From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Euclidean construction of ellipse's foci
Date: 8 Dec 1995 06:18:36 GMT
In article ,
Nelson G. Rich wrote:
>Is it possible to locate the foci of a given elllipse using
>Euclidean tools only? Is so, how?
Well, I see others have had the same response I had: tell me where one
axis is and I'll be happy to locate the foci. Otherwise...
Well, otherwise I shall pout and throw a tantrum. Take your ellipse and go!
Take it -- except for five points. For five points determine the ellipse,
so that at least in principle I must "know" the foci if I am given these
points.
Indeed, if the points are (x1, y1), ..., (x5, y5) and the ellipse has
the form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, then when I substitute in
the five points' coordinates, I get 5 linear equations in these six
variables, which (if the points are in general position at least) allows
a determination of A through F, now unique up to constant multiples.
(If you pull a fast one and leave me five points _not_ on any ellipse,
I'll notice it when this quadratic leads to another kind of conic.)
Well, this leads to some singularly unattractive formulae, at least in
the general case. I'm sure more pleasant descriptions are available at
least if the points are not chosen at whim but rather according to
some pattern. I tried something of this sort; the calculations are
attached at the end -- really, there's nothing instructive about the
answers.
But it _is_ clear that the coordinates of the center of the ellipse
are rational functions of the coordinates of the five points. In addition,
the slopes of the axes (which go through that center) satisfy a quadratic
equation whose coefficients are rational in these coordinates. Since
the arithmetic operations + - * / and even sqrt can be accomplished
with ruler an compass, this means that yes (in principle) you can
find the foci with traditional Euclidean tools.
Back in the century in which emperors and presidents gave mathematical
proofs there was so much collective understanding of Euclidean constructions
that someone would have leapt in to demonstrate the elegant geometrical
construction (perhaps generalizing the construction of the center of a
circle from three points on it). But that was then, this is now, and I'm
afraid no slick answer may be forthcoming.
dave
Here's my attempt at "efficient" calculations.
Note that if I draw a line through a point P on the ellipse, that line
must meet the ellipse in precisely one more point (we are solving a
quadratic f(P + t*v)=0 which already has a root at t=0; I will ignore the
degenerate case of a line tangent at P). So I proceed as follows.
Given five points on the
ellipse, draw coordinates so that one point is the origin and another
is the point (1,0). Then repeatedly follow perpendiculars to new
points on the curve, to find points (1,0), (0,0), (0, y0), (x0,y0), and
(x0,y1) on the ellipse. If I have done this right, the equation containing
these five points is
0 = y0 y1 (x^2-x) + (x0-1) y[ -y1 x + x0 y - x0 y0 ], i.e.
0 = y0y1 x^2 - y1(x0-1) xy + x0(x0-1) y^2 - y0y1 x - x0y0(x0-1) y
However you find the equation of your ellipse, you may then find the center
without much trouble. If the ellipse is determined with a quadratic f(x,y)=0,
then that center is the location of the absolute min/max of f, which
we find by setting the partials equal to zero:
0 = 2A x + B y + D
0 = B x + 2C y + E
so that the center is at (x,y) = (BE-2CD, BD-2AE)/(4AC-B^2). In the notation
above this is ( x0y0(x0-1)+2x0y0, y0y1+2y0x0y0)/(4y0x0-y1(x0-1)).
Then you need the direction of the foci from here. This you can do with
Lagrange multipliers: if you translate the midpoint to the origin (so
that the ellipse is just 0 = Ax^2+Bxy+Cy^2+F') then the ends of the
axes are the points which minimize or maximize x^2+y^2. The Lagrange
multiplier condition which results is (2x)(Bx+2Cy)=(2y)(2Ax+By), or
B x^2 + 2(C-A) xy - B y^2=0, so that the ends of the axes lie on the
lines y - c x=0 where c is either of the roots to X^2 - 2(C-A)/B X - 1=0.
In particular, if you draw the lines of slopes c through the midpoint,
these will be the axes of the ellipse.
At this point you have the original ellipse with, now, the axes added.
As several other posters have pointed out, the construction of the foci
is trivial.
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Euclidean construction of ellipse's foci -- Got it now.
Date: 10 Dec 1995 15:32:58 GMT
In article ,
Nelson G. Rich wrote:
>Is it possible to locate the foci of a given elllipse using
>Euclidean tools only? Is so, how?
