Ellipse: geometric proof of the equivalence of two definitions

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 (P) are such that the sum of the distances from P to two fixed points (the foci, F_1 and F_2) is constant. That is

\overline {PF_1} + \overline {PF_2} = \text{const} = 2a (where a is the semi-major axis of the ellipse)

B) An ellipse is a plane curve whose points (P) are such that the ratio of the distance of P from a fixed point (focus) F_1 and from a fixed line (directrix) d_1 is constant. That is, calling H_1 the perpendicular projection of P on d_1, it is

\overline{PF_1}/ \overline{PH_1}= \text{const}=e (where 0<e<1)

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.

equivalence of the two foci and focus-directrix definitions

fig.1: equivalence of the “two foci” and “focus-directrix” definitions

I) First part: A) \rightarrow B)

Given the locus of points for which is constant the sum of the distances from two fixed points F_1 and F_2 then for any point P belonging to the locus it is \overline{PF_1}=e\cdot \overline{PH_1}, where \overline{PH_1} is the distance from P and an appropriate fixed line (directrix) perpendicular to the line joining F_1 and F_2.

Demonstration

Let it be

    \[\overline{F_1F_2}=2c; \; \; \overline{PF_1}+\overline{PF_2}=2a; \; \; c/a=e; \; \; \overline{PF_1}=f_1; \; \; \overline{PF_2}=f_2\]

First, consider the triangle F_1F_2P, and construct the circumscribed circle around it.

The perpendicular bisector of the segment F_1F_2 meets the circle (on the arc \stackrel {\frown} {F_1F_2} not including P) in V . The quadrilateral VF_1F_2P is then a cyclic quadrilateral.

Let’s call N the intersection of the line PV with the line joining F_1 and F_2, and we’ll call \overline{NF_1}=m_1 and \overline{NF_2}=m_2.

We also define H_1 and H_2 as the points where the line through P and parallel to F_1F_2 meets the lines VF_1 and VF_2 respectively. We’ll then call \overline{PH_1}=d_1; \; \; \overline{PH_2}=d_2

Since \overline{F_1V}=\overline{F_2V}, it follows, for the chord properties, that the segment PV bisects the angle F_1PF_2, so that F_1\widehat{P}N=N\widehat{P}F_2=F_1\widehat{F_2}V=\alpha

Applying the angle bisector theorem to the triangle F_1PF_2 we have

    \[f_1:m_1=f_2:m_2\]

It’s also

    \[(f_1+f_2):f_1=(m_1+m_2):m_1 \; \; \rightarrow \; \; 2a:f_1=2c:m_1 \; \; \rightarrow \; \; m_1/f_1=c/a\]

and

    \[(f_1+f_2):f_2=(m_1+m_2):m_2 \; \; \rightarrow \; \; 2a:f_2=2c:m_2 \; \; \rightarrow \; \; m_2/f_2=c/a\]

The triangles H_1PF_1 and F_1PN are similar (they have the same angle \alpha and two alternate interior angles). So, also the triangles H_2PF_2 and F_2PN are similar.

Thence it is

    \[m_1:f_1=f_1:d_1=c/a\]

and

    \[m_2:f_2=f_2:d_2=c/a\]

whence

    \[f_1:d_1=f_2:d_2=c/a \; \rightarrow \; f_1=(c/a) \cdot d_1; \; \; \;  \;  \;  f_2=(c/a) \cdot d_2\]

That is

    \[\overline{PF_1}=e\cdot \overline{PH_1}; \; \;  \;  \;  \; \overline{PF_2}=e\cdot \overline{PH_2}\]

To complete the demonstration we must show that the points H_1 and H_2 lie on two parallel lines that have a certain constant distance, independently of the choice of the point P 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 H_1H_2 is, by construction, parallel to the focal axis F_1F_2 and that the two points H_1 and H_2 are symmetric with respect to the perpendicular bisector of the segment F_1F_2.

