1050. The extremities of a diameter which terminate in the curve are called the vertices of that diameter. Thus the points V and Q are the vertices of the diameter VQ. 1061. A straight line drawn from any point of a diameter parallel to a tangent at either tremity of the diameter to meet the curves is called an ordinate to the two abscissas. Thus PM, being parallel to a tangent at V, is an ordinate to the two abscissas VP, PQ. 1052. If a diameter be drawn through the centre parallel to a tangent at the extremity of another diameter, these two diameters are called conjugate diameters. Thus VQ and RS are conjugate diameters. 1063. A third proportional to any diameter and its conjugate is called the parameter or latus rectum. 1054. The points in the axis where the ordinate is equal to the semi-parameter are called the foci. 1065. THEOREM I. In the ellipsis the squares of the ordinates of an aris are to each other as the rectangles of their abscissas. Let AVB (fig. 408.) be a plane passing through the axis of the cone, and AEB another section of the cone perpendicular to the plane of the former ; AB the axis of the elliptic section, and PM, HI ordinates perpendicular to it; then it will be PM2 : HI2:: AP × PB: AH × HB. For through the ordinates PM, HI draw the circular sections KML, MIN parallel to the base of the cone, having KL, MN for their diameters, to which PM, HI are ordinates as well as to the axis of the ellipse. Now, in the similar triangles APL, AHN, AP: PL:: AH; HN, And in BPK, BHM, Taking the rectangles of the corresponding terms, AP BP: PL x PK:: AH BH: HN x HM. By the property of the circle, Coroll. 1. If C be the centre of the figure, AP × PB= CA2 – CP2, and AH × HB = CA-CH2. Therefore PM2: HI2:: CA2-CP2: CA2-CH2. For AP CA-CP, and PB= CA+CP: consequently AP PB (CA-CP) (CA+CP) = CA2-CP2; and in the Baine manner it is evident that AH × HB=(CA+CH)(CA — CH) = CA2 — CH2. Coroll. 2. If the point P coincide with the middle point C of the semi-major axis, PM will become equal to CE, and CP will vanish; we shall therefore have PM2: HI2:: CA2-CP2: CA2-CH2 Now CE2: HI2:: CA2; CA2- CH2, or CA2 x HI2=CE2(CA2 — CH2). 1066. THEOREM II. In every ellipsis the square of the major axis is to the square of the Bigr axis as the rectangle of the abscissas is to the square of their ordinate. Let AB (fig. 409.) be the major axis, DE the minor axis, C the centre, PM and HI ordinates to the axis AB; then will CA2: CE2:: AP x PB: PM2. X For since by Theor. I., PM2 : HI2:: AP × PB : AH × HB; and if A Coroll. 1. CE? PM2: CE2:: AP x PB: CA2; And, alternately, CA2: CE:: AP × PB:: PM2, 1 E M Hence, if we divide the two first terms of the analogy by AC, it will be CA: CA: AP × PB: PM2. But by the definition of parameter, AB: DE::DE: pa 1067. THEOREM III. In every ellipsis, the square of the minor axis is to the square of t major axis as the difference of the squares of half the minor aris and the distance of an ordinate from the centre on the minor axis to the square of that ordinate. E M Draw MQ (fig. 410.) parallel to AB, meeting CE in Q; then will A For by Cor. 2. Theor. II., CA2: CAo-CP2:: CE2: PM2; Therefore, by division, CA2: CP2:: CE2: CE2-PM2. Therefore, since CQ=PM and CP=QM; CAo: QM2:: CE2; CE2- CQ2. Coroll. 1. If a circle be described on each axis as a diameter, one being inscribed with the ellipse, and the other circumscribed about it, then an ordinate in the circle will be to the corresponding ordinate in the ellipsis as the axis belonging to this ordinate is to the axis belonging to the other; that is, CA: CE:: PG: PM, and CE CA::pg: pM; and since CA2: CE2:: AP × PB: PM2, and because AP × PB=PG2; CA2: CE2:: PG2: PM2, G E P g R B IC D Fig. 411. In the same manner it may be shown that CE: CA::pg : pM, or, alternatel CACE::pM: pg; therefore, by equality, PG: PM::pM: pg, or PG: Cp::CP : p therefore CgG is a continued straight line. Coroll. 2. Hence, also, as the ellipsis and circle are made up of the same number corresponding ordinates, which are all in the same proportion as the two axes. it follow that the area of the whole circle and of the ellipsis, as also of any like parts of them, a in the same ratio, or as the square of the diameter to the rectangle of the two axes; that the area of the two circles and of the ellipsis are as the square of each axis and t rectangle of the two; and therefore the ellipsis is a mean proportional between the tv circles. Coroll. 