Ha! I did it. It turns out to have been child's play. :-)
Below is the Euclidean construction of the foci of an ellipse. I will
describe the procedure first and then show why it works. You'll
probably want to sketch this yourself because it's a little lengthy
and there's no way I'm going to attempt an ascii portrait.
Begin with an ellipse. Pick any three points A, B, C on it and
construct the following points and lines. You might find it easier to
draw the following if A is "between" B and C.
First we will draw one line passing through the center of the ellipse.
Let L be the line bisecting AB through the midpoint P0 of AB.
Draw a line CD parallel to AB and passing through C; let D be the
point where meets the ellipse again.
Bisect the segment AC and find the intersection P1 of this bisector and L.
Bisect the segment BD and find the intersection P2 of this bisector and L.
Mark off a segment of P0,Q on L congruent to CD.
Draw a parallel to AB from Q and mark off a(n oriented) segment QR on it,
congruent to P1,P2.
Draw the line L' joining P0 and R.
Next, repeat this construction, drawing a parallel to AC through B
and extending to meet the ellipse at another point E.
Continue with the construction to get another line L'' which passes
through the midpoint of AC.
Let O be the intersection of L' and L''. This is the center of the ellipse.
Now we find the axes and foci.
Draw a circle at O with radius OA (say). Let the points where it
meets the ellipse be A, X, Y, Z (in circular order).
Bisect AX at U1; bisect AZ at U2.
Draw the lines O-U1 and O-U2; these are the axes of the ellipse.
Let W1 and W2 be points on the ellipse and on the lines O-U1 and O-U2
respectively.
Without loss of generality, assume segment O-W1 is longer than O-W2.
Draw the circle of radius O-W1 at W2.
The intersection of this circle with line O-U1 are the foci.
(This procedure is moderately difficult to draw but even harder to describe
verbally! Believe me, it's a lot simpler than the methods I had considered.)
Now let's try to figure out why this works. After P0 and L have
been drawn, we can impose coordinates so that the line AB is the
x-axis, L is the y-axis, and A-P0 is the unit of length. In these
coordinates it is clear that P0=(0,0), A=(1,0), and B=(-1,0).
The ellipse may be described in this coordinate system with a quadratic
equation in x and y. Since A and B are on it, the equation must
be (a multiple of) an equation of the form
0 = (x^2-1) + y (bx+cy+e)
for some b, c, and e. Now that coordinates have been introduced, we may
compute the coordinates (xi,yi) of any point on the ellipse, but we don't
know what b, c, and e are. On the other hand, each such point gives a
linear equation in b, c, and e, namely
b xi + c yi + e = (1-xi^2)/yi.
Each point P on the ellipse also determines a circle A B P. We used
in our construction the usual determination of the center of a circle from
three points on it. In the present coordinates, the center must be a
point (0,r), so with (+-1, 0) on the circle already, we see it must
have the equation x^2+y^2-2ry=1. If the point P=(xi,yi) is on this
circle, we have (1-xi^2)/yi = yi - 2r. Combined with the previous paragraph,
this shows that each point P on the ellipse helps determine b, c, and e
by giving a linear equation they must satisfy:
b xi + c yi + e = yi - 2 ri
(where (0, ri) is the center of the circle A B P).
So if our initial point C has coordinates (x1, y1), and the center P1
of its circle has coordinates (0, r1), then we have
b x1 + c y1 + e = y1 - 2 r1.
By construction D has coordinates of the form (x2,y1). The center P2
of circle A B D lies at some other point P2=(0,r2), so that this
point D yields an equation
b x2 + c y1 + e = y1 - 2 r2.
Subtracting this one from the equation for C leaves
2(r1-r2) + b(x1-x2) = 0
which I interpret to mean (2, b) is perpendicular to (r1-r2, x1-x2).
So we construct Q to be the point with coordinates (r1-r2,0), and
then R is the point (r1-r2,x1-x2). This makes L' the line with
equation 2x+by=0.
Now, the center of the ellipse is the unique critical point of the defining
quadratic. With the ellipse given by 0 = x^2-1 + y(bx+cy+e), we see
this critical point is determined to be the point (x,y) satisfying both
0 = 2x + by and
-e = bx + 2c y.
In particular, the center lies on the line L' we have just constructed.
