
The Focus of an Ellipse

An ellipse has the property that any ray coming from one of its foci is
reflected to the other focus. This is occasionally observed in
elliptical rooms with hard walls, in which someone standing at one
focus and whispering can be heard clearly by someone standing at the
other focus, even though they're inaudible nearly everyplace else in
the room.
On this page, we'll show that this is true, by looking
at several triangles in a magnified image of part of an ellipse. We'll
also find an equation for an ellipse, and just for amusement we'll show
that an "infinitely stretched" ellipse turns into a parabola.
Definition of an Ellipse
An
ellipse has two foci. The sum of the distances from any point on
the ellipse to the two foci is the same for every point on the ellipse.
In
figure 1, we show an ellipse
in which the foci are 1.7 units apart, and in which the sum of the
distances to the two foci is 2 for every point on the ellipse.
Figure 1  An Ellipse, with
foci at +/ 0.87, sum of radii = 2:
Proof that Rays from One Focus Travel to the Other Focus
Just looking at
figure 1,
this assertion certainly seems plausible; for a ray following any of
the blue lines, the angle of incidence and the angle of reflection
where it hits the ellipse look equal. That's what we'll now show
is true.
Figure 2 shows a highly
magnified view of two points on the ellipse,
P1 and
P2,
which are extremely close together. They are assumed to be so
close together, and the region in figure 2 is so small, that rays
b1
and
b2 which run to focus
f_{2} are
(nearly)
parallel, and rays
a1 and
a2 which run
to focus
f_{1}
are also (nearly) parallel. A careful examination of the picture
demonstrates the proof; we shall now go over it in words.
Figure 2  An infinitesimal
segment of an ellipse, with rays to foci shown:
Rays
a1 and
b1 run from point
P1 to the two
foci, and rays
a2 and
b2 run from point
P2 to
the two foci
. By definition of the ellipse, the sum of
the lengths of rays
a1 and
b1 must
equal the
sum of rays
a2 and
b2. Ray
b2 is ε units
longer than ray
b1; therefore, if
a2+
b2 is to
equal
a1+
b1, it must be the case that ray
a2 is
ε units
shorter than ray
a1. And that is the key
to the proof!
Look at triangles
ABC and
EDC. For added clarity,
these are shown highlighted in
figure 3 and
figure 4. They are each right triangles.
Side
D and side
B are equal (each is ε units
long), and side
C
is the hypotenuse of both triangles. We must, therefore, have the
length of side
A equal to the length of side
E, as
well; the triangles are mirror images of each other.
Since the two triangles are identical (save that one is flipped), we
must have angle α equal to angle β. Now, since
b1 is
parallel to
b2, it must strike this tiny,
nearly straight
section of the ellipse at the same angle as
b2,
so we must also have angle α equal to angle γ. But then angle β
must equal angle γ as well. But then, a beam coming from focus 2
which strikes the ellipse at
P1 will come in along ray
b1;
it will be reflected with an angle of reflection equal to the
angle of incidence, which means it will have to be reflected along ray
a1,
which leads to focus 1. And that is what was to be shown.
Figure 3  Ellipse segment,
with triangle ABC highlighted:
Figure 4  Ellipse segment,
with triangle EDC highlighted:
The Limit of a Stretched
Ellipse is a Parabola
Rotate the ellipse so it's vertical, with focus
f_{2}
above, and focus
f_{1} below (this
just makes this argument a little easier to state). Send
focus
f_{2} up to infinity. Now,
lines which come from
f_{2} must all
go
straight down. So, all lines arriving parallel to
the axis will be reflected to focus
f_{1},
and the figure must be a parabola, as we saw
here on the
parabola focus page. Q.E.D.
That may have been a little too concise, so here's a longer version
that argues directly from the definition of a parabola:
Suppose
we make the foci of an ellipse arbitrarily far apart. At the same
time, suppose we keep the sum of the distances from points on the
ellipse to the foci just slightly longer than the distance between the
foci. If the distance between the foci is 2
f, let's set
the sum of the distances to the foci from any point on the ellipse to 2
(f+
h).
And now, let's shift the ellipse so that one end of it just
touches the origin, with its long axis extending to the right along the
X axis. What will it look like?
The "close" focus,
f_{1}, must be
h
units from the origin. Let's draw a line, parallel to the Y axis,
h units from the origin, on the
opposite side of the
Y axis from
f_{1}. Call it the
directrix.
Now, the origin, which lies on the ellipse, is
h units
from the directrix, and
h units from
f_{1}.
As we move along the ellipse to points that are farther from
f_{1},
they must grow
closer to
f_{2},
which we have assumed is arbitrarily far away. Because
f_{2}
is so far away, all lines to it will appear parallel. So, a point
on the ellipse which is
h+δ units from
f_{1}
must be δ units farther to the right  or δ units from the Y axis, or
h+δ
units from the directrix. That means the figure is, by definition, a
parabola.
This
visual argument could be shown more clearly with a picture but I
haven't drawn one for it (as yet). I'll also show this
symbolically,
below.
An Equation for an Ellipse
We've done everything so far just using the definition of an ellipse.
Let's find an equation for one.
Referring back to
figure 1, if each
focus is
f units from the origin, then the distance from
a point on the ellipse to focus
f_{1} must be
and the distance from a point on the ellipse to focus
f_{2}_{}
must be
If we let the sum of the distances to the foci total
2r, then
we have
or
Squaring both sides, expanding (x+f)
^{2} and (xf)
^{2}
on the left and right respectively, and canceling common terms, we
obtain,
Dividing through by 4 and then squaring yet again, we obtain,
We expand (xf)
^{2} and cancel common terms,
Dividing through by r
^{2} and collecting terms,
Since we must have r > f, we see that the coefficient on
x^{2}
must be between zero and one, and the term on the right can be any
positive value.
^{}
Limit of a Stretched Ellipse
Above
we gave a visual argument (sans picture) to show a stretched ellipse
turns into a parabola. Here we'll do it symbolically.
Let's
change coordinates so that the left end of the ellipse just touches the
Y axis; then we'll stretch it to the right. So, let:
Substituting into (
8) we get:
Multiplying through by r
^{2}/(r
^{2}f
^{2})
and multiplying out the square:
Collecting terms:
Now we're going to set:
If we keep δ fixed, then focus
f_{1} will
remain fixed δ units to the right of the Y axis even as we vary
f,
and the left end of the ellipse will remain at the Y axis even as we
vary
f. Substituting (
13)
into (
12), we obtain:
and, in the limit as
f goes to infinity,
which was to be shown (compare with the
parabola equation we found
previously).
Page created on 11/6/06, and last updated on 11/15/06