Furthermore, the distance \overline{H_1H_2}=d_1+d_2 is constant.
In fact the proportion f_1:f_2=d_1:d_2 can be rewritten as

    \[(d_1+d_2):d_2=(f_1+f_2):f_2\]

and since f_1+f_2=2a, d_2/f_2=a/c and d_1+d_2=\overline{H_1H_2} we have

    \[(d_1+d_2)=2a \cdot d_2/f_2 \rightarrow (d_1+d_2)=2a \cdot (a/c)\rightarrow\overline{H_1H_2}=2a \cdot (a/c)\]

So the distance \overline{H_1H_2} doesn’t depend on the choice of the point P on the ellipse and the distance of the directrix line from the vertical axis of the ellipse is a^2/c=a/e.

This completes the demonstration that \overline{PF_1}=e\cdot \overline{PH_1} (or \overline{PF_2}=e\cdot \overline{PH_2} ) and that \overline{PH_1} (or \overline{PH_2}) is the distance from P to an appropriate fixed line (directrix) perpendicular to the line joining F_1 and F_2 as for the initial statement1.

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 F_1F_2 and from the fact that the distances  \overline{H_1H_2} are constant. Only a straight vertical line satisfies both these conditions.

II) Second part: B) \rightarrow A)

Given the locus of points for which the ratio between the distances from a fixed point F_1 (focus) and a fixed line (directrix) has a constant value e with 0<e<1, then

II.1) There exists a second focus F_2 and a second directrix d_2 for which the same relation  between the distances from P apply, That is  \overline{PF_2}/\overline{Pd_2}=e= \text{const}.

II.2) For any point P belonging to the locus the sum of the distances of P to the foci F_1 and F_2 has a constant value.

Demonstration of II.1

Basically, starting from the focus F_1, the directrix d_1, a point P belonging to the locus, and calling H_1 the perpendicular projection of P on the directrix it’s possible to build the triangle PF_1H_1, and the similar triangle PF_1N where the side F_1N is parallel to PH_1 and with the angle F_1 \overhat{P} N congruent to the angle F_1 \overhat{H_1}P. The lines PN and H_1F_1 meets at a point V. The second focus F_2, on the line FN, is such that the line PV bisects the angle F_2\widehat{P}F_1. The line VF_2 intersects the line H_1P in the point H_2 that defines the position of the second directrix d_2 (that is the line through H_2 parallel to d_1).

ellipse-construction-part2

fig. 2: ellipse construction from the “focus-directrix” definition (partial)

With this construction we can show that the triangle VF_1F_2 is isosceles.

In fact, the triangles H_1PF_1 and PF_1N are similar (by construction).
The triangles PF_2N and F_1VN are similar, since they have two congruent angles: F_1 \widehat NV = F_2 \widehat NP (opposite angles) and N \widehat F_1V=N \widehat P F_2 (N \widehat F_1 V=P \widehat H_1F_1=N \widehat P F_2).
Then it is \overline{PN}:m_2=m_1:\overline{NV}.

Also  the triangles PF_1N and NF_2V are similar, for the SAS similarity criterion, since they have a congruent angle, P \widehat NF_1=F_2 \widehat NV (opposite angles) and the two sides adjacent to that angle have lengths in the same ratio. In fact, as shown above, it is \overline{PN}:m_2=m_1:\overline{NV} (or, equivalently, \overline{PN}:m_1=m_2:\overline{NV}).

Then V \widehat F_1N=F_1 \widehat P N=F_1 \widehat P N=N \widehat F_2V and this proves the fact that the triangle F_1VF_2 is isosceles.
Thence \overline{F_1V}=\overline{F_2V} and the point V belongs to the perpendicular bisector of the segment F_1F_2.
Furthermore it can be easily proved that F_1 \widehat PF_2+F_1 \widehat VF_2=F_1 \widehat PF_2+F_1 \widehat V F_2=180^\circ and that imply that the quadrilateral PF_1VF_2 is a cyclic quadrilateral and that the point V belongs to the circumscribed circle of the triangle F_1PF_2.
With the construction of the point H_2 we can see that also the triangles  NPF_2 and PF_2H_2 are similar (for they have the congruent alternate interior angles P \widehat F_2 N=F_2 \widehat P H_2 and the congruent angles N \widehat P F_2=P \widehat H_2 F_2).