By repeating the whole construction starting with the midpoint of AC,
we create another line L'' containing the center, so the center O is just
the intersection of L' and L''.
(You may think of L' as a geometric construction of the coefficient b.
It turns out not to be too difficult to compute c and e too, using
the circle ABC: it meets the ellipse at a fourth point F whose
equation when subtracted from that of C gives an easy relation between
b and c. But using the resulting values of b, c, e is a little
cumbersome. That's why I took a more geometric perspective, using two
lines each of which is "natural" in a different cooridinate system,
rather than forcing a second line in the first coordinate system.)
Now that we have the center, the rest of the problem is easy. By
translating coordinates to O, we have a quadratic whose critical point
is at the origin, and so is of the form a x^2+ b xy + c y^2 = 1.
By rotating the coordinate system, too, we may assume b=0. The
circle through A of course has equation x^2+y^2=r for some r,
so that the intersection points are of the form (+-x0, +-y0) in
these coordinates. The segments joining consecutive ones will be
vertical and horizontal, with bisectors U1=(x0,0) and U2=(0,y0) (say).
It follows that the lines O-U1 and O-U2 are the coordinate axes,
which are in the present coordinates the same as the axes of the ellipse.
Then W1=(w1,0) and W2=(0,w2) where a w1^2=1 and b w2^2 = 1.
Now, the foci F1 and F2 will be on the major axis. If a < b, this
is the horizontal axis O-U1. Let their coordinates be (+-w3,0). The
defining property of the foci is that the sum of the distances
F1-P + F2-P is constant as P ranges over the ellipse. In particular,
using P=W1, we see this constant is (w1-w3)+(w1+w3)=2 w1, ie., twice
the distance O-W1. But then using P=W2 we see the constant is twice
the distance W2-F1, so that the lengths W2-F1 and O-W1 are equal.
Thus F1 (and F2) will lie on the last circle drawn, as well as on
the major axis O-W1.
Commentary: The construction is pretty clear (I think) after the center
is found. The construction of the center is more complex but is similar
to what is done to find the center of a circle: take two points on the
circle, and draw the line from their midpoint to the center; repeat
with another pair of points and intersect. Unfortunately for us, the
line to be drawn in an ellipse won't be perpendicular to the segment
joining the two points. We need two additional points to calculate how
much to bend the perpendicular at the midpoint so it does meet the
center. But in retrospect, it's not all that hard to carry out that
calculation geometrically.
Notice that I found the center with 5 points on the ellipse. This is the
minimum number of points needed to uniquely determine an ellipse, so it
is perhaps not too surprising that 5 points are needed to find the center.
In principle it should be possible to find the axes, too, using only these
five points, rather than drawing the last circle; but that seems unnecessarily
complicated. Likewise, it is possible to find the foci from _any_ five
points on the ellipse, as I indicated in a previous post, but that looks
to be tremendously complex.
I should caution that I have been a little cavalier in giving the
recipe for construction. When computing the point R it is necessary
to mark off segments away from P0. Since I haven't specified the
_directions_ away from P0 and Q in which to mark off the points
Q and R respectively, this leads to four possibilities for R, that is,
two possible lines L', only one of which, of course, passes through O.
It is possible to trace through the supporting material with coordinates
and decide which line is correct, but I think in practice it will be
easier to eyeball the right line, for most choices of A, B, and C.
Likewise, I am glossing over certain degenerate cases, in which for
example A is already on one of the axes of the ellipse (so that the
circle of radius OA meets the ellipse only twice). If in any case
it's not immediately clear how to patch the construction details, one
may always start over with a different choice of A, B, and C. I hope
this is satisfactory.
dave
==============================================================================
From: jrr@atml.co.uk (John Rickard)
Newsgroups: sci.math
Subject: Re: Euclidean construction of ellipse's foci
Date: 10 Dec 1995 18:02:14 GMT
Nelson G. Rich (ngrich@naz.edu) wrote:
: Is it possible to locate the foci of a given elllipse using
: Euclidean tools only? Is so, how?
I've found a construction.
First step: The centre.
Choose A, B, C on the ellipse. Draw l through C parallel to AB; let
D be the other point of intersection of l with the ellipse.
(Special case: if l is a tangent -- intersects the ellipse only at
C -- then let D = C.) Let L and M be the midpoints of AB and CD.
