# Geometric proof of equivalence between two constructs of ellipse

Pretending that we don't know any analytic geometry and trigonometry.

Consider the following two constructs of an ellipse, where admittedly the second one is an ad-hoc construct for the ellipse parametrization that is more naturally done algebraically.

1. Two-foci construct: given two fixed points that shall be called the focal points, construct the locus of the points which sum of distances to the foci is constant.

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1. Two-circle construct: given two concentric circles of different radii and an arbitrary infinite line $$L_1$$ that passes through the center (of the circles).
• Make a line $$L_2$$ that also passes through the center and is perpendicular to $$L_1$$.
• Make a ray emitting from the center that intersects with the larger circle at point $$P_1$$ and intersects with the smaller circle at $$P_2$$.
• Find the perpendicular foot $$Q_1$$ on $$L_1$$ such that $$\overline{P_1 Q_1} \perp L_1$$
• Find $$Q_2$$ along $$\overline{P_1 Q_1}$$ such that $$\overline{P_2 Q_2} \perp \overline{P_1 Q_1}$$.
• The locus of $$Q_2$$ for all rays sweeping a complete revolution is the desired.

Is there a geometric proof that the above two constructs are equivalent (when the given foci and given concentric circles "match")?

Algebraically this is standard, however, I find it difficult to geometrically map the locus of constant-distance-to-two-foci to the locus from the concentric circles.

I don't mind if the mapping involves tools beyond compass and straightedge, as long as it's geometric in nature. Any pointer is appreciated.

• Seems that if definition 1 ('two-foci construct') is to define a single ellipse & not an indefinite number of ellipses, it should say the sum of distances to the foci for points on the curve is not just 'constant', but 'a specified constant' (say $2a$), and the distance between the foci another specified constant < $2a$, say $2ea$. Then to define the same ellipse by the two-circle method, the larger circle needs to be stated to have diameter $2a$, and the smaller circle diameter $2a * sqrt(1-e^2)$. Feb 26, 2019 at 21:19
• @terry-s Sorry for the late response. Yeah that's a good point. I originally framed it that way just to "pretend" that one cannot be sure how the ellipses of the two definitions are going to match, leaving the possibility of a "discovery" of the matching conditions and "why" the two definitions are equivalent. Mar 11, 2019 at 5:31

Here is my answer using some ellipse geometric properties (location of foci and combined lengths of side elevations of constructing triangle for method $$2$$).

This answer is a little unsatisfying, as it amounts to: The two geometric constructions are equivalent, because the relations describing them are algebraically equivalent. I believe it's possible to remove some of the algebra, but I haven't found the cleanest way to do it ... yet!

Let circles $$\alpha$$ and $$\beta$$ have common center $$O$$ and respective radii $$a$$ and $$b$$. Let a variable ray from $$O$$ meet these circles at $$A$$ and $$B$$, and let the projections of $$A$$ and $$B$$ onto "horizontal" and "vertical" diameter-lines be $$X$$ and $$Y$$, and let the projection lines meet at $$P$$.

Defining $$x := |OX|$$ and $$y:=|OY|$$, the similarity of $$\triangle OAX$$ and $$\triangle BOY$$, and the right-triangle-ness of $$\triangle OAX$$, imply $$\frac{|AX|}{|OA|} = \frac{|OY|}{|OB|} \quad\to\quad a^2 y^2 = b^2\;|AX|^2 \quad\to\quad a^2 y^2 = b^2 \left( a^2 - x^2 \right) \tag{1}$$

Of course, $$(1)$$ is equivalent to the "standard form" of the ellipse equation: $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \tag{2}$$ Since we "know" $$(2)$$ derives from the focus construction, we could say we're done. But I want to investigate $$(1)$$ a little more.

Introduce $$c$$ such that $$a^2 = b^2 + c^2$$, and define $$e := c/a$$. Then $$b^2 = a^2(1-e^2)$$, and we can write $$(1)$$ as

\begin{align} y^2 &= \left(1-e^2\right)\left(a^2-x^2\right)\tag{3} \\[6pt] &= (a+ex)^2-(ea+x)^2 \tag{4} \end{align} Defining $$z := ex$$, we can write $$(4)$$ as $$(a+z)^2 = y^2 + ( c+x )^2 \tag{5}$$

We interpret $$(5)$$ geometrically by introducing a third circle, $$\gamma$$, centered at $$O$$ and having radius $$c$$. Let $$\gamma$$ meet the variable ray at $$C$$ and the "horizontal" diameter-line at $$C_{+}$$ and $$C_{-}$$. Also, let $$Z$$ be the projection of $$C$$ onto that diameter. (While we're at it, let's say that the diamter meets $$\alpha$$ at $$A_{+}$$ and $$A_{-}$$.)

Now, $$a=|OA_{+}|$$ and $$c=|OC_{+}|$$, while proportionality tells us that $$z = |OZ|$$. We also have $$y = |PX|,\qquad a+z = |A_{+}Z|,\qquad c+x = |C_{+}X| \tag{6}$$ so that $$(5)$$ implies, by way of right triangle $$\triangle PXC_{+}$$, $$|A_{+}Z|^2 = |PX|^2 + |C_{+}X|^2 = |PC_{+}|^2 \quad\to\quad |A_{+}Z|=|PC_{+}| \quad\left(\text{likewise,}\; |A_{-}Z| = |PC_{-}|\right) \tag{7}$$

The reader may notice that @PhilH's expression for what he calls "$$a$$" amounts to the same observation. Even so, it's interesting (to me) to formulate things thusly:

Circle $$\gamma$$ meets the variable ray at a point whose projection onto the "horizontal" diameter of $$\alpha$$ divides that diameter into precisely the segments needed to join the foci to $$P$$.

I have no doubt that this interpretation exists in the literature; however, I only recently realized it, myself. Be that as it may ... We have recaptured the fact that the sum of the focus-to-$$P$$ distances is constant, namely the diameter of $$\alpha$$. $$\square$$

As I mentioned, I believe some of the algebra can be removed and replaced with more geometry. For instance, if we read equation $$(5)$$ as $$y^2 = \left(\;(a+z)+(c+x)\;\right)\cdot\left(\;(a+z)-(c+x)\;\right) \tag{5a}$$ then we see that $$y$$ is the geometric mean of lengths $$a+z+c+x$$ and $$a+z-c-x$$; moreover, the arithmetic mean (aka, the average) of those lengths is $$a+z$$. These values and relations are geometrically (ahem) mean-ingful. (The "geometric mean" link shows how they feature in a classic construction involving a right angle inscribed in a semi-circle.) I have a way to deduce the relations (mostly) geometrically from the two/three-circle construction, but it's currently a little messier than the already-messy algebraic route. If (when?) I find a tidier argument, I'll update this answer.

• Thank you so much for the detailed answer! You are very humble to say that it is unsatisfying (for a specific goal), while the post shows great creative thinking. Some parts actually gave me new ideas I'd like to try. I'll let you know if I get anything worthy. Nov 5, 2018 at 21:31