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rallel to PU (II. 29.), because QR and QX are bisected in N and C respectively (17.); and RX is an ordinate to the diameter D Z, (17. Cor. 2.) because RY is to Y X as QC to CX, that is, in a ratio of equality. From this reciprocal relation such diameters PU and DZ are called conjugate diameters; and the diameter which is conjugate to the transverse axis (and therefore (13. Cor. 1.) perpendicular to that axis) is called the conjugate axis of the ellipse, for, being perpendicular to its ordinates, it divides the figure symmetrically, and therefore is a second axis of the figure.

In the hyperbola, which, as we have seen, although so different in form, is very like the ellipse in its properties, there are no diameters, properly speaking, except such as lie in the angle made by the asymptotes. Let us, however, define the conjugate diameter of any diameter PU to be a straight line DZ, which is drawn through the centre C parallel to the ordinates of PU, is bisected in the centre, and is such that CD is to CP as the square of the ordinate QN to the rectangle under the abscissæ PN, NU. Such a straight line D Z will, it is evident, as in the ellipse, bisect all straight lines which are drawn parallel to the diameter PU, and terminated by the hyperbola. But there

The conjugate axis of the ellipse being always less than the principal or transverse axis, the former is frequently called the minor axis, and the latter the major axis of the ellipse.

There is, however, no other straight line which divides the figure symmetrically, that is, no third axis. For, if CQ be joined in the first figure of prop. 17, and if PC P' be supposed to represent the transverse axis, then if it were possible that C Q could divide the figure symmetrically, or (which is the same thing) bisect its ordinates at right angles, CQT would be a right angle, and, consequently, because QN is perpendicular to CT, CQ would be a mean proportional between CN and C T, (II. 34. Cor.) and therefore equal to C P (18.), so that Q N2 would be to C P2--C N2 or P N X N P' in a ratio of equality, and consequently (19.) the square of every other semi-ordinate of the axis would be equal to the rectangle under its abscissæ, and the figure would be a circle, not an ellipse.

is yet a more striking analogy, to which we are led by observing that, in the ellipse, the curve in which the extremity of a conjugate diameter so defined lies, is a part of the ellipse itself, or, as we may consider it, an ellipse having the same centre and axes with the given ellipse.

It might be expected that the same, or something similar, would obtain in the hyperbola; and such, we are about to demonstrate, is the case; viz. that "the locus of the extremities of all the conjugate diameters of a given hyperbola is an hyperbola which has the same centre and the same axes as the first."

To demonstrate this:

Let C B be drawn parallel to the tangent at the principal vertex A, and therefore perpendicular to the transverse axis

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A'CA (13. Cor. 1.): let P be any point in the hyperbola, and draw PM likewise parallel to the tangent at A, to meet CA produced in M, so that PM is a semiordinate to the transverse axis A'CA (17. and def. 17.); take CB such that CB2 shall be to CA as PM to AM x MA', and make CB' equal to CB, so that, according to the above definition, B B' is the conjugate axis of the hyperbola. Let PT be drawn touching the hyperbola in P to meet CA in T; through A draw A Q parallel to PT, and therefore (def. 17.) an ordinate to the diameter PU, by which it is consequently bisected (17.) in the point of intersection N; through C draw CD parallel to P T, and take CD such that CD shall be to CP2 as QN to PN × NU, and make C Z equal to CD, so that DZ is the diameter which is conjugate to the diameter C P. The points D, Z shall lie in an hyperbola which has BB' for its transverse axis and A A' for its conjugate axis.

From D draw DE perpendicular to CB produced. Then, because the sides of the triangles CDE, PTM are parallel, each to each, those triangles are similar (I. 18.): therefore, CE: PM :: CD: PT (II. 31.), and, consequently

(II. 37. Cor. 4.), CE

2

PM:: CD

2

PT. Now, CD is to PT in a ratio which is compounded of the ratios of C D to QN, and Q N° or NA to PT. And because, by supposition, CD CP2 :: Q N2: PN × NU or (I. 34.) CN2 CP alternando C D2 QN:: CP: CN-CP; also, because NA is parallel to PT, N A2 : PT:: CN2: CP (II. 30. Cor. 2. and II. 37. Cor. 4.); therefore, the ratio which is compounded of the ratios of CD to Q N2 and NA to PT2 is

the same with the ratio which is compounded of the ratios of CP to CNCP and CN to CP2, the same, that is, with the ratio of C N to CN-CP that is, again, since (II. 29.) CN : CP :: CA CT, the same with the ratio

of CA to C ACT (II. 37. Cor. 4. and II. 20. Cor. 1.). Therefore, CE: PM: CA2: CA2 - C T2 (a).

