# Dandelin Spheres Theory

Dandelin Spheres
Field: Geometry
Image Created By: Hollister (Hop) David
Website: Hop's Gallery

Dandelin Spheres

This image shows a head floating on the ocean surface with a funny cone-shaped hat. A round fish is kissing the ocean surface under the water in the hat. The hat intersects the ocean surface in a "Conic Section", which in the image is an ellipse.

# Basic Description

Mathematicians see this image more from a mathematical perspective. We can view the water level inside the cone as being an ellipse created by the intersection of a cone and a plane, and the fish and the man's head as being spheres that are tangent to the cone and to the ellipse.

In geometry, the Dandelin spheres are one or two spheres that are tangent both to a plane and to a cone that intersects the plane. The intersection of the cone and the plane is a conic section, and the point at which either sphere touches the plane is a focus of the conic section, so the Dandelin spheres are also sometimes called focal spheres. See Figure 1.

In addition, Dandelin Spheres provide a way of looking at the three-dimensional definition of a Conic Section, and relationg it to a two-dimensional definition in terms of foci and focal constants.

Since the Dandelin Spheres are created by conic sections, the types of Dandelin Spheres can be categorized by different types of conic sections:

## History

Dandelin Spheres are named after the Belgian mathematician and military engineer Germinal Pierre Dandelin (1794–1847). Adolphe Quetelet is sometimes given partial credit as well. Conic sections are defined by the intersection of a plane with a cone. Their more useful definitions, however, are those from plane geometry. Dandelin published an article, in which he showed the importance of the intersections created by spheres tangent to the cone and a conic section. He also gave an elegant proof that the spheres are tangent to the conic section at its foci.

The Dandelin spheres can be used to prove at least two important theorems. Both of those theorems were known for centuries before Dandelin, but he made it easier to prove them.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Geometry, Algebra and Basic Knowledge of Conics Sections

## Two Theorems

Sum of Distances to Foci Property, the first theorem proven by Dandelin Sphe [...]

## Two Theorems

Sum of Distances to Foci Property, the first theorem proven by Dandelin Spheres is that a closed conic section (i.e. an ellipse) is the of points such that the sum of the distances to two fixed points (the foci) is constant. This was known to Ancient Greek mathematicians such as Apollonius of Perga, but the Dandelin spheres facilitate the proof.
Focus-Directrix Property, the second theorem proven by Dandelin Spheres is that for any conic section, the distance from a fixed point (the focus) is proportional to the distance from a fixed line (the directrix), the constant of proportionality being called the eccentricity. Again, this theorem was known to the Ancient Greeks, such as Pappus of Alexandria, but the Dandelin spheres facilitate the proof.
The directrix of a conic section can be found using Dandelin's construction. Each Dandelin sphere intersects the cone at a circle; let both of these circles define their own planes. The intersections of these two planes with the conic section's plane will be two parallel lines; these lines are the directrices of the conic section. However, a parabola has only one Dandelin sphere, and thus has only one directrix.
Using the Dandelin Spheres, it can be proved that any conic section is the locus of points for which the distance from a point (focus) is proportional to the distance from the directrix. Neither Dandelin nor Quetelet used the Dandelin spheres to prove the focus-directrix property. The first to do so was apparently Pierce Morton in 1829. The focus-directrix property is essential to proving that astronomical objects move along conic sections around the Sun.
Germinal Pierre Dandelin employs spheres inscribed in a cone which touch the intersecting plane in two points, the foci of the conic section. In what follows, both of Dandelin's proofs are presented.
This will be shown in Explore Different Dandelin Spheres.

## Explore Different Dandelin Spheres

Explore Different Dandelin Spheres, we will start from prove the two definitions with different Dandelin Spheres.

We can use the Dandelin Spheres Theory to help to prove that when using a plane to cut a cone, the plane creates a conic section. The theory can also help to prove that the intersections of the spheres with the plane are foci. All we need to do is to show that the conic section under consideration satisfies the definition of a circle, an ellipse, a hyperbola, or a parabola.

Use this Applet to play with Dandelin Spheres.

