important theorem: The product of the perpendiculars let fall from any point of a conic on two opposite sides of an inscribed quadrilateral is in a constant ratio to the product of the perpendiculars let fall on the other two sides.

From this property we at once infer, that the anharmonic ratio of a pencil, whose sides pass through four fixed points of a conic, and whose vertex is any variable point of it, is constant. For the perpendicular

but the right-hand member of this equation is constant, while the left-hand member is the anharmonic ratio of the pencil OA, OB, OC, OD.

The consequences of this theorem are so numerous and important that we shall devote a section of another chapter to develope them more fully.

260. If S=0 be the equation to a circle, then (Art. 90) S is the square of the tangent from any point xy to the circle; hence S- kaß=0 (the equation of a conic whose chords of intersection with the circle are a and B) expresses that the locus of a point, such that the square of the tangent from it to a fixed circle is in a constant ratio to the product of its distances from two fixed lines, is a conic passing through the four points in which the fixed lines intersect the circle.

This theorem is equally true whatever be the magnitude of the circle, and whether the right lines meet the circle in real or imaginary points; thus, for example, if the circle be infinitely small, the locus of a point, the square of whose distance from a fixed point is in a constant ratio to the product of its distances from

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 coordi nates 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 sine+b sin"0

where, by Art. 134, c' is the result of substituting in the equation the coordinates 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, q, and we take on it a point 0, 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 coordinates of the centre shall be unity. Let t', t" be the lengths of the tangents, and let S' be the result of substituting the coordinates of their intersection; then

t'2: b'2 :: S': 1, t"2: b"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. 154,

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 curvature may be deduced if in the last example we suppose the two tangents to coincide, in which case the diameter of the circle becomes the radius of curvature (see Art. 398); 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' = 2b′2. For if S' be the result of substituting in the equation the coordinates of the middle point of the chord, 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 B2 = b2S', w = pS', w' = 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 first two conics,

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

Now, on subtracting these equations, we find Ľ3 – 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. t, 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 - Atzt, must be identical with c12 — λc22. Now c12-Ac2 = 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 tits, t2t, and tit, tatą. For t1t2- λtзt1 = 0 must denote a locus passing through these points.

Ex. 3. If 2a, 26, 27, 28 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+ M3, S+N2; 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+ L2, 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

tt - c2 = 0, tt - c2 = 0, tt-c” = 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 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

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) cominon 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 tittats 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.