either one or two of them may be at an infinite distance; the fifth point 0, to which the pencil is drawn, may be also either at an infinite distance, or may coincide with one of the four points, in which latter case one of the legs of the pencil will be the tangent at that point; then, again, we may measure the anharmonic ratio of the pencil by the segments on any line drawn across it, which we may, if we please, draw parallel to one of the legs of the pencil, so as to reduce the anharmonic ratio to a simple ratio.

The following examples being intended as a practical exercise to the student in developing the consequences of this theorem, we shall merely state the points whence the pencil is drawn, the line on which the ratio is measured, and the resulting theorem, recommending to the reader a closer examination of the manner in which each theorem is inferred from the general principle.

We use the abbreviation {0.ABCD} to denote the anharmonic ratio of the pencil OA, OB, OC, OD.

Let these ratios be estimated by the segments on the line CD; let the tangents at A, B meet CD in the points T, T'', and let the chord

that is, if any chord CD meet two tangents in T, T'', and their chord of contact in K,

KC.KT'.TD = KD.TK. T'C.

(The reader must be careful, in this and the following examples, to take the points of the pencil in the same order on both sides of the equation. Thus, on the lefthand side of this equation we took K second, because it answers to the leg OB of the pencil; on the right hand we take K first, because it answers to the leg OA).

Ex. 2. Let T and T' coincide, then

KC.TD-KD.TC,

or, any chord through the intersection of two tangents is cut harmonically by the chord of contact.

Ex. 3. Let 7'' be at an infinite distance, or the secant CD drawn parallel to PT', and it will be found that the ratio will reduce to

TK2= TC.TD.

Ex. 4. Let one of the points be at an infinite distance then {0. ABC ∞ } is constant. Let this ratio be estimated on the line Co. Let the lines AO, BO cut C∞ Ca

in b; then the ratio of the pencil will reduce to сь ; and we learn, that if two fixed points, A, B, on a hyperbola or parabola, be joined to any variable point 0,

and the joining line meet a fixed parallel to an asymptote (if the curve be a hyperbola), or a diameter (if the curve be a parabola), in a, b, then the ratio Ca: Cb will be constant.

Ex. 5. If the same ratio be estimated on any other parallel line, lines inflected from any three fixed points to a variable point, on a hyperbola or parabola, cut a fixed parallel to an asymptote or diameter, so that ab: ac is constant.

Ex. 6. It follows from Ex. 4, that if the lines joining A, B to any fourth point O' meet C∞ in a', b', we must have

Now let us suppose the point C to be also at an infinite distance, the line C∞ becomes an asymptote, the ratio ab: a'b' becomes one of equality, and lines joining two fixed points to any variable point on the hyperbola intercept on either asymptote a constant portion (p. 182).

Let these ratios be estimated on C∞o; then if the tangents at A, B, cut C∞ in a, b, and the chord of contact AB in K, we have

a C K

B

(observing the caution in Ex. 1). Or, if any parallel to an asymptote of a hyperbola, or a diameter of a parabola, cut two tangents and their chord of contact, the intercept from the curve to the chord is a geometric mean between the intercepts from the curve to the tangents. Or, conversely, if a line ab, parallel to a given one, meet the sides of a triangle in the points a, b, K, and there be taken on it a point C such that CK2 Ca.Cb, the locus of C will be a parabola, if Cb be parallel to the bisector of the base of the triangle (Art. 211), but otherwise a hyperbola, to an asymptote of which ab is parallel.

=

Ex. 8. Let two of the fixed points be at infinity,

{∞.AB ∞ ∞'} = {∞'.AB ∞ ∞'};

the lines∞∞, ∞∞', are the two asymptotes, while ∞ ∞ is altogether at infinity. Let these ratios be estimated on the diameter OA; let

this line meet the parallels to the asymptotes B ∞, B ∞',

in a and a'; then the ratios become

Or,

ОА Oa' Oa OA' parallels to the asymptotes through any point on a hyperbola cut any semi-diameter, so that it is a mean proportional between the segments on it from the centre.

Hence, conversely, if through a fixed point O a line be drawn cutting two fixed lines, Ba, Ba', and a point A taken on it so that OA is a mean between Oa, Oa', the locus of A is a hyperbola, of which O is the centre, and Ba, Ba', parallel to the asymptotes.

a

B

Ob'
(0 being

Оа

=

Ob Oa'

Let the segments be measured on the asymptotes, and we have the centre), or the rectangle under parallels to the asymptotes through any point on the curve is constant (we invert the second ratio for the reason given in Ex. 1).

327. We next examine some particular cases of the anharmonic property of the tangents to a conic (Art. 275).

Ex. 1. This property assumes a very simple form, if the curve be a parabola,

for one tangent to a para

Ex. 2. Let two of the four tangents to an ellipse or hyperbola be parallel to each other, and let the variable tangent coincide alter

nately with each of the parallel tangents. In the

Hence the rectangle Ab. Db' is constant.

It may be deduced from the anharmonic pro

perty of the points of a conic, that if the lines joining any point on the curve 0 to A, D, meet the parallel tangents in the points b, b', then the rectangle Ab. Db' will be constant.

328. We now proceed to give some examples of problems easily solved by the help of the anharmonic properties of conics.

Ex. 1. To prove Mac Laurin's method of generating conic sections (p. 236), viz.To find the locus of the vertex V of a triangle whose sides pass through the points A, B, C, and whose base angles move on the fixed lines Oa, Ob.