### Spheres Tangent To A Circle

More information about Dandelin Spheres that are tangent to a circle.

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In Figure 2, when the plane intersects all generators of the cone, it is possible to inscribe two spheres which will touch the conical surface and the plane.

The upper sphere centered at point $C$ touches the cone surface in a circle $k$ and the plane at a point $F$. The lower sphere centered at point $C'$ touches the cone surface in a circle $k'$ and also touches the plane at the point $F$.

The arbitrary chosen generating line $TT'$ intersects the circle $k$ at a point $T$, the circle $k'$ at a point $T'$, and the intersection curve at a point $P$.

We see that points $T$ and $F$ are the tangency points of the upper sphere, and points $T'$ and $F$ are the tangency points of the lower sphere.

Focus-Directrix Property

In this part we will show how Dandelin proved the Focus-Directrix Property:

Eq. 1         $TP = PF$.

By using the same method, we can get:

Eq. 2         $T' P = PF$.

Detail of proof for the circle.

The image Proof of the Sphere shows a sphere centered at point $O$.
Proof of the Sphere
Line $PA$ is tangent to the sphere at point $A$, and line $PB$ is tangent to the sphere at point $B$. Because the lines are tangent to the sphere, $\angle OAP$ and $\angle OBP$ are right angles. The two triangles $\triangle OAP$ and $\triangle OBP$ share the side $OP$. Sides $OA$ and $OB$ are radii of the sphere, so $OA=OB$. Therefore, the two triangles are congruent triangles.

Thus, $AP=BP$.

Imagine all such lines that start at P and touch the sphere. All these lines will form a conical cap of the sphere. It is then easy to see that the distances from $P$ to all the points of contact at the sphere are equal. This means $AP=BP$.

Sum of Distances to Foci Property

Prove: When either of the two right circular conical surfaces are intersected with the plane perpendicular to the axis of the cone, the resulting intersection is a circle.

From Eq. 1 and Eq. 2, we can get:

Eq. 3         $TP + T' P = TT' = 2 PF$.

Since point $P$ is an arbitrary point on the conic section, and the length of $TT'$ is constant, $PF$ must be constant. Since $PF$ is the distance from a point to the edge of the conic section, we know that the conic section has a constant radius, and is therefore a circle."

We can also see that the circle is a special case of the ellipse where the plane is perpendicular to the axis of the cone.

### Spheres Tangent To An Ellipse

More information about Dandelin Spheres that are tangent to an ellipse.

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In Figure 3, when a plane intersects all generators of the cone, it is possible to inscribe two spheres which will touch the conical surface and the plane.

The upper sphere $G_1$ touches the cone surface in a circle $k_1$ and the plane at a point $F_1$. The lower sphere$G_2$ touches the cone surface in a circle $k_2$ and the plane at a point $F_2$.

The arbitrary chosen generating line $S P_2$ intersects the circle $k_1$ at a point $P_1$, the circle $k_2$ at a point $P_2$ and the intersection curve $E$ at a point $P$.

We see that points $P_1$ and $F_1$ are tangency points of the upper sphere and points $P_2$ and $F_2$ are tangency points of the lower sphere.

Sum of Distances to Foci Property

Inscribe two spheres $G_1$ and $G_2$ inside the cone in such a way that they are tangent to both the cone and the plane containing the conic section.

Since sphere $G_1$ is tangent to the cone and the intersection $k_1$ of the sphere and the cone is a circle containing point $P_1$, line $P_1 P$ is tangent to sphere $G_1$. The sphere is also tangent to the conic section at point $F_1$, so the line $F_1 P$ on the conic section is also tangent to the sphere. Because of the proof we gave above in Proof of the Sphere, two tangent line sections of the same sphere are equal.

Eq. 1         $P_1 P = F_1 P$.

Using the same method, we can get:

Eq. 2         $P_2 P = F_2 P$.

Focus-Directrix Property

Prove: When the cutting plane is inclined to the axis of the cone at a greater angle than that made by the generating segment or generator, the resulting curve is an ellipse.

