through L draw GH parallel to A B cutting the opposite side BC of the rectangle in M, and through the point O draw KI parallel to AD or BC cutting the opposite side DC in N. In NK or NK produced, make NQ r N equal to NC, and join CG. ; draw QR parallel to GH cutting CB or CB produced in R; make EH and EG each equal to QC, as “t L RTH also EI and EK each equal to PC; then will GH be the major axis and KI the minor axis of the ellipsis required. A. Ho B The demonstration of this method, in which the line QK has Fig. 429. nothing to do with the construction, is as follows : — 1083. The direction of a plane cutting a cone, which produces the form called the hyperbola, has been already described; its most useful properties will form the subject of the following theorems, which we shall preface with a few definitions : — 1. The primary axis of an hyperbola is called the transverse aris. 2. A straight line drawn through the centre of an hyperbola and terminated at each extremity by the opposite curves is called a diameter. 3. The extremities of a diameter terminated by the two opposite curves are called the vertices of that diameter. 4. A straight line drawn from any point of a diameter to meet the curve parallel to a tangent at the extremity of that diameter is called an ordinate to the two abscissas. . A straight line which is bisected at right angles by the transverse axis in its centre, and which is a fourth proportional to the mean of the two abscissas, their ordinate, and the transverse axis, is called the conjugate aris. 6. A straight line which is a third proportional to the transverse and conjugate axis is called the latus rectum or parameter. Q’ 7. The two points in the transverse axis cut by ordinates which are equal to the semi-parameter are called the foci. 1084. ThroReM I. In the hyperbola the squares of the ordinates of the transverse aris are to each other as the rectangular of their abscissas. Let QVN (fig. 430.) be a section of the cone passing along the axis VD, the line of section of the directing plane, HB the line of axis of the cutting plane, the directing and cutting plane being perpendicular to the plane QVN. Let the cone be cut by two planes R perpendicular to the axis passing through the two points P, H, meeting the plane of section in the lines PM, HI, which are ordinates to the circles and to the figure of the section, of the same time. Q By the similar triangles APL and AHN, AP : PL::AH : HN; Fig. 450 And by the similar triangles BPK and BHQ, BP: PK:: BH : HQ. g-i-xù. Therefore, taking the rectangles of the corresponding terms, AP x BP: PL x PK:: A H x 1085. TheoreM II. In the hyperbola, as the square of the transverse aris is to the square of the conjugate aris, so is the rectangle of the abscissas to the square of their ordinate. Let AB (jig. 431.) be the transverse axis, GE the conjugate axis, G—io-e C being the centre of the opposite curves; also let HI and PM be or- dinates as before; then will AB2 : GE2::PA X PB : PM2, Coroll. Hence A B* : GE*::CP2-CA3 : PM3 (fig. 432.). For let the cutting plane of the opposite hyperbola intersect two circles parallel to the base in HI and PM, and let the cone be cut by another plane parallel to the base, passing through the centre C of the transverse axis, and let mn be the diameter of the circle made by the plane QVN. Then ACm, APK are similar, and AC : Cm :: A P : PK. And as BCn, BPL are similar, BC : Cn :: BP : PL. Therefore, taking the rectangles of the corresponding terms, BC x A C : Cn x Cm :: BP × AP : PL x PK. But BC = AC ; Cm x Cn = Ct?; and PL x PK = PM2. Therefore AC3 : Ct? :: A P × B P : PM2. Though Ct is not in the same plane, it is what is usually called the semi-conjugate axis, and it agrees with what has been demonstrated in the first part of this proposition. 1086. TheoreM III. In the hyperbola, the square of the semi- Fig. 432. conjugate aris is to the square of the semi-transverse aris as the sum of the squares of the semi-conjugate aris and of the ordinate parallel to it is to the square of the abscissas. Let AB (fig. 433.) be the transverse axis, GE the conjugate, C the cen- \ / In the same manner we find fM = coor, CA; And, subtracting the upper equation from the lower, fM-FM = 2CA. Coroll. 1. Hence is derived the common method of describing the hyperbolic curve mechanically. Thus : — In the transverse axis AB produced (fig. 435.), take the foci F, f, and any point I in the straight line AB so produced. Then, with the radii AI, BI, and the A a centre F, f, describe arcs intersecting each other; call the points of intersection E, then E will be a point in the curve; with the same distances another point on the other side of the axis may be found. In like manner, by taking any other points I, we may find two more points, one on each side of the axis, and thus continue till a sufficient number of points be found to describe the curve by hand. By the same process, we may also describe the opposite hyperbolas. Coroll. 