1071. THEOREM VII. In the ellipsis, half the major axis is a mean proportional between the distance of the centre and an ordinate, and the distance between the centre and the intersection of a tangent to the vertex of that ordinate. To the major axis draw the ordinates PM (fig. 416.) and HI, and the minor axis CE. Draw MN perpendicular to HI. Through the two points I, M draw MT, IT, meeting the major axis produced in T; then will CT: CA:: CA: CP. For, IE M N B A Р HC By Cor. 1. Theor. VI., CE2 CA2::(IH+HN)IN: (PC+ CH)HP; Fig. 416. Therefore, by equality, PM2.. CAo— CP2::(IH+HN)IN : (PC + CH)HP. By similar triangles, INM, MPT; IN: NM or PH:: PM: PT or CT-CP. Therefore, taking the rectangles of the extremes and means of the two last equations, and throwing out the common factors, they will be converted to the equation PM(CT-CP)(CP+CH)=(CA2— CP2)(IH + HN). But when HI and PM coincide, HI and HN will become equal to PM, and CH will become equal to CP; therefore, substituting in the equation 2CP for CP+ PH, and 2PM for IH+HN, and throwing out the common factors and the common terms, we have CT. CP CA or CT: CA:: CA: CP. Coroll. 1. Since CT is always a third proportional to CP and CA, if the points P, A, B remain fixed, the point T will be the same; and therefore the tangents which are drawn from the point M, which is the intersection of PQ, and the curve, will meet the curve in the point T in every ellipsis described on the same axis AB. Coroll. 2. When the outer ellipsis AQB, by enlarging, becomes a circle, draw QT perpendicular to CQ, and joining TM, then TM will be a tangent to the ellipsis at MM. Coroll. 3. Hence, if it were required to draw a tangent from a given point Tin the prolongation of the major axis to the ellipsis AEB, it will be found thus: - · On AB describe the semicircle AQB. Draw a tangent TQ to the circle, and draw the ordinate PQ intersecting the curve AEB of the ellipsis in the point M; join TM; then TM is the tangent required. This method of drawing a tangent is extremely useful in practice. 1072. THEOREM VIII. Four perpendiculars to the major axis intercepted by it and a tangent will be proportionals when the first and last have one of their extremities in the vertices, the second in the point of contact, and the third in the centre. Let the four perpendiculars be AD, PM, CE, BP, of which AD and BF have their extremities in the vertices A and B, the second in the point of contact M, and the third in the centre C; then will AD: PM::CE: BF. For, by Theor. VII., TC: AC:: AC: CP; M E B By division, That is, By composition, Therefore Fig. 417. TC-AC CA-CP::TC: AC or CB; TA: TA+ AP::TC: TC + CB: TA: TP::TC: TB. But by the similar triangles TAD, TPM, TCE, and TBF, the sides TA, TP, TC, and TB are proportionals to the four perpendiculars AD, PM, CE, and BF; therefore AD: PM::CE: BF. Coroll. 1. If AM and CF be joined, the triangles TAM and TCF will be similar. For by similar triangles, the sides TD, TM, TE, TF are in the same proportion as the sides TA, TP, TC, TB. Therefore TD: TM::TE: TF; Alternately, TD:TE::TM:TF: but TAD is similar to TCE; Therefore, by equality, AP: PM:: CB: BF. Coroll. 3. If AF be drawn cutting PM in I, then will PI be equal to the half of PM. For, since AP: PM:: CB: BF, and, by the similar triangles API, ABF, Therefore PM: PI:: CB: AB. But CB is the half of AB; therefore, also, PI is the half of PM. 1073. THEOREM IX. If two lines be drawn from the foci of an ellipse to any point in the curve, these two lines will make equal angles with a tangent passing through that point. Let TM (fig. 418.) be a tangent touching the curve at the point M, and let F, ƒ be the two foci; join FM, ƒM, then will the angle FMT be equal to the angle ƒMR. For draw the ordinate PM, and draw ƒR parallel to FM, then will the triangles TFM and T TfR be similar; and by Cor. Theor. VII., G D R AFPN C ƒ B By division and composition, CT-CF : CT+ CF::FM: 2CA-FM; That is, TF: Tƒ:: FM : ƒM. By the similar triangles TFM, TƒR; TF: Tƒ:: FM : ƒR. It therefore appears that ƒM is equal to ƒR, therefore the angle fMR is equal to the angle fRM: but because FM and ƒR are parallel lines, the angle FMT is equal to the angle fRM; therefore the angle FMT is equal to the angle ƒMR. Coroll. 1. Hence a line drawn perpendicular to a tangent through the point of contact will bisect the angle FMf, or the opposite angle DMG. For let MN be perpendicular to the tangent TR. Then, because the angle NMT and NMR are right angles, they are equal to one another; and since the angles FMT and ƒMR are also equal to one another, the remaining angles NMF and NMƒ are equal to one another. Again, because the opposite angles FMN and IMG are equal to one another, and the opposite angle ƒMN and IMD are equal to one another; therefore the straight line MI, which is the line MN produced, will also bisect the angle DMG. Coroll. 