Every time I try to get some deeper insight about Newton’s gravitational law I stumble upon the geometrical properties of the ellipse.

After many years of these *strange and challenging encounters* I really think that the ellipse is a rich and wonderful trove of geometrical nuances and subtleties.

The geometric definition of an ellipse can be given with **two alternative but equivalent statements**:

**A**) An ellipse is a plane curve whose points () are such that the sum of the distances from to two fixed points (the foci, and ) is constant. That is

(where is the semi-major axis of the ellipse)

**B**) An ellipse is a plane curve whose points () are such that the ratio of the distance of from a fixed point (focus) and from a fixed line (directrix) is constant. That is, calling the perpendicular projection of on , it is

(where )

The equivalence of the two definitions can be elegantly proved in *space geometry (3D)* with the *Dandelin spheres*, and using the fact that the ellipse is a *conic section*.

Another possible road is to use the analytic geometry in a Cartesian frame of reference.

Anyway I have been for a long time looking for a geometric proof of the equivalence of the two definitions using pure **plane geometry** (in 2D) using the classic procedures of Euclidean geometry.

Since I couldn’t find any proof of this kind in the web, some years ago (in 2008) I tried to devise one by myself.

By pure chance I’ve rediscovered in these days (October 2016) that old demonstration of several years ago and I’ve found it quite interesting and potentially rich of clues for other possible geometric investigations, although in need of some refinement.

So, here is a possible (and tentative)** pure geometric demonstration using plain plane geometry** showing the equivalence of the two definitions of the ellipse.

It is essentially unchanged with respect to the old version, but I have added a few more details to smooth out some passages.

I’m fully aware this demonstration is much more complex than the one with the Dandelin spheres, and that it could be improved and maybe made simpler, but it temporarily meets my needs for a proof living in plane (2D) geometry (until I find something better than this).

The first part **I)** is aimed to prove that definition **A)** implies definition **B)** while the second part **II)** (just sketched as it is mostly a reversion of the steps followed in the first part) is aimed to prove that definition **B)** implies **A)**.

In a third part **III)** I’ve included a geometric construction of the ellipse, taken from a classic textbook.

## I) First part: A) B)

Given the locus of points for which is constant the sum of the distances from two fixed points and then for any point belonging to the locus it is , where is the distance from and an appropriate fixed line (directrix) perpendicular to the line joining and .

*Demonstration*

Let it be

First, consider the triangle , and construct the circumscribed circle around it.

The perpendicular bisector of the segment meets the circle (on the arc not including ) in . The quadrilateral is then a cyclic quadrilateral.

Let’s call the intersection of the line with the line joining and , and we’ll call and .

We also define and as the points where the line through and parallel to meets the lines and respectively. We’ll then call

Since , it follows, for the chord properties, that the segment bisects the angle , so that

Applying the angle bisector theorem to the triangle we have

It’s also

and

The triangles and are similar (they have the same angle and two alternate interior angles). So, also the triangles and are similar.

Thence it is

and

whence

That is

To complete the demonstration we must show that the points and lie on two parallel lines that have a certain constant distance, independently of the choice of the point of the ellipse.

These two lines are the two alternative choices for the directrix, at the right or left side of the segment joining the two foci.

First, it can be observed that the line segment is, by construction, parallel to the focal axis and that the two points and are symmetric with respect to the perpendicular bisector of the segment .

Furthermore, the distance is constant.

In fact the proportion can be rewritten as

and since , and we have

So the distance doesn’t depend on the choice of the point on the ellipse and the distance of the directrix line from the vertical axis of the ellipse is .

This completes the demonstration that (or ) and that (or ) is the distance from to an appropriate fixed line (directrix) perpendicular to the line joining and as for the initial statement^{1}.

^{1} That the directrix is a straight line perpendicular to the focal axis follows from the fact that the second directrix is symmetric to the first with respect of the perpendicular bisector of the segment and from the fact that the distances are constant. Only a straight vertical line satisfies both these conditions.