Again, because PM is a semiordinate of the diameter A A' to which B B' is conjugate, P M2: BC2:: AM × MA' or C M2 CA: CA. Therefore, combining this with the proportion (a), ex æquo perturbato, CE: BC:: C M2 CA CA2 CT: and, consequently, because CM, CA, CT are proportionals (18.), C E : BC2:: CM CA2 (II. 22.). And hence, dividendo, CE2 С В2 : С В2 :: CM CA2: CA2 (b).

But, because the triangles CDE, PTM are similar, DE: M T2 :: CE2 : PM2 (II. 31. and II. 37. Cor. 4.); therefore, by the proportion (a), DE2 is to M T as CA to C A2-C T2, that is, since CM, CA, CT are proportionals (18.), as C M to CM-CT or MT (II. 37. Cor. 2. and II. 20. Cor. 1.), or as CM MT to MT (II. 35.); therefore, DE2 is equal to C MX MT, or, since MA, MT, M C, MA' are proportionals (II. 47.), to A M× MA' (II. 38.), that is (I. 34.) to CM - CA. Therefore, by the proportion (b), DE2 : CA2:: CE2-CB2: CB2; and alter nando (II. 19.) DE: CE2 - CB2 or (I. 34.) BE× EB' :: CA: CB, which shows that the point D is in an hyperbola which has the transverse axis B B' and the conjugate axis A A'. (19.) And, because CZ is equal to CD, and that the diameters of an hyperbola are bisected by the centre, the point Z is in the same hyperbola.

This hyperbola, which has for its transverse axis the conjugate axis of the given hyperbola, is called the conjugate hyperbola. Thus the two hyperbolas are

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point d within or without the circle which is the projection of the conic section (12.): also any straight line passing through the former may, in like manner, be transferred to a corresponding straight line passing through the latter, and the tangents at the points in which the conic section is cut by the former straight line to tangents of the circle (6.) at the points in which it is cut by the latter, and the point P in which the tangents of the conic section intersect one another to a corresponding point p in which the tangents of the circle intersect one another. But the points p lie in a straight line (III. 53.) because they are the intersections of tangents to a circle at the extremities of chords passing through the same point. Therefore, the projections of the points P are such that their projections lie in a straight line, that is, they also lie in a straight line (2.).

Again, in the circle, any straight line which passes through the projection of the point taken, is divided harmonically by the circumference and the straight line which is the locus of p (III. 53. Cor.); and lines harmonically divided are the projections of other lines which are likewise harmonically divided (II. 49.); therefore, also, in the conic section, any straight line which passes through the point taken is divided harmonically by the conic section and the straight line, which is the locus of P. Therefore, &c.

PROP. 21.

In every conic section APQ, if in the right cone of which it is a section there be inscribed a sphere which touches the plane of the conic section in a point S, and the conical surface in a circle, the plane of which is produced to cut the plane of the conic section in a straight line RX; the distances SP and PR of any point P in the conic section from the point S, and the straight line RX, shall be to one another in a constant ratio.*

Let VO be the axis of the cone, and AM the axis of the conic section, so

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that the plane VA MA' passes through V O, and is perpendicular to the plane of the conic section APQ (13. and def. 13.): then, because the axis V O makes equal angles with the slant sides VA, VA', if the angle VAM be bisected by a straight line cutting VO in O, the point O will be the centre of a circle touching the three straight lines VA, V A', and AM (III. 59.); and, if OB, OS be drawn perpendicular to V A', A M respectively, it will touch VA' in the point B, and AM in the point S. Therefore, if the half of this circle, which is upon the same side of VO with the tangent V B, be made to revolve

