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Subject: "Brief elllipse/conic section proof?"     Previous Topic | Next Topic
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ossipoff
Member since Nov-7-03
Nov-07-03, 11:35 AM (EST)
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"Brief elllipse/conic section proof?"
 
  
I've seen the proof, using 2 inscribed spheres, that when a plane intersects a cone in a way that the intersection is a closed curve, the intersection will be an ellipse.

But aren't there other demonstations of that fact? Is there a briefer proof, or one that is naturally and obviously motivated? Of course analytic geometry could prove it too, but that would be a long brute-force demonstration.

The reason why I ask is that sometimes things can be demonstrated in ways that are brief, and naturally motivated. For instance, there was an article in _Scientific American_ showing several brief, simple, and natural demonstrations of the Pythagorean theorem, demonstrations much simpler than the standard one.

Another request:

How about a brief and natural demonstration that when lines are drawn from a point to all the points on a circle, the surface formed by the lines is an elliptical cone. (If the circle is perpendicular to the line connecting its center to the point,then of course the elliptical cone is a circular cone).

Mike Ossipoff


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Vladimir
Member since Jun-22-03
Nov-26-03, 01:41 PM (EST)
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1. "RE: Brief conic section proof?"
In response to message #0
 
   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.

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

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Please solution
guest
Feb-28-06, 10:13 AM (EST)
 
2. "Thankyou"
In response to message #1
 
   ABC is acute-angled and is not isosceles. The bisector of the acute angle between the altitudes from A and C meets AB at P and BC at Q. The angle bisector of B meets the line joining HN at R, where H is the orthocenter and N is the midpoint of AC. Show that BPRQ is cyclic


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Bruce5
guest
Sep-23-07, 10:58 PM (EST)
 
3. "RE: Brief conic section proof?"
In response to message #1
 
   I enjoyed your brief conic section proof.

Instead of intersecting a right circular cone with an oblique plane,

an elliptical right cone could be cut by an oblique planc.

I haven't been able to prove whether the resulting section is an ellipse, or an egg-shaped oval, or some toher quadric surface.

Could you help me get going on a proof? Thanks.

Bruce Petrovics


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alexbadmin
Charter Member
2081 posts
Sep-24-07, 00:31 AM (EST)
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4. "RE: Brief conic section proof?"
In response to message #3
 
   The proof is the same as Vladimir's. You have an equation of a plane and a coordinate system in that plane. All that you have to do is to write an equation of a straight cone with an ellipric base. Just add a coefficient to the equation of the circular cone.


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