two fixed lines, is a conic section; and the fixed lines may be considered as chords of imaginary intersection of the conic with an infinitely small circle whose centre is the fixed point.

261. Similar inferences can be drawn from the equation S-ka 0, where S is a circle. We learn that the locus of a point, such that the tangent from it to a fixed circle is in a constant ratio to its distance from a fixed line, is a conic touching the circle at the two points where the fixed line meets it; or, conversely, that if a circle have double contact with a conic, the tangent drawn to the circle from any point on the conic is in a constant ratio to the perpendicular from the point on the chord of contact.

In the particular case where the circle is infinitely small, we obtain the fundamental property of the focus and directrix, and we infer that the focus of any conic may be considered as an infinitely small circle, touching the conic in two imaginary points situated on the directrix.

262. In general, if in the equation of any conic the co-ordinates of any point be substituted, the result will be proportional to the rectangle under the segments of a chord drawn through the point parallel to a given line.*

For (Art. 148) this rectangle

=

a cos20+2h cose sin 0 + b sin20"

where, by Art. 134, c' is the result of substituting in the equation the co-ordinates of the point; if, therefore, the angle be constant, this rectangle will be proportional to c'.

Ex. 1. If two conics have double contact, the square of the perpendicular from any point of one upon the chord of contact, is in a constant ratio to the rectangle under the segments of that perpendicular made by the other.

Ex. 2. If a line parallel to a given one meets two conics in the points P, Q, P, I, and we take on it a point O, such that the rectangle OP.OQ may be to Op.Oq in a constant ratio, the locus of O is a conic through the points of intersection of the given conics.

Ex. 3. The diameter of the circle circumscribing the triangle formed by two b'b" tangents to a central conic and their chord of contact is ; where b', b" are the p

semi-diameters parallel to the tangents, and p is the perpendicular from the centre on the chord of contact. [Mr. Burnside].

*This is equally true for curves of any degree.

It will be convenient to suppose the equation divided by such a constant, that the result of substituting the co-ordinates of the centre shall be unity. Let t', t" be the lengths of the tangents, and let S' be the result of substituting the co-ordinates of their intersection; then

t'2: 6'2 :: S': 1, t''2: 6'2 :: S': 1.

But also if be the perpendicular on the chord of contact from the vertex of the triangle, it is easy to see, attending to the remark, Note, p. 149,

But the left-hand side of this equation, by Elementary Geometry, represents the diameter of the circle circumscribing the triangle.

Ex. 4. The expression (Art. 242) for the radius of the osculating circle may be deduced from the last example by supposing the two tangents to coincide; or also from the following theorem due to Mr. Roberts: If n, n' be the lengths of two intersecting normals; p, p' the corresponding central perpendiculars on tangents; b' the semi-diameter parallel to the chord joining the two points on the curve, then np + n'p' = 26′2. For if S' be the result of substituting in the equation the co-ordinates of the middle point of the chord, w, w' the perpendiculars from that point on the tangents, and 2ẞ the length of the chord, then it can be proved, as in the last example, that ẞ2 = 6'2S', w = pS', ' = p'S', and it is very easy to see that nw + n'w' = 2p2.

263. If two conics have each double contact with a third, their chords of contact with the third conic, and a pair of their chords of intersection with each other, will all pass through the same point, and will form a harmonic pencil.

Let the equation of the third conic be S=0, and those of the other two conics,

S+L=0, S+ M2=0.

Now, on subtracting these equations, we find L' - M2 = 0, which represents a pair of chords of intersection (L+ M=0) passing through the intersection of the chords of contact (L and M), and forming a harmonic pencil with them (Art. 57).

Ex. 1. The chords of contact of two conics with their common tangents pass through the intersection of a pair of their common chords. This is a particular case of the preceding, S being supposed to reduce to two right lines.

Ex. 2. The diagonals of any inscribed, and of the corresponding circumscribed quadrilateral, pass through the same point, and form a harmonic pencil. This is also a particular case of the preceding, S being any conic, and S+ L2, S + M2 being supposed to reduce to right lines. The proof may also be stated thus: Let t1, t2, C1; t3, t4, C2 be two pairs of tangents and the corresponding chords of contact. In other words, c1, c2 are diagonals of the corresponding inscribed quadrilateral. Then the equation of S may be written in either of the forms

The second equation must therefore be identical with the first, or can only differ from it by a constant multiplier. Hence t1t2- Att must be identical with c12 - λc22. Now cλc2 = 0 represents a pair of right lines passing through the intersection of C1, C2, and harmonically conjugate with them; and the equivalent form shows that these right lines join the points t1tз, t2t and t1t, tatз. For tit2- λtzt = 0 must denote a locus passing through these points.

Ex. 3. If 2a, 26, 27, 26 be the eccentric angles of four points on a central conic, form the equation of the diagonals of the quadrilateral formed by their tangents. Here we have

Hence reasoning, as in the last example, we find for the equations of the diagonals

264. If three conics have each double contact with a fourth, six of their chords of intersection will pass three by three through the same points, thus forming the sides and diagonals of a quadrilateral.

