The Focus of a Hyperbola
can be considered as an ellipse
turned inside out. Like the ellipse, it has two foci; however,
in the distances to the two foci is fixed for all points on the
hyperbola. For an ellipse, of course, it's the sum of the
distances which is fixed. If a hyperbola is "stretched" to the
limit, it turns into a parabola, as does the ellipse; but for the
hyperbola as we've drawn it here, it's the lower focus which goes to
infinity (to form an
upward-facing parabola), while for a similarly oriented ellipse, it's
the upper focus (again, if you want form an upward-facing parabola).
To turn an ellipse into a hyperbola, then, we can send its upper
infinity, and then retrieve a new lower focus from -infinity. At
the boundary between the two, the figure is a parabola, which could be
said to have three foci, two of which are infinitely distant (but we
normally say, instead, that it has a directrix and just one focus).
In figure 1
we show a hyperbola with foci on the Y axis at +/- 2 units from the
origin, for which the difference in the distances to the two foci is 2
units. We've shown both "branches" of the hyperbola, though on
the rest of this page we'll be concerned only with the upper branch.
a parabola or ellipse, a hyperbola has its own "focus property":
All incident rays which are directed at the lower focus and which
hit the upper branch will be reflected to the upper focus, instead.
Conversely, rays which come from the first focus and strike the
hyperbola will be reflected along a path that makes them appear to have
come from the second
focus; in this sense, f2
is an "image" of f1
We shall prove that fact on this page, and then we'll show that
an infinitely stretched hyperbola really does turn into a parabola, and
go on to derive an equation for a hyperbola, and finally demonstrate
that the familiar y=1/x is really a hyperbola.
Figure 1 -- Hyperbola, foci
4 units apart, difference=2
Proof that Light Directed to
f2 is Reflected to f1
In figure 2
, we show two points on a
which are infinitesimally close together, in a greatly magnified image.
The proof is contained entirely in the picture; we will describe
the details in words, below.
Figure 2 -- An infinitesimal
segment of a hyperbola, greatly magnified:
From point P1
, we show rays a1
lead to focus 1 and focus 2, respectively. From point P2
we show rays a2
which lead to focus 1 and focus
2, respectively. P1
are assumed to be so
close together that rays a1
parallel, and rays b1
are (nearly) parallel.
By the definition of a hyperbola, the difference in length between ray a1
and ray b1
must be equal to the difference in length between
and ray b2
. Ray b2
is ε units
longer than ray b1
. Therefore, if the differences are to
be equal, ray a2
must also be ε units longer than ray a1
That is the key to the proof!
Look at triangles ABC
in figure 1. They're right triangles, they each have side C as
the hypotenuse, and sides A and D are equal (each is ε units long).
Therefore the two triangles are identical (save that one's
flipped over). Therefore, angles α and β must be equal.
The line marked "Incident Light" is the extension of ray b2
through the hyperbola, and the angle it forms with the hyperbola must,
of course, be equal to the angle b2
forms with the hyperbola.
So, angles β and γ must be equal.
But then angles α and γ must also be equal. Light coming to point
which was directly at focus f2
must come in along the "Incident Light" line, with incident angle γ.
Its angle of reflection must equal its angle of incidence; but
then, it must be reflected along ray a2
. Ray a2
leads to focus f1
, which is what was to be
that the argument given here was identical to the argument
we used on the parabola focus
page. The only significant difference is that figure 2
on the parabola page
was shown rotated, so that lines B1
run straight down to the directrix. They could just as well be
running down to a lower focal point which was located very far away; we
discuss this at a bit more length in the next section.
Stretching a Hyperbola Makes
Move focus f2
down to -infinity.
Now, lines which are directed toward f2
must all go straight down
. So, all lines arriving
parallel to the axis will be reflected to focus f1
and the figure must be a parabola, as we saw here
parabola focus page. Q.E.D.
That may have been too concise. So here's a version with a few
steps, which proves the claim directly from the definitions:
Shift everything in figure 1
down a bit so
that the upper branch of the hyperbola touches the origin. Place f1
units above the X axis, and fix the difference in the distance to the
two foci at 2
-δ) units, where the distance between the
foci is 2f
Now, send the lower focus arbitrarily far away. The
hyperbola will still just touch the origin, but what happens at nearby
points on the hyperbola?
As we move away from the origin, and consequently farther from f1
we must also move farther from f2
order to keep the difference in the distances fixed. But since
is arbitrarily far away, all lines
run (almost exactly) straight
. Draw a line parallel to the X axis, and δ units
below the origin; call it the directrix
. Lines leading
are all (almost exactly) perpendicular to the directrix. A point
on the hyperbola which is ε units farther from f1
and consequently ε units farther from f2
must also be ε units farther from the directrix. So, the
difference in distances between a point on the hyperbola and f1
and the directrix must also be the same for all points on the
hyperbola. By the difference in the distances for the point at
the origin is zero (it's δ units from f1
and from the directrix). So, all points on the hyperbola are
equidistant from f1
directrix, which is the definition of a parabola. And that is
what was to be shown.
proof would benefit enormously from a picture, but I haven't drawn one
for it (as yet). A little farther down the page we'll do this over symbolically
An Equation for a Hyperbola
So far we've just worked directly with the definition of a hyperbola.
Let's find an equation for one.
Given a particular point on the hyperbola
define the following:
We then have:
Moving the second root to the right, squaring, and eliminating common
terms, we obtain:
Multiplying out the squares, we obtain:
to the other side and squaring again, we obtain:
Multiplying everything out, eliminating common terms, and collecting
what's left, we get:
This is still pretty ugly. Observing that we must always have r
< 2f, we make the following substitution:
which, with a little rearrangement, leads to:
A Symbolic Proof that a
Stretched Hyperbola is a Parabola
The form of (8
isn't useful for taking a limit as the lower focus goes to
infinity, because the hyperbola heads off to infinity, too. We
want a formula which keeps the upper branch of the hyperbola touching
the X axis as we "stretch" it. To obtain an easily recognizable form,
we'd also like the "end" of the hyperbola to be touching the origin.
To that end, we'll first define the following:
the distances to f1
as above. We'll then shift the axes up a bit, so that the foci fall as:
We can then immediately see that:
Squaring both sides and dividing by 2, we obtain the following mess:
Multiplying out and canceling common terms, we obtain:
Now we divide through by f
and take the limit as f
goes to infinity:
Collecting all but the square root on the left, we square it again:
Multiply out the squares:
And we're done:
(Compare with the formula for a parabola
we found on the parabola focus page.)
What About y=1/x ?
also a hyperbola; it's just rotated relative to the hyperbolas we've
been looking at. We'll start with the usual formula:
and we'll rotate the axes to the right 45 degrees, which is an angle of
The rotation matrix is:
We obtain the following substitutions:
Plugging them into (19
) we obtain:
Looking back at equation (8
), we see that it
matches equation (23) with the following values:
and this is, indeed, a hyperbola.
Page created on 11/7/06 and last updated on 11/15/06