PF1H1, and the similar triangle PF1N. The lines PN and H1F1 meets at a point V. The second focus F2 is such that the line PV bisects the angle F2PF1.
    Having built the same figure as in the demonstration of the first part it follows that the focus F2 is the second focus and the line VF2 inrtersects the line H1P in the point H2 that defines the position of the second directrix.

    Demonstration of 2)
    Since f1:d1=f2:d2=e, it follows that f1=e*d1 and f2=e*d2 and summing f1+f2=e*(d1+d2)=e*H1H2. So the sum of the distances of P to the foci F1 and F2 has a constant value.
     

    Attached Files:

  7. Aug 7, 2008 #6

    tiny-tim

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    splendid!

    Hi lemma28! :smile:
    Quite right! :biggrin:
    Your proof is really neat!

    I particularly like the way you proved the angles are the same. :smile:

    (I've spent some time trying to find a shortcut to Part I that doesn't involve the circle, but I can't find one … so your idea of using the circle is most helpful … how did you come up with it?

    My only objection to Part I is that I think you laboured the ratios a bit. Otherwise it's excellent. :smile:

    Part II seems fine to me … the diagram works in reverse … I'd just leave it at that. Nice statement of the Part II theorem, also.)

    Cool diagram, too! :cool:

    (Have you done diagrams for hyperbola and parabola? The hyperbola proof is probably the same. The parabola diagram is presumably going to be a bit trickier … though the proof is trivial, if you're allowed to subtract infinities. :wink:)

    Final suggestion: Can you prove … same way … that line PV is perpendicular to the tangent to the ellipse?

    With our without that, since I've not seen anything like this before, I think the proof and diagram are good enough to send to one of the more amateur-friendly maths journals! :smile:
     
  8. Aug 7, 2008 #7
    Thanks again tiny-tim.

    I must admit that it was hard work to get to the circle. I've tried several other directions before.... It's very time consuming to get to the right path in geometry (unless you have some sort of genial mind of course, which is not my case... :grumpy:)

    I've just discovered GeoGebra. Very good program to make some geometry analysis and diagrams :cool:

    Don't provoke me! :wink:. I've absolute no intention of being absorbed again in the conics, where everything seems easy and turns out not to be. I'm content with the ellipse, for the time being.

    That would be equivalent to proving the reflection property of the ellipse. I've already seen that demonstration somewhere before. If I remember well it goes like assuming that the perpendicular to PV has a second different point of intersection with the ellipse and then proving that this second point cannot lie on the elllipse since it wouldn't respect the definition (Reductio ad absurdum).
     
  9. Aug 8, 2008 #8
    There are two problems with Part One:

    1. The line from which H1 and H2 are derived, is not itself defined. This makes any claims about H1 and H2 -- particularly that the triangles H1PF1 and F1PN are similar -- suspect.

    2. The point V is defined as the intersection of a circle and a nontangential line -- giving us a choice of two possible points. Which one do you choose? Or does it not matter?
     
  10. Aug 8, 2008 #9

    tiny-tim

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    Hi Doodle Bob! :smile:

    lemma28 defines H1 and H2 as the points where the "horizontal" line through P meets VF1 and VF2.

    And the other "possible point" on the circle is P itself.

    So its all ok! :smile:
     
  11. Aug 8, 2008 #10

    No tiny-tim, I think Doodle Bob was right in pointing out that V was not properly defined.
    Originally it was defined as the intersection between the perpendicular bisector of the segment F1F2 and the circle.
    After Doodle Bob remark it should be corrected as "intersection between the perpendicular bisector of the segment F1F2 and the circle on the arc F1F2 not including P. Otherwise the point N goes outside the segment F1F2 and everything goes astray.
    After that we can define H1 and H2 as "the points where the "horizontal" line through P meets VF1 and VF2".
    Note that the distance of the line horizontal H1H2 from the horizontal axis F1F2 changes with P.

    I think with these adjustments the demonstration still holds.
    Thanks Doodle Bob for your remark.
     
  12. Aug 8, 2008 #11

    tiny-tim

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    Hi lemma28! :smile:
    Or more simply as the intersection of the bisector of F1PF2 with the circle … since equal angles are subtended by equal arcs or chords, that proves that VF1 = VF2. :smile:

    I must admit that I took one look at the diagram today, and automatically assumed that that was how it had been defined, without checking. :redface:

    (btw, I had earlier tried to use that bisector to define N, and to then proceed without using the circle at all, and eventually had to admit that your idea of using the circle was best, since it's a clever way of proving that the other relevant angles are equal. :wink:)

    Though I suspect your original proof works for either choice of V anyway, with the other choice just needing a larger piece of paper! :smile:
     
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