2. Because coor is a fourth proportional, CA, CF, CP CA : CF::CP : CA 4 FM. 1089. ThroREM VI. As the square of the semi-transterse aris is to the square of the semi-conjugate, so is the difference of the squares of any two abscissas to the difference of the squares of their ordinates. CA2 : CE*:: CP2 – CA2 : PM2 (fig. 436.), By Theor. II. to ; cf. off Too' iii. ) Therefore, by CH2 – CA2 : CP2 – CA2 :: HI2 : PM2 or last equations, and neglecting the common factors, it will be PM(CP for I H + HN, and neglecting the common factors and common terms, / ti the points P and A to remain constant, the point T will also remain constant; therefore all the tangents will meet in the point T which are drawn from the ex tremity of the ordinate M of every hyperbola described on the same N axis A.B. 1091. Theon EM VIII. Four perpendiculars to the transverse aris in- N • tercepted by it and a tangent, will be proportionals when the first and last d § hare one of their ertremities in each rerter, the second in the point of con- But by the similar triangles TAD, TPM, TCE, and TBF, the sides AT, PT, CT, and BT are proportional to the four perpendiculars AD, PM, CE, BF, Therefore AP : PM :: CE : BF. 1092. Theon EM IX. The two radius rectors meeting the curre in the same point will make equal angles with a tangent passing through that point. (Fig. 439.) a And by the similar triangles TFM, Tf R, FT : f'T:: FM : frt. Therefore f R is equal to fM; consequently the angle f RM is equal to the angle fMR: and because flt is parallel to fM, the angle FMT is equal to the angle f RM ; therefore the angle FMT is equal to the angle f RM. 1093. ProbleM. I. To describe an hyperbola by means of the end of a ruler moveable on a pin F (fig 440.) fired in a plane, with one end of a string fired to a point E in the same plane, and the other extremity of the string fastened to the other end C of the ruler, the point C of the ruler being moved towards G in that plane. While the ruler is moving, a point D being made to slide along the edge of the ruler, kept close to the string so as to keep each of the parts CD, DE of the string taught, the point D will describe the curve of an hyperbola. \ If the end of the ruler at F (fig. 441.) be made moveable about the point E, and the string be fixed in F and to the end C of the ruler, as before, another curve may be described in the same manner, which is called the opposite hyperbola : the points E and F, about which the ruler is made to revolve, are the foci. There are many occasions in which the use of this conic section occurs in architectural details. For instance, the profiles of many of the Grecian mouldings are hyperbolic ; and in conical roofs the forms are by intersections such that the student should be well acquainted with the methods of describing it. 1094. PRob. II. Given the diameter AB, the abscissa BC, and the double ordinate DE in position and H-----, magnitude, to describe the hyperbola. (Fig. 442.) Fig. 440. Fig. 441. Through B draw FG parallel to DE, and draw DF and EG parallel to AB. Divide DF and DC each into the same number of equal parts, and from the points of division in B F draw lines to B, also from the points of division in DC draw straight lines to A; then through the points of intersection found by the lines drawn through the corresponding points draw the curve DB. In like manner the curve EB may be drawn so that DBE will form the curve on each side of the diameter AB. If the point A be considered as the vertex, the opposite hyperbola HAI may be described in the same manner, and thus the two curves formed by cutting the opposite cones by the same plane will be found. By the theorists, the hyperbola has been considered a proper figure of equilibrium for an arch whose office is to support a load which is greatest at the middle of the arch, and diminishes towards the abutments. This, however, is matter of consideration for another part of this work. of The PARAbol A. 1095. Definitions.—1. The parameter of the axis of a parabola is a third proportional to the abscissa and its ordinate. 2. The focus is that point in the axis where the ordinate is equal to the semi-parameter. 3. The diameter is a line within the curve terminated thereby, and is parallel to the axis. 4. An ordinate to any diameter is a line contained by the curve and that diameter parallel to a tangent at the extremity of the diameter. 1096. ThroneM. I. In the parabola, the abscissas are proportional to the squares of their ordinates. Let QVN (fig. 443.) be a section of the cone passing along the axis, and let the directrix RX pass through the point Q perpendicular to QN, and let the V parabolic section be ADI meeting the base QIND of the cone in the line DI, and the diameter QN in the point H ; also let KML be a section of the cone parallel to the base QIN intersecting the plane VQN in the line KL, and the section ADI in PM. Let P be the point of concourse of the three planes QVN, KML, AHI, and let H be the point of concourse of the three planes QVN, KML, AHI ; then, because the planes VRX and ADI are parallel, and p the plane AQN is perpendicular to the plane ARX, the plane ADI is also perpendicular to the plane AQN. Again, because the plane R QIN is perpendicular to the plane QVI, and the plane KML is NT parallel to the plane QIN, the plane KML is perpendicular to the 9 “HoP plane QVN : therefore the common sections PM and HI are perpendicular to the plane AQN ; and because the plane KML is pa- *... 443. rallel to the plane QIN; and these two planes are intersected by the plane QVN, their common sections KL and QN are parallel. Also, since PM and HI are each perpendicular to the plane QVN, and since KL is the common section of the planes QVN, KML, and QN in the common section of the planes QVN, QIN; therefore PM and HI are perpendicular respectively to KI and QN. Consequently AP : AH; : PM2 : HIQ. For, by the similar triangles APL, AHN, AP : AH :: PL: HN, Or A P : AH :: K P × PL : K P × H.N. 1097. Theon EM II. As the parameter of the aris is to the sum of any two ordinates, so is the difference of these ordinates to the difference of their abscissas. Multiplying the first of these equations by AP and the second by AH, Coroll. Hence, because Px KM = GK X KI; the focus is equal to one fourth of the parameter. Let LG (fig. 445.) be a double ordinate passing through the focus, then LG is the parameter. For by the definition of parameter A F : FG:: FG : P=2FG. Therefore 2AF = FG= }LG; Consequently AF = | LG. |