
The Focus of a Parabola

A parabolic mirror focuses light. What does that mean? It
means that all
rays which run parallel to the parabola's axis which hit the face of
the parabola will be reflected directly to the focus. But why
does this happen?
On this page we provide a simple proof of this fact, just by examining
an infinitesimal piece of a parabola and looking at some triangles.
Definition of a Parabola
(Note  A parabolic mirror is
actually a "paraboloid", which just means that a vertical "slice"
through the mirror's surface has the shape of a parabola. On this
page I'm rather sloppily using the term "parabola" for the 2
dimensional curve which really is a parabola, and for the 3 dimensional
surface of a mirror, which really is a paraboloid. Hopefully this won't
cause any confusion.)
A "parabola" is the set of all points which are equidistant from a
point, called the
focus, and a line, called the
directrix.
Later on we'll show that this leads directly to the usual formula for a
gardenvariety parabola, y=x
^{2}, but for now we're going to
work directly with the definition. In
figure
1
we've shown a
portion of a parabola, with some distances marked off to illustrate
this. This particular parabola has its focus located at (0,0.25),
with its directrix running 1/4 unit below the X axis. We've shown
the distances from the directrix and focus to three points on the
parabola: the points at 0.25, 0.5, and 1.25 units from the focus and
directrix.
(Note
that a "paraboloid", such as the surface of a telescope mirror, is
defined identically, save that the directrix is extended to a plane.)
Figure 1  A Parabola
Proof that the Parabola Brings Light to a
Focus
We'll now look at an infinitesimal segment of the parabola. In
figure 2 we've shown a highly magnified view of
two points on the parabola, marked
P1 and
P2, along
with the lines leading from the focus and directrix to them. The
segment shown in the picture is so small, and
P1 and
P2
are so close together, that the lines from the focus to
P1 and
P2
are (almost exactly)
parallel. (The lines perpendicular
to the directrix which lead to
P1 and
P2 are, of
course, necessarily parallel.)
The proof is contained entirely in the picture. We will, however,
discuss it in a bit more detail, below.
Figure 2:
Lines
A1 and
B1 lead from point
P1 to the focus
and directrix, respectively. Since
P1 is on the parabola,
lines
A1 and
B1 must be the
same length.
Just a little farther along the parabola we have marked point
P2.
We've drawn line
A2 from
P2 to the focus, and
we've drawn line
B2 straight down to the directrix. As
mentioned,
P1 and
P2 are actually so close together
that
A1 and
A2 are (essentially) parallel.
Segment
b runs perpendicularly from the end of
A1
to
A2.
A2 is longer than
A1 by the piece
extending past segment
b, marked "
a"; it
is ε units long.
Segment
d runs perpendicularly from the end of
B1
to
B2, and
e is the "extension" of
B2
versus
B1. Since
P2 must also be equidistant from
the directrix and the focus, segment
e must also be ε
units long. That is the key to the proof!
Triangles
abc and
edc are right triangles and two of
their sides (
a and
e, and the hypotenuse,
which is
c for each of them) are certainly the same
length. So, the third pair of sides,
b and
d,
must also be the same length. So, if lines
A1 and
A2
are δ units apart, then
B2 and
B2 must also be δ units
apart.
So, triangles
abc and
edc
are just mirror images, so angles θ and φ must be identical.
Angles φ and γ are also identical, since they're opposite angles
of two intersecting lines. But then angles θ and γ must also be
identical.
A light ray coming straight down from the top of the page (parallel to
the parabola's axis), along line
B2, strikes the parabola at
angle γ. It's naturally reflected at
the same angle 
so, since γ = θ, the ray will head down line
A2,
straight to the focus, as was to be shown.
If it Brings Parallel Rays
to a Focus, Must it Be a Parabola?
Yes, absolutely ...
if it's continuous.
Pick
a focus, and pick a point on an arbitrary curve. If, at that
point, incoming light is reflected to the focus, then the curve must
have the same slope as a parabola at that point. That is, its
derivative must match that of a parabola which passes through that
point. If its derivative matches that of a parabola at
every
point, then, since a function is the integral of its derivative, it
must
be a parabola.
There's
a caveat here, though, which is that this argument only works if the
curve is continuous. A discontinuous curve could take a finite
number of "jumps", and still bring incoming rays to a focus nearly
everywhere. That's the principle on which Fresnel lenses and
mirrors are based.
On the other hand, camera lenses which do not behave as ideal parabolic
lenses also don't "really" focus light to a
point.
That is, parallel rays are brought together into a tiny disk at
the film plane (the "blur disk"), but not a single point. It's a
tradeoff: Parabolic mirrors (and lenses) focus onaxis rays
perfectly (within the limits set by diffraction), but do a poor job of
focusing items which are offaxis. Camera lenses are optimized to
bring rays from many angles to a "pretty good" focus.
The Parabola as a Limit
As we can see on the
ellipse focus
and
hyperbola focus
pages, a parabola can be considered as the limit of a "stretched
ellipse", or as the limit of a "stretched hyperbola". It can also
be considered as the figure which stands midway between an ellipse and
hyperbola. It has a single focal point and a focal
line,
but the directrix could also be considered to represent either an
"ellipselike" focal point at
or
a "hyperbolalike" focal point at
,
or perhaps both.
The Equation of a Parabola
Just
for completeness, we'll now go ahead and derive the simplest of the
parabola equations, to show that the definition we're talking about is
really the usual y=x
^{2} thing.
In
figure 3,
we've shown the parabola again, with a number of distances marked.
We'll use those to find the equation which describes the parabola.
Figure 3  Parabola
with distances marked:
Since,
by definition, the distance from a point on the parabola to the focus
equals the distance from that point to the directrix, the two distances
marked "
D" must be identical. We can identify a right
triangle in the upper left of
figure 3 bounded
by yf, x, and D, and we
can see on the green line that y+f must be D. So, we can read off:
Squaring (2) and equating the D
^{2} terms, we obtain
Multiplying it out and collecting terms we see,
and, when f = 1/4, this is just the familiar y=x
^{2}.
Page created 11/2/06, and last modified 11/15/06