COR. It is evident that [PQRP']=[P'Q'R'P].

28. PROP. Any two conjugate points form, with the two foci, an harmonic range.

Let K1, K ̧ be the foci, then

K1P=p−k, K‚P=p+k,

kc

and K,P. K,P = (k) (k + p) = (p2 − k') ;

Ρ

:. K‚P.K ̧P' = K1P'. K„P,

- k2);

or the four points in question form a harmonic range.

Conversely, if there be a system of pairs of points in a straight line, such that each pair forms, with two given points, an harmonic range, the aggregate of the pairs of points will form a system in involution, of which the two given points are the foci.

29. A system of straight lines, intersecting in a point, may be treated in the same manner as a system of points lying in a straight line, the sine of the angle between any two lines taking the place of the mutual distance of two points. From the proposition, proved in Art. 20, it will follow that, if a system of straight lines in involution be cut by a transversal, the points of section will also be in involution.

33

CHAPTER II.

SPECIAL FORMS OF THE EQUATION OF THE SECOND DEGREE.

1. WE now proceed to the discussion of the curve represented by the equation of the second degree. We shall first prove that every curve, represented by such an equation, is what is commonly called a conic section; and then, before proceeding further with the consideration of the general equation, shall investigate the nature of the curve corresponding to certain special forms of the equation.

PROP. Every curve represented by an equation of the second degree is cut by a straight line in two points, real, coincident, or imaginary.

The general equation of the second degree is represented by

ua2 + vß2 + wy2+ Qu'ßy + 2v'ya + 2w'aß = 0.

To find where the curve, of which this is the equation, is cut by the straight line

la + mB + ny = 0,

we may eliminate a between the two equations. This will give us a quadratic for the determination of to each of the

B

, γ

two values of this ratio, real, equal, or imaginary, one value of a will correspond; whence it appears that the straight line and the curve cut one another in two real, coincident, or imaginary points.

Hence, the curve is of the same nature as that represented by the equation of the second degree in Cartesian co-ordinates, and is, therefore, a conic section.

2. We shall now inquire what are the relations of the conic section to the triangle of reference, when certain relations exist among the coefficients of the equation.

First, suppose u, v, w, all = 0.

The equation then assumes the form

u'By + v'ya + w'aß = 0,

which we shall write

λβγ + μγα + ναβ = 0.

Now, if in this equation we put a = 0, it reduces itself to λβγ = 0,

which requires either that ẞ=0, or that y = 0.

It hence appears that the curve passes through two of the angular points (B, C) of the triangle of reference. It may similarly be shewn to pass through the third. Hence the equation

λβη + μγα + ναβ = 0,

represents a conic, described about the triangle of reference.

3. Let us now inquire how the line

If in the equation of the conic we put +=0, or,

which is the same thing, uy+v8 = 0, it reduces to λBy=0.

μ

ע

which it meets the lines = 0, y=0; but these two points coincide, since the line in question evidently passes through the point of intersection of B=0 and y=0. Hence the straight line and the conic meet one another in coincident points, that is, they touch one another at the point A.

Similarly, the equations of the tangents at B and C are

4. To determine the position of the centre of the conic.

Through the angular points A, B, C of the triangle of reference draw the tangents EAF, FBD, DCE.

Fig. 14.

Bisect

AC, AB respectively in H, I, join EH, FI, and produce them to intersect in O. Then, since every straight line drawn through the intersection of two tangents so as to bisect their chord of contact passes also through the centre, O will be the centre of the conic.

COR. We may hence deduce the relation which must hold between λ, μ, v, in order that the conic may be a parabola. For, since the centre of a parabola is at an infinite distance, its co-ordinates will satisfy the equation

aa+bB+cy=0.

We hence obtain the following equation:

λ2a2 +μ2b2 + v2c2 — 2μvbc — 2vλca — 2λμab = 0,