Since

Eq. 1         $P_1 P = F_1 P$
Eq. 2         $P_2 P = F_2 P$,
and Points $P$, $P_1$, and $P_2$ are on one line,
Using Eq. 1 $+$ Eq. 2, we can get:
Eq. 3 $P_1 P + P_2 P = F_1 P + F_2 P$.
Eq. 4 $P_1 P_2 = F_1 P + F_2 P$.

$P_1 P_2$ is the line section on the cone between circle $k_1$ and circle $k_2$.

1. Spokes from the upper sphere’s hat band are all equal because they are tangent lines sharing a far point.
2. Spokes from the lower sphere’s belt are all equal length for the same reason.
3. Take away hat spokes from belt spokes and you're left with lamp shade spokes which are all equal.

$P_1 P_2$ is constant, and $F_1 P + F_2 P$ is constant too.

Since the intersection curve is the locus of points in the plane for which the sum of distances from the fixed points $F_1$ and $F_2$ is constant, the intersection curve $E$ is an ellipse.

The sum of the distances from any point $P$ on the ellipse to those two foci is constant and equal to the major diameter, $P F_1 + P F_2 = 2 a$. Each of these two points is called a focus of the ellipse.

By the definition of an ellipse, any two points $F_1$ and $F_2$ that make $P_1 P_2 = P F_1 + P F_2$ constant for all points $P$ on the ellipse are the foci of the ellipse.

### Spheres Tangent To A Hyperbola

More information about Dandelin Spheres that are tangent to a hyperbola.

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When the intersecting plane is inclined to the vertical axis at a smaller angle than that of the generator of the cone, the plane $\pi$ cuts both cones creating a hyperbola, which is shown in Figure 4.

Inscribed spheres centered at $C'$ and $C$ touch the plane from the same side at points $F'$ and $F$, and the cone surface at circles $k'$ and $k$.

The generator intersects the circles $k'$ and $k$ at points $T'$ and $T$, and the intersection curve at the point $P$.

By rotating the generator around the vertex $V$ by 360 degrees, the point $P$ will move around and trace both branches of the hyperbola.

While rotating, the generator will coincide with the plane two times and then will have common points with the curve only at infinity.

Sum of Distances to Foci Property

Insert the upper sphere in the upper of the cone so that the sphere is tangent to the cone and to the upper part of the plane containing the conic section. Insert the lower sphere in the lower nappe of the cone so that the sphere is tangent to the cone and to the lower part of the plane containing the conic section.

Point $P$ is an arbitrary point on the conic section. Line $T' P$ goes through the two spheres, and intersects the upper sphere at point $T'$. The upper sphere intersects the plane containing the conic section at point $F'$, and intersects the cone in a circle $k'$. The lower sphere intersects the plane containing the conic section at point $F$, and intersects the cone in a circle $k$. Using the same technique shown in Proof of the Sphere, since Points $T'$ and $F'$ are on the sphere centered at $C'$, we can get:

Eq. 1         $F' P=T' P$.

Since Points $T$ and $F$ are on the sphere centered at $C$,

Eq. 2         $F P=T P$.

Focus-Directrix Property

Prove: When the cutting plane is inclined to the axis at a smaller angle than the generator of the cone, the resulting curve is a hyperbola.

Since the planes of circles $k'$ and $k$ are parallel, all generating segments from $k'$ to $k$ are of equal length.

Since

Eq. 1         $F' P=T' P$
and
Eq. 2        $F P=T P$,
and points $P$, $T'$, and $T$ are on one line,
Using Eq. 2 $-$ Eq. 1, we can get:
Eq. 3         $|F P - F' P|=|T P - T' P|$,
Eq. 4         $T' T = |F P - F' P|$.

$T' T$ is the line section on the cone between circle $k'$ and circle $k$.

$T' T$ is constant, so $|F P - F' P|$ is constant too.

Since the intersection curve is the locus of points in the plane for which the difference of the distances from the two fixed points $F'$ and $F$ is constant, the conic section curve is a hyperbola.

A hyperbola may be defined as the locus of points where the absolute value of the difference of the distances to the two foci is a constant equal to $2a$, the distance between its two vertices.

