Collingwood <1>
Sep 20/44
My dear Sir
I had got quite wrong about the Cone – of course it is possible that a cone having for example a hyperbolic base may yet admit of a reentering section, viz: a circle – for since a section of a right cone may be any conic section (and ∴ an hyperbola) conversely an oblique cone having an hyperbola for its base may be in fact no other than a right cone with a circular base – Aliter <2> – Since any conic § may be cut out of some right cone or other by some plane or other therefore some right cone necessarily exists which can stand (obliquely) on any given conic § as a base.
As regards the proposition that every cone of the second order is necessarily identical with some right cone having an ellipse for its base (or with some oblique cone having a circle for its base – for the expressions are convertible) I suppose the proof will run thus:
1. By the reasoning in my last, the general equations of every cone of the second order must be of the form
c = Ax2 + By2 + Cz2 + pxy + qxz + ryz
For if x & y be changed in any given ratio z must change in the same ratio which excludes all the non homogeneous terms
2. Change coordinates retaining the Vertex as their axis
or change x into α x + βy + γz
y – α′x + β′y + γ′z
z – α″x + β″y+ γ″z
Then since x2 + y2 + z2 remains unchanged this gives 6 equations if conditions
α 2 + α′2 + α″2 = 1, &c
αβ + α′ β′ + α″ β″ = 0, &c
and three of the 9 constants remain arbitrary.
3. Now this substitution gives an equation of form
Kx2 + Ly2 + Mz2 = kxy + lxz + myz = 0
when K L M k l m are homogeneous functions of α β γ α′ β′ γ′ &c of two dimensions, with A,B,C p, q, r for coefficients – and of which three are undetermined and may therefore be assumed to satisfy the 3 equations
k = 0 l = 0 m = 0
4. This done the equation of our cone becomes
0 = Kx2 + Ly2 + Mz2
Now I shall assume that it is possible so to satisfy the equations k = 0 l = 0 m = 0 as to give α, β, γ &c real values – which perhaps it would not be impracticable to shew.
5. Let the Cone be cut by 3 planes perpendicular to the x, y, z, at distances = 1 and the conic sections respectively produced will have for equations
0 = K + Ly2 + Mz2 [illegible deletion] (1)
0 = L + Kx2 + Mz2 [illegible deletion] (2)
0 = M + Kx2 + Ly2 [illegible deletion] (3)
6. Now if K, L, M be all + or all − the cone is impossible. But by hypothesis it is a real cone. Ergo K L M cannot be all + or all −
7. Therefore one must be + and two −
or one − and two +
Take any one as K + and L, M −
Then equation (1) is the equation of an ellipse about its center
Take any one as K − and L, M each + and again also (1) is as aforesaid
And in both cases the other two are equations of hyperbolas also about the center.
The whole onus of the proof then lies in shewing that k = 0 l = 0 m = 0 may be satisfied by real values – But they are ugly complicated functions and do not seem very easy to deal with
Your demonstration of the theorem about [mathematical equation] is simple and straitforward, but it appears to me not absolutely free from all prior reasoning or general theory for you refer to certain preunderstood properties of “Symmetricals” as per general theory.
I have not time now (having to draw up the Magnetic report for the British Association) <3> to go further into the question about cones – but it has just occurred to me that it would be simpler in point of form of equations to set the problem on its head and from a general assumed equation
0 = Kx2 + Ly2 + Mz2
to shew that by a possible choice of coordinates an equally general equation of the form
0 = Ax2 + By2 + Cz2 + pxy + qxz + ryz
may be derived.
I remain my dear Sir yours very try
J. F.W. Herschel
Envelope:
H. F. Talbot Esq31 Sackville St
London
Notes:
1. Hawkhurst, Kent.
2. in another way
3. ‘Report of a Committee … Appointed to Conduct the Co-operation of the British Association in the System of Simultaneous Magnetical and Meteorological Observations’, British Association for the Advancement of Science Report for 1843, 1844, pp.60–113.