CHAPTER II

PROJECTIVE CONICS

THE solutions and demonstrations given in this chapter are, generally speaking, of a dual character, based respectively on Newton's Principia and the principles of modern projective geometry.

7. PROBLEM.-Given five points SADBS (Fig. 13) on a conic, to construct the curve.

We shall first give the modern solution of this problem and then deduce the proof from a general theorem given by Newton in his Principia, so as to illustrate the intimate connexion existing between Newton and modern projective geometry, which so far seems to have escaped observation.

SOLUTION I. Take any two of the given five points, such as S and S1, as poles of projection. Through a third A draw any two lines u and u,. From pole S, project the remaining two points B and D by rays S,B and SD intersecting line u, in the points b, and d1. Similarly from the other pole S draw rays to B and D intersecting the line u in b and d. Join bb, and dd, meeting in the centre of perspectivity S, of the lines u and u,. Then to find a sixth. point on the curve, draw any ray through S, intersecting u and u, in k and k, and project k from S and k, from S, by lines meeting in K, a point on the required curve.

PROOF.-Newton has solved this problem for a particular case of which we will now give a short account, and thence deduce the more general method described above.

Let ABCD (Fig. 14) be a quadrangle inscribed in an ellipse, P a point upon the ellipse; then, if PS and PQ be two chords meeting the sides of the quadrangle in S and will be constant.

T, R and Q respectively, the ratio

PR. PQ

PS. PT

CASE I. For, first, let PR and PQ be parallel to the side AC, PS and PT parallel to the adjacent side AB, the opposite sides AC and BD being parallel to each other. Then the line bisecting the sides AC and BD will be a diameter of the ellipse which will bisect RQ in O, and PO

will be the ordinate of P parallel to the axis conjugate of this diameter. Produce PO to K, making OK = OP; then K will be a point on the ellipse.

Hence (Drew, ch. ii.,

Prop. XXIX.) if x be the centre of the ellipse and xz be parallel to PO, we have

PQ. QK

AQ. QB = (27),

a constant ratio for all positions of P, provided that, as assumed, PQ and PS are drawn parallel to the adjacent sides of the quadrangle. But

therefore

CASE II.

When BD, and AC are not parallel, draw

BD parallel to AC meeting the conic in D and the line. Join CD, intersecting PQ in R and D,N

PST in T.

parallel to PQ in M. Then by similar triangles,

whence, by multiplication,

PQ. RR D1N. D1M

PS. TT AN. BN

(3).

But, by Case I., since D, is a point on the ellipse and similarly situated with respect to M and N as P is with respect to R and Q,

and the lines R'T' and RT are parallel; so that, in order to construct the ellipse, all that is required is to divide the two lines PT and PQ by a series of parallel lines meeting them in points T and R which, being projected from poles B and C respectively, will intersect by means of their corresponding rays in points upon the required conic. Such is Newton's method, which we will now proceed to extend to the case illustrated in Fig. 13. It will be observed that when, as in Fig. 14, the lines PT and PR are drawn parallel to the adjacent sides AC and AB, the rays R'T', RT are parallel, and the centre of perspective of the punctuated lines PT and PR lies at an infinite distance. When, however, the lines PT and PR are shifted into the positions PT' and PR' (Fig. 15) being no longer parallel to the adjacent sides branching from A, the points D, E, O on the ellipse are projected from B upon line PT' in d,e,o,, and from pole C upon the line PR' in d'e'o,'. Now, in order to demonstrate the method given at the beginning of this article, it is necessary to shew that the lines joining d, and d', e, and e', o, and o,', all meet in one and the same point