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Given a circle, it's easy to contruct its center.

The question is: given an ellipse, draw the foci.

I don't know whether it's possible to do this using only straightedge and compass.

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I'm assuming you are given the major and minor axes of the ellipse. Hint: what is the distance from a foci of the ellipse to the "top" of the ellipse? –  David Mitra Dec 21 '12 at 16:24
    
No, you're only given the ellipse's line, the same way as in the case of the circle. –  Sgernesto Dec 21 '12 at 16:26
    
What do you mean by that? Do you mean you're given the locus of points on the ellipse? –  David Mitra Dec 21 '12 at 16:27
    
Yes, only that. –  Sgernesto Dec 21 '12 at 16:29
    
Oh yes, of course; sorry for being dense... –  David Mitra Dec 21 '12 at 16:44

2 Answers 2

Draw two perpendicular coordinate axes. Choose somehow $6$ points on the ellipse ($5$ are enough). (The "somehow" is there because for classic constructibility, all points constructed have to be explicit. Here we will have to select at least one "random" point.)

By projecting these $6$ points on the axes, we can construct the $x$ and $y$-coordinates of our $6$ points. With straightedge and compass, any required squares and products of these coordinates can be constructed.

The ellipse has equation of shape $ax^2+bxy+cy^2+dx+ey+f$ for some constants $a$ to $f$. Thus the constants satisfy a system of $6$ linear equations in $6$ unknowns, with constructible coefficients. Solving the system involves only arithmetical operations, which can be done by straightedge and compass.

Now the ellipse can be put in standard form using arithmetical operations and square root. And as has been pointed out by rschwieb, once that has been done the foci can be constructed by straightedge and compass.

Remark: This settles the question of whether the job can, in principle, be done. There remains the task of doing it in a geometrically nice way. The above recipe for a construction gives no information about that.

There does remain something interesting about the construction. It is one of many instances where a geometric problem is solved using coordinatization. Because the elementary theory of real-closed fields is decidable, most geometric problems of the classical kind can be answered, albeit clumsily, by a computer program.

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To simplify matters slightly, we can choose our coordinates so $(0,0)$ and $(1,0)$ are two arbitrary points on the ellipse. Then we know $f=0$ and $a+d=0$. Since the equations are homogeneous in $a,b,c,d,e,f$, we can assume $a=1$ and $d=-1$ (we can't have $a=d=0$ since there are only two points of the ellipse on the $x$ axis). That leaves $b,c,e$ to be determined from three equations in three unknowns. –  Robert Israel Dec 21 '12 at 21:33

Let $\lambda$ a given ellipse, let's draw its foci.

See the figure below: enter image description here

  1. Draw two parallel chords $AB$ and $DC$.
  2. Find out $M$ and $N$ (midpoints of $AB$ and $DC$ respectively).
  3. Draw the straightline $r$ through $M$ and $N$.

  4. Let $\{E, E'\} = r \cap \lambda$, find out $O$ the midpont of $EE'$.

  5. Draw an arc $\mu$ centered at $O$ and radius $AO$, such that $\{A, A'\}= \mu \cap \lambda$.
  6. Draw the straightline $s$ through $A$ and $A'$.
  7. Draw a straightline $t$ such that $O \in t$ and $t\parallel s$.
  8. Draw $u$ (the perpendicular bisector of $AA'$).
  9. Let $\{G, G'\} = u \cap \lambda$, and $\{J, J'\} = t \cap \lambda$, $GG'$ and $JJ'$ are the axes of symmetry of $\lambda$. Now draw an arc $\nu$ centered at $G$ and radius $JO$, such that $\{F1, F2\}= \nu \cap t$. $F1$ and $F2$ are the desired foci of $\lambda$.
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