Since $F P - F' P$ is constant, Point $F'$ and point $F$ are the foci of the hyperbola.

### Sphere Tangent To A Parabola

More information about Dandelin Spheres that are tangent to a parabola.

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See Figure 5. An inscribed sphere, centered at $C$, touches the plane at point $F$ and the cone surface at circle $k$.

The generator intersects the circle $k$ at point $T$, and the intersection curve at point $P$.

By rotating the generator around the vertex $V$ by 360 degrees, the point $P$ will move around the conic section.

While rotating, the generator will coincide with the plane two times and then will have common points with the curve only at infinity.

Sum of Distances to Foci Property

The sphere is centered at point $C$ situated below the cone vertex $V$. The sphere touches the surface of the cone from the inside in a circle $k$ and it also touches the intersecting plane at a point $F$. In addition to the intersecting plane there is also another plane displayed on the image. It is the plane in which circle $k$ is embedded. The intersection of this plane with the intersection plane is a line $p$.

$P$ is an arbitrary point on the conic section.

Point $T$ is the intersection point of the line $PV$ with circle $k$. The point $Q$ is the perpendicular projection of point $P$ to the plane of circle $k$. The point $S$ is the foot of a perpendicular from point $P$ to line $p$.

This creates two triangles: $\triangle PQT$ and $\triangle PQS$.

As we showed in Proof of the circle, because points $F$ and point $T$ are on the sphere, we know that:

Eq. 1         $|PF| = |PT|$.

Focus-Directrix Property

Prove: If the cutting plane is parallel to any generator of one of the cones, then the intersection curve is the parabola.

This is different from the previous cases, in that we cannot use Eq. 1 $|PF| = |PT|$ directly to get the answer. If we want to prove that the conic section is a parabola, we need to prove that $|PT| = |PS|$.

Lines $PT$ and $PS$ are hypotenuses of two right triangles $\triangle PQT$ and $\triangle PQS$. These two triangles are congruent since they have one identical side $PQ$ and two congruent angles.

Their right angles are congruent and also $\angle PTQ$ is congruent to $\angle PSQ$.

Click to see why the two angles are congruent

Why are the two last angles congruent?

If one wants parabola to emerge as a conic section, the intersecting plane has to be very special. Its inclination angle to the horizontal plane has to be the same as the inclination angle of any line lying on the conical surface going thru its vertex $V$.

The angles PTQ and PSQ are equal to that angle.

Since the two triangles are congruent, we can make the conclusion that:

Eq. 2         $|PF| = |PT| = |PS| =$ the distance from point $P$ to the line $p$.

It means that the distance of arbitrary point $P$ of the investigated curve from point $F$ is equal to its distance from line $p$.

Therefore, the investigated curve is a parabola and line $p$ is its directrix.

There is another way to prove the sphere touches the parabola at the focus.

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Sum of Distances to Foci Property

See Figure 6. As the generator approaches the position needed to be parallel to the intersection plane, the point $P$ moves far away from $F$. This shows the basic property of the parabola that the line at infinity is a tangent.

Since the segments $PF$ and $PT$ belong to tangents drawn from $P$ to the sphere, we can get:

Eq. 1         $PT = PF$.

Since the planes of the circles $k$ and $k'$ are parallel to each other and perpendicular to the section through the cone axis, and since the intersection plane is parallel to the slanting edge $VB$, the intersection $d$, of planes $E$ and $K$, is also perpendicular to the section through the cone axis.

$PN$ is perpendicular from $P$ to the line $d$. Thus,

Eq. 2         $PN = BA = PT$, or
Eq. 3         $PF = PN$.

Focus-Directrix Property

Prove: If the cutting plane is parallel to any generator of one of the cones, then the intersection curve is the parabola.

We get Eq. 2 and Eq. 3 from the above proof. Therefore, for any point $P$ on the intersection curve the distance from the fixed point $F$ is the same as it is from the fixed line $d$, proving that the intersection curve is the parabola.

