= AE = SCHOLIUM 1. The property established in this proposition suggests the following simple method of constructing a parabola whose principal parameter is given :Make AF one-fourth of the parameter (Prop. 1., Cor. 2), and in EAF produced, take any point M; with F as centre and radius EM, describe a circle, intersecting a perpendicular PMP' to EAF produced, in P and P'; then will P and P' be two points in the curve. Other points are found in a similar way. SCHOLIUM 2. The property itself is the determining or constant ratio, for the parabola, to which reference is made in Def. 15. PROPOSITION III. If a line be drawn to bisect the angle made by two lines, one of which is drawn from a point in the parabola to the focus, and the other from the same point perpendicular to the directrix, that line will be a tangent to the parabola at that point. Let P be a point in the parabola, F the focus, EX the directrix, perpendicular to which PX is drawn; then the line PT drawn to bisect the angle FPX, will be a tangent at P. For since P is a point in the curve, TP meets the curve; and if it be not a tangent, it will meet the curve again in some other point K. Then, also, K is a point in the parabola. Join KF, and draw KL perpendicular to the directrix, and join XK. = M N Then because K is a point in the curve, and KL is perpendicular to the directrix, KF KL (Prop. II.); and since by the same Prop., XP = PF, and by hypothesis the angle XPK = FPK, and KP common to the triangles XPK, FPK, the base FK, therefore, is equal to KX. But it has been shown that FK KL, and therefore KL KX, the less to the greater, which is impossible. Whence the point K is not in the parabola. And in the same way it may be shown that no other point of PT is in the parabola except P. Hence (Def. 9) PT is a tangent to the parabola. = = = = = COR. 1. Since the line XP is parallel to FT, the angle XPT PTF; but by hypothesis, XPT FPT. Hence FTP FPT, and FT = FP; that is, the point of contact of a tangent to a parabola and the point of intersection of the tangent and axis produced, are equally distant from the focus. EA + AM = = COR. 2. Draw the ordinate PM; then AF + AT FT = FP = PX = EM = AF + AM. Hence AT = AM; that is, the subtangent of a parabola is bisected in the vertex (Def. 11). COR. 3. Draw the normal PN. Then (Def. 12), TPN is a right angle, and therefore (Euc. VI. 8), TM: MP :: MP: MN, TM.MN = MP2. But (Prop. 1., Cor. 1), p. AM P.AM TM. MN. And since by last Cor., = TM VOL. II. = = or MP2; hence 2 AM, conse Q = quently MN p.; that is, in the parabola, the subnormal is equal to half the parameter of the axis. = COR. 4. From the focus F draw FH perpendicular to the tangent PT, and join A and H. Then because FT FP (Cor. 1), and MA = AT (Cor. 2), the triangles PTM and HAT are similar, since the perpendicular FH evidently bisects PT. Wherefore AH is the tangent at the vertex A. Hence, a perpendicular from the focus of a parabola on any tangent intersects that tangent in the tangent from the vertex. COR 5. Also by Euc. vI. 8, FT: FH:: FH: FA; that is, if a straight line be drawn from the focus of a parabola perpendicular to the tangent at any point, it will be a mean proportional between the distances of the focus from the vertex and the intersection of the tangent and axis. SCHOLIUM 1. From the property of tangents to the parabola established in this proposition, the term focus is given to the point F. For by a property of optics, rays of light which proceed parallel to the axis. of a parabola and fall on a polished surface whose figure is that produced by the revolution of the parabola about its axis, are reflected to this point. SCHOLIUM 2. It would be easy to extend this series of properties of the parabola if the limits of this course admitted of such extension. All those properties relative to the principal axis and its coordinates hold for any diameter and its coordinates. The following is one of the most important of these. PROPOSITION IV. The abscissas of any diameter of a parabola are to one another as the squares of their ordinates. Let EH be any ordinate to the diameter CV, so that it is parallel to the tangent CT at the point C; CM the corresponding abscissa; AI the axis; and F the focus of a parabola. From E and C draw EG, CD perpendicular to the axis, and from the focus F draw FY perpendicular to the tangent CT; also join FC, and produce EG to meet CV in S. Then it will be obvious by Prop. M V E III., Cor. 4, that the triangles FCY, ESM, are equiangular; hence EM2: ES2:: FC2: FY2. But (Prop. III., Cor. 5), since FT = FC, FC: FY:: FY: FA, or FC: FA:: FC2: FY2. Wherefore, EM: ES:: FC: FA. Again, by the similar triangles CTD, ESM, ES: SM:: CD: DT:: CD2: CD. DT. And since (Prop. 1., Cors. 1, 2), CD 4 AF. AD, and (Prop. III., Cor. 2), DT=2 AD; we have = ES: SM:: 4 AF. AD: 2 CD. AD:: 4 AF: 2 CD, or, 4 AF.SM=2 CD.ES . . . (2). But (Prop. 1., Cor. 4), 4 AF. CS CD2 - EG2 = (CD + EG) (CD — EG) =(CD+ EG) ES . . Wherefore, taking the difference of (2) and (3), we get = 4 AF. MC, or ES2 = 4 AF. MC. (CD - EG) ES Consequently the proportion (1) becomes EM2: 4 AF. MC::FC: FA, or, EM2 = 4 FC. MC. Hence for the same diameter of a parabola, any two abscissas are to one another, as the squares of the corresponding ordinates. This is the general case of the theorem established in Prop. 1, THE ELLIPSE. PROPOSITION I. The squares of the ordinates of the transverse diameter of an ellipse are to one another as the rectangles of their abscissas. As