3. Draw MQ parallel to GC, meeting ED in Q; then will QM=CGCA and let R be the point where QM cuts AB; then, because QMGC is a parallelograr QM is equal to CG=CE; and therefore, since QM is equal to CA, half the major ax and RM CE, half the minor axis QR is the difference of the two semi-axes, and hen we have a method of describing the ellipsis. This is the principle of the trammel, so wi known among workmen. If we conceive it to move in the line DE, and the point R in the line AB, while t point M is carried from A, towards E, B, D, until it return to A, the point M will in i progress describe the curve of an ellipsis. 1068. THEOREM IV. The square of the distance of the foci from the centre of an ellipsis equal to the difference of the square of the semi-axes. E G A Let AB (fig. 412.) be the major axis, C the centre, F the focus, and FG the semi-par meter; then will CE2-CA2 – CF2. For draw CE perpendicular to AB, and join FE. By Cor. 2. Th. II., CA: CE :: CA2CF2 FG2, and the parameter FG is a third proportional to CA, CE; therefore CA2: CE2:: CE2: FG, and as in the two analogies the first, second, and fourth terms are identical, the third terms are equal; consequently Coroll. 1. CE CA- CF2. Fig. 412. Coroll. 2. The two semi-axes and the distance of the focus from the centre are the sid of a right-angled triangle CFE, and the distance FE from the focus to the extremity the minor axis is equal to CA or CB, or to half the major axis. Coroil. 3. The minor axis CE is a mean proportional between the two segments of t axis on each side of the focus. For CE-CA2 — CF2 = (CA + CF) × (CA — CF). 1069. THEOREM V. In an ellipsis, the sum of the lines drawn from the foci to any point the curve is equal to the major axis. Let the points F, f(fig. 413.) be the two foci, and M a point in the curve; join FM and ƒM, then will AB-2 CA = FM +ƒ M. By Cor. 2. Th. II., CA2: CE2:: CA2-CP2: PM2, And by taking the rectangle of the extremes and means, and dividing the equation CA, the result is And since CF2. CP2 FM2 CA-2CF.CP+ CA2 Extracting the root from each number, FM-CA CF.CP In the same manner it may be shown that FM-CA + these is FM+ƒM=2CA. CF.CP Coroll. 1. A line drawn from a focus to a point in the curve is called a radius vector, and the difference between either radius vector and half the major axis is equal to half the difference between the radius vectors. For, since Coroll. 2. Because CF::CP: CA-ƒM. Coroll. 3. Hence the difference between the major axis and one of the radius vectors gives the other radius vector. For, since FM+ Mƒ=2CA ; Therefore FM=2CA-Mƒ. A H E h B C Coroll. 4. Hence is derived the common method of describing an ellipsis mechanically, by a thread or by points, thus: - Find the foci Fƒ (fig. 414.), and in the axis AB assume any point G; then with the radius AG from the point F as a centre describe two arcs H, H, one on each side of the axis; and with the same radius from the point ƒ describe two other arcs h, A one on each side of the major axis Again, with the distance GB from the point ƒ describe two arcs, one on each side of the axis, intersecting the arcs HH in the points HH; and with the same radius from the point ƒ describe two other arcs, one on each side of the axis, intersecting the arcs described at h. h in the point h, h. find as many points as we please; and a sufficient number being found, the curve will be formed by tracing it through all the points so determined. 1070. THEOREM VI. The square of half the major axis is to the square of half the minor aris as the difference of the squares of the distances of any two ordinates from the centre to the difference of the squares of the ordinates themselves. Let PM and HI (fig. 415.) be ordinates to the major axis AB; draw MN parallel to AB, meeting HI in the point N; then will PM HN, and MN=PH, and the property to be demonstrated is thus expressed CA: CE2:: CP2-CH2: HJ2-HNo. Or by producing HI to meet the curve in the point K, and making CQ=CP, the property to be proved will be By Cor. 2. Theor. II. Therefore But, by division, Alternately, CA2-CH2: CA2-CP2::HI2: PM2 or HN2; And, since we have above, CA2- CH2; HI2::CA2: CE, But since And since CP?