Equivalence between ellipse definitions

Question

In a YouTube video by MathyJaphy, the creator uses a geometric\kinematic way to describe the focus and directrix definition of the ellipse (or, equivalently, the slice-of-a-cone definition). He even created a desmos file if anyone wants to play with it a bit.

When the moving line goes faster than the expanding circle, it draws an ellipse. I understand why it's equivalent to the conic section definition of the ellipse (or the focus and directrix one), but I can't see in a straightforward way why it's gives the same result as the two foci definition, or the squashed circle one (or for that matter - why the resulting shape is symmetric).

Does anyone have a simple explanation why they're all the same? Ideally, it would be with as little algebra as possible, and without using Dandelin spheres (at least, without introducing a third space dimension). Any insight is appreciated.

Answer 1

We need first of all a nice lemma:

In a right triangle $ABC$, let $E$ be a point on hypotenuse $AB$ and $D$ a point on the extension of $BA$ after $A$ (see figure). Then $\angle DCA=\angle ACE$ if and only if $DA:EA=DB:EB$.

Proof. If $\angle DCA=\angle ACE$ then by the angle bisector theorem (in triangle $DCE$) we have $$DA:AE=DC:CE.$$ On the other hand, $BC$ turns out to be the bisector of exterior angle $\angle ECF$ (because $\angle ECB=\angle BCF=90°-\angle ACE$) and by the exterior angle bisector theorem we get $$BD:BE=DC:CE.$$ Comparing the above results we thus find: $$DA:EA=DB:EB,$$ as it was to be proved. The converse is obvious by the unicity of the construction and can be easily proved by RAA. $\square$

Let's see now how one can construct points on an ellipse, given a focus $S$, its directrix $XM$ (with $SX\perp XM$, see figure below) and its eccentricity $e<1$. We first construct the major axis $AA'$ of the ellipse, and then pairs of points, symmetric about the perpendicular bisector of $AA'$.

On segment $SX$ we can in fact construct a point $A$, such that $SA:AX=e$. And on the extension of $XS$ we can construct another point $A'$, such that $SA':A'X=e$. Points $A$ and $A'$ belong then to the ellipse and are by definition the endpoints of its major axis.

Take now any point $M$ on the directrix. Through $M$ and $S$ draw line $MYSY'$, meeting lines $AY$ and $A'Y'$, drawn at right angles to $AA'$, at points $Y$ and $Y'$. On $YY'$ as diameter describe a circle and draw line $MPP'$ parallel to $AA'$, intersecting the circle at points $P$ and $P'$. Then $P$ and $P'$ lie on the ellipse.

Proof. We have $SY : YM = SA : AX$ and $SY': Y'M = SA' : A'X$, whence:
$$SY : YM = SY': Y'M.$$ But $YPY'$ is a right triangle and we can apply the lemma proved above, obtaining that $PY$ is the bisector of $\angle MPS$. From the angle bisector theorem we then have: $$ PS:PM=SY:YM=SA:AX=e, $$ hence $P$ lies on the ellipse. We can repeat the same reasoning for right triangle $YP'Y'$, to prove that $P'$ is also on the ellipse. $\square$

By construction, $P$ and $P'$ are symmetric about the perpendicular bisector $OC$ of $AA'$ (the projection of the midpoint $O$ of $YY'$ on line $AA'$ is in fact the midpoint $C$ of $AA'$).

It follows that we could consider another focus $S'$ and another directrix $X'M'$, reflections of $S$ and $XM$ about $OC$, and we would get the same ellipse.

Finally, let's prove that $SP+S'P=AA'$ for any point $P$ of the ellipse. From $$SA':A'X=SA:AX$$ we get $$SA':SA=A'X:AX \quad\text{and}\quad (SA'+SA):SA=(A'X+AX):AX,$$ that is: $$AA':SA=XX':AX, \quad\text{or}\quad AA':XX'=SA:AX.$$ From that and $SP:PM=SA:AX$ one then gets $$SP:PM=AA':XX'.$$ In the same way one obtains $$S'P:PM'=AA':XX'$$ and combining the last two equalities we find $$(SP+S'P):(PM+PM')=AA':XX'.$$ But $PM+PM'=MM'=XX'$, hence: $$SP+S'P=AA'.$$

Here's an animation made with GeoGebra, showing how points $P$ and $P'$ trace out the ellipse.