From the " Transactions of the Cambridge Philo

sophical Society," Vol. III. No. VIII.

together with VB about the axis VO of the cone, it will generate a sphere (IV. def. 21.) which touches the conical surface (in which V B always lies) in the circle BDE (IV. 3. Cor. 2.), generated by the point B. And the same sphere will touch the plane A PQ in the point S (IV.8.); for OS, being drawn in the plane VAM perpendicular to A M, which is the common section of the plane V AM with the plane A P Q to which it is perpendicular, is perpendicular to the plane APQ (IV. 18.). It is supposed, therefore, that the plane of the circle BDE is produced to meet the plane A PQ in the line RX; and it is required to show that, if from any point P of the conic section, PR is drawn perpendicular to R X, and SP is joined, SP shall be to PR in a constant ratio.

Through V draw V L parallel to AM (I. 48.), and, since V L so drawn is in the plane V A M (I. def. 12. and IV. 1.) let it meet the straight line EB, in which the plane of the circle is cut by the plane V A M, or E B produced, in L; join VP, and let it cut the circumference B D E in D, and join LD, D R. Then, because the plane BDE of the circle is perpendicular to the axis V O (11.), and consequently to the plane V AM which passes through V O (IV. 18.), and that the plane APQ of the conic section is perpendicular to the same plane V A M, the common section RX is perpendicular to the same plane (IV. 18. Cor. 2.), and therefore, also, to the line X AM which meets it in that plane (IV. def. 1.). But RX is also perpendicular to RP. Therefore RP is parallel to XAM (I. 14.), that is (IV. 6.), to V L. Therefore the points L, D, R are in the plane of the parallels V L, RP; but they are, also, in the plane of the circle B DE; therefore they are in the common section of these two planes, that is, in a straight line (IV. 2.), and LD, DR are in one and the same straight line. Again, because the plane VPS cuts the sphere in a circle (VI. 1.), and that the straight lines PD, PS meet this circle and do not cut it, they touch it in the points D, S respectively, and consequently, PD is equal to PS (III. 2. Cor. 3.). But, because V L is parallel to R P, the triangles VDL, PDR are similar (I. 15.). Therefore, PD is to PR as V D to V L (II. 31.). Therefore, since SP is equal to PD, and V B to VD, SP is to PR as V B to V L, that is, in a constant ratio.

Therefore, &c.

Cor. 1. If the conic section be an ellipse or an hyperbola, a second circle may be described in the angle AV B or in the angle vertical to it, touching the straight lines VA, VA', and A A', and accordingly, a second sphere inscribed in the cone touching the plane of the ellipse or hyperbola in a second point S of the axis, and the surface of the ccne in a second circle B'D' E'. And if the plane of this circle be produced to cut the plane of the ellipse or hyperbola in a line R' X', it may be shown, as in the proposition, that the distances S'P and PR' of any point P, from the point S' and the line R'X', are to one another in the constant ratio of V B' to V L', that is, in the same constant ratio as before, of V B to V L (II. 29.).

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also C S', CA', C X', are proportionals. For, since (by the proposition) SA is to AX as SA' to A'X, the straight lines A'X, A' A, and A'S, are in harmonical progression (II. 45. Cor.); and consequently, the mean A A' being bisected in C, C S, CA, and C X are proportionals (II. 46.); and the like demonstration applies to CS', CA' and C X'.

Cor. 3. The constant ratio of SP to PR is the same with that of C S to CA, or (which is the same) of CS' to C A'. For A being a point of the ellipse or hyperbola, SA is to A X in the constant ratio; and because C S, CA, C X are proportionals, C S is to CA as SA to AX (II. 22.) And for the like reason CS' is to CA' as S' A' to A'X', that is, likewise in the constant ratio.

Cor. 4. In the ellipse SP+ PS' = A A'; and in the hyperbola SP-PS' =AA'. For, since SP is to PR in the constant ratio of CS to CA, or of CA to CX, or again (II. 23.), of A A' to X X', and that S' P is to PR' in the same ratio, SPS'P is to PR+PR' in the same ratio of A A' to XX' (II. 23. and II. 22.). But in the ellipse PR +PR is equal to R R', and RR' is equal to XX' (I. 22.); therefore (II. 18. Cor.), SP+ S'Pis equal to A A'. And, in the hyperbola, PR ~ PR' is equal to R R', and R R' is equal to X X' (I. 22.); therefore (II. 18. Cor.) SP~ S'P is equal to A A'.