Let the conics be

S+L=0, S+ M2=0, S+N" = 0.

By the last Article the chords will be

L-M=0, M-N=0, N-L=0;

L+M=0, M+N=0, N-L=0;
L+M=0, M-N=0, N+L=0;

L-M=0, M+N=0, N+L=0.

As in the last Article, we may deduce hence many particular theorems, by supposing one or more of the conics to break up into right lines. Thus, for example, if S break up into right lines, it represents two common tangents to S+M", S+N"; and if I denote any right line through the intersection of those common tangents, then S+L also breaks up into right lines, and represents any two right lines passing through the intersection of the common tangents. Hence, if through the intersection of the common tangents of two conics we draw any pair of right lines, the chords of each conic joining the extremities of those lines will meet on one of the common chords of the conics. This is the

extension of Art. 116. Or, again, tangents at the extremities of either of these right lines will meet on one of the common chords.

265. If S+L3, S+ M2, S + N2, all break up into pairs of right lines, they will form a hexagon circumscribing S, the chords of intersection will be diagonals of that hexagon, and we get Brianchon's theorem: "The three opposite diagonals of every hexagon circumscribing a conic intersect in a point." By the opposite diagonals we mean (if the sides of the hexagon be numbered 1, 2, 3, 4, 5, 6) the lines joining (1, 2) to (4, 5), (2, 3) to (5, 6), and (3, 4) to (6, 1); and by changing the order in which we take the sides, we may consider the same lines as forming a number (sixty) of different hexagons, for each of which the present theorem is true. The proof may also be stated as in Ex. 2, Art. 263. If

2

2

2

tt-c2=0, tt-c2=0, tt-c2 = 0,

be equivalent forms of the equation of S, then c1 = c2 = c ̧ represents three intersecting diagonals.*

266. If three conic sections have one chord common to all, their three other common chords will pass through the same point.

Let the equation of one be S=0, and of the common chord L=0, then the equations of the other two are of the form

S+LM=0, S+LN=0,

which must have, for their intersection with each other, L(M-N)=0;

but M-N is a line passing through the point (MN).

According to the remark in Art. 257, this is only an extension of the theorem (Art. 108), that the radical axes of three circles meet in a point. For three circles have one chord (the line at infinity) common to all, and the radical axes are their other common chords.

* Mr. Todhunter has with justice objected to this proof, that since no rule is given which of the diagonals of t1ttat is c1 = + c2, all that is in strictness proved is that the lines joining (1, 2) to (4, 5) and (2, 3) to (5, 6) intersect either on the line joining (3, 4) to (6, 1), or on that joining (1, 3) to (4, 6). But if the latter were the case the triangles 123, 456 would be homologous (see Ex. 3, p. 59), and therefore the intersections 14, 25, 36 on a right line; and if we suppose five of these tangents fixed, the sixth instead of touching a conic would pass through a fixed point.

H II

The theorem of Art. 264 may be considered as a still further extension of the same theorem, and three conics which have each double contact with a fourth may be considered as having four radical centres, through each of which pass three of their common chords.

The theorem of this Article may, as in Art. 108, be otherwise enunciated: Given four points on a conic section, its chord of intersection with a fixed conic passing through two of these points will pass through a fixed point.

Ex. 1. If through one of the points of intersection of two conics we draw any line meeting the conics in the points P, p, and through any other point of intersection B a line meeting the conics in the points Q, q, then the lines PQ, pq, will meet on CD, the other chord of intersection. This is got by supposing one of the conics to reduce to the pair of lines OA, OB.

Ex. 2. If two right lines, drawn through the point of contact of two conics, meet the curves in points P, P, Q, q, then the chords PQ, pq, will meet on the chord of intersection of the conics.

A

B

This is also a particular case of a theorem given in Art. 264, since one intersection of common tangents to two conics which touch, reduces to the point of contact (Cor., Art. 117).

267. The equation of a conic circumscribing a quadrilateral (ay=kBd) furnishes us with a proof of "Pascal's theorem," that the three intersections of the opposite sides of any hexagon inscribed in a conic section are in one right line.

Let the vertices be abcdef, and let ab=0 denote the equation of the line joining the points a, b; then, since the conic circumscribes the quadrilateral abcd, its equation must be capable of being put into the form

ab.cd-bc.ad= 0.

But since it also circumscribes the quadrilateral defa, the same equation must be capable of being expressed in the form

de.fa-ef.ad = 0.

From the identity of these expressions, we have

ab.cd-de.fa= (bc — ef) ad.

Hence we learn that the left-hand side of this equation (which from its form represents a figure circumscribing the quadrilateral formed by the lines ab, de, cd, af) is resolvable into two factors, which must therefore represent the diagonals of that quadrilateral. But ad is evidently the diagonal which joins the vertices