respect to the circle. Hence the cross ratio of the pencil formed by four intersecting straight lines is equal to that of the pencil subtended at the centre of the circle by their poles, and therefore equal to the cross ratio of the range formed by their poles.

322. The cross ratio of the pencil formed by joining any point on a conic to four fixed points is constant, and is equal to that of the range in which the tangents at those points are cut by any tangent.

Since the cross ratios of pencils and ranges are unaltered by projection, we need only prove the proposition for a circle.

Let A, B, C, D be four fixed points on a circle; let P be any other point on the circle, and let the tangent at P meet the tangents at A, B, C, D in the points A', B, C, D'.

Then, if O be the centre of the circle, OA' is perpendicular to PA, OB′ to PB, OC′ to PC, and OD' to PD. Hence

{A'B'C'D'} = 0 {A'B'C'D'} = P {ABCD}.

But the angles APB, BPC, CPD are constant, since A, B, C, D are fixed points.

Therefore

{A'B'C'D'}=P{ABCD} = const.

If Q be any point which is not on the circle, Q{ABCD} cannot be equal to P {ABCD}; this is seen at once if we take P such that APQ is a straight line, and consider the

ANHARMONIC PROPERTIES OF CONICS.

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ranges made on BC by the two pencils. Hence we have the following converse proposition.

If a point P move so that the cross ratio of the pencil formed by joining it to four fixed points A, B, C, D, is constant; P will describe a conic passing through A, B, C, D.

Ex. 1. The four extremities of two conjugate chords of a conic subtend a harmonic pencil at any point on the curve.

Let the chords be AC, BD; let E be the pole of BD, and let F be the point of intersection of AC, BD. The four points subtend, at all points on the curve, pencils of equal cross ratio. Take a point indefinitely near to D; then the pencil is D{ABCE}. But the range A, B, C, E is harmonic, which proves the proposition.

Ex. 2. If two triangles circumscribe a conic, their six angular points are on another conic.

Let ABC, A'B'C' be the two triangles. Let B'C' cut AB, AC in E', D', and let BC cut A'B', A'C' in E, D. Then the ranges made on the four tangents AB, AC, A'B', A'C' by the two tangents BC, B'C' are equal.

The proposition may also be proved by projecting B, C into the circular points at infinity; the conic is thus projected into a parabola, of which A is the focus; and it is known that the circle circumscribing A'B'C' will pass through A.

323. DEF. Ranges and pencils are said to be homographic when every four constituents of the one, and the corresponding four constituents of the other, have equal cross ratios.

Another definition of homographic ranges or pencils is the following:-two ranges or pencils are said to be homographic which are so connected that to each point or line of the one system corresponds one, and only one, point of the other.

To show that this definition of homographic ranges is equivalent to the former, let the distances, measured from fixed points, of any two corresponding points of the two

systems be x, y; then we must have an equation of the form

The proposition follows from the fact that the cross ratio of every four points of the one system, namely

Ex. 1. The points of intersection of corresponding lines of two homographic pencils describe a conic.

Let P, Q, R, S be four of the points of intersection, and O, O' the vertices of the pencils.

Then_O{PQRS}=0'{PQRS}; therefore [Art. 322] O, O', P, Q, R, S are on a conic. But five points are sufficient to determine a conic; hence the conic through O, O' and any three of the intersections will pass through every other intersection.

Ex. 2. The lines joining corresponding points of two homographic ranges envelope a conic.

Let a, b, c, d be any four of the points of one system, and a', b', c', be the corresponding points of the other system. Then aa', bb', cc', dd are cut by the fixed lines in ranges of equal cross ratio. Hence a conic will touch the fixed lines, and also aa', bb', cc', dď. But five tangents are sufficient to determine a conic; hence the conic which touches the fixed lines, and three of the lines joining corresponding points of the ranges, will touch all the others.

Ex. 3. Two angles PAQ, PBQ of constant magnitude move about fixed points A, B, and the point P describes a straight line; shew that Q describes a conic through A, B. [Newton.]

Corresponding to one position of AQ, there is one, and only one, position of BQ. Hence, from Ex. 1, the locus of Q is a conic.

Ex. 4. The three sides of a triangle pass through fixed points, and the extremities of its base lie on two fixed straight lines; shew that its vertex describes a conic. [Maclaurin.]

Let A, B, C be the three fixed points, and let Oa, Oa' be the two fixed straight lines. Suppose triangles drawn as in the figure.

Then the ranges {abcd...} and {a'b'c'd'...} are homographic. Therefore the pencils B {abcd.....} and C {a'b'c'd'...} are homographic.

Ex. 5. If all the sides of a polygon pass through fixed points, and all the angular points but one move on fixed straight lines; the remaining angular point will describe a conic.

Ex. 6. A, A' are fixed points on a conic, and from A and A' pairs of tangents are drawn to any confocal conic, which meet the original conic in C, D and C', D′; shew that the locus of the point of intersection of CD and C'D' is a conic.

The tangents from A to a confocal are equally inclined to the tangent at ▲ [Art. 228, Cor. 3], therefore the chord CD cuts the tangent at A in some fixed point O [Art. 195, Ex. 2]. So also C'D' passes through a fixed point O'. Now if we draw any line OCD through O, one confocal, and only one, will touch the lines AC, AD; and the tangents from A to this confocal will determine C' and D', so that corresponding to any position of OCD there is one, and only one, position of O'C'D'. The locus of the intersection is therefore a conic from Ex. 1.

Ex. 7. If AOA', BOB', COC', DOD'... be chords of a conic, and P any point on the curve, then will the pencils P{ABCD...} and P{A'B'C'D'...} be homographic.

Project the conic into a circle having O for centre.

Ex. 8. If there are two systems of points on a conic which subtend homographic pencils at any point on the curve, the lines joining corresponding points of the two systems will envelope a conic having double contact with the original conic.

Let A, B, C, D..., and A', B′, C', D'... be the two systems of points. Project AA', BB', CC' into equal chords of a circle [Art. 319, Ex. 5]; let P, P' be any pair of corresponding points, and O any point on the circle; then we have O{ABCP}=0{A'B'C'P'}. Hence PP' is equal to AA', and therefore the envelope of PP' is a concentric circle.

Ex. 9. If a polygon be inscribed in a conic, and all its sides but one pass through fixed points, the envelope of that side will be a conic.

This follows from Ex. 7 and Ex. 8.

324. Any two lines at right angles to one another, and the lines through their intersection and the circular points at infinity, form a harmonic pencil.

Let the two lines at right angles to one another be xy= 0, then the lines to the circular points at infinity will be given by x2+y2=0. By Art. 58 these two pairs of lines are harmonically conjugate.

We may also shew that two lines which are inclined at any constant angle, and the lines to the circular points at infinity, form a pencil of constant cross ratio.

Ex. The locus of the point of intersection of two tangents to a conic which divide a given line AB harmonically is a conic through A, B, and the envelope of the chord of contact is a conic which touches the tangents to the original conic from A, B.

Project A, B into the circular points at infinity and the proposition becomes; the locus of the point of intersection of two tangents to a conic which are at right angles to one another is a circle; and the envelope of the chord of contact is a confocal conic.

325. The following are additional examples of the methods of reciprocation and projection.