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This gives, by the result of Art. 8,

Nb+ McLc + Na = Ma + Lb.

To solve these equations, put each member equal to r, we then get

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Adding together the last two of these equations, and subtracting the first, we get

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Similar expressions being obtained for M and N, we see that

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It may similarly be proved that the escribed circles, of which the centres are respectively given by

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We may remark that, at every point in the circle which touches BC externally, a is essentially negative, so that the form (-a) represents a real quantity. Similarly the appearance of (—), (—y) in the equations of the other two escribed circles may be accounted for.

11. The next form of the general equation of the second degree which we propose to consider is that in which u', v', w', the respective coefficients of 2ẞy, 2ya, 2aß, are all = 0. The equation then assumes the form

ua2 + vß2 + wy2 = 0.

We observe in the first place, that if this equation represent a real conic, the coefficients of a2, B2, y2, cannot be all of the same sign. Suppose the coefficient of a to be of a different sign from the other two, then writing, for convenience of future investigations, L2, - M2, - N2 for u, v, w respectively, our equation assumes the form

L2a2 — M2ß2 — N2y2 = 0.

12. We have now to enquire how this conic is related to the triangle of reference.

Putting B=0, we get

La+Ny.

The interpretation of this equation is, that the two straight lines drawn from B to the points in which the conic is cut by CA, form, with BC, BA, an harmonic pencil.

It may similarly be proved that the two straight lines drawn from C to the points in which the conic is cut by AB, form, with CA, CB, an harmonic pencil.

If we put a =0, we get

MB= + V(− 1) Ng

shewing that BC cuts the conic in two imaginary points. The analytical condition of harmonic section is, however, satisfied here also.

13. We may next investigate the equations of the tangents drawn through the points A, B, C.

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If in the equation of the conic we put La Ny, we get B=0, shewing that the straight line La-Ny = 0 meets the conic in two coincident points, and, therefore, touches it.

Similarly

La + Ny = 0, La-MB-0, La+MB=0,

are tangents to the conic.

The tangents to the conic drawn through A would be analytically represented by the equations

MB=√(−1) Ny, MB=-√(−1) Ny,

which shew that these tangents are imaginary, or that the point A lies within the concavity of the conic.

14. Since the two tangents drawn through B meet the conic in points situated in the line CA, it follows that CA is the chord of contact of tangents to the conic drawn through B, or that CA is the polar of B, and B the pole of CA with respect to the conic. Similarly, C, AB, stand to one another in the relation of pole and polar.

Again, since the pole of AB is the point C, and the pole of AC is the point B, it follows that the line joining Band C is the polar of the point of intersection of AB, AC, i. e. that A is the pole of BC, and BC the polar of A.

We come then to this conclusion, that when an equation of the second degree does not involve the terms ẞy, ya, aß, the conic represented by it is so related to the triangle of

reference, that each side of the triangle is the polar, with respect to the conic, of the opposite angular point*.

This is expressed by saying that the triangle is self-conjugate with respect to the conic; or that the three angular points of the triangle form a conjugate triad.

The geometrical properties of the conic having been thus established, we shall, in future investigations, write for the sake of symmetry of form, L instead of L2, so that the equation of the conic will be written

L2a2 + M2ß2 + N2y2 = 0.

It must here be borne in mind that one of the three quantities L, M, N is essentially imaginary.

15. Any two conic sections represented by such equations as

L'a2+M2ß2 + N2y2 = 0,

L'2a2+M'2ß2+N'y2=0,

have important relations to one another, which we proceed to consider.

They will of course intersect in four points, which may be real or imaginary. We will first suppose them real, and represent them by the letters P, Q, R, S.

Now the locus of the equation

(L3M" — L'M2) ß2 + (L3N"2 – L'2N3) y2 = 0

passes through all the points P, Q, R, S; and, since it may be resolved into linear factors, represents two straight lines. Suppose them to be PQ and RS. The intersection of these two straight lines is given by the equations

* If the coefficients of B2 and y2 be equal, and the triangle of reference be right-angled at A, the form of the equation shews that A will be a focus of the conic, and BC the corresponding directrix.

(L2M" — L2M2) * B = (L'N2 – L2N') Y,

(L3M'2 — L'2M2) 1 B = — (L'2N2 – L3N'2) Y,

ß:

which evidently give ẞ=0, y=0.

Hence PQ, RS intersect in A.

Similarly, PR, QS intersect in B, and PS, QR intersect in C. Hence, the angular points of the triangle of reference coincide with the intersections of the line joining each pair of points of intersection of the conics with the line joining the other pair. Hence also, if any number of conic sections be described about the same quadrangle*, and the diagonals of that quadrangle intersect in A, while the sides produced intersect in B and C, then A, B, C form, with respect to each of the circumscribing conics, a conjugate triad. The points A, B, C may themselves be called vertices of the quadrangle, or of the system of circumscribing conics.

It will be seen, from the preceding investigation, that any two conics which intersect in four real points can be reduced, by a proper choice of the triangle of reference, to the form

L2a2 + M2ß2 + N2y2 = 0.

The same reduction may also be effected in every case with the reservation that if two of the points of intersection of the conics be real and two imaginary, then two of the angular points of the triangle of reference (or vertices) will be imaginary and the remaining one real. If all the points of intersection be imaginary, the vertices of the conics will be all real. This we shall prove hereafter.

16. To find the condition that a given straight line may

touch the conic.

Let the equation of the straight line be

la+mß + ny = 0.

* I employ the term quadrangle in preference to quadrilateral, considering a quadrangle as a figure primarily determined by four points, a quadrilateral by four indefinite straight lines.

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