One of the major activities of mathematicians and scientists of all kinds is learning, and attempting to prove, how and why things work.

Since Ancient Greece at the beginning of the millennium, and likely well before, the properties of conic sections have been studied. One that has been explored extensively is the "Focal Property", which we have referred to in nearly every lesson in this section.

There are a number of mathematical proofs of the focal property, but trying to explain such proofs to someone who does not dedicate his or her life to mathematics is difficult. The Dandelin Spheres are one solution.

### Conic Sections and Dandelin Spheres

#### Dandelin Spheres and the Equation of an Ellipse

It wasn’t until 1822 that the French mathematician Germinal Dandelin thought of this very clever construction. Dandelin found a way to find the foci and prove the focal property in one fell swoop.

Take the conic section in question. Then choose a sphere that is just the right size so that when it’s dropped into the conic, it touches the intersecting plane, as well as being snug against the cone on all sides. If you prefer, you can think of the sphere as a perfectly round balloon that is blown up until it “just fits” inside the cone, still touching the plane. Then do the same on the *other* side of the plane. After we’ve drawn both of these spheres we have this picture:

or

These spheres are often called “Dandelin spheres”, named after their discoverer. It turns out that not only is our shape an ellipse (which, like all ellipses satisfies the focal property), but these spheres touch the ellipse exactly at the two foci. To see this, consider this geometric argument.

The first thing to notice is that the circles \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*} shown on the diagram below, where each sphere lies snug against the cone, lie in parallel planes to one another. In particular, each line passing through these circles and the vertex of the cone, such as the line \begin{align*}l\end{align*} drawn below, cuts off equal segments between the two circles. Let's call \begin{align*}d\end{align*} the shortest distance along the cone between circles \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*}. This can also be thought of as the shortest distance between \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*} that passes through the vertex of the cone.

The next thing to remember is a property of tangents to spheres that you may have learned in geometry. If two segments are drawn between a point and a sphere, and if the line containing each segment is tangent to the sphere, then the two segments are equal. In the diagram below, \begin{align*}AB=AC\end{align*}. (This follows from the fact that tangents are perpendicular to the radii of a sphere and that two congruent triangles are formed in this configuration.)

Now consider the point \begin{align*}P\end{align*} on the ellipse drawn below. Let \begin{align*}\overline{QR}\end{align*} be the segment of length \begin{align*}d\end{align*} between \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*} that passes through \begin{align*}P\end{align*}. The distances between the two foci are marked \begin{align*}d_1\end{align*} and \begin{align*}d_2\end{align*}. But \begin{align*}d_1 =RP\end{align*} and \begin{align*}d_2 =PQ\end{align*} by the property of tangents to spheres discussed above. So \begin{align*}d_1 + d_2 = RP + PQ = QR = d\end{align*}. And this sum will always equal \begin{align*}d\end{align*}, no matter what point \begin{align*}P\end{align*} on the ellipse is chosen. So this proves the focal property of ellipses: that the sum of the distances between any point on the ellipse and the two foci is constant.

#### Dandelin and the Parabola

Like the ellipse, the parabola has a focal property. And, also like the ellipse, a construction similar to Dandelin’s with the spheres can show us what it is. Dandelin himself didn’t prove the focal property for parabolas that we are about to discuss, but Pierce Morton used a sphere construction similar to Dandelin’s to prove the focal property of parabolas in 1829. We’ll look at Morton’s argument here.

In contrast with the argument we made for the ellipse, for a parabola we can only fit one tangent sphere inside the cone. That is, only one sphere can be tangent to both the cone and the cutting plane. In the diagram below, the sphere fits underneath the cutting plane, but there is no room for a sphere to lie on top of the cutting plane and still be tangent to the cone.

As with the ellipse, the point where the sphere intersects the plane is called a **focus**. But because there is only one sphere in this construction, and this is related to the fact that a parabola has only one focus. The other geometric object of interest is called the **directrix**. This is the line that results from the intersection between the cutting plane and the plane that contains the circle of contact between the sphere and the cone. In the diagram below, the directrix is labeled \begin{align*}l\end{align*} and is found by intersecting the plane defined by circle \begin{align*}C\end{align*} and the cutting plane (the planes are shown in dashed lines for clarity). Finally, we will call the angle between the planes \begin{align*}\theta\end{align*}.

