Wherefore, taking the difference of (2) and (3), we get (3) (CD - EG) ES = 4 AF. MC, or ES2 = 4 AF. MC. Consequently the proportion (1) becomes EM: 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, IIIi, 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 triangles AFL, AHN, and KFB, MHB, K AF: AH:: FL: HN, and FB: HB:: KF: MH. FL.KF = FG, and HN. MH = HI'. Consequently, AF. FB: AH.HB:: FG: HI2; 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 A K AD. DB: AC:: DE2: aC, or AD. DB: DE:: AC: aC :: AB2 : ab2. 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:: AB: al. But by the proposition, AB2: al2:: AD·DB: DE2. 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. 111. 35, AD.DB=DF. Ilence by the proposition, DF2: DE2 :: AC: aC2, 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 or 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 Prop. 11. and Euc. 11. 5, CA: Ca2 : : AF. FB: EF2:: CA CF: EF. But because F is the focus, EF = p (Def. 14), and therefore by Def. 13 AC: aC :: aС2: ¦ p2 or EF2. Hence, CA - CF a2, or CF2 = CA aC2; that is, the square of M P And similarly for the 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. = = CA - = Ca', or CF2 + Ca2 = = Fa. COR. 1. By the proposition CF CA'. But CF + Ca2 Wherefore, the distance of either extremity of the conjugate diameter from the focus is equal to the semi-transverse diameter. CA2 - CF2 = COR. 2. Since Ca2 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 f 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:: CM: CP; hence since (Prop. III.), CA - CF = Ca3, CA: Ca2:: CM2: CM2 CP2. But (Prop. 11.), CA: Ca2 :: CA2 CM: MD2. CM2:: CM2 CP2: MD2. CA: CM :: CM2 CP? + MD® : CM CP2, or CA: CM2 CP+MD:: CM2: CM2 Now (Euc. 11. 12) FD2 = FC2 + CD2 2 FC. CM; but as has been shown, = == = CA Ca3. CD2 Pa2 PC2 + Ca; also (Constr.), CA: CF :: CM: CP, or FC. CM - CA. CP, and (Prop. 111.), FC* Wherefore FD2 = (CA PC), or FD In a similar way it is shown that BP = ƒD. = AP, and ƒ D = CA CP --- = AP. = COR. 1. Since FD BP; hence FD + ƒD AB; that is, the sum of the two lines drawn from the foci to any point in the ellipse is equal to the transverse diameter. COR. 2. With F as centre and ƒ D radius, describe a circle to intersect the ellipse in D', below AB; then it follows from the proposition that the distance of D' from f is equal to FD. Wherefore if the points F, D' and ƒ, D' be joined, the figure FDf D' will be a parallelogram; and, therefore, since the diagonals of a parallelogram bisect one another, DD' passes through C the middle of Ff; that is, DD' is a diameter (Def. 7), hence all diameters of the ellipse are bisected in the centre. SCHOLIUM. From this proposition is deduced a very simple method of tracing an ellipse by points. For in the transverse axis AB, take the foci F, f, and any point P; with both centres F, f and radii AP, BP, describe circles; these will intersect in four points, which are evidently four points in the curve. Proceed similarly with another point taken in AB, and get four other points in the curve, and so on to any extent. PROPOSITION V. The distance of the directrix from any point in the ellipse has to the distance of the focus from the same point a constant ratio. L Let NL be the directrix with respect to the focus F, and N'L' that with respect to f; then the ratio of FD to DL, or the ratio of ƒ D to DL', is equal to the constant ratio of CF to CA (C being the centre), wherever the point D is taken in the curve. Take, in the axis produced, CF : CA: CA: CN, and draw NL perpendicular to CN; then NL is the directrix (Def. 15). N CP M Also take, as in Prop. iv., CP a fourth proportional to CA, CF, and CM, the distance of the ordinate DM from the centre. Then we have CF CA: CA: CN, and CF : CA :: CP : CM. Hence, CA CN :: CP: CM, or CA: CP:: CN : CM. Wherefore we readily get by composition, AP CA: MN: CN, or since AP MN = = FD, by Prop. iv., and DL; FD: DL :: CA: CN:: CF: CA. Similarly, ƒD: DL' :: CF: CA. SCHOLIUM. The constant ratio investigated in this proposition is the property referred to in Def. 15, with respect to the ellipse. PROPOSITION VI. The straight line which bisects the angle adjacent to that which is contained by two straight lines drawn from any point in the ellipse to the foci is a tangent to the curve at that point. To E any point in the ellipse, let the straight lines FE, ƒE, be drawn from the foci; then the straight line EP which bisects the angle ƒEG which is adjacent to FEf, is a tangent to the curve. = = P H B Take T any other point in PE produced; and make EG Ef, and join Tf, TG, and fG, meeting EP in P. Because Ef= EG, and EP is common to the triangles EƒP, EGP, and the angles ƒEP, GEP are equal, these triangles are equal, and ƒP = PG ; and hence fT TG. Wherefore FT +ƒT FTTG; but FT + TG is greater than FG. that is, greater than FE+ Ef, or AB. Consequently FT+ƒT is greater than AB. But if T be a point in the curve, FT +ƒT = AB (Prop. Iv., Cor. 1), which is impossible; and therefore T is not in the curve. In the same way it may be shown that no other point in PET but E is in the curve; that is, PET is a tangent at the point E. COR. 1. Since the angle ƒEP GEP, it must also be equal to FET which is the vertical angle of GEP. Hence a tangent to the ellipse makes equal angles with the straight lines drawn from the point of contact to the foci. = = COR. 2. Produce EC to meet the curve again in E', and draw the tangent HE'; then EE' is a diameter (Def. 7), and it is bisected (Prop. IV., Cor. 2), in C. The line Ff is also bisected in C; hence FEƒE is a parallelogram, and the angle FEƒ FEf. Now (Cor. 1) since the angle TEF = PEf, each of these angles is half the supplement of FEƒ; and for the same reason the angle HE'ƒ is half the supplement of FE'f; and since FEƒ FEf, the angle TEF HEF. Also, because EF is parallel to E'f, the angle FEC = CEf; consequently the whole angle TEE' HE'E, and therefore PET is parallel to HE'; that is, the tangents at the opposite extremities of any diameter are parallel to one another. The convex of this is also true, viz., that parallel tangents are at the extremities of the same diameter. = = - = = COR. 3. Let f P be a perpendicular from the focus fon the tangent at the point E; and let PG = Pf, as before, so that FEG is a straight line equal to AB; join CP. Then because FC = ƒC, and ƒ PPG, the triangles FG f, CP f, are similar. Whence FG: CP:: Ff: Cf::2:1; hence CP FGAB CB. Wherefore the point P is in a circle described on AB as a diameter. Similarly, the perpendicular Fp on the same tangent meets that tangent in the circle on AB. Consequently, the perpendiculars drawn from the foci on any tangent to the ellipse meet that tangent in a circle described on the transverse diameter. COR. 4. Since the tangent at E' (Cor. 2) is parallel to that at E, the perpendicular from fon the tangent at E' will meet that tangent |