_CH2=(CP-CH)CP+CH)= PH × QH, Coroll. 1. Hence half the major axis is to half the minor axis, or the major axis is to the minor axis, as the difference of the squares of any two ordinates from the centre is to the rectangle of the two parts of the double ordinate, which is the greatest made of the sum and difference of the two semiordinates. For KN=HK + HN= HI+HN, which is the m of the two ordinates, and NI= HI-HN, which is the difference of the two ordinates. Coroll. 2. Hence, because CP2- CH2 = (CP — CH)(CP + CH), and since HI-HN 2 = (HI−HN)(HI+HN), and because CP-CH=PH_and_ HI-HIN=NI; therefore CA: CE::(CP + CH)PH : (HI + IN)NI. X 1071. THEOREM VII. In the ellipsis, half the major axis is a mean proportional betw 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, Q I E 4 PH C Fig. 416. By Cor. 1. Theor. VI., CE2 : CA2::(III + HN)IN : (PC + CH)HP; By Cor. 2. Th. II., CE2: CA2:: PM2; CAo— CP2; Therefore, by equality, PM2: CA2 – 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, a throwing out the common factors, they will be converted to the equation PM(CT-CP)(CP+CH)=(CAo — CP2)(IH+HN). But when HI and PM coincide, HI and HN will become equal to PM, and CH ▾ become equal to CP; therefore, substituting in the equation 2CP for CP+ PH, and 21 for IH+HN, and throwing out the common factors and the common terms, we have CT. CP-CA2 or CT: CA::CA: CP. Coroll. 1. Since CT is always a third proportional to CP and CA, if the points P, A, remain fixed, the point T will be the same; and therefore the tangents which are dra from the point M, which is the intersection of PQ and the curve, will meet in the point in every ellipsis described on the same axis AB. Coroll. 2. When the outer ellipsis AQB, by enlarging, becomes a circle, draw QT p pendicular to CQ, and joining TM, then TM will be a tangent to the ellipsis at M. Coroll. 3. Hence, if it were required to draw a tangent from a given point T in the p longation of the major axis to the ellipsis AEB, it will be found thus: Ou AB descr the semicircle AQB. Draw a tangent TQ to the circle, and draw the ordinate PQ int secting the curve AEB of the ellipsis in the point M; join TM; then TM is the tang 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 t. gent 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, BF, 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 Cir then will AD: PM::CE: BF. For, by Theor. VII., TC: AC:: AC: CP; E M D P C 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, : 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 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 P 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 eurre, 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, f be the two foci; join FM. ƒM, then will the angle FMT be equal to the age ƒMR. For draw the ordinate PM, and draw parallel to FM, then will the triangles TFM and T IfR be similar; and by Cor. Theor. VII., By Cor. 2. Theor. V., Therefore, by equality, CA: CP:: CT: CA; G D CA CP::CF: CA-FM; APPN By division and composition, CT− CF : CT+CF::FM: 2CA-FM; By the similar triangles TFM, TƒR; TF: Tƒ::FM: ƒR. It therefore appears that ƒM is equal to fR, therefore the angle ƒMR is equal to the angle RM: but because FM and ƒR are parallel lines, the angle FMT is equal to the angle fRM; therefore che angle FMT is equal to the angle ƒ MR. 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 pro duced, 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 AMI 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 mother. ་ 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 asual 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 fom 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 ƒM 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. 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 H and PM ordinates to the axis, and draw half the minor Taxis 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 K E M D H B P C Fig. 419. G rectangle CIDM will be equal to the rectangle made of the two axes AB and GE. |