a relation which is satisfied by the co-ordinates of every line passing through the point P. This equation therefore represents the point P.

14. In this system, as in that described in the former part of the present chapter, an equation represents the curve, the co-cordinates of whose tangents satisfy it, and an equation of the nth degree will therefore represent a curve of the nth class.

15. If the perpendicular OQ let fall from O on the straight line HK (fig. 23) be denoted by p, and the angle QOX by p, we shall have

and a point will then be represented by the equation

a cos +b sin & = p ;

an equation which, if a+b be put = c2, and

We thus obtain a method of representing curves by a relation between the perpendicular from a fixed point on the tangent and the inclination of that perpendicular to a fixed straight line. These may be called the tangential polar coordinates of the curve. This method will be found discussed in the Quarterly Journal of Pure and Applied Mathematics. Vol. I. p. 210.

EXAMPLES.

1. Prove that the distance between the points aέ+ by = 1, a'§ + b'ŋ= 1, is {(a' — a)2 + (b′ — b)2}1.

2. Prove that the cosine of the angle between the lines (§, n),

3. Prove that the distance from the point (a + bŋ = 1) to the line (§, n,) is (a§, + bn, − 1) (§, 2 + n‚3)−§.

2

4. Prove that the equation & + n2 + 2P§ + 2Qn+ R= 0, represents a conic, of which the focus is the origin.

What are the co-ordinates of its directrix? What is its eccentricity, and what its latus-rectum ?

Prove that the equation p=a+c cos represents a circle; and determine the radius of the circle.

6. Prove that the evolute of the ellipse a3ç3 + b2n2 = 1 is represented by the equation

145

CHAPTER VIII.

ON THE INTERSECTION OF CONICS, AND ON PROJECTIONS.

1. We shall here say a few words on the subject of the intersection of two conics, as an acquaintance with this branch of the subject will be useful in future investigations.

Since every conic is represented by an equation of the second degree, it follows that any two conics intersect in four points, which may be (1) all real, (2) two real and two imaginary, or (3) all imaginary.

2. Through these four points of intersection three pairs of straight lines can be drawn. If the four points be called P, Q, R, S, the pairs of straight lines will be PQ and RS, PR and QS, PS and QR. If PQ and RS intersect in L, PR and QS in M, PS and QR in N, the points L, M, N are called (see Art. 15, Chap. II.) the vertices of the quadrangle PQRS. Also the three points L, M, N will form, with respect to every conic passing through the points P, Q, R, S, a conjugate triad; and therefore, each of them will have the same polar with respect to all such conics.

3. The equations of the pairs of lines PQ, RS, &c. (the sides and diagonals of the quadrangle) may be found as follows. Let the equations of the conics be

$ (a, B, y) = ua2 + vß2 + wy2 + 2u'By + 2v'ya +2w'aß = 0 ..... (1), ¥ (a, B, y) = pa2 + qB2 + ry2 + 2p'By + 2q'ya + 2r'aß = 0 ..... (2)

then every conic passing through their four points of intersection will be represented by an equation of the form

$ (α, B, y) + k↓ (2, B, y) = 0

(3). ·

If the left-hand member of this equation break up into two factors, the conic degenerates into two straight lines, real or imaginary. The condition that this should happen is

a cubic for the determination of k, of which the roots are either all real, or one real and two imaginary. If the roots be all real, the vertices of the quadrangle, which will be the centres of the several conics included in the form (3), will be all real. If one root only be real, then one vertex only of the quadrangle will be real. We proceed to consider how the reality of the vertices L, M, N depends upon that of the points P, Q, R, S.

4. First, suppose all the four points P, Q, R, S to be real, then it is clear that all the vertices will be real.

5. Next, let two of the points, P, Q, for example, be real, and R, S, imaginary.

Then, the line PR can have no other real point but P. For, if it had, it would itself become a real line, and we should have a real line cutting a real conic in one real and one imaginary point, which is impossible.

Hence the point M, which lies on PR, is imaginary. Similarly the line PS, and the point N, which lies on it, are imaginary. The real vertex must therefore be L, which lies on PQ.

We may observe that the line RS will be real. For the two lines PQ, RS, considered as one locus, will be represented by equation (3) when for k is substituted the real value corresponding to the point L. Hence the form of the expression & (a, B, y) + k ↓ (a, B, y) answering to PQ, contains one real linear factor, and the other linear factor, which answers to RS, will therefore also be real.

VERTICES OF A QUADRANGLE.

147

6. Thirdly, let all the four points of intersection be imaginary. Then the three vertices will all be real,

For, by what has been shewn above, one vertex is necessarily so. Take this as the angular point A of the triangle of reference, and let its polar with respect to the two conics be taken for the side BC.

The point B being chosen arbitrarily, let its polar with respect to one of the conics be taken as AC, Then this conic may be represented by the equation

a2 + vß2 + wy2 = 0

Let the other be represented by

..(1).

a2+qß2 + ry2+2p'By = 0 ....................... (2).

Since the four points of intersection are imaginary, the roots of the quadratic

(qv) B® + (r− 2) y2+2p' By = 0

Now let (0, g, h) be the co-ordinates of either vertex. Then, since it has the same polar with respect to both conics, the equations

vgB+ why = 0,

(gg+p'h) B+ (p'g+rh) y = 0

will represent the same straight line, hence

The two values of, given by this equation, will deter

mine the vertices. Now the roots of this equation are real or imaginary, as

(qw - rv)2 + 4vwp0.
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