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.


  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.

  • $\begingroup$ 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) $. $\endgroup$
    – terry-s
    Feb 26, 2019 at 21:19
  • $\begingroup$ @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. $\endgroup$ Mar 11, 2019 at 5:31

2 Answers 2


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

enter image description here


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!

We'll start with the two-circles construction:

enter image description here

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 $A_{-}$.)

enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ Nov 5, 2018 at 21:31

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