112 GIVEN CONJUGATE DIAMETERS. First Method. By continuous motion (fig. 62), From C draw Ca perpendicular to CD and through P draw Pa parallel to CD meeting Ca in a. On CP make Cd=CD and on Ca make Cp=CP. Through a draw ab parallel to pd meeting CP in b. If a triangle equal and similar to the triangle ab°C be moved round so that the angle a is always on the diameter DCD, and the angle b on PCP,, the angle C will be on the curve. The most convenient way of proceeding practically is to cut a strip of paper of breadth equal to the perpendicular distance between C and ab. The points a and b can then be marked off on one edge (as at a,b,) and the point C on the other edge (as at C). The slip can easily be adjusted in any number of positions and the corresponding positions of C, marked. Any number of points on the curve may thus be determined*. 1 Second Method (fig. 63). Draw PM, PM, parallel to CD and DMM parallel to CP meeting PM in M and PM, in M ̧. Divide MD into any number of equal parts 1, 2, 3... and CD into the same number of equal parts. Then lines drawn from P to any of the points on MD intersect lines drawn from P, through the corresponding points on CD in points on the curve, and thus any number of points in the quadrant PD can be determined. * I am indebted to Prof. Minchin for this construction. 1 Similarly if MD be divided into any number of equal parts 1', 2', 3'... and CD into the same number 1, 2, 3... lines drawn from P, to the points on MD intersect lines drawn from P to the corresponding points on CD in points on the required curve, and thus the quadrant DP, can be determined. When half the curve is drawn the remainder can be put in by symmetry, since every diameter is bisected by the centre; thus if QCQ, be drawn and C'Q, be made equal to CQ, Q, will be a point on the curve and similarly for any other points on the semi-ellipse P, D1, P,. 1 Third Method (fig. 64). PCP,, DCD, are the given conjugate diameters. Draw PM, P,M, parallel to CD and MDM, parallel to CP meeting PM in M and P ̧M, in M ̧ 1° Draw the line PD and take on it any number of points 1, 2, 3... Draw the lines la, 2b, 3c... parallel to CD meeting DM in a, b, c.......; and the lines M1, M2, M13..... meeting MP in a', b', c'........... Then the lines aa', bb', cc'... will be tangents to the curve, which must be drawn in to touch these lines, so giving the quadrant PD. A similar construction will give a second quadrant DP1, and the remaining semi-ellipse can of course be put in similarly or by drawing any number of diameters. 2 3 3 2 3 Fourth Method (fig. 64). Draw PM,, P,M.E parallel to CD, and M ̧Ð ̧M ̧ parallel to PP, meeting PM, in M, and PM ̧ in M ̧. Make M ̧E=P1M, and divide PM, into any number of equal parts as at 1, 2, 3.... Draw E1, E2, E3... cutting D ̧M ̧ in f, g, h 3 1 3 2 114 GIVEN AXIS AND POINT. respectively. Then the lines joining corresponding points on PM, and MD1, as ƒ3, g2 and so on, will be tangents to the curve, which must therefore be drawn touching these lines. 2 Similarly for the remaining quadrants, or as before, when half the ellipse is obtained the other can be put in by symmetry. To draw a tangent at any point of an ellipse having a given pair of conjugate diameters. Let Q (fig. 64) be the point, PCP1, DCD, the given conjugate diameters. Draw QN parallel to CP meeting CD, in N, so that CN is the abscissa and QN the ordinate of Q referred to the given conjugate diameters as axes. Make Cn on CP, equal to CN and Cd= CD, and draw through d a line dT parallel to nD, cutting CD1 in T. The line QT will be the tangent at Q, for by similar triangles 1 CT: Cd :: CD ̧ : Cn, i.e. CT: CD, CD, CN (Prop. 3, p. 107). PROBLEM 64. To describe an ellipse, one axis and a point (1) on the curve being given (Fig. 55). The axis is of course given in direction and magnitude, and this really involves the centre of the curve and the position of the other axis. First, suppose the major axis A4, given. Bisect it in Cand draw BCB, perpendicular to AA,. From P with AC as radius 1 mark the point a on BB, and draw Pa cutting AA, in b. Pb will be the length of the semi-minor axis, which can therefore be marked off from C to B and B1. Second, if the minor axis BB, is given. Bisect it in C, through C draw ACA, perpendicular to BB1. From P with radius BC mark off the point b on AA, and draw Pb, producing it to meet BB, in a. Then Pa will be the length of the semi-major axis, which can be set off from C to A and A ̧. The construction is obvious from the original method of drawing the curve. PROBLEM 65. To describe an ellipse, an axis ACA, and a tangent Tt being given (Fig. 65). T, t are the points in which the given tangent cuts the axes. Take CN on CA, a third proportional to CT, CA (i.e. on CB make Ca=CA, draw An parallel to Ta, cutting CB in n, and make CN=C'n). Then N is the foot of the ordinate of the point of contact of the given tangent (Prop. 3, p. 107), and therefore by drawing NP perpendicular to CA meeting the tangent in P, a point on the curve is determined and the problem reduces to Problem 64. 116 GIVEN CENTRE, 2 POINTS & DIRECTIONS OF CONJ. DIAM“. PROBLEM 66. To describe an ellipse, the directions of a pair of conjugate diameters CA, CB, a tangent PT and its point of contact P being given (Fig. 65). In the figure the given conjugate diameters are the axes, but the construction holds in any case. Through P draw PN parallel to CB meeting CA in N. Take CA a mean proportional between CN and CT, which determines the length of the semi-diameter CA. Similarly determine the length CB. PROBLEM 67. To describe an ellipse, the centre (C), two points on the curve (P and Q), and the directions of a pair of conjugate diameters (CA, CB) being given. The lengths CA, CB are not given (Fig. 66). From P and Q draw PM, QN parallel to BC meeting CA in M and N. [In order that the problem may be possible, if PM is 1 1 less than QN, CM must be greater than CN.] Produce PM to E and P, making MP, equal to MP; P, will evidently be a point on the curve. Similarly, drawing EQnQ, parallel to CA meeting PM in E and CB in n, and making nQ1 = nQ, Q1 will be a point on the curve. Through M draw MX parallel to PQ and MY parallel to P1Q, meeting EQ, in X and Y respectively. Find ND a mean proportional between EX and EY and set it up from Non a perpendicular to CA. [The mean proportional may con 1 |