In the above diagram, we have labeled the point where the sphere contacts the cutting plane with \begin{align*}F\end{align*}, and we’ll call that point the focus of the parabola. Suppose \begin{align*}P\end{align*} is an arbitrarily chosen point on the parabola. Then, let \begin{align*}Q\end{align*} be the point on circle \begin{align*}c\end{align*} such that \begin{align*}\overline{P Q}\end{align*} is tangent to the sphere. In other words \begin{align*}Q\end{align*} is chosen so that \begin{align*}\overline{P Q}\end{align*} lies on the cone itself. Let \begin{align*}L\end{align*} be the point on the directrix \begin{align*}l\end{align*} such that \begin{align*}\overline{P L}\end{align*} is perpendicular to \begin{align*}l\end{align*}. Then \begin{align*}PF = PQ\end{align*} since both segments are tangents to the sphere from the same point \begin{align*}P\end{align*}. We can also show that \begin{align*}PQ=PL\end{align*}. This follows from the fact that the cutting plane is parallel to one side of the cone. Consider the point \begin{align*}P'\end{align*} that is the projection of \begin{align*}P\end{align*} onto the plane containing circle \begin{align*}C\end{align*}. Then \begin{align*}\angle \ PP'Q\end{align*} and \begin{align*}\angle \ PP'L\end{align*} are both right angles by the definition of a projection. \begin{align*}\angle \ PQP'\end{align*} and \begin{align*}\angle \ PLP'\end{align*} are both equal to the angle \begin{align*}90 - \theta\end{align*} , where \begin{align*}\theta\end{align*} is the angle defined above, because the cutting plane and the cone both have an angle of \begin{align*}\theta\end{align*} with the horizon. Since they also share a side, triangles \begin{align*}\triangle PQP'\end{align*} and \begin{align*}\triangle PLP'\end{align*} are congruent by \begin{align*}AAS\end{align*}. So the corresponding sides \begin{align*}\overline{P Q}\end{align*} and \begin{align*}\overline{P L}\end{align*} are congruent. By the transitive property we have \begin{align*}PF=PL\end{align*}, so the distance between the point \begin{align*}P\end{align*} on the parabola to the focus is the same as the distance between \begin{align*}P\end{align*} and the directrix \begin{align*}l\end{align*}. We have just proven the focus-directrix property of parabolas.

#### Dandelin Spheres and Hyperbolas

To prove the focal property of hyperbolas, we examine Dandelin’s sphere construction. Unlike the construction for ellipses, which used two spheres on one side of the cone, and the sphere construction for parabolas, which used one sphere on one side of the cone, this construction uses two spheres, one on each side of the cone. As with the ellipse construction, each sphere touches the plane at one of the foci of the hyperbola. And as with the argument for the elliptical focal property, the argument uses the fact that tangents from a common point to a sphere are equal.

In the above diagram, suppose \begin{align*}P\end{align*} is an arbitrary point on the hyperbola. We would like to examine the difference \begin{align*}PF_2 - PF_1\end{align*}. Let \begin{align*}C_1\end{align*} be the point on the upper sphere that lies on the line between \begin{align*}P\end{align*} and the vertex of the cone. Let \begin{align*}C_2\end{align*} be the point on the upper sphere when this line is extended (so \begin{align*}P\end{align*}, \begin{align*}C_1\end{align*}, and \begin{align*}C_2\end{align*} are all on the same line and \begin{align*}PC_1 + C_1 C_2 = PC_2\end{align*} and the cone By the common tangent property, \begin{align*}PF_1 = PC_1\end{align*} and \begin{align*}PF_2 = PC_2\end{align*} for some points \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*} on the circles where the spheres meet the cone. So \begin{align*}PF_2 - PF_1 = PC_2 - PC_1 = (PC_1 + C_1 C_2) - PC_1 = C_1 C_2\end{align*}. But \begin{align*}C_1 C_2\end{align*} is the distance along the cone between the two circles of tangency and is constant regardless of the choice of \begin{align*}C_1\end{align*} and \begin{align*}C_2\end{align*}. So the difference \begin{align*}PF_2 - PF_1\end{align*} is constant.

### Examples

#### Example 1

What would the Dandelin spheres look like if used to explore a circle?

Since the spheres touch the figure within just at the foci, and since a circle is an ellipse with both foci at the same point, the spheres would sit directly above each other, and would touch the circle at the center point on both sides.

#### Example 2

Conceptually speaking, why is there only one Dandelin Sphere used in the proof of a parabola?

As we learned in the proof for an ellipse, the Dandelin sphere must be tangent to both the cone, e.g., the sphere touches the cone all the way around in a circle, and the plane, where the tangency is a single point. In the case of a parabola, the sphere below the plane "fits" just fine, but a sphere above the plane would "sit on" the plane, and not touch the cone all the way around. The figure only allows for one sphere to be tangent.

#### Example 3

Explain why for any positive number \begin{align*}b\end{align*} and \begin{align*}a\end{align*}, there exists a \begin{align*}c\end{align*} such that \begin{align*}b^2=c^2-a^2\end{align*}. Let \begin{align*}c = \sqrt{a^2+b^2}\end{align*}.

Since \begin{align*}a^2 + b^2\end{align*} is always positive for positive \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, this number is always defined. Geometrically, let \begin{align*}c\end{align*} be the hypotenuse of a right triangle with side lengths \begin{align*}a\end{align*} and \begin{align*}b\end{align*}.

### Review

- Who was the first mathematician who conceptualized dandelin spheres? When?
- What is proved by dandelin spheres?
- How do you identify the foci of an ellipse using dandelin spheres?
- If two tangents are drawn from a single point to a sphere, what can you say about the line segments formed?
- How do the tangents relate to a radius of a sphere?
- Describe the focal property of ellipses.
- How and when did Germinal Dandelin prove the focal property for parabolas?
- What is the line that results from the intersection between the cutting plane and the plane that contains the circle of contact between the sphere and cone?
- What is defined by the point where the sphere intersects the cutting plane?
- Which construction uses two spheres in a single cone?
- Which construction uses one sphere in a single cone?
- Which construction uses two spheres and two cones?

Identify the parts listed on the diagram as specified below:

- Directrix Line - Small Sphere
- Directrix Line - Large Sphere
- Focus - Small Sphere
- Focus - Large Sphere
- Vertex - Small Sphere
- Vertex - Large Sphere
- Directrix Plane - Small Sphere
- Directrix Plane - Large Sphere
- Cutting Plane
- What conic section is illustrated here?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.8.