Ellipse: geometric equivalence of two definitions

[lemma28]

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 http://gallica.bnf.fr/ark:/12148/bpt6k99630k/f240" seems interesting).

Any help from a true geometry lover will be appreciated.

 

[tiny-tim]

Hi lemma28!

Any help from a true geometry lover will be appreciated.

ooh! me! I luurve geometry! :!)

Use Dandelin spheres … see the PF Library entry "https://www.physicsforums.com/library.php?do=view_item&itemid=98"c".

 

[HallsofIvy]

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.

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.

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 http://gallica.bnf.fr/ark:/12148/bpt6k99630k/f240" seems interesting).

Any help from a true geometry lover will be appreciated.

 

[lemma28]

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.

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.
...

 

[lemma28]

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.

 

[tiny-tim]

splendid!

Hi lemma28!

I was not completely satisfied since I looked for demonstrations using "plane" geometry.

Quite right!

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.

Your proof is really neat!

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

(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.

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!

(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. )

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! ​

 

[lemma28]

Thanks again tiny-tim.

(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?

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... )

Cool diagram, too!

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

(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. )

Don't provoke me! . 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.

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

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).

 

[Doodle Bob]

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?

 

[tiny-tim]

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?

Hi Doodle Bob!

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!

 

[lemma28]

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?

Hi Doodle Bob!

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!

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.

 

[tiny-tim]

Hi lemma28!

… 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. …

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.

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.

(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. )

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