If we slice through a cone, depending on the angle of the cut, the
edges will form a circle, ellipse, parabola, or hyperbola (figure 1
On this page, we'll discuss the shape each cut appears to have,
simply from an inspection of the cone and the way the lines pass
through it, and then we'll use a little algebra to prove that the
sections really do have the claimed forms.
that each of the conic sections has its own "focus property", different
for each. The focus properties are shown pictorially on our parabola focus
, ellipse focus
, and hyperbola focus
pages, where we also
derive equations for each of those figures starting with the
Figure 1 -- Cone, with
This is really pretty obvious! If
you cut through the cone parallel to the base the result is a circle;
it must be true simply from the way we construct a cone.
This one may not be at all obvious! If we cut through the cone
at an angle, certainly the section formed should have an oval
shape -- but, since the cone is "fatter" farther from the point, one
might expect the resulting figure to be egg-shaped, rather than a
It can help to realize that, when one cuts through the cone this way,
the "fattest" part of the cut is not
on the centerline of the cone. The line of the cut is descending
as it's crossing the cone. The cone can be viewed as being made
up of stacked circles. As the cut crosses the centerline, it also
crosses the widest chord of the circle through which it's cutting.
But since that is exactly the widest point, moving an
infinitesimal amount away from it to either side won't change the
length of the chord -- this is just the principle on which finding
maxima and minima is based. But as the cut crosses the
centerline, it's descending at a nonzero (linear) rate, and
diameter of the cone is increasing at a nonzero (linear) rate.
So, we would expect the diameter of the cut section to be increasing
at the moment when the cut crosses the centerline. This is indeed
the case, as we shall see later.
So, the section isn't egg-shaped -- but it also isn't
symmetric about the centerline of the cone.
If we cut exactly parallel
to the side of the cone, we
certainly won't get an ellipse. (In figure 1
the line labeled "parabola" should run along a cut which is exactly
parallel to the left edge of the cone. However, it was drawn
"free-hand" so it may not be quite
will remain at a fixed distance from the edge of the cone. If we
imagine the cut as it goes far down the cone, we realize that there's
no limit to how far apart the lines may become. But at the same
time, it seems that they must be descending ever more rapidly down the
cone's sides: as the cone becomes very very fat, the cut begins
to look more and more like we're just peeling off a "strip of bark"
from the cone, and the edges would seem to grow nearly parallel. Since
they never cease to grow farther apart, however, they're certainly not
approaching vertical asymptotes, like the curve of a tangent function.
In fact, it appears that they never approach any asymptotes.
This certainly describes a parabola, and we'll see later that this is
exactly what this curve is. (para
bola == the cut which is para
to the side of the cone.)
If we cut even more steeply, so that the slice is not parallel to the
side, we form a hyperbola. (hyper
=> more; a hyperbola
is cut more steeply than a parabola.)
primary trait of a hyperbola is that the sides approach asymptotes.
For any hyperbola, we can find two lines (the "asymptotes") such
that, if we go far enough out on the legs, the sides of the curve
approach arbitrarily closely to them. We can see that the edges
of any cut which is steeper than the parabola will approach asymptotes:
First, if we cut straight down, as the hyperbola in figure
appears to be, this property is obvious: as we go arbitrarily far
down the cone, the cone swells out on either side of the cut, and the
cut is left traveling almost "dead center". The radius of
curvature of a circle about the cone increases arbitrarily as we
descend, and the distance between the two sides of the cut becomes
almost the same as the diameter of the circle; eventually the hyperbola
will look almost as though we cut straight down through the point ...
which would certainly just produce two straight lines.
If we cut down from the point
at an angle, then the fraction of the way from the center of the cone
to its edge at which the cut is located would be fixed, all the way
down. The edges would be straight lines, which would separate
from each other at a smaller angle than the "sides" of the cone.
If, however, we start our cut some distance to one side of the
point, it will still eventually
find itself about the same
fraction of the distance from the center of the circle to the edge as
if we had cut from the point
... and as we descend, the "initial aberration" we provided by starting
the cut to one side of the point will make less and less difference.
In short, the edges will again approach asymptotes.
We'll prove later that this curve really is a hyperbola.
Proof, Part I -- Preliminaries, and Proof that a Parallel Cut
yields a Parabola
In figure 2
, we show a flattened view of the
cone from the side, with the "cut" line passing through it. Note
that the "cut line
" is really a cut plane
viewed edge-on. We have marked the included angle θ between
the edge of the cone and the centerline, and the angle φ between the
cut line and the centerline.