Bcause of the definition of parabola, given a point (the focus) and a corresponding line (the directrix) on the plane, the locus of points in that plane that are equidistant from them is a parabola. The line perpendicular to the directrix and passing through the focus.

Since $PN = BA = PT$ and $PF = PN$,

we can draw the conclusion that the tangent point is the focus of the parabola.

# Applied Dandelin Spheres

Applied Dandelin Spheres is always related to conic sections, and the most useful way to facilitate some previous complicated proofs, and even related [...]

## Conic Section Eccentricity

The four defining conditions above can be combined into one condition that depends on a fixed point $F$ (the focus), a line $L$ (the directrix) not containing $F$ and a nonnegative real number $e$ (the eccentricity). The corresponding conic section consists of the locus of all points whose distance to $F$ equals $e$ times their distance to L. For $0 < e < 1$ we obtain an ellipse, for $e = 1$ we get a parabola, and for $e > 1$ we get a hyperbola.

Explore eccentricity of different conic sections

Ellipse's Eccentricity

Drop a hinge line straight off of the hat brim to a point on ellipse. Hinge line is a leg of both the blue and the red right triangles. It forms a constant angle with the cone wall (on which the blue hypotenuse rests) and a constant angle with the cutting plane (on which the red hypotenuse rests). The blue hypotenuse is the point's distance to focus. The red hypotenuse is thepoint's distance to directrix line. Since the triangles remain the same shape and share a leg, the ratio of the blue hypotenuse to the red hypotenuse remains constant. This constant is called the ellipse's eccentricity.

An ellipse's eccentricity is between 0 and 1. A circle's eccentricity is called zero although its directrix line is not defined.

Hyperbola's Eccentricity

As with the ellipse, extend a perpendicular hinge from the hat brim to a point on the hyperbola. The hinge is a shared leg of 2 right triangles. The blue hypotenuse is the distance to the focus and the red hypotenuse is the distance to the directrix line. As with the ellipse, the ratio of the blue to the red hypotenuse is constant and this constant is the hyperbola's eccentricity.

A Hyperbola's eccentricity is always greater than 1.

Parabola's Eccentricity

Dropping the hinge from the hat brim we can see the blue hypotenuse is the same as the red hypotenuse. Distance to directrix $=$ Distance to focus

and so the parabola's eccentricity is exactly 1.

## Newton's Astronomy Theory

Dandelin Spheres help to prove that all astronomical objects move along conic sections move around the Sun.

In A simple Cartesian treatment of planetary motion by Andrew T Hyman in 1992, the author illustrates taht two famous theorems are proved in his book in a much simplier manner. It is proved that planets pursuing Keplerian trajectories have accelerations which conform to Neton's central $\frac{1}{R^2}$ equation. Then it is proved that, conversely, planetary orbits must be Keplerian if Neton's central $\frac{1}{R^2}$ equation holds true.

In the book, Hyman talks that Kepler introduced the first two laws in his 1609 Astronomia Nova. The third or "harmonic" las was suggested in his 1619 Harmonice Mundi and is often stated in terms of the length $a$ of the semimajor axis ($a$ is half the orbit's greatest width).

The discovery of Kepler's laws was the greatest advance since Aristarchus deduced nineteen centuries earlier that planets circle the Sun (see Heath 1981).

Recall that ellipses are the closed curves formed by intersecting a circular cone and a plane. The ancient Greeks proved (see Heath 1981) that, everywhere along an ellipse, the distance to a point (the focus) divided by the distance to a line (the directrix) is a constant eccentricity $e$. A beautiful Proof of this focus-directrix Property was devised in 1822 by G P Dandelin, for both open $e \geqslant 1$ and closed $0 \leqslant e < 1$ conic sections (see Shenk 1977 or Tho- mas and Finney 1984).

This is a perfect example that Dandelin Spheres Theory helps to facilitate some complicated theorems.

## Interesting Math problems

The Senior Second Round of the 2006 Harmony SA Mathematics Olympiad contained an interesting problem in Question 20. For example, consider . If $AE = 3$, $DE = 5$, and $CE = 7$, then $BF$ equals:

(a) 3.6

(b) 4.0

(c) 4.2

(d) 4.5

(e) 5.0

Before reading further the reader is encouraged to perhaps stop and first try to solve this problem before reading further.

