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Conic Sections and Dandelin's Spheres
(A note to myself: Hidden code -- do not delete! --G.J.)

INSTRUCTIONS: click the "Open in Cocalc" link on the upper right to make the 3d graphics display. You may have to refresh your browser once or twice to load everything. Keep scrolling down until you reach the last frame, which says "The End."

Conic Sections and Dandelin's Spheres

A conic section, or "conic", is a curve obtained by intersecting a plane with a right circular cone. Let α\alpha be the angle between the cone's axis and any line that lies in the cone (a "generator"), and let β\beta be the angle between the cone's axis and the plane. (The plane is viewed "edge-on" in the figure.)

The shape of the conic depends on these angles, and whether or not the plane intersects the vertex of the cone.

If the plane intersects the vertex the conic is "singular"; it is an isolated point if α<β\alpha < \beta, two intersecting lines if α>β\alpha > \beta, or one line "counted twice" if α=β\alpha = \beta.

If the plane does not intersect the vertex the conic is "smooth". A smooth conic is

  1. an ellipse if α<β\alpha < \beta,

  2. a hyperbola if α>β\alpha > \beta, or

  3. a parabola if α=β\alpha = \beta.

I'll concentrate on the smooth conics. Below are some pictures. To rotate a picture select it then move the mouse holding down the left mouse button. Enlarge the picture with the mouse wheel on a PC, or "shrink or enlarge" finger motions on a mac. All the 3d pictures work better if one rotates and enlarges them.

Ellipse
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Parabola
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Hyperbola
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Focus and Directrix

The ellipse has two special points called foci. Dandelin constructed them by placing spheres inside the cone so that each sphere is tangent to the plane at a point, and tangent to the cone along a circle that is perpendicular to the axis of the cone. (To see that this always possible, put very small spheres inside the cone then inflate them until they are in the right position.) The foci are the points where the spheres touch the plane.

Foci of ellipse
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The foci of the ellipse have a special distance-to-foci property:

The sum of the distances from the foci to the point PP is the same for every point PP on the ellipse.

In the figure the purple length plus the green length is constant.

The hyperbola also has two foci, constructed in a similar way using Dandelin's spheres.

Foci of hyperbola
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The foci of the hyperbola have a special distance-to-foci property:

The difference of the distances from the foci to the point PP is the same for every point PP on the hyperbola.

In the figure, the purple length minus the green length is constant

The parabola has one special point, its focus, plus a directrix, which is a special line. As before, Dandelin constructed them by placing a sphere inside the cone, tangent to both the plane and the cone. The sphere is tangent to the plane at the focus and tangent to the cone along a circle. This circle lies in a plane that is perpendicular to the axis of the cone -- the green plane in the figure. The intersection of this plane with the plane containing the parabola is the directrix.

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The parabola's focus and directrix have a distance-to-focus property

The distance from the focus to PP equals the distance from PP to the directrix, for every point PP on the parabola.

In the figure the purple length equals the green length. Of course the distance to the directrix is measured along a perpendicular to the directrix.

The ancients knew these distance-to-foci properties and an easy corollary:

Corollary. Tangent lines and Angle bisectors:

Conic-section-shaped mirrors reflect light (sound, radio waves, ...) in a special way.

Ellipse: light (black line) leaving one focus reflects into the other focus.

Hyperbola: light (black line) traveling toward one focus reflects toward the other focus.

Parabola: light (black line) perpendicular to the directrix (green line) reflects into the focus

The reflection properties follow immediately from the fact that the tangent lines (blue) bisect the obvious angles in the following figures.

Proof that the angle bisector is tangent to the ellipse

The proof is based on the distance-to-focus property of the ellipse, which will be proved below. The proofs for the other conics are similar.

It's enough to show that the angle bisector (blue line) intersects the ellipse only at one point, PP.

Let QPQ\neq P be any other point on the blue line. Extend the segment PF2\overline{PF_2} to EF2\overline{EF_2} so the lengths of the green segments EF2=PF1EF_2 = PF_1 are equal and E,P,F2E,P,F_2 are collinear. The blue line bisects EPF1\angle EPF_1 so triangles ΔEPQΔF1PQ\Delta EPQ \cong \Delta F_1PQ are congruent (side-angle-side). QQ cannot be on the ellipse because the sum of the distances

F1Q+QF2=F2Q+QE>F2E=F1P+PE=F1P+PF2. F_1Q + QF_2 = F_2Q + QE > F_2E = F_1P + PE = F_1P + PF_2.

are not equal.

Finally ...

Dandelin's lovely proof of the Distance to Foci properties of conics.

Here is our hero, Germinal Pierre Dandelin (1794-1847), Belgian soldier and professor of engineering.

His proof uses a simple property of circles and tangent lines.

If PA\overline{PA} and PB\overline{PB} are tangent to a circle at AA and BB then the lengths

PA=PBPA = PB

are equal.

The red lengths are equal.

The same is true of spheres: if PA\overline{PA} and PB\overline{PB} are tangent to a sphere at AA and BB then

PA=PB.PA = PB.
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Proof of the distance-to-foci properties

Ellipse:

In the figure the yellow point is on the ellipse. The red points are foci. Each line segment (red or purple) connecting the yellow point to a focus is tangent to the corresponding sphere at the focus because the segment lies in the plane and the plane is tangent to the sphere at the focus. The spheres are tangent to the cone along the green and purple circles. The green and purple segments connecting the yellow point to the circles lie in the cone. The cone is tangent to the spheres, so the segments connecting the yellow point to the circles are tangent to the spheres at the green and purple points where the segments meet the circles. Therefore, because they are tangent to the same sphere, both green segments have the same length, and both purple segments have the same length.

Thus the purple length plus the green length is the same as the distance in the figure between the green point on the green circle and the purple point on the purple circle, along the line segment connecting these points in the cone. The cone, spheres, and circles are rotationally symmetric around the cone's axis, so the length of this line segment is the same no matter where the yellow point is located on the ellipse. In other words,

(purple length) + (green length) is constant.

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Hyperbola:

The proof is similar to the proof for the ellipse. In the figure the red points are foci, the yellow point is on the hyperbola, each spheres is tangent to the cone along a green or purple circle. The green segments connecting the yellow point to the focus and green circle have equal length. The purple segment connecting yellow point to the other focus has the same length as the green-and-purple segment connecting the yellow point to the purple point on the purple circle.
The difference in the distances, yellow point-on-hyperbola to foci, equals the length of the segment, green point to purple point, connecting the circles.

This length is just the distance between the circles, measured along a line in the cone, so it doesn't depend on the location of the yellow point.

Thus the difference beween the distances to the foci does not depend on where the yellow point is located on the hyperbola.

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Let

α\alpha = angle between the cone's axis and any cone generator

"green" = the green segment connecting parabola to circle

"black" = black segment, "blue" = segment

  1. "black" is parallel to the axis of the cone
  2. "green" lies on a generator of the cone, so
  3. α=(green, black)\alpha = \angle(\text{green, black})
  4. the plane is parallel to one of the cone's tangent planes, so
  5. "blue" is parallel to the generator in that tangent plane, so
  6. α=(blue, black)\alpha = \angle(\text{blue, black})

Thus their lengths (blue), (green), (black) satisfy

(blue)cos(α)\cos(\alpha) = (black) = (green)cos(α)0\cos(\alpha) \neq 0

Conclusion:

(blue) = (green)

A picture showing the blue and black lines translated to the x-z plane so the black line coincides with the cone's axis. The blue plane is parallel to the tangent plane along the purple generator.

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