Cor. 5. In the ellipse S P is less than PR; in the hyperbola SP is greater than PR; and in the parabola SP is equal to PR.

Scholium.

The points S and S' are called the foci, and the straight lines R X, R'X' the directrix-es, of the conic section: the ellipse and hyperbola having two foci at equal distances from the centre upon either side of it, and two directrix-es; the parabola one focus and one directrix only. From the simple properties which have just been demonstrated with regard to these remarkable points, viz., that

1. In the ellipse SP+S'P=AA'. 2. In the hyperbola S P~S' P= A A'. 3. In the parabola SP=PR, others of very considerable imporThe three conic tance are derived. sections are, indeed, commonly defined by these properties, and from these, by help of the theorems of Plane Geometry, all other properties are derived in order.

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230

PART III.-Plane Sections of the oblique cone, of the right cylinder, and of the oblique cylinder. PROP. 22.

Every section of an oblique cone which is made by a plane parallel to its base, is a circle having its centre in the axis of the cone.

See the demonstration of Prop. 11.
PROP. 23.

In an oblique cone, if V A, V A' are`
the slant sides in which the surface of
the cone is cut by a plane passing
through the aris V O perpendicular to
the base, and if the cone be cut by a
plane PQR which is perpendicular to
the plane VA A', and is inclined to
either of the slant sides V A at the same
angle at which the base is inclined to the
other slant side VA', the section made
by the plane PQR shall be a circle:
in other words, every subcontrary sec-
tion of an oblique cone is a circle.
Let the plane PQR
cut the plane V A A'
in the straight line
PP'; take any point
Q in the curve PQR,
and through Q draw
the plane A QA'R
parallel to the base of
the cone, and let this
plane cut the plane p

A

R

Q

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straight line QR, which is drawn per-
pendicular to it from a point Q of the
section P Q R, and that the square Q No
of the half of such straight line is equal
to the rectangle PN x Ñ P', under the
segments of PP', the section P Q R is a
circle, having PP' for its diameter (III.
3. and III. 20.).
Therefore, &c.

PROP. 24.

Every plane section of an oblique cone which neither passes through the vertex, nor is parallel to the base, nor is subcontrary, is a conic section, that is, either an ellipse or a parabola or an hyperbola.

VAA' in the straight line A'A', and the
plane PQR in the straight line QR
(IV. 2.); then A QA' is a circle having
its centre O in the axis of the cone (22.).
And, because the planes A Q A', P Q R
are each of them perpendicular to the
plane V A A', their common section QR
is perpendicular to the same plane (ÏV.
18.Cor. 2.), and consequently (IV.def. 1.)
to the straight lines AÂ', PP' which meet
it in that plane. Now, because, by the
supposition, the planes PQR, A Q A'
are equally inclined to the straight lines
VA, VA respectively, the angles of
inclination V P'P and V A' A are equal
to one another, and (I. 3.) the vertical
angles PN A' and A N P' are likewise
équal; therefore the triangles PN A',
AN P' are similar, and (II. 31.) PN ·
is to N A' as AN to N P', and con-
sequently (II. 38.), PN × N P' is equal
to AN X N A'. But, because A A' is
the diameter of the circle A QA', and is
perpendicular to the chord QR at the
point N, QR is bisected in N (III. 3.),
and QN is equal to AN × NA' (III. 20.).
Therefore Q N2 is equal to PN ×N P'.

Therefore, because P P' bisects every

Let V be the vertex, and VO the axis of an oblique cone, and let it be cut by any plane PQR: through V draw the straight line VD parallel both to the base of the cone and to the plane PQR (IV. 40. Cor.): through any point in the base of the cone draw a straight line parallel to VD, and (III. 57.) a straight line touching the base parallel to this straight line or (IV. 6.) to V D; and let VP be the slant side of the cone which passes through the point of con

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