## II) Second part: B) A)

Given the locus of points for which the ratio between the distances from a fixed point (focus) and a fixed line (directrix) has a constant value with , then

**II.1**) There exists a second focus and a second directrix for which the same relation between the distances from apply, That is .

**II.2**) For any point belonging to the locus the sum of the distances of to the foci and has a constant value.

*Demonstration of II.1*

Basically, starting from the focus , the directrix , a point belonging to the locus, and calling the perpendicular projection of on the directrix it’s possible to build the triangle , and the similar triangle where the side is parallel to and with the angle congruent to the angle . The lines and meets at a point . The second focus , on the line , is such that the line bisects the angle . The line intersects the line in the point that defines the position of the second directrix (that is the line through parallel to ).

With this construction we can show that the triangle is isosceles.

In fact, the triangles and are similar (by construction).

The triangles and are similar, since they have two congruent angles: (opposite angles) and ().

Then it is .

Also the triangles and are similar, for the *SAS similarity criterion*, since they have a congruent angle, (opposite angles) and the two sides adjacent to that angle have lengths in the same ratio. In fact, as shown above, it is (or, equivalently, ).

Then and this proves the fact that the triangle is isosceles.

Thence and the point belongs to the perpendicular bisector of the segment .

Furthermore it can be easily proved that and that imply that the quadrilateral is a cyclic quadrilateral and that the point belongs to the circumscribed circle of the triangle .

With the construction of the point we can see that also the triangles and are similar (for they have the congruent alternate interior angles and the congruent angles ).

Here is an updated figure representing the geometric properties found so far, highlighting the similarity of the triangles , the similarity of the triangles and the congruent angles.

For the angle bisector theorem applied to the triangle , with the angle in bisected in the two congruent angles and , it is

(a)

For the similitude of the triangles and it is

(b)

For the similitude of the triangles and it is

(c)

Putting together (a) with (b) and (c) it is

that is

*Demonstration of II.2*

Since

it follows that

and

Summing the terms it is

so that the sum of the distances of to the foci and has a constant value.

Q.E.D.

## III) A geometric construction of the ellipse using definition B)

To enrich the geometrical investigation of the ellipse from the point of view of the definition **B)** here is a geometrical construction of the points of the ellipse.

It is based on the classic textbook

*W. H. Besant – Conic Sections Treated Geometrically – 1895 – *chapter 3 (see the References at the bottom for a link to the free ebook).

I’ve only changed the naming of the points and lines and have given some more detail to some of the passages.

With this construction, starting from a given directrix and a given focus and calling the perpendicular projection of on , each point of the ellipse must satisfy the condition

(d)

where

**Construction**

Let be the focus, the directrix, the line through perpendicular to and the intersection between and .

Divide at the point in the given ratio ().

The point is one point of the ellipse since it complies with the definition (d). In particular it is one of the ellipse vertices.

Construct another point on satisfying the proportion .

is another point of the ellipse and another of its vertices.

Choose some point anywhere on the directrix , join it with (line ) and draw, from another line such that it forms with the line the same angle as the one that forms with the horizontal line .

Join the points and to form the line and call the intersection point between the lines and .

Draw from a line parallel to the horizontal line . Call its intersection with the directrix and its intersection with the line .

Since by construction it is and since (alternate interior angles) it follows that and the triangle is isoceles.

Then it is .

Considering the parallel lines and , cut by the lines , and drawn from the common point , for Thales’ *intercept theorem* it is

and putting in the congruence between and it is also

Then the the point belongs to the locus of points (ellipse) for which it is .

Changing the choice of the point on will result in a different point but this will nonetheless belong to the same ellipse.

## References

https://en.wikipedia.org/wiki/Ellipse

https://en.wikipedia.org/wiki/Dandelin_spheres

https://en.wikipedia.org/wiki/Conic_section

https://www.physicsforums.com/threads/ellipse-geometric-equivalence-of-two-definitions.247417/

W. H. Besant – Conic Sections Treated Geometrically (1895) – (free ebook in Project Gutenberg)