through L draw GH parallel to AB cutting the opposite side BC of the rectangle in M, and through the point O draw K1 parallel to AD or BC cutting the opposite side DC in N. In NK or NK produced, make NQ equal to NC, and join CQ; draw QR parallel to GH cutting CB or CB produced in R; make EH and EG each equal to QC, as also El and EK each equal to PC; then will GH be the major axis and KI the minor axis of the ellipsis required. The demonstration of this method, in which the line QK has nothing to do with the construction, is as follows: C P L E M Q 10 B K Fig. 429. H By the similar triangles CPM and CQR, we have CP: CM:: CQ: CR. But EN is equal to MC, and EM equal to NC; Whence EI EH::MC: CN. But since the wholes are as the halves, we shall have KI: GH:: BC: CD. This problem is useful in its application to architecture about domes and pendentives, as well as in the construction of spheroidal ceilings and other details. OF THE HYPERBOLA. 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 axis. 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. 5. 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 axis. 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 cal ed the foci. 1084. THEOREM I. In the hyperbola the squares of the ordinates of the transverse axis are to each other us the rectangles 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 perpendi cular to the plane QVN. Let the cone be cut by two planes 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. R DN H Fig. 430. By the similar triangles APL and AHN, AP: PL::AH: HN; X But in the circle, PL x PK = PM2, and HN × HQ= HI2; Therefore Or, alternately, PM2: HI2:: AP: PB: AH: BH. 1085. THEOREM II. In the hyperbola, as the square of the transverse axis is to the square of the conjugate axis, so is the rectangle of the abscissas to the square of their ordinate. Let AB (fig. 431.) be the transverse axis, GE the conjugate axis, C being the centre of the opposite curves; also let HI and PM be ordinates as before; then will Or AB GE: PA x PB: PM By Theor. I., PA × PB : HA × HB:: PM2 ; HI2; G F. 431. For let the cutting plane Coroll. Hence AB2: GE2:: CP2-CA2: PM2 (fig. 432.). of the opposite hyperbola intersect two circles parallel to the base in III 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 such plane. Then ACm, APK are similar, and AC: Cm :: AP: PK. And as BCn, BPL are similar, BC Cn: BP: PL. Therefore, taking the rectangles of the corresponding terms, BC AC: Cn × Cm:: BP x AP: PL × PK. X But Though Ct is not in the same plane, it is what is usually called the seini-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 semiconjugate axis is to the square of the semi-transverse axis 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 centre of the figure, and PM an ordinate, then will GE2: AB2:: CE2+ PM2 CP2. For, by Theor. II., CE2: CA2:: PM2: CP2 — CA2, This demonstration may be also applied to what are called conjugate hyperbolas. 1087. THEOREM IV. In the hyperbola, the square of the distance of the focus from the centre is equal to the sum of the squares of the semi-axes. Let AB (fig. 434.) be the transverse axis, CE the semi-conjugate. AB, produced within the curve each way, let F be one focus; and ƒ the other, and let FG be the semi-parameter then CF2= CA2 + CE2. G C In Coroll. 1. The two semi-axes, and the distance of the focus from the centre, are the sides of a right-angled triangle CEA, of which the distance AE is the distance of the focus from the centre. Coroll. 2. The conjugate axis CE is a mean proportional between FA and FB, or between ƒB and ƒA, for CE2= CF2-CA=(CF+ CA) × (CF-CA) = BF × AF. 1088. THEOREM V. The difference of the radius vectors is equal to the transverse axis. (fig. 435.) That is, fM-FM-AB=2CA=2CB. And, subtracting the upper equation from the lower, ƒM- 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 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 CFX CP CA: CF:: CP: CA+ FM. is a fourth proportional to CA, CF CP, 1089. THEOREM VI. As the square of the semi-transverse axis is to the square of the semi-conjugate, so is the difference of the squares of any two obscissas to the difference of the squares of their ordinates. CA2: CE2:: CP2-CA2: PM2 (fig. 436.), By Theor. II., {CA Therefore, by equality, But Therefore CH2-CA2: CP2 — CA2:: HI: PM2 or CH2-CA2 : CH2-CP2:: HI2; HI2- HN2; CA: CE2:: CH2-CP2; HI2- HN2. H Fig. 436. E Coroll. 1. If IH be produced to K, and CQ be made equal to CP, then will CH2CP (CH+CP)(CH-CP)=(CP+CH)PH; and HI2-HN2 = (HI+HN) (HI— HN)=(HI+HN)NI. Therefore the analogy resulting becomes CA2: CE2::(CP+ CH)PH: (HI + HN)NI. So that the square of the transverse axis is to the square of the conjugate, or the square of the semi-transverse is to the square of the semi-conjugate, as the rectangle of the sum and difference of the two ordinates from the centre is to the rectangle of the sum and difference of these ordinates. 