Then LM passes throught the centre of the ellipse, and the centre is
midway between the points of intersection of LM with the ellipse.
(For this obviously works if the "ellipse" is a circle, and all the
steps are affine-invariant.)
Second step: The axes.
Draw a circle with centre at the centre of the ellipse, passing
through a point on the ellipse. This circle will intersect the
ellipse in four points, which are the vertices of a rectangle with
sides parallel to the axes (special cases of two or infinitely many
points of intersection left to the reader). The sense of the
intersections shows which is the major axis and which the minor.
(Or just compare the lengths of the axes!)
Third step: The foci.
If F and G are the foci, then for any P on the ellipse PF + PG
equals the length of the major axis. Let P be an end of the minor
axis; by symmetry PF = PG, so PF = PG = half length of major axis.
Draw a circle with centre P and radius equal to half the length of
the major axis; this intersects the major axis at the foci.
--
John Rickard
==============================================================================
From: kubo@abel.harvard.edu (Tal Kubo)
Newsgroups: sci.math
Subject: Re: Euclidean construction of ellipse's foci
Date: 11 Dec 1995 07:55:30 GMT
In article <4af7b6$kr8@gatekeeper.atml.co.uk>,
John Rickard wrote:
>
>First step: The centre.
>
> Choose A, B, C on the ellipse. Draw l through C parallel to AB; let
> D be the other point of intersection of l with the ellipse.
> (Special case: if l is a tangent -- intersects the ellipse only at
> C -- then let D = C.) Let L and M be the midpoints of AB and CD.
> Then LM passes throught the centre of the ellipse, and the centre is
> midway between the points of intersection of LM with the ellipse.
This is the degree 2 case of "Newton's theorem on secants":
the locus of the average of the N points of intersection of a
plane algebraic curve of degree N with lines in a fixed direction,
is a straight line.
>
> (For this obviously works if the "ellipse" is a circle, and all the
> steps are affine-invariant.)
Likewise, whenever the curve is symmetric around a point, the "Newton
line" in any direction must pass through the center.
==============================================================================
Date: Tue, 12 Dec 95 12:01:14 CST
From: rusin (Dave Rusin)
To: fc3a501@GEOMAT.math.uni-hamburg.de
Subject: Re: Euclidean construction of ellipse's foci -- Got it now.
Newsgroups: sci.math
Yes, I thought about this. I can carry out geometrically all the
algebraic computations, and I don't even need circles to find the
center (except to be able to draw parallels to lines and to be able to
mark off congruent segments). I think you do need to draw at least one
circle in order to find the axes (equivalently, the foci) once the
center is known.
But the attempt to carry out geometrically the algebraic computations
is incredibly convoluted and boring.
It's not even clear that one can easily find the foci if one is able to
consult the ellipse 5 times. That is, in the construction I posted (and
in the shorter version someone else posted) we pick 3 points on the
ellipse arbitrarily, then consult a fourth point found by intersecting the
ellipse with a line. I had to find one more point on the ellipse to
find the center; the other poster used two more.
Once the center is known, the construction we used to find the axes
required two more points on the ellipse. Moreover, we assumed we could
find an intersection of the ellipse with each axis.
So the constructions presented already require at least four more points
on the ellipse than are strictly speaking necessary.
If you want to look for a nicer construction, this would be the first step:
to find a construction which only uses 5 points on the ellipse. The more
general problem, of constructing the foci from 5 _arbitrary_ points on
the ellipse, seems even more formidable.
dave
==============================================================================
From: jons5451@telcel.net.ve
Newsgroups: sci.math
Subject: Re: Ellipse problem
Date: Sun, 01 Mar 1998 07:50:03 -0600
In article <6d9fhj$2g7$1@gannett.math.niu.edu>,
rusin@vesuvius.math.niu.edu (Dave Rusin) wrote:
>
> In article <34F70BC3.7ACCB378@online.no>,
> Alf Jacob Munthe wrote:
> >Given an ellipse on a paper, is it a geometrical way to find the focal
> >points or the long and short axis?
>
> We had this problem once before. Check out
> http://www.math.niu.edu/~rusin/known-math/index/51M04.html
> and scan for "ellipse". I think you'll want to skip over the responses
> by someone named "rusin"; the others are much clearer.
In the last sentence of that document you wrote:
> The more general problem, of constructing the foci from 5 _arbitrary_
> points on the ellipse, seems even more formidable.