Let us suppose four such triangles drawn, then since the pencil {C.aa'a"a""} is the same pencil as {C.bb'b''b'''}, we have

V' v'

Vm

A

B

Or the reasoning may be stated thus: The systems of lines through A, and through B, being both homographic with the system through C, are homographic with each other; and therefore (Art. 297) the locus of the intersection of correspond

ing lines is a conic through A and B. The following examples are, in like manner, illustrations of the application of this principle of Art. 297.

Ex. 2. M. Chasles has showed that the same demonstration will hold if the side ab, instead of passing through the fixed point C, touch any conic which touches Oa, Ob; for then any four positions of the base cut Oa, Ob, so that

{aa'a"a""} = {bb'b''b'''} (Art. 275),

and the rest of the proof proceeds the same as before.

Ex. 3. Newton's method of generating conic sections:-Two angles of constant magnitude move about fixed points P, Q;

the intersection of two of their sides traverses the right line AA'; then the locus of V, the intersection of their other two sides, will be a conic passing through P, Q.

For, as before, take four positions of the angles, then

{P.AA'A"A""} = {Q. AA'A"A""};

but (P.AA'A"A""} = {P.VV'V"V""},

{Q.AA'A"A""} = {Q.VV'V"V""},

P

since the angles of the pencils are the same; therefore

Α Α' Α" Α"

{P.VV'V"V"} = {Q.VV'V"V""} ;

and, therefore, as before, the locus of V"" is a conic through P, Q, V, V', V".

Ex. 4. M. Chasles has extended this method of generating conic sections, by supposing the point A, instead of moving on a right line, to move on any conic passing through the points P, Q; for we shall still have

{P.AA'A"A""} = {Q.AA'A"A""}. ·

Ex. 5. The demonstration would be the same if, in place of the angles APV, AQV being constant, APV and AQV cut off constant intercepts each on one of two fixed lines, for we should then prove the pencil

{P.AA'A"A""} = {P.VV'V"V""},

because both pencils cut off intercepts of the same length on a fixed line.

Thus, also, given base of a triangle and the intercept made by the sides on any fixed line, we can prove that the locus of vertex is a conic section.

Ex. 6. We may also extend Ex. 1, by supposing the extremities of the line ab to move on any conic section passing through the points AB, for, taking four positions of the triangle, we have, by Art. 276,

and the rest of the proof proceeds as before.

Ex. 7. The base of a triangle passes through C, the intersection of common tangents to two conic sections; the extremities of the base ab lie one on each of the conic sections, while the sides pass through fixed points A, B, one on each of the conics: the locus of the vertex is a conic through A, B.

The proof proceeds exactly as before, depending now on the second theorem proved, Art. 276. We may mention that this theorem of Art. 276 admits of a simple geometrical proof. Let the pencil {0. ABCD} be drawn from points corresponding to {o.abed}. Now, the lines OA, oa, intersect at r on one of the common chords of the conics; in like manner, BO, bo, intersect in 'on the same chord, &c.; hence {rr'""} measures the anharmonic ratio of both these pencils.

Ex. 8. In Ex. 6 the base, instead of passing through a fixed point C, may be supposed to touch a conic having double contact with the given conic (see Art. 276).

Ex. 9. If a polygon be inscribed in a conic, all whose sides but one pass through fixed points, the envelope of that side will be a conic having double contact with the given one.

For, take any four positions of the polygon, then, if a, b, c, &c. be the vertices of the polygon, we have

{aa'a"a""} = {bb′b"b""} = {cc°c"c""}, &c.

The problem is, therefore, reduced to that of Art. 277,-"Given three pairs of points, aa'a", dd'd", to find the envelope of a""d"", such that

{aa'a"a""} = {dd'd"d""'}.”

Ex. 10. To inscribe a polygon in a conic section, all whose sides pass through fixed points.

If we assume any point (a) at random on the conic for the vertex of the polygon, and form a polygon whose sides pass through the given points, the point z, where the last side meets the conic, will not, in general, coincide with a. If we make four such attempts to inscribe the polygon, we must have, as in the last example,

{aa'a"a""} = {zz'z'z'''}.

Now, if the last attempt were successful, the point a"" would coincide with z""', and the problem is reduced to,-" Given three pairs of points, aa'a", zz'z", to find a point K such that

{Kaa'a"} = {Kzz'z"},"

Now if we make az"a'za"z' the vertices of an inscribed hexagon (in the order here given, taking an a and z alternately, and so that az, a'z', a"z", may be opposite vertices), then either of the points in which the line joining the intersections of opposite sides meets the conic may be taken for the point K. For, in the figure, the points ACE are aa'a", DFB are zz'z"; and if we take the sides in the order ABCDEF, L, M, N are the intersections of opposite sides. Now, since {KPNL} measures both {D.KACE} and {A.KDFB}, we have

{KACE} = {KDFB}. Q.E.D.*

K

M

It is easy to see, from the last example, that K is a point of contact of a conic having double con

B

tact with the given conic, to which az, a'z', a"z" are tangents, and that we have therefore just given the solution of the question, "To describe a conic touching three given lines, and having double contact with a given conic."

Ex. 11. The anharmonic property affords also a simple proof of Pascal's theorem, alluded to in the last example.

We have {E.CDFB} = {A.CDFB), Now, if we examine the segments made by the first pencil on BC, and by the second on DC, we have

{CRMB} = {CDNs}.

This construction for inscribing a polygon in a conic is due to M. Poncelet (Traité des Propriétés Projectives, p. 351). The demonstration here used is Mr. Townsend's. It shows that Poncelet's construction will equally solve the problem, "To inscribe a polygon in a conic, each of whose sides shall touch a conic having double contact with the given conic." The conics touched by the sides may be all different.

PP