LAST EDITED ON Nov-27-03 AT 12:29 PM (EST)
 
Hello Mike,

Analytical geometry usually means a brute force, but it does not take too long to prove that when a plane intersects a cone in such a way that the intersection is a closed curve, the intersection is an ellipse.

Place the xyz coordinate system so that the cone axis is parallel to the z-axis and the cone vertex is on the x-axis. Moreover, place the y-axis into the intersecting plane. The equations of the cone surface and of the intersecting plane are then:

cone: (x - p)2 + y2 = z·tana
plane: z = x·tanq

where p > 0 is the distance of the cone vertex from the origin, 0 < a < p/2 is the cone half-angle, and -p/2 < q £ p/2 is the angle between the intersecting plane and the xy-plane. The conic section will obviously be a closed curve iff |q| < p/2 - a.

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3fc46e495eea98f3.gif

Rotate the coordinate system around the y-axis by the angle q in order to make the intersecting plane coincident with the x'y'-plane in the rotated coordinate system x'y'z'. The rotation is expressed by the equations (they can be also expressed in a matrix form):

x' = x·cosq + z·sinq
y' = y
z' = -x·sinq + z·cosq

The inverse rotation is expressed by the equations (replace q by -q):

x = x'·cosq - z'·sinq
y = y'
z = x'·sinq + z'·cosq

Equation of the intersecting plane in the rotated coordinate system is

x'·sinq + z'·cosq = (x'·cosq - z'·sinq)·tanq
x'·sinq + z'·cosq = x'·sinq - z'·sin2q/cosq
z'·(sin2q + cos2q)/cosq = z'/cosq = 0
z' = 0

as expected. Equation of the conic surface in the rotated coordinate system is

(x'·cosq - z'·sinq - p)2 + y'2 = (x'·sinq + z'·cosq)·tana

We can get the equation of the conic section by substituting z' = 0 (equation of the intersecting plane) into the last equation of the conic surface:

(x'·cosq - p)2 + y'2 = x'·sinq·tana
x'2·cos2q - 2x'·cosq·(p + 1/2·tanq·tana) + p2 + y'2 = 0
{x'·cosq - (p + 1/2·tanq·tana)}2 - (p + 1/2·tanq·tana)2 + p2 + y'2 = 0
cos2q·{x' - (p + 1/2·tanq·tana)/cosq}2 + y'2 = (p + 1/4·tanq·tana)·tanq·tana

Denote q = (p + 1/2·tanq·tana)/cosq. Then

cos2q·(x' - q)2 + y'2 = (q·cosq + p)·(q·cosq - p) = q2·cos2q - p2

Denote further

b2 = q2·cos2q - p2 > 0
a2 = b2/cos2q > b2

The equation of the conic section then becomes

(x' - q)2/a2 + y'2/b2 = 1

which is clearly equation of an ellipse in the x'y'-plane (identical with the intersection plane) with the center at the point {q, 0} and with the primary and secondary axes a and b along the x'- and y'-axes, respectively.

Compare this with the 2 inscribed spheres proof: Place 2 inscribed spheres inside the cone, one above and the other below the intersecting plane. The 2 spheres touch the conic surface from inside at 2 circles, planes of which are perpendicular to the cone axis. The 2 spheres also touch the intersecting plane at the points E and F, respectively. Consider an arbitrary point P on the conic section. Put a cone element (a line of the conic surface through the cone vertex) through the point P. The cone element intersects the 2 circles of tangency at the points A and B, respectively. The distance between the points A and B does not depend on the point P: AB = AP + BP = const. Since the lines AP and EP are both tangents to the first sphere from the point P, AP = EP. Since the lines BP and FP are both tangents to the second sphere from the point P, BP = FP. The 2 lines EP and FP are both in the intersecting plane and EP + FP = AB = const. Locus of the points (P) with a constant sum of distances (AB) from 2 fixed points (E and F) is an ellipse. This is a brief, completely natural, and beautiful proof.

Regards, Vladimir