If I remember well this was a standard problem in a course about
projective geometry that I took some (well, many) years ago. The
solution was more or less as follows:
1) Given the points M, N, P, Q and R use Pascal's theorem (limit case)
to construct the tangent m to the ellipse at M (the rule suffices).
2) Draw a circle c tangent to m at M. This circle is homologous to the
ellipse, with M as center of homology (this last word is used here with
the meaning it has in the context of projective geometry, of course).
3) Let N', P' and Q' be the (second) points where MN, MP and MQ meet
the circle c, respectively. The axis e of the homology is determined by
the points NP.N'P' and PQ.P'Q'.
4) Let S be the point where P'Q' intersect a parallel to PQ by M. The
parallel i' to the axis e by S' is the "limit line", i.e. the homologous
of the line i at infinity. Let X' be the intersection of m and i'. With
center X' and radius X'M intersect i' at points Y' and Z'. Since the angle
Y'MZ' is right, MY' and MZ' give the directions of the principal diameters.
5) Let O' be the pole of the line i' with respect to c. Let the line O'Y'
meet c at points A' and B', and let O'Z' meet c at C'and D'. The points A,
B, C and D homologous of A', B', C' and D' are the vertices of the ellipse.
The point O = AB.CD (homologous of O') is the center. The foci are easily
found: if AB is the major axis draw a circle with center C and radius OA
and intersect it with AB.
In other post cited in the same document you say:
>Back in the century in which emperors and presidents gave mathematical
>proofs there was so much collective understanding of Euclidean constructions
>that someone would have leapt in to demonstrate the elegant geometrical
>construction
It seems that I'm older than I thought :-(
José H. Nieto
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==============================================================================
From: jhnieto@my-dejanews.com
Newsgroups: sci.math
Subject: Re: finding the focci of an ellipse
Date: Fri, 07 Aug 1998 00:55:16 GMT
In article <35C9F411.6525@_._>,
Dave Conant <_@_._> wrote:
> Let's say you have a accurately drawn ellipse, or better still, one
> drawn in a cad package. Is there a way to find the focci using only
> a strait edge and compass?
>
five points are all you need to determine completely the ellipse.
The foci may be found as follows (this is only a sketch, look up the
details in a good ol' book on projective geometry).
1) Given the points M, N, P, Q and R use Pascal's theorem (limit
case) to construct the tangent m to the ellipse at M.
2) Draw a circle c tangent to m at M. This circle is homologous to
the ellipse, with M as center of homology (this last word is used
here with the meaning it has in the context of projective geometry).
3) Let N', P' and Q' be the (second) points where MN, MP and MQ meet
the circle c, respectively. The axis e of the homology is determined
by the points NP.N'P' and PQ.P'Q'.
4) Let S be the point where P'Q' intersect a parallel to PQ by M.
The parallel i' to the axis e by S' is the "limit line", i.e. the
homologous of the line i at infinity. Let X' be the intersection of
m and i'. With center X' and radius X'M intersect i' at points Y'
and Z'. Since the angle Y'MZ' is right, MY' and MZ' give the
directions of the principal diameters.
5) Let O' be the pole of the line i' with respect to c. Let the line
O'Y' meet c at points A' and B', and let O'Z' meet c at C'and D'. The
points A, B, C and D homologous of A', B', C' and D' are the vertices
of the ellipse. The point O = AB.CD (homologous of O') is the center.
The foci are now easily found: if AB is the major axis draw a circle
with center C and radius OA and intersect it with AB.
Best regards,
Jose H. Nieto
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==============================================================================
==============================================================================
From: azeynel@hotmail.com
Newsgroups: sci.math
Subject: Re: Ellipse problem
Date: Sun, 08 Mar 1998 17:27:16 -0600
In article <34F70BC3.7ACCB378@online.no>,
Alf Jacob Munthe wrote:
>
> Given an ellipse on a paper, is it a geometrical way to find the focal
> points or the long and short axis?
>
> Alf
> I saw this in an old geometry textbook. Totally impractical, but it seems
correct:
Take a hollow cone and cut it to appropriate angle to get the desired
ellipse. Place a sphere between the vertex and the ellipse touching the
ellipse at point S. Now place another sphere under the ellipse this time
touching it at point H. (Of course spheres snugly fit inside the cone). The
points S and H are the focii.
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