We have coordinates ξ and y
in the plane of the cut.
Given a particular point, P
, on the edge of the
cut, ξ is its distance from the centerline of the cone
measured along the centerline of the cut
, and y
is the perpendicular distance from the centerline of the cut
. Note that P
is at the edge
of the cone -- not
in the plane of the picture in figure
has coordinates (ξ,y
) in the
cut plane. x
is the perpendicular distance from the
centerline of the cone
to the point (ξ,0
) in the plane
of the cut. We wish to find an equation relating ξ and
Figure 2 -- Cone with
cut, flattened, seen from the side:
This should be clearer in figure
below, which shows the cone from the top
; the distances x,
, and r
, shown in figure 3
, relate points all
of which lie in a plane h
units from the top of the cone.
The cut crosses the centerline of the cone a distance h0
from the top of the cone, and the point P
on the cut edge lies distance h
from the top of the
cone. In the plane of figure
, the point P
may be seen to lie x
units from the centerline of the cone, and figure 3
we can see it lies y
units from the centerline of the cut. h
the top, the cone has radius r
. Since the cone's
units from the top is a circle, we can see
immediately that x2
Figure 3 -- Cone with cut,
seen from the top:
hope I haven't confused you totally with these attempts at describing
what we see in the images. The situation is simple enough in 3
dimensions; the difficulty is in producing a clear picture using only two
dimensions! But let us proceed.
From figures 1 and 2 we can read off the following:
and as already noted we can see that:
) with the formulas from (1
) for x
Rearranging a little,
And now we need to stop and take a good look at the second term in (4
) before we do anything more to it. If,
similar to the way we have drawn figure 1
> θ, then we also have
and so the second term must be positive.
If φ = π/2, then cos(φ)=0, sin(φ)=1, and we obtain a circle, just
as we would expect.
If, on the other hand, we have φ = θ then the second term in (4
) will be zero
. In that case we
can replace φ with θ everywhere to obtain,
which is certainly the equation for a parabola, as we saw here
, on the parabola focus page.
So, if the cut is parallel to the side of the cone, the edge
does, indeed, form a parabola.
Proof that φ > θ => we obtain an Ellipse
When φ > θ, the second term in (4
) is positive.
To reduce the clutter we make the following substitutions:
Note that k1
are both positive, and,
< 1. We pull out the
coefficient on ξ2
complete the square,
shift the origin by making the following substitution,
and finally obtain:
Looking back at (6
), we see that that the
coefficient on ζ2
is between 0 and 1, and the term on the
right is positive. Comparing this with equation 8
on the ellipse focus
page, we see that this is, indeed, the equation for an ellipse.
Note that the substitution in (9
) is necessary
exactly because the minor axis of the ellipse does not touch the
centerline of the cone.
Proof that φ < θ => we obtain a Hyperbola
Looking back at (6
) we see that, if φ < θ, then
must be between 0 and
The right hand side of (10
), on the other
hand, is positive, of any magnitude. Comparing this with equation 8
on our hyperbola
focus page, we see that in this case (10
indeed, the equation for a hyperbola.
Cuts which Don't Cross the Centerline of the Cone
As it happens, the cut labeled "Hyperbola" in figure
appears to go straight down, and never cross the centerline at all.
We have not yet proven that such cuts, nor cuts which angle
"outward" and hence also avoid the centerline, also produce hyperbolas.
We need to change our analysis slightly, and we need a new
picture. In figure 4
, we've turned the
cut so it never crosses the center line. We've kept the same
meanings for the symbols used in figure 2
and figure 3
, but we've added a new one: x0
is the horizontal distance to the start of the cut.
Figure 4 -- Cone, side view,
with cut that doesn't cross the center line:
From figure 4
, we can read off:
Plugging the values for r
) into (2
) we obtain:
Multiplying out and collecting terms we obtain:
that we must have φ < θ (or else the "cut" misses the cone
entirely). By inspection, then, we see that both coefficients in
(13) are negative
. To reduce clutter we'll make the
Note that both k3
positive and unbounded.
We'll now complete the square and collect terms to obtain:
We change variables to shift the origin:
and finally arrive at:
The coefficients are positive and unbounded. Comparing this with equation 8
on our hyperbola
focus page, we see that it is, indeed, the equation for a hyperbola.
Page created on 11/10/06