2. The tangent will bisect the angle formed by one of the radius vectors, and the prolongation of the other. For prolong FM to G. Then, because the angles RMN and RMI are right angles, they are equal to one another; and because the angles NMƒ and IMD are equal to one another, the remaining angles RMG and RMƒ are equal to one another. Scholium. Hence we have an easy method of drawing a tangent to any given point M in the curve, or of drawing a perpendicular through a given point in the curve, which is the usual mode of drawing the joints for masonic arches. Thus, in order to draw the line IM perpendicular to the curve: produce FM to G, and ƒM to D, and draw MI bisecting the angle DMG; then IM will be perpendicular to the tangent TR, and consequently to the curve. As in optics the angle of incidence is always found equal to the angle of reflection, it appears that the foundation of that law follows from this theorem; for rays of light issuing from one focus, and meeting the curve in any point, will be reflected into lines drawn from these points to the other focus: thus the ray fM is reflected into MF: and this is the reason why the points Fƒ are called foci, or burning points. In like manner, a sound in one focus is reflected in the other focus. 1074. THEOREM X. Every parallelogram which has its sides parallel to two conjugate diameters and circumscribes an ellipsis is equal to the rectangle of the two axes. K E M Let CM and CI (fig. 419.) be two semi conjugate diameters. Complete the parallelogram CIDM. Produce CA and MD to meet in T, and let AT meet DI in t. Draw IH and PM ordinates to the axis, and draw half the minor axis CE. Produce DM to K, and draw CK perpendicular to DK then will the parallelogram CIDM be equal to the rectangle, whose sides are CA and CE; or four times the rectangle CIDM will be equal to the rectangle made of the two axes AB and GE. By Cor. Theor. VII., Therefore By the similar triangles CtI, TCM, By equality, therefore, { CA: CT:: CP:.CA, Ct: CT:: CI: TM; By the similar triangles CIH, TMP, CI: TM::CH: PT; Therefore, by equality, Consequently But by Theor. VII., CH: PT:: CP: CH. CP x CT-CA2; Fig. 419. G B But by the similar triangles HIC, KCT, HI: CI:: CK: CT; Therefore Consequently CE: CI:: CK: CA: CE x CACI x CK. The ellipsis is of so frequent occurrence in architectural works, that an acquaintance with all the properties of the curve, and the modes of describing it, is of great importance to the architect. Excepting the circle, which may be called an ellipsis in which the two foci coincide, it is the most generally employed curve in architecture. 1075. PROBLEM I. To describe an ellipsis. F 3 2 D G Let two pins at E and F (fig. 420.) be fixed in a plane within a string whose ends are made fast at C. If the point C be drawn equally tight while it is moved forward in the plane till it returns to the place from which it commenced, it will describe an ellipsis. 1076. PROB. II. The two diameters AB and ED of an ellipse being given in position and magnitude, to describe the curve through points. Fig. 420. A 123 C E Let the two diameters cut each other at C (fig. 421.). Draw AF and BG parallel to ED. Divide AC and AF each into the same number of equal parts, and draw lines, as in the figure, through the points of division; viz. those from the line AF to the point D, and the lines through AC to the point E; then through the points of intersection of the corresponding lines draw the curve AD, and in the same manner find the curve BD; then ADB will be the semi-ellipsis. D It is evident that the same method also extends to a circle by making CD equal to CA; (fig. 422.); and it appears that the two lines forming any point of the curve to be drawn will make a right angle with each other. For these lines terminate at the extremities of the diameter ED, and the point of concourse being in the curve, the angle made by them must be a right angle; that is, the angle EAD, or EhD, or EiD, or EkD, is a right angle: and from this property we have the following method of drawing the segment of a circle through points found in the curve. Fig. 422. Thus, let AB be the chord, and CD be the versed sine of an arc of a circle, to describe the arc. Through D draw HI (fig. 423.) parallel to AB; join AD and DB; draw AH perpendicular to AD, and BI perpendicular to BD; divide AC and HD each into the same number of equal parts, and join the corresponding points; divide AF into the same number of equal parts, and through the points of division draw lines to D, and through the corresponding points where these lines meet the former draw a curve AD. In the same manner the other half BD may be drawn. 1077. PROB. III. A diameter KH of an ellipsis being given, and an ordinate DL, to find the limits of the other conjugate diameter. H IF 2 3 D Bisect KH in I (fig. 424.), through I draw EA parallel to DL, and draw DC and KB perpendicular to EA; from the point L with the distance K describe an arc cutting EA at F; join LF, and produce LF to C; make IE and IA each equal to LG; then will EA be a diameter conjugate to KH. 1078. PROB. IV. A diameter KH and an ordinate DL of an ellipsis being given, to describe the curve. (See fig. 424.) Find the limits E and A of the other conjugate diameter by the preceding construction. Produce KB to q, and make Kq equal to IA or IE, and through the centre I of the curve and the point q, draw the straight line MN. Then, suppose the straight line KBq to be an inflexible rod, having the point B marked upon it. Move the rod round, so that the point q on the rod may be in the line MN, while the point B is in the line EA; then, at any instant of the motion, the place of the point K on the plane whereon the figure is to be drawn may be marked; the points thus found will be in the curve. Instead of a rod, a slip of paper may be used, and in some cases a rod with adjustible points to slide in a cross groove, and a sliding head