SA Mathematics Olympiad provides the answer is (c).

By looking at the angles we see that triangles...

By looking at the angles we see that triangles $\triangle AED$ and $\triangle BFA$ are similar, as are triangles $\triangle CFB$ and $\triangle DEC$.

Therefore,

$\frac{BF}{AF} = \frac{AE}{DE}= \frac{3}{5}$

and

$\frac{BF}{CF} = \frac{CE}{DE} = \frac{7}{5}$

so

$\frac{CF}{AF} = \frac{3}{7}$.

But

$CF + AF = CE + AE = 10$

so

$AF = 3$

and

$AF = 7$.

Therefore,

$BF = \frac{BF}{AF} * AF = \frac{3}{5} * 7 = 4.2$.

However, the problem and solution masks an interesting underlying theorem, Dandelin Spheres. Since there are right angles at $F$ and $E$, we can draw two circles centred at $B$ and $D$ respectively as shown in Figure7. Assume the two circles are of different size with circle $B$ smaller than circle $D$. Then draw tangents from $A$ and $C$ to circles $B$ and $D$ respectively.

This is therefore another good illustrative example, as will be discussed further in De Villiers, of where a 2D result can be proved much more easily by considering it as a special case of a 3D result!

# Why It's Interesting

A video helps to understand the Dandelin Spheres.

This video shows the effect of light, which causes the ball's shadow and creates different Dandelin Spheres.

Other interesting Dandelin Spheres images from Hop's Gallery

# Teaching Materials

There are currently no teaching materials for this page. Add teaching materials.

# References

[1] Heath, Thomas. A History of Greek Mathematics, Clarendon Press, 1921.

[2] Kendig, Keith. Conics, Cambridge University Press, 2005.

[3] Taylor, Charles. An Introduction to the Ancient and Modern Geometry of Conics, Deighton, Bell and co., 1881.

[4] Hyman, Andrew. A Simple Cartesian Treatment of Planetary Motion, European Journal of Physics, Vol. 14, 1993.

[5] De Villiers, Michael. Further reflection on a SA Mathematics Olympiad Problem, Teaching & Learning Mathematics, No. 4, Feb 2007, page. 25-27.

[6] Wikipedia. (n.d.). Dandelin spheres. Retrieved from http://en.wikipedia.org/wiki/Dandelin_spheres

[7] Tuleja, Slavomir and Hanc, Jozef. Java applet JDandelin. 2002. Retrieved from http://www.liceomendrisio.ch/~marsan/matematica/materiale_vario/coniche/Dandelin/JDandelinEn.htm

[8] Dandelin Spheres. From MedicBD Health. Retrieved from http://www.health.medicbd.com/wiki/Dandelin_spheres#Proof_that_the_curve_has_constant_sum_of_distances_to_foci

[9] Dandelin’s Spheres - proof of conic sections focal properties. From Nabla. Retrieved from http://www.nabla.hr/Z_IntermediateAlgebraConicsFamilyOfSimilarlyShapedCurve_2.htm

[10] Conics, a family of similarly shaped curves - properties of conics. From Nabla. Retrieved from http://www.nabla.hr/Z_CollegeConicsFamilyOfSimilarlyShapedCurveProperties_2.htm

[12] Wikipedia. (n.d.). Hyperbola - Conic section. Retrieved from http://en.wikipedia.org/wiki/Hyperbola#Conic_section

[13] Wikipedia. (n.d.). Conic section. Retrieved from http://en.wikipedia.org/wiki/Conic_section

[14] Wikipedia. (n.d.). Parabola - Derivation of the focus. Retrieved from http://en.wikipedia.org/wiki/Parabola#Derivation_of_the_focus

[15] David, Hollister. (n.d.). Dandelin Spheres. From Hop’s Gallery. Retrieved from http://clowder.net/hop/Dandelin/Dandelin.html

Part of Circle proofs.
Need to add some in applied dandelin spheres.
Work on the References!

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