1090. THEOREM VII. If a tangent and an ordinate be drawn from any point in an hyperbola to meet the transverse axis, the semi-transverse axis will be a mean proportional between the distances of the two intersections from the centre. CE2: CA2::(IH+HN)IN:: (PC+ CH)HP, And by Theor. I., CE2 : CA2::PM2; CP2– CA2; For (fig. 437.) By equality, PM2 CP2 - CA2:: (IH + HN) IN: (PC + And by similar triangles ÎNM, MPT, IN: NM or PH::PM: PT or CP-CT. Therefore, taking the rectangles of the extremes and means of the two last equations, and neglecting the common factors, it will be PM(CP - CT)(CP+CH)=(CP2- CA2)(IH + HN); but when IH and PM coincide, IH and HN each become equal to PM, and CH equal to CP: therefore in the equation substitute 2CP for CP+ CH, and 2PM for IH+HN, and neglecting the common factors and common terms, the result is CT.CP= CA2, or CT: CA:: CA: CP. Coroll. Since CT is always a third proportional to CP, CA; suppose Fig. 437. E 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 extremity of the ordinate M of every hyperbola described on the same axis AB. 1091. THEOREM VIII. Four perpendiculars to the transverse axis intercepted by it and a tangent, will be proportionals when the first and last have one of their extremities in each vertex, the second in the point of contact, and the third in the centre. Let the four perpendiculars he AD, PM, CE, BF (fig. 438.), whereof AD and BF have their extremities in the vertices A and B, and the second in the point of contact M of the tangent and the curve, and the third in the centre C. B 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 AD: PM::CE: BF. 1092. THEOREM IX. equal angles with a tangent For, by Theor. VII., By Cor. 2. Theor. V., The two radius vectors meeting the curve in the same point will make passing through that point. (Fig. 439.) CA: CP:: CT: CA; CA CP:: CF: CA+ FM; By division and composition, CF-CT: CF + CT:: FM : 2CA And by the similar triangles TFM, TƒR, FT : ƒT::FM : ƒR. Therefore ƒR is equal to ƒM; consequently the angle fRM is equal to the angle fMR: and because ƒR is parallel to fM, the angle FMT is equal to the angle fRM; therefore the angle FMT is equal to the angle ƒRM. 1093. PROBLEM 1. To describe an hyperbola by means of the end of a ruler moveable on a pin F (fig 440.) fixed in a plane, with one end of a string fixed 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 R along the edge of the ruler, kept close to the string so as to keep each of the parts CD, DE of the string stretched, 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 mould. ings 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 C magnitude, to describe the hyperbola. (Fig. 442.) G Fig. 440. H Fig. 441. F 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 BF 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. F D A 2 3 C OF THE PARABOLA. 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. THEOREM I. ordinates. In the parabola, the abscissas are proportional to the squares of their V K M 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 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 I be the point of concourse of the three planes QVN, KML, AHI; then, because the planes VRX and ADI are parallel, and the plane VQN is perpendicular to the plane VRX, the plane ADI is also perpendicular to the plane VQN. Again, because the plane R QIN is perpendicular to the plane QVN, and the plane KML is parallel to the plane QIN, the plane KML is perpendicular to the plane QVN; therefore the common sections PM and HI are perpendicular to the plane VQN; and because the plane KML is parallel 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 KL and QN. Consequently AP: AH:: PM2 : HI2 For, by the similar triangles APL, AHN, AP : AH::PL; HN, Q H X Fig. 443. Therefore Therefore, by substitution, AP: AH:: PM2 : HIo, Coroll. By the definition of the parameter, which we shall call P, PM2 AP PM:: PM : P=PA' And P x AP=PM2, or P x AH = HIo. Therefore PPM::PM: AP, or P: HI::HI: AH. 1097. THEOREM II. As the parameter of the axis is to the sum of any two ordinates, so i' the difference of these ordinates to the difference of their abscissas. That is, PHI + PM :: HI−PM:AH-AP. Multiplying the first of these equations by AP and the second by AII, H K Fig. 414 Subtract the corresponding numbers of the first equation, and P(AH – AP) = HI2— PM2. But the difference of two squares is equal to a rectangle under the sum and difference of their sides. Or, by drawing KM parallel to AH, we have GK PM+ HI, and KI = HI-PM; and since PH AH-AP; P: GK::KI: PH, or KM. And since X So that any diameter MK is as the rectangle of the segments GK, KI of the double ordinate GI. From this a simple method has been used of finding points in the curve, so as to describe it. 1098. THEOREM III. The distance between the vertex of the curve and Fig. 445. 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 AF: FG::FG : P=2FG Therefore 2AF=FG=4LG; Consequently AF = {LG. |