ᏃᎡ (eq. 3). Thus, in ZS being given in direction by the ratio e Fig. 63, set off s"" per- ZS and = TZ n, being the same value for any point on the line TZ; for, if Z' be such a point and Z'S' be drawn perpen dicularly to QS', we have Z'S' = n. Hence, from equation (4), ZA n Therefore, in order to find Z, take any point Z' in TZ, n from which with radius Z'A': TZ' draw an arc meeting = TA produced in A'. Then from A draw AZ parallel to A'Z' intersecting TZ in the required point Z. COR. I.—If AZ and BZ be equal, the point Z will lie on the perpendicular bisecting AB. COR. II.-If AZ = BZ = CZ, the point Z will lie at the intersection of the two perpendiculars traversing respectively the points of bisection of AB and AC; and, therefore, at the centre of the circle passing through A, B, and C. COR. III.—If the sums were given instead of the differences, the points A, B, C would be foci of two ellipses, whose major axes would be BZ + AZ and AZ + CZ respectively, and whose ellipticities e= BZ+AZ would be greater than unity. and e' = AZ+CZ 45. PROBLEM.-Given the focus S, the tangent TP at P a PN M S R Fig. 64. point on the conic, together with a second point P' on the curve; to find the second focus H (Fig. 64). Let fall the perpendicular ST upon the tangent TP and produce it to V, making TV = ST. Join and produce VP, and join SP and SP'. Then and SP major axis - HP SP' major axis - HP'. Thus, by subtraction, we find that the difference HP HP is equal to the known difference SP' - SP. Hence H will lie at the intersection of VH with the hyperbola whose foci coincide with P and P' and whose ellipticity is SP'-SP e= PP' 46. PROBLEM.-Given the focus S and three points B, C, D on the conic, to construct the curve (Fig. 65). K H F Α E Fig. 65. Join BC and CD, producing them to E and F so that Then draw SG perpendicular to the line EF, and set off the points A and a on SG, making so that A and a are the vertices and Aa the major axis of the conic which, according as GA> or = or <SA, will be an ellipse, parabola, or hyperbola. which shews that EF is the directrix of the curve. 47. PROBLEM. To find the position of a body moving in a given parabolic curve at any assigned time, assuming the focal radius vector of the body to describe equal areas in equal times. Let (Fig. 66) GS be made equal to SA, S being the focus and A the vertex of the parabola. Erect AGS Fig. 66. the perpendicular GH, making area described GH=3M, where M= 4AS and about H as a centre with radius HS describe a circle meeting the curve in the required position P of the body at the given time. PROOF. We have by construction AG2+GH2=HP2=(AO - AG)2+(GH – PO)2 wherefore But =(AO2+PO2 - 2GAO) – 2GH. PO + (AG2+GH2); 2GH. PO=AO2+ PO2 - 2GAO. 2GAO=1.4SAO=1PO2, ... 2GH. PO=AO2+ PO2, where AO2=AO. PO2 4AS Hence, multiplying both sides by 2 AS 3' PO' we have the area described in given time is ASP. COR. I. When the body is at P' perpendicularly over AS S, the assigned area is AS; M'AS÷ 4AS = 3 At any other time t, the assigned area GH': = 3M' = AS. directly proportional to t or ASP = nt, where n is area described per unit of time, we have d (3 nt 3 n dt GH=2()-1 dt 4 AS 4 a an expression for the velocity of the point H. = COR. III. Since AH HP, the perpendicular from H upon AP bisects the latter. Hence we can find the time for any arc AP by joining AP, and erecting a perpendicular at its middle point to meet GH in H, and then using the proportion given in Cor. I. 48. PROBLEM.-To find the position of a body moving in an ellipse at any assigned time. Produce the major axis OA (Fig. 67) to G, making OA OG trochoid traced by the point A, when the outer circle rolls on line GX. Then t being the assigned time, T the period of revolution, make GK of such a length that |