Question

in Conic section was discovered by Menaechmus when he want to build tomb for his son. He used the concept of " doubling the cube". pls clarify if this is true and im curious pls put whole story on how concept of doubling the cube got involved with building a tomb for his son? and how it created the conic section with intercsection of a plane and a double-napped cone and created 4 basic conics (circle,ellipse,parabola,hyperbola) and the degenerate conics? how are they found and how is the idea originated?

history/discovery of conic section

Answer 1

According to tradition, the idea of the conic sections arose out of the exploration of the problem of "doubling the cube". This problem, and the accompanying story, is presented in a letter from Eratosthenes of Cyrene to King Ptolemy Euergetes, which has come down to us as quoted by Eutocius in his commentary on Archimedes' On the Sphere and Cylinder, which appears in Heath. Eratosthenes told the king that the legendary King Minos wished to build a tomb for Glaucus and felt that its current dimensions - one hundred feet on each side - were inadequate.

Too small thy plan to bound a royal tomb.

Let it be double; yet of its fair form

Fail not, but haste to double every side.

(Heath, 1961, p. xvii)

Clearly, doubling every side will increase the volume by a factor of eight, not by the desired factor of two. Mathematicians worked diligently on this problem, but were having tremendous difficulty in solving it. A breakthrough of a kind occurred when Hippocrates of Chios reduced the problem to the equivalent problem of "two mean proportionals", though this formulation turned out to be no easier to handle than the previous one (Heath, 1961, p. xviii). Eratosthenes continued by mentioning the Delians, who had an interest in exactly the same problem of "doubling a cube". When they called upon geometers at Plato's Academy in Athens for a solution, two geometers found answers to the equivalent mean proportions problem. Archytas of Tarentum used "half- cylinders", and Eudoxus used "curved lines". These solutions, however, only gave demonstrations of the existence of the desired number as a geometrical quantity, but they could not actually construct the mean proportion mechanically, so they did not reach the point of practical application until Menaechmus, who he achieved it with considerable difficulty (Heath, 1961, pp. xvii-xviii).

The mention above of Hippocrates' mean proportions is of interest. What this means is that, given two lengths a and b, we find x and y such that a:x::x:y and x:y::y::b, or in modern notation a/x=x/y=y/b; if we denote this ratio by r, then r^3 = (a/x)(x/y)(y/b)=a/b, and as Hippocrates noted, that if the segment a is twice as long as the segment b, then the doubling of the cube would be solved using the length r. Needless to say, he did not have an algebraic notation able to support the argument in the form we have given it, and had to argue directly.

Continuing our analysis in modern terms, the continued proportion above yields the three following equations which we now know to describe conics:

• (i) x² = ay
• (ii) y² = bx
• (iii) xy = ab

Again, without using algebraic notation, the Greeks - specifically, Menaechmus - arrived at the consideration of the same three curves, which they described geometrically.

Menaechmus was a pupil of Eudoxus, a contemporary of Plato (Heath, 1921, p. 251). Much of what we know about Menaechmus's work comes to us from the commentaries of Eutocius, a Greek scholar who discussed the works of many mathematicians of his and earlier times, including Menaechmus, Archimedes, and Apollonius. In his solutions, Menaechmus essentially finds the intersection of (ii) and (iii) (see Solution 1, below), and then, alternatively, the intersection of (i) and (ii) (see Solution 2, below). Menaechmus's proof deals with the general case of the mean proportions. Once we have this, we can take the special case a=2b to double the cube. Before giving these two solutions, it should be noted that Menaechmus did not use the terms "parabola" and "hyperbola" - these terms are due to Apollonius. Instead, he called a parabola a "section of a right-angled cone", and a hyperbola a "section of an obtuse-angled cone" (Heath, 1921, p. 111).

Solution 1:

y M P B Z х ΤΑ

• Let AO, AB be two given straight lines such that AO > AB and let them form a right angle at O.
• Suppose the problem solved and let the two mean proportionals be OM measured along BO produced and ON measured along AO produced. (Heath, 1921, p. 253).
• Complete the rectangle OMPN.
• Because AO : OM = OM : ON = ON : OB, we have by cross multiplication the following relations:
• (1) OB.OM = ON² = PM² [the "." refers to multiplication], such that P lies on a parabola which has O for its vertex, OM for its axis, and OB for its latus rectum.
• (2) AO.OB = OM.ON = PN.PM, such that P lies on a hyperbola with O as its center, and OM and ON as its asymptotes.
• To find the point P, we must construct the parabola in (1) and the hyperbola in (2), and once we do so, the intersection of the two solves the problem, for AO : PN = PN : PM = PM : OB.

Solution 2:

у M Р B N х ТА

• Suppose AO and AB are given and the problem to be solved as in the first two steps to Solution 1.
• Again, we have AO : OM = OM : ON = ON : OB, giving us
• (1) as in solution 1, the relation OB.OM = ON² = PM ², such that P lies on a parabola which has O for its vertex, OM for its axis, and OB for its latus rectum.
• (2) the relation AO.ON = OM² = PN², such that P lies on a parabola which has O for its vertex, ON for its axis, and OA for its latus rectum.
• To find the point P, we must construct the two parabolas described in (1) and in (2). The intersection gives us the point P such that AO PN = PN : PM = PM : OB