in its point of intersection H with the circle on AB, so that fP, f II, are in the same straight line. Wherefore (Eu“. 111. 35), A f. f B = Pf.f H. But the triangles Fp E, fH E' are evidently equal in all respects, and therefore fH = Fp; consequently Af.fB = Pf.F p. Whence, since the rectangle under Af and fB (Prop. III., Cor. 2) is equal to the square of the semi-conjugate diameter, the rectangle under the perpendiculars from the foci of an ellipse upon any tangent is equal to the square of the semi-conjugatc diameter. Cor. 5. A perpendicular to the transverse axis at either of its ectremities is a tangent to the ellipse; also a perpendicular to the conjugate axis at either of its extremities is a tangent to the curve, PROPOSITION VII. If a tangent and an ordinate be drawn from any point in the ellipse to meet the transverse axis, the semi-transverse axis will be a mean proportional between the distances of the two intersections from the centre. hence FT + Tf:FT – Tf::FE + Ef: FE - Ef, (1). But FE’ – E f' = FD - f Do, or (FE + Ef) (FE – Ef) = (FD + fD) (FD - fD); that is, since FE + Efi= 2 BC, D + f D=Ff, and FD-fD=2 CD, Ff:FE – Ef:: 2 BC: 2 CD. (2) Wherefore we readily get from (1) and (2), 2 CT: 2 CB :: 2 CB : 2 CD, or CT : CB :: CB : CD; that is, the semi-transverse CB is a mean proportional between CT and CD. PROPOSITION VIII. The straight line drawn from either, focus of an ellipse to the intersec tion of two tangents to the curre makes equal angles with the straight lines drawn from the same focus to the points of contact. Let TD, TE, tangents to an ellipse at the points D and E, intersect in T; draw to the focus F the lines DF, EF, TF; also draw FG, FH perpendicular to the tangents DT, ET, to meet the lines f D, FE, drawn from the other focus in K and L; and join TK, TL. Then because the angle FDG = KDG (Prop. vi.), and the angles at G right angles, the triangles KGD, FGD, are equal, and therefore KD DF. Wherefore Kf = KD + D f = FD + Df = AB; and similarly, lof = AB. = AB. Whence Kf = L f. The triangles FGT, KGT, are also evidently equal in all respects, and therefore KT = TF. For the same reason, LT TF; whence LT = TK. It will H hence be obvious that the triangles TK f and TL f are equal in all respects, and therefore the angle Df T = L ft. Also if perpendiculars be drawn from f on the tangents produced, it may be shown in a similar way that the angle TFD = TFE. Consequently, the straight line FT drawn from the focus F makes equal angles with the lines FD, FE, drawn from F to the points of contact Ď and E; and also f T makes equal angles with f D, J E, the corresponding lines and angles with respect to the other focus f. Cor. 1. It is shown in the proposition that TK TL; and in a similar way it might be shown that perpendiculars from T on FD, FE, are equal; wherefore, perpendiculars drawn from the intersection of two tangents to an ellipse on the lines drawn from either focus through the points of contact are equal. Cor. 2. Let the other lines be drawn as in the figure. Then hecause FII = HL, and FC = 7C (C being the centre), the triangles TEF, TCF, ECF, are respectively equal to TEL, TCf, ECf. Wherefore the triangle TEC is one-half of the triangle Tlf. For the same reason TCD is one-half of TKf. But it has been shown in the proposition, that TLj' = TKf; therefore TEC = TDC. Whence the perpendiculars from D and E on TC are equal (the perpendiculars are not given in the fig.), and therefore DM ME; that is, the line drawn from the intersection of two tangents to an ellipse to the centre bisects the line which joins the points of contact. PROPOSITION IX. Any diameter of an ellipse bisects all its own ordinates. Let WS be a tangent to an ellipse at the point V; VA a diameter through the same point; and ED any ordinate parallel to WS (Def. 10); then the diameter V A bisects ED in the point M. For let C be the centre of the ellipse, and W, S, the points in the tangent at V, in which tangents at Danıl E meet the tangent WS. Join SC, WC, cutting the chords DV, EV, in G, H, and the chord DE in Q and R. Draw T D M also through the centre C the line LK parallel to DE, and from V, VN parallel to CS. Now because the lines WS, DE, LK are parallel, the triangles VWG, SVH, are respectively similar to DGQ, REH; and since (Prop. viii., Cor. 2), VG = GD, and VH = HE, those triangles are respectively equal in all respects. Hence VW = DQ, and SV = ER. Again, by the similar triangles LCS, RES, CL : RE:: CS : SR. Also by the similar triangles CWK, QWD, CW:WQ :: CK: QD. But CS : SR :: CW:WQ. Hence CL : RE:: CK : QD. Now, by Prop. VIII., Cor. 2, the line which passes through C, the centre, and the intersection of the tangents DK, EL (we do not know yet that this point of intersection is in VA), bisects the line DE, which