Suppose A infinitely distant, then we may put in (1) xa for a, where a is infinitely great, and, omitting all but the higher terms, the equation of the cylinder formed by tangent lines parallel to a is (Sppp-1) Sapa (Sppa)2 = 0.

254.] To study the nature of the surface more closely, let us find the locus of the middle points of a system of parallel chords.

Let them be parallel to a, then, if a be the vector of the middle point of one of them, w+xa and -xa are simultaneous values of p which ought to satisfy (1) of § 252.

ρ

The latter equation shews that the locus of the extremity of ✩, the middle point of a chord parallel to a, is a plane through the centre, whose normal is pa; that is, a plane parallel to the tangent plane at the point where OA cuts the surface. And (d.) shews that this relation is reciprocal-so that if ß be any value of ☎, i. e. be any vector in the plane (1), a will be a vector in a diametral plane which bisects all chords parallel to B. The equations of these planes are

δπφα = 0,
Sapa

δαφβ

Sw&B = 0,

so that if V. papßy (suppose) is their line of intersection, we have

and (1) gives

=

Sypa 0 Sapy,
= =

Sypẞ0 Spoy,
=0= δβφγ,

δβφα = 0 = δαφβ.

(2)

Hence there is an infinite number of sets of three vectors a, ß, y, such that all chords parallel to any one are bisected by the diametral plane containing the other two.

255.] It is evident from § 23 that any vector may be expressed as a linear function of any three others not in the same plane, let then p = xa+yẞ+zy,

where, by last section, Sapß Sẞpa = 0,

And let

=

so that a, ẞ, and y are vector conjugate semi-diameters of the surface we are engaged on.

Substituting the above value of p in the equation of the surface, and attending to the equations in a, ẞ, y and to (a.), (b.), and (d.), we have Spopp S(xa+yẞ+zy) p(xa+y3+2y),

=

= x2 + y2+z2 = 1.

To transform this equation to Cartesian coordinates, we notice that a is the ratio which the projection of p on a bears to a itself, &c. If therefore we take the conjugate diameters as axes of §, n, S, and their lengths as a, b, c, the above equation becomes at once

£2 n2 x2
+ + = 1,
a2 72 c2

the ordinary equation of the ellipsoid referred to conjugate diameters. 256.] If we write y2 instead of p, these equations assume an interesting form. We take for granted, what we shall afterwards prove, that this halving or extracting the root of the vector function is lawful, and that the new linear and vector function has the same properties (a.), (b.), (c.), (d.) (§ 251) as the old. The equation. of the surface now becomes

If we compare this with the equation of the unit-sphere

Tp=1,

we see at once the analogy between the two surfaces. The sphere can be changed into the ellipsoid, or vice versa, by a linear deformation of each vector, the operator being the function or its inverse. See the Chapter on Kinematics.

257.] Equations (2) § 254 now become

Say23 = 0 = Stayß, &c.,

(1)

so that ya, ẞ, y, the vectors of the unit-sphere which correspond to semi-conjugate diameters of the ellipsoid, form a rectangular system. We may remark here, that, as the equation of the ellipsoid referred to its principal axes is a case of § 255, we may now suppose i, j, and 2:2 y2 22 + + a2 62

k to have these directions, and the equation is which, in quaternions, is

1,

We here tacitly assume the existence of such axes, but in all cases,

by the help of Hamilton's method, developed in Chapter V, we at once arrive at the cubic equation which gives them.

It is evident from the last-written equation that

which latter may be easily proved by shewing that √2p = — Op.

And this expression enables us to verify the assertion of last section about the properties of y.

As Sip-x, &c., x, y, z being the Cartesian cöordinates referred to the principal axes, we have now the means of at once transforming any quaternion result connected with the ellipsoid into the ordinary one.

258.] Before proceeding to other forms of the equation of the ellipsoid, we may use those already given in solving a few problems. Find the locus of a point when the perpendicular from the centre on its polar plane is of constant length.

If be the vector of the point, the polar plane is

and the length of the perpendicular from O is Hence the required locus is

a concentric ellipsoid, with its axes in the same direction as those of the first. By § 257 its Cartesian equation is

x2 y2 22
+ + = C2.
64

a1

c4

259.] Find the locus of a point whose distance from a given point is always in a given ratio to its distance from a given line.

Let p=xß be the given line, and A(OA = a) the given point, and let Saß 0. Then for any one of the required points

=

a surface of the second order, which may be written p2 - 2 Sap+ a2 = e2 (S2ßp— B2 p2).

Let the centre be at d, and make it the origin, then

p2+2 Sp (d-a)+(d—a)2 = e2 {S2.ẞ (p+8)-B2 (p+8)2},

and, that the first power of p may disappear,

(d—a) = e2 (BSẞd — ß2¿),

d)

a linear equation for d. To solve it, note that Saß 0, operate by

S.B and we get (1-e2ß2 + e2ß2)Sßò = Sßd = 0.

=

which shews that it belongs to a surface of revolution (of the second order) whose axis is parallel to B, as its intersection with a plane SBp=a, perpendicular to that axis, lies also on the sphere

In fact, if the point be the focus of any meridian section of an oblate spheroid, the line is the directrix of the same.

260.] A sphere, passing through the centre of an ellipsoid, is cut by a series of spheres whose centres are on the ellipsoid and which pass through the centre thereof; find the envelop of the planes of intersection.

Let (p-a)2= a2 be the first sphere, i.e.

261.] From a point in the outer of two concentric ellipsoids a tangent cone is drawn to the inner, find the envelop of the plane of contact. If Sapa 1 be the outer, and Spyp 1 be the inner, and

=

being any two self-conjugate linear and vector functions, the plane of contact is

Soup = 1.

Hence, for the envelop, Sap = 0, 1
Sw'pw = 0, S

another concentric ellipsoid, as y1y is a linear and vector function =x suppose; so that the equation may be written

262.] Find the locus of intersection of tangent planes at the extremities of conjugate diameters.

If a, ß, y be the vector semi-diameters, the planes are

Hence -ψπδ.ψαψβψγ = ψπ = ψα+β+ψy, by § 92, therefore

since ya, ẞ, y form a rectangular system of unit-vectors. This may also evidently be written

shewing that the locus is similar and similarly situated to the given ellipsoid, but larger in the ratio √3:1.

263.] Find the locus of the intersection of three spheres whose diameters are semi-conjugate diameters of an ellipsoid.

If a be one of the semi-conjugate diameters

with similar equations in ẞ and y. Hence, by § 92,

↓1pS.↓a¥ß↓y = −4−1p = p2(Ya+¥ß+¥v),

This is Fresnel's Surface of Elasticity in the Undulatory Theory. 264.] Before going farther we may prove some useful properties of the function in the form we are at present using-viz.