in the parabola, let AVB be the transverse plane, AGIB the sectional plane; AB the transverse axis, which in this case (Def. 5) meets the cone in two points A and B; KGL, MIN, any two sections of the cone perpendicular to its axis. Also let KL, MN, be the intersections of KGL, MIN, with the transverse plane, and GFg, IHi, those with the sectional plane. Then as in the case of the parabola (Prop. 1.), KL, MN are diameters of the circles KGL, MIN, and the transverse diameter AB bisects Gg, Ii, perpendicularly in F and H. Hence by the similar a triangles AFL, AHN, and KFB, MHB, M AF: AH:: FL:HN, and FB:HB::KF: MH. Consequently, AF. FB: AH.HB:: FG: HI; HI2. A that is, the squares of the ordinates FG, HI, are to one another as the rectangles of their corresponding abscissas. It will be obvious that the ellipse is composed of one single and continuous curve, and that the transverse diameter divides it into two equal branches, symmetrically situated on each side of this axis. PROPOSITION II. If any point be taken in an ellipse, the rectangle of the abscissas belonging to that point is to the square of the corresponding ordinate as the square of the transverse diameter is to the square of its conjugate. Let E be any point in the ellipse AaBb, AB the transverse diameter, and ab its conjugate (Def. 6); then if ED be drawn perpendicular to AB, the rectangle under AD, DB will be to the square of the ordinate ED as the square of AB is to the square of ab. For by Prop. I., AD. DB: AC.CB:: DE2: aC2. But because C is the centre, AC = CB (Def. 6); hence AD.DB: AC:: DE2: aC2, or AD. DB: DE3:: AC2: aC :: AB3 : ab3. COR. 1. Take p a third proportional to AB and ab; so that AB: ab:: ab: p, then by the property of duplicate ratio, AB:p:: AB2: ab2. But by the proposition, AB : ab2::AD·DB: DE3. Hence AB: p::AD. DB: DE; and similarly for any other point in the curve. This is the property relative to the ellipse, to which reference is made in Def. 13. The quantity p is constant when AB and ab are given, and is called the parameter of the transverse. Whence this property, the transverse diameter of an ellipse is to its parameter as the rectangle under the abscissas of any point in the curve is to the square of the ordinate of that point. COR. 2. Produce the ordinate DE to meet the circle described on the axis AB of the ellipse in F. Then by Euc. III. 35, AD.DB = DF2. Hence by the proposition, DF2: DE2 :: AC2: aC, or DF: DE :: AC: aC. A similar property is easily shown to belong to the circle described on ab as a diameter. Hence, if two circles be described, one on each principal diameter of an ellipse, an ordinate in either circle is to the corresponding ordinate of the ellipse as the diameter to which this ordinate is drawn is to the other diameter. COR. 3. Ordinates to either conjugate diameter which intercept equal segments of that diameter from the centre are equal to one another, and conversely, equal ordinates intercept equal segments of the diameter from the centre. SCHOLIUM 1. By Def. 14, the focus of the ellipse is that point in the transverse diameter at which the double ordinate is equal to the parameter p (Cor. 1) of that diameter; and as two such points are evidently on each side of the centre, it will be obvious that in the ellipse there are two foci at equal distances from the centre. For a similar reason, there will be two directrices (Def. 15). SCHOLIUM 2. By means of the property in Cor. 2, the curve of an ellipse whose transverse axes are given, may be constructed by points. For if on AB, one of the axes, a circle be described, and in it any ordinate DF be drawn; then since AC: aC:: DF: DE, a point E can be found in the ellipse; and similarly for other points. PROPOSITION III. In the ellipse, the square of the distance of the focus from the centre is equal to the difference of the squares of the semi-axes. Let AB be the transverse diameter of an ellipse, ab its conjugate, F and f the foci, C the centre, and EF an ordinate at the focus F. Then by = = CA ac2; that is, the square of MP α CF, the distance between the centre and focus F, is equal to the difference of the squares of the semi-axes AC, aC. distance Cf. = = And similarly for the = COR. 1. By the proposition CF2 = CA Ca2, or CF2 + Ca2 CA2. But CF+ Ca2 Fa; hence CA Fa2, or CA = Fa. Wherefore, the distance of either extremity of the conjugate diameter from the focus is equal to the semi-transverse diameter. = = COR. 2. Since Cas CA - CF AF. FB; hence, the rectangle under the segments of the transverse diameter made by one of the foci is equal to the square of the semi-conjugate axis. PROPOSITION IV. If in the ellipse a fourth proportional be taken to the semi-transverse axis, the eccentricity, and the distance of an ordinate to any point in the curve (all estimated from the centre along the transverse axis); then the extremity of this fourth proportional will divide the transverse axis into two parts which are equal respectively to the focal distances of that point. (See the Fig. of last Prop.) Let D be any point in the ellipse Aa Bb, F and ƒ the foci, and CP a fourth proportional to CA, CF and CM (DM being the ordinate to the point D); then the segments AP, BP of the axis AB will be respectively equal to the focal distances FD, ƒD, drawn from the foci to the paint D. For by construction, CA: CF :: CM3: CP; hence since (Prop. III.), CA2 — CF2 = Ca3, CA: Ca :: CM2: CM2 CP2. But (Prop. II.), CA: Ca2:: CA2 CM2: MD2. CA: CM2:: CM2 CP2+ MD2: CM2 |