The Dance of the Foci David Seppala-Holtzman St. Joseph’s College faculty.sjcny.edu/~holtzman

Introduction Take a cone and choose a point on its axis other than its vertex Construct a line, perpendicular to the axis, through this point Take the set of all planes that contain this line The intersection of all these planes with the cone will yield all the conic sections

This is Very Old News This has been known since the time of Apollonius (262 – 190 B.C.) Euclid Apollonius √ Pythagoras

New News What is not well known is the trajectories of the foci of these conics while the plane is rotating about this line Of particular interest will be how their paths articulate with one another as we pass from ellipse to parabola to hyperbola We have some good news to report on this front

The Plan We shall derive two pairs of parametric equations which trace out the trajectories of the two foci of the ellipse and the hyperbola We shall show how these trajectories articulate smoothly with the single focus of the parabola at the point of transition

The Set-up (1) We begin with a right-circular cone with vertex at the point (0,0,-1) and whose sides make an angle of φ (0 < φ < π/2) with the horizontal. We shall keep φ fixed throughout The equation of this cone is:

The Set-up (2) We will consider the set of planes that contain the x-axis These “cutting planes” will be parameterized by the angle, θ, that they make with the horizontal These planes will be given by the equations:

Consider the Ellipse

Ellipse (1) Here we have taken θ to be in the range 0 < θ < φ Let us take a closer look at the situation by removing the cone and plane and drawing the major axis of the ellipse along with its center, vertices and foci

Ellipse (2)

Ellipse (3) It is a simple matter to find the coordinates of the vertices. These are simply the points of intersection of the cutting plane, the cone and the yz-plane:

Ellipse (4) The center of the ellipse is easily derived from the coordinates of the vertices Let the horizontal coordinates of the vertices be denoted by y1 and y2. Then the coordinates of the center are:

Ellipse (5) This much is straight forward But how are we to derive the coordinates of the foci? Let us consider the diagram on the right:

Ellipse (6) In the diagram on the previous slide, we have drawn a horizontal line segment through the center of the ellipse and connected it to the vertices and foci with vertical line segments This generates two pairs of similar triangles: the larger ones have hypotenuses of length a and the smaller ones have hypotenuses of length c where a and c are the traditional symbols for the semi-major axis and the distance from the center to the foci, respectively

Ellipse (7) In other words, the ratio of the small triangles to the large ones is e = c/a, the eccentricity of the ellipse The key to our derivation is that one can also express the eccentricity in terms of the angles involved:

Ellipse (8) Using the eccentricity, we can calculate the horizontal and vertical displacement of the foci from the center of the ellipse as a function of θ (recall that φ is fixed) Compiling all of this and simplifying yields the four parametric equations on the next slide

The 4 Parametric Equations

Ellipse (9) The space coordinates of the left-hand focus of the ellipse are: (0, S(θ), T(θ)) The space coordinates of the right-hand focus of the ellipse are: (0, U(θ), V(θ))

Consider the Hyperbola

Hyperbola (1) Here we have selected a value of θ in the range φ < θ < π/2 The vertices and the center of the hyperbola are computed precisely as in the elliptical case Also, as in the elliptical case, we add several line segments that produce similar triangles relating the vertices and foci and the eccentricity

Hyperbola (2)

Hyperbola (3) We use the same method of derivation as in the elliptical case and, unsurprisingly, the same four parametric equations result The space coordinates of the upper focus of the hyperbola are: (0, S(θ), T(θ)) The space coordinates of the lower focus of the hyperbola are: (0, U(θ), V(θ))

Consider the Parabola

Parabola (1) In the parabolic case θ = φ Our 4 parametric equations are undefined in this case In addition, the methods that we previously employed will not work here A new approach is called for

Parabola (2) Our plan is to impose a new coordinate system, (x’, y’), on the cutting plane, one in which the equation for the parabola will be in standard form:

Parabola (3) In the standard equation on the previous slide, (h, k) gives the coordinates (in the x’y’-system) of the vertex of the parabola The value p in this equation denotes the distance from the vertex to the focus This will allow us to determine the location of the focus Consider the following coordinate system for the cutting plane:

Parabola (4)

Parabola (5) In the previous slide, we gave the cutting plane the following coordinate system: x’ = x y’ = y sec(φ) These were the two red lines in the diagram

Parabola (6) Intersecting the cone with the appropriate cutting plane and then translating into the x’y’-coordinate system yields the following equation for the parabola in standard form:

Parabola (7) From the previous equation, we deduce that p = cos2(φ)/2 sin(φ) This information, in turn, allows us to give the space coordinates of the focus of the parabola:

Parabola (8) Of immediate concern is how this location for the focus of the parabola articulates with the left-hand focus of the ellipse and the upper focus of the hyperbola We have good news:

Parabola (9) Thus, the sole focus of the parabola is precisely the point toward which the left-hand focus of the ellipse and the upper focus of the hyperbola were converging

The Dance of Foci Although we have only calculated the trajectories of the foci of the conics in the range 0 < θ < π/2, we see that our work is done The second half, π/2 < θ < π, is merely the mirror image (about the z-axis) of the first half The geometry of the situation makes this obvious but it is nice to see that our parametric equations corroborate this. Observe:

The Dance Part 1

The Dance Part 1 In the previous slide, we let θ range from 0 to φ. The blue curve (S, T) gave us the left-hand focus of the ellipse and the green (U, V) gave us the right-hand focus The red circle locates the single focus of the parabola, the point to which the blue curve is converging

The Dance Part 2

The Dance Part 2 In the previous slide we let θ range from 0, through φ and on to π/2 After passing through the focus of the parabola, the blue curve (S, T) became the upper focus of the hyperbola The green curve (U, V) escaped to infinity in the first part of the dance and now returns from negative infinity to trace out the lower focus of the hyperbola

The Dance Part 3

The Dance Part 3 In the previous slide, we let θ continue on to π – φ where the conic is a parabola once more The blue curve (S, T) now traces out the lower focus of the hyperbola and the green (U, V) gives the upper As in the first part of the dance, the upper focus converges on the parabola’s focus

(S, T) in Summary The blue curve represents (S(θ), T(θ)). It gives, in turn, the left focus of the ellipse, the focus of the parabola (as a limiting value in the third quadrant), the upper focus of the hyperbola, the vertex of the cone, the lower focus of the hyperbola, and, after going off to infinity, the left focus of the ellipse

(U, V) in Summary The green curve represents (U(θ), V(θ)). It gives, in turn, the right focus of the ellipse, the escape to infinity, the lower focus of the hyperbola, the vertex of the cone, the upper focus of the hyperbola, the focus of the parabola in the fourth quadrant (as a limiting value) and the right focus of the ellipse

The Dance in its Entirety