Bt2 The secant joining the points where t has the values t and ť is represented by the equation ßt2 Bt2 -at (§ 30) 2 2. ť 2 Put ľ=t, and write x for x (t' – t) [which may have any value] and the equation of the tangent at the point (t) is p= att +x (a + Bt). 2 Bt2 + 2 2 2 = or the intercept of the tangent on the diameter is – the abscissa of the point of contact. Otherwise: the tangent is parallel to the vector a+ßt or Bt2 at+ ßt2 or att or OQ+OP. But TP = TO + OP, hence To = OQ. (g.) Since the equation of any tangent to the parabola is B1 p = att +x (a + ßt), let us find the tangents which can be drawn from a given point. Let the vector of the point be p=pa+qß (24). Since the tangent is to pass through this point, we have, as conditions to determine t and X, t +* = P, t2 2 by equating respectively the coefficients of a and B. Hence t=p+p2 – 29. Thus, in general, two tangents can be drawn from a given point. These coincide if p2 = 2; that is, if the vector of the point from which they are to be drawn is p=pa+aß = pat B, 2 i.e. if the point lies on the parabola. They are imaginary if 29>p, i.e. if the point be pa p=pat +r) B, go being positive. Such a point is evidently within the curve, as at = = + xt = 9; 2p2 R, where OQ = B, QP = pa, PR = r. 2 2 (h.) Calling the values of t for the two tangents found in (9) ty and t, respectively, it is obvious that the vector joining the points of contact is 31,2 – ata 2 which is parallel to ti + t2 B ; 2 or, by the values of tq and t, in (9), a+pB. Its direction, therefore, does not depend on q. In words, If pairs of tangents be drann to a parabola from points of a diameter produced, the chords of contact are parallel to the tangent at the vertex of the diameter. This is also proved by a former result, for we must have OT for each tangent equal to QO. (i.) The equation of the chord of contact, for the point whose vector is p=pa +9B, is thus p= at + Bt? +y(a +PB). 2 Suppose this to pass always through the point whose vector is p=aa+b3. Then we must have tity ti+y = a, = b ty=p+vp2 – 2 pa +26. Comparing this with the expression in (g), we have q=pa-6; that is, the point from which the tangents are drawn has the vector p=pa+(pa-6) B =-6B+p(a+aß), a straight line (§ 28 (1)). The mere form of this expression contains the proof of the usual properties of the pole and polar in the parabola ; but, for the sake of the beginner, we adopt a simpler, though equally general, process. Suppose a = 0. This merely restricts the pole to the particular diameter to which we have referred the parabola. Then the pole is , where р b3; and the polar is the line T'U, for which p= --bB+pa. 2 or Hence the polar of any point is parallel to the tangent at the extremity of the diameter on which the point lies, and its intersection with that diameter is as far beyond the vertex as the pole is within, and vice versá. (3.) As another example let us prove the following theorem. If a triangle be inscribed in a parabola, the three points in which the sides are met by tangents at the angles lie in a straight line. Since 0 is any point of the curve, we may take it as one corner of the triangle. Let t and t, determine the others. Then, if W1, W2, Wg represent the vectors of the points of intersection of the tangents with the sides, we easily find W3 = ti+t tı– 2 W3 = 0. 2 2 These values give 2t, 2t-ti W2 tty Also = 0 identically. ti t Hence, by g 30, the proposition is proved. (k.) Other interesting examples of this method of treating curves will, of course, suggest themselves to the student. Thus p = a cost+ß sint p= ax+BV1 - x2 represents an ellipse, of which the given vectors a and ß are semiconjugate diameters. В Again, p= a tan x + ß cotx t evidently represents a hyperbola referred to its asymptotes. But, so far as we have yet gone with the explanation of the calculus, as we are not prepared to determine the lengths or inclinations of vectors, we can investigate only a very small class of the properties of curves, represented by such equations as those above written. or p= att + or р (2.) We may now, in extension of the statement in § 29, make the obvious remark that ρ = Σφα is the equation of a curve in space, if the numbers P1, P2, &c. are functions of one indeterminate. In such a case the equation is sometimes written p= $(t). But, if P1, P2, &c. be functions of two indeterminates, the locus of the extremity of p is a surface; whose equation is sometimes written p= $(t, u). (m.) Thus the equation p= a cost+ß sin t+yt belongs to a helix. Again, p=pa+B+ry with a condition of the form apa + bq2 + cr2 = 1 belongs to a central surface of the second order, of which a, b, y are the directions of conjugate diameters. If a, b, c be all positive, the surface is an ellipsoid. 32.] In Example (f) above we performed an operation equivalent to the differentiation of a vector with reference to a single numerical variable of which it was given as an explicit function. As this process is of very great use, especially in quaternion investigations connected with the motion of a particle or point; and as it will afford us an opportunity of making a preliminary step towards overcoming the novel difficulties which arise in quaternion differentiation; we will devote a few sections to a more careful exposition of it. 33.] It is a striking circumstance, when we consider the way in which Newton's original methods in the Differential Calculus have been decried, to find that Hamilton was obliged to employ them, and not the more modern forms, in order to overcome the characteristic difficulties of quaternion differentiation. Such a thing as a differential coefficient has absolutely no meaning in quaternions, except in those special cases in which we are dealing with degraded quaternions, such as numbers, Cartesian coordinates, &c. But a quaternion expression has always a differential, which is, simply, what Newton called a fluxion. As with the Laws of Motion, the basis of Dynamics, so with the foundations of the Differential Calculus; we are gradually coming to the conclusion that Newton's system is the best after all, с p= att dp= (a - )at. = From the very nature of the question it is obvious that the length of dp must in this case be ds. This remark is of importance, as we shall see later; and it may therefore be useful to obtain afresh the above result without any reference to time or velocity. 39.] Following strictly the process of Newton's VIIth Lemma, let us describe on Pq2 an arc similar to PQ2, and so on. Then obviously, as the subdivision of ds is carried farther, the new arc (whose length is always ds) more and more nearly coincides with the line which expresses the corresponding approximation to dp. 40.] As a final example let us take the hyperbola B t Here B dt. 12 This shews that the tangent is parallel to the vector B at t In words, if the vector (from the centre) of a point in a hyperbola be one diagonal of a parallelogram, two of whose sides coincide with the asymptotes, the other diagonal is parallel to the tangent at the point. 41.] Let us reverse this question, and seek the envelope of a line which cuts off from two fixed axes a triangle of constant area. If the axes be in the directions of a and B, the intercepts may B evidently be written at and Hence the equation of the line is t (30) ß t dp = 0. = B ß 0 dæ. Hence (1 — 2) dt - t dx = 0, da and redt t2 This gives = {a--(2 + a)}dt+ -at) dc * o (i – at t = 0. t * We are not here to equate to zero the coefficients of dt and dx; for we must remember that this equation is of the form 0 = pa+qB, where and q are numbers; and that, so long as a and B are actual and non-parallel vectors, the existence of such an equation requires р p = 0, 9 = 0. |