Here is an updated figure representing the geometric properties found so far, highlighting the similarity of the triangles tr1,\;tr1',\;tr1'', the similarity of the triangles tr2,\;tr2',\;tr2'' and the congruent angles.

Ellipse construction from focus-directrix

fig. 3: ellipse construction from focus-directrix (complete)

For the angle bisector theorem applied to the triangle PF_1F_2, with the angle in \widehat P bisected in the two congruent angles F_1 \widehat{P} N and N \widehat{P} F_2, it is

(a)   \[f_1:m_1=f_2:m_2 \]

For the similitude of the triangles H_1PF_1 and PF_1N it is

(b)   \[m_1:f_1=f_1:h_1 \]

For the similitude of the triangles H_2PF_2 and PF_2N it is

(c)   \[m_2:f_2=f_2:h_2 \]

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

    \[f_1:h_1=f_2:h_2\]

that  is

    \[\overline{PF_1}/\overline{PH_1}= \overline{PF_2}/\overline{PH_2}=e\]

Demonstration of II.2

Since

    \[f_1:d_1=f_2:d_2=e\]

it follows that

    \[f_1=e\cdot d_1\]

and

    \[f_2=e\cdot d_2\]

Summing the terms it is

    \[f_1+f_2=e \cdot (d_1+d_2)=e \cdot \overline{H_1H_2}\]

so that the sum of the distances of P to the foci F_1 and F_2 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 P 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 d_1 and a given focus F_1 and calling H_1 the perpendicular projection of P on d_1, each point P of the ellipse must satisfy the condition

(d)   \[\overline{PF_1}/ \overline{PH_1}= \text{const}=e \]

where 0<e<1

Geometric construction of the ellipse by W.H. Besant

fig. 4: geometric construction of the ellipse by W.H. Besant

Construction
Let F_1 be the focus, d_1 the directrix, a the line through F_1 perpendicular to d_1 and K_1 the intersection between a and d_1.

Divide F_1K_1 at the point A_1 in the given ratio (\overline{F_1A_1}/\overline{A_1K_1}=e).

The point A_1 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 A_2 on a satisfying the proportion \overline{F_1A_2}:\overline{K_1A_2}=\overline{F_1A_1}:\overline{K_1A_1}.
A_2 is another point of the ellipse and another of its vertices.

Choose some point E anywhere on the directrix d_1, join it with F_1 (line u) and draw, from F_1 another line s such that it forms with the line EF_1 the same angle as the one that EF_1 forms with the horizontal line a.
Join the points E and A_1 to form the line r and call P the intersection point between the lines s and r.
Draw from P a line h parallel to the horizontal line a. Call H_1 its intersection with the directrix d_1 and L its intersection with the line EF_1.
Since by construction it is L\widehat{F_1}A_2=L\widehat{F_1}P and since L\widehat{F_1}A_2N=P\widehat{L}F_1 (alternate interior angles) it follows that L\widehat{F_1}P=P\widehat{L}F_1 and the triangle PLF_1 is isoceles.
Then it is \overline{PL}=\overline{PF_1}.
Considering the parallel lines h and a, cut by the lines r, u and d_1 drawn from the common point E, for Thales’ intercept theorem it is

    \[\overline{A_1F_1}:\overline{A_1K_1}=\overline{PL}:\overline{PH_1}=e\]

and putting in the congruence between \overline{PL} and \overline{PF_1} it is also

    \[\overline{A_1F_1}:\overline{A_1K_1}=\overline{PF_1}:\overline{PH_1}=e\]

Then the the point P belongs to the locus of points (ellipse) for which it is \overline{PF_1}:\overline{PH_1}=e= \text{const}.
Changing the choice of the point E on d_1 will result in a different point P 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)

Last edited 11 oct 2016