for a pencil is convenient; and such an instrument is called a trammel. D K 9 When the diameters KH and EA (fig. 425.) are at right angles to each other, the straight line Kq coincides with the diameter KH, and consequently the line MN, on which the point q of the inflexible line Kq moves, will also fall upon the diameter KH. Therefore in this case nothing more is required to find the limits of the other diameter, than to take the half diameters IK, KH of the given diameters, and from the extremity L with that distance describe an arc cutting the unlimited diameter in the point F; then drawing LF, and producing it to q, and making IE and IA each equal to qL, EA will be the other diameter; and since the two diameters are at right angles to each other, they are the two axes given in position and magnitude, and thus the curve may be described as before. Н Fig. 425. A method of describing the curve from any two conjugate diameters is occasionally of considerable use, and particularly so in perspective. For, in every representation of a circle in perspective, a diameter and a double ordinate may be determined by making one of the diameters of the original circle perpendicular to the plane of the picture and the other parallel to it; and then the representation of the diameter of the original circle, which is perpendicular to the intersecting line, will be a diameter of the ellipsis, which is the representation of that circle; and the representation of the diameter of the circle which is parallel to the intersecting line will become a double ordinate to the diameter of the ellipsis which is the perspective representation of the circle. 1079. PROB. V. Through two given points A and B to describe an ellipsis, the centre C being given in position and the greater axis being given in magnitude only. About the centre C (fig. 426.) with a radius equal to half the greater axis describe a circle HEDG; join AC and BC; draw AD perpendicular to AC, and BE perpendicular to BC, cutting the circumference in the points D and E; draw also BF parallel to AC, and find BF, which is a fourth proportional to AD, AC, and BE; through the point F and the centre C draw FG to cut the circle in H and G, and GH is the major axis of the ellipsis. By drawing an ordinate Bq, the curve may be described by the preceding problem, having the axis GH and the ordinate Bq. G D Fig. 426. E 1080. PROB. VI. Through a given point in the major axis of a given ellipsis to describe another similar ellipsis which shall have the same centre and its major axis on the same straight line as that of the given ellipsis. A C K H B E Let ACBD (fig. 427.) be the given ellipsis, having AB for its major axis and CD for its minor axis, which are both given in position and magnitude. It is required to draw a similar ellipsis through the point G in the major axis AG. Draw BK perpendicular and CK parallel to AB, and join KE. Again, draw GL perpendicular to AB cutting EK at L, and draw LH parallel to AB cutting CD in H. On the axis CD make EI equal to EH, and on the axis AB make EF equal to EG. Then, having the major axis AB, and the minor axis FG, the ellipsis FIGH may be described, and when drawn, it will be similar to the given ellipsis ADBC. Fig. 127. 1081. PROB. VII. Through any given point p, within the curve of a given ellipsis to describe another ellipsis which shall be similar and concentric to the given one. U Let C (fig. 428.) be its centre. Draw the straight line CpP, cutting the curve of the given ellipsis in P. In such curve take any other number of points Q, R, S, &c., and join QC, RC, SC, &c.; join PQ and draw pq parallel thereto cutting qC at q: join PR and draw pr parallel to PR, cutting RC at r; join PS and draw ps parallel to P PS cutting SC in s. The whole being completed, and the curve p, s, t, u drawn through the points p, q, r, s, &c., the figure will be similar and concentric to the given ellipse P, S, T, U; or when the points at the extremities for one half of the curve have been drawn, the other half may be found by producing the diameter to the opposite side, and making the part produced equal to the other part. R Fig. 428. 1082. PROB, VIII. About a given rectangle ABCD to describe an ellipsis which shall have its major and minor axes respectively parallel to the sides of the rectangle and its centre in the points of intersection of the two diagonals. Bisect the sides AD and AB (fig. 429.) of the rectangle respectively at L and O; through L draw GH parallel to AB 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 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 EI 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: I N P H L E Q IM 10 B Fig. 129. K By the similar triangles CPM and COR, 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 called the foci. 1084. THEOREM I. In the hyperbola the squares of the ordinates of the transverse axis 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 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 K H Fig. 430. By the similar triangles APL and AHN, AP PL:: AH: HN; But in the circle, PL x PK = PM2, and HN × HQ=HI2; Or, alternately, PM2: HI2:: AP: PB: AH: BH. 1085. THEOREM II. axis is to the square of the conjugate axis, so is the rectangle of the abscissas to the square of their ordinate. In the hyperbola, as the square of the transverse 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 |