Spheres in a Cone; or, Proving the Conic SectionsAuthor(s): DAVID ATKINSONSource: The Mathematics Teacher, Vol. 80, No. 3 (MARCH 1987), pp. 182-184Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27965315 .

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Spheres in a Cone ; or,

Proving the Conic Sections By DAVID ATKINSON, Olivet Nazarene University, Kankakee, IL 60901

Anyone

who has taught the conic sec

tions has no doubt used a model of a

right circular cone that comes apart to dem

onstrate the elliptical, parabolic, and hy

perbolic cross sections obtained when the cone is cut by a plane at various angles. This result is believable but certainly not

intuitively obvious, yet it is almost never

proved. In fact many of us who teach the

conic sections have never seen a proof of

this result. A truly elegant proof of the conic sec

tions was done in the early nineteenth cen

tury by the Belgian mathematicians Germi nal Dandelin (1794-1847) and Adolphe Que telet (1796-1874). The proof is done by in

serting one sphere (parabola) or two

spheres (ellipse and hyperbola) into the cone so that each sphere is tangent to the cone in a circle and tangent to the cutting

Fig. 1. Dandelin's spheres?elliptic case

182

plane at a point. The spheres are referred to as Dandelin spheres.

The proof that follows is for the ellipse. We use the common definition of an ellipse as the locus of points in a plane, the sum of whose distances from two fixed points in the plane (foci) is a constant. As a bonus this proof also determines the directrix lines and relates them to the foci and eccen

tricity, which is always between zero and one for an ellipse.

We begin with one nappe of a right circular cone (see fig. 1) with vertex V. We

shall show that an ellipse is produced when the cutting plane a intersects only one

nappe of the cone and is neither perpen dicular to the cone's axis nor parallel to a

generator. (A generator is any line on the cone passing through the vertex.)

Imagine inserting two small inflatable

V

Fig. 2. Dandelin's spheres?elliptic case with direc

trices

Mathematics Teacher

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spheres into the cone, one above the plane a

and one below. Then imagine inflating each

sphere until the upper sphere is tangent to

the cone at circle c? and to the plane a at

point Fl and until the lower sphere is

tangent to the cone at circle c2 and to the

plane a at point F2. We claim that Fx and

F2 are the foci of the ellipse. Let be a representative point on the

curve of intersection of plane ? and the cone. Draw the generator line VP intersect

ing circle cl at Al and circle c2 at A2. Also

draw segments PFl and PF2. Since these

segments are in plane a, they are tangent to

their respective spheres at Fl and F2. Also

because is a generator line it is tangent to the spheres at A and A2 , as shown. Now

PF1=PA1 and PF2 =

PA2 because tangents to a sphere from an external point are

equal. Adding these equalities, we get

PF1 + PF2 = PA1 + PA2.

But

PAl + PA2 =

A,A2 =

kx,

a constant independent of P. (The circles c1 and c2 determine equal segments on any

generator.) The statement

PF1 + PF2 =

kl9

a constant, is precisely our definition of an

ellipse! When studying conies we learn that cor

responding to each focus is a line called a

directrix with the property that the dis tance from any point on the conic to a focus over the distance from the point to the cor

responding directrix is a constant called

the eccentricity, denoted by e. We now

locate the directrix lines and show that

they have this property. Planes ? and ?2 (see fig. 2) are the

planes of the circles of tangency, c1 and c2 ,

respectively. Since these planes are each

perpendicular to the axis of the cone, they are parallel, and so the lines of intersec

tion, and l2, of these planes with the cut

ting plane a are parallel. These lines are the

directrix lines of the ellipse !

Proceed by labeling the points of inter

section of the line through the foci and the

March 1987 -

lines Zx and l2 as ?1 and B2, respectively. We draw the segment from perpendicular to /1 at Di and extend this segment from

meeting l2 at D2. This segment will be per

pendicular to l2 at D2, since lines and l2 are parallel.

Since segments AlA2 and DlD2 inter sect at P, they determine a plane that cuts

the parallel planes ?, and ?2 in the parallel lines AiDl and A2D2 (not shown). It readi

ly follows that triangle PA1Dl is similar to

triangle PA2 D2 . Using corresponding sides

of these triangles, we obtain

PA, _PA2 PD, PD2'

Substituting the equations PAl =

PFi and

PA 2 = PF2 obtained previously yields

PFt PF2

PD,

"

PD2 '

This ratio is precisely the eccentricity, and it only remains to be shown that this ratio is constant for all points on the ellipse.

From the preceding proportion we have

PF2 PD2 PFl

~

PD, '

Adding one to each side of this proportion yields

PFt + PF2 PD, + PD2 PFX

= PD,

It was shown that PFy + PF2 = (a con

stant). Since P, Dlf and D2 are collinear by construction, then

PD1 + PD2 =

DlD2

is the distance between the parallel lines l? and l2 and is therefore a constant, k2. From

the preceding proportions we obtain

PF2 PF, PFt + PF2 kx PD2

~

PD1

~

PD, + PD2

"

k2 9

which is a constant, the eccentricity. To show that the eccentricity, e, is less

than one for the ellipse, we consider the case when is at a vertex of the ellipse. (The vertices are the points of intersection

:-183

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of the line FtF2 and the ellipse.) Suppose that is at the vertex between Ft and so that Dx coincides with Bx and D2 coin cides with B2. It is then apparent that

PF2 < PD2, and so

PF2

PD2

In most analytic developments of the el

lipse, it is given (but not proved) that the

eccentricity is the ratio of the distance from the center (midpoint of F^J to a focus over the distance from the center to a vertex. We leave the proof as an exercise and give the hint to consider the case when

is at a vertex. In conclusion observe that when the cut

ting plane is parallel to exactly one gener ator of the cone, a parabola is produced and that when it intersects both nappes of the cone (but does not contain the cone's

vertex), a hyperbola is determined. (Stu dents often erroneously conclude from models that the cutting plane must be

parallel to the cone's axis to produce a hy perbola.) The Dandelin-sphere method works in both cases, and the proof is similar to that of the ellipse. Figures 3 and 4 can be

copied and used for these proofs. In figure 3

plane a is parallel to the generator VA of

Fig. 3. Dandelin's spheres?parabolic case

Fig. 4. Dandelin's spheres?hyperbolic case

the cone. is a representative point on the curve of intersection of plane a and the cone. Assume that the curve shown through A and is the circle through whose

plane is perpendicular to the axis of the cone. Try doing these proofs and perhaps challenge some superior students to try them.

BIBLIOGRAPHY

Courant, Richard, and Herbert Robbins. What is Math ematics? New York: Oxford University Press, 1941.

Eves, Howard. A Survey of Geometry, Vol. 1. Boston:

Allyn & Bacon, 1963.

Partridge, Alden R. "Ellipses from a Circular and

Spherical Point of View." Two-Year College Math ematics Journal 14 (November 1983) :436-38. ?

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