joins the points of contact. Hence because LK is parallel to the line DE, LK is bisected in C, and therefore LC = CK. Wherefore by the last proportion, RE = QD. But it has been shown that RE SV, and QD = VW; whence SV = VW. Consequently, RM MQ, and therefore EM = MD; that is, the diameter VA bisects any of its ordinates ED in the point M. Cor. 1. It was proved in Prop. VII., Cor. 2, that C the centre, M the middle of ED, and the intersection of the tangents, were in the same straight line ; and it has now been shown that C, M, V are in the same straight line. Wherefore the tangents at the extremities of any chord of an ellipse meet in the diameter of that chord. Cor. 2. Since SH is parallel to NV (Constr.), and EH = HV, therefore ES = SN. But by the parallels NV, SC, and SV, EM, we have CV: CT :: SN : ST, and SE: ST :: MV : VT; and since ES SN, we have CV: CT :: MV : VT. This gives, by alteration and division, CM: CV::CV : CT. Hence, if a tangent and ordinate be drawn to meet any diameter of an ellipse, the semi-diameter will be a mean proportional between the distances of the two intersections from the centre. This generalizes the property established in Prop. vii. Cor. 3. All the ordinates to the same diameter of an ellipse are parallel to one another. Cor. 4. A straight line which bisects two parallel chords of an ellipse, and terminates in the curve, is a diameter. PROPOSITION X. The square of any diameter has to the square of its conjugate the same ratio that the rectangle under the abscissas of any point has to the square of the corresponding ordinate. Let DE, GH be two conjugate diameters (Def. 10), and PQ an ordinate to the former, corresponding to the point P in the curve. Also let the tangent at P meet the conjugate diameters in S and T. Then by Cor. 2 of last Prop., and the property of duplicate ratio, CQ: CD :: CD: CS, or CQ*: CD' :: CQ : CS. Hence, by division and similar triangles, remembering that QP = RC, we have CD – CQ' : CD? :: QS : CS :: RC: CT. But by Prop. ix., Cor. 2, and duplicate ratio, RC: CT:: RC: CH®; wherefore CD – CQ* : CD: :: RCR : CHR. Now because C is the middle of ED, CD – CQ* = EQ .QD; consequently EQ. QD : QP :: CDP : CHP :: EDP : HG'; that is, the square of the diameter CĎ is to the square of its conjugate HG as the rectangle under the abscissas EQ, QD, is to the square of the ordinate PQ. This is the general case of the property established in Prop. II. Cor. 1. The rectangles of the abscissas of any diameter are as the squares of the corresponding ordinates. THE HYPERBOLA. N M PROPOSITION I. The squares of the ordinates of the trunsverse diameter are to one another as the rectangles of their abscissas. As in the parabola and ellipse, let AVB be the transverse plane; AGIB the hyperbolic sectional plane which meets both sheets of the cone (Def. 3); AB the transverse axis which meets the two sheets of the cone in 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 it may be shown, as in the parabola (Prop. I.), that KL, MN are diameters of the circles KGL, MIN, and that the transverse diameter AB bisects Gg, li, perpendicularly in F and H. Hence by the similar triangles AFL, AHN, and KFB, MHB, AF : AH :: FL : HN, and FB: HB :: KF : MH. Whence AF. FB: AH . HB :: FL . KF : HN. HM. But by Euc. III. 35, FL. KF FGʻ; and HN. HM HI'. Wherefore, AF. FB : AH.HB :: FG': HI'; that is, the squares K of the ordinates FG, HI, are to one another as the rectangles of their corresponding abscissas. The complete hyperbola, therefore, is situated on both sheets of the cone, and is composed of two separate and infinitely extended branches (Def. 1). Those branches, moreover, are divided by the transverse diameter into two parts, which are symmetrical with respect to that diameter. PROPOSITION II. The rectangle under the abscissas of any point in the hyperbola is to the square of the corresponding ordinate as the square of the transverse diameter is to the square of its conjugate. (See the Fig. of last Prop.) Let a circular section of the cone meet the middle of the diameter AB in C; and let Ca be a tangent from C to this circle. Then Ca is the conjugate axis to AB. Let WT be the diameter of this circular section; then by the similar triangles AFL, ATC, and FBK, BWC, we have AF : AC :: FL: TC, and FB : BC :: FK: WC. Hence, AF. FB : AC.CB :: FL. FK: TC . WC. But since AC CB, AC. CB AC“, also (Euc, III, 35, 36) FL. FK = FG ’ and TC. WC = Ca'. Wherefore AC?: Ca:: AF. FB : FG”, or AB' : 4Ca' :: AF.FB:FG; that is, the square of the transverse diameter is to the square of its conjugate as the rectangle under the abscissas of the point G is to the square of the ordinate of the same point. Cor. Take p, a third proportional to AB and 2Ca; then by the property of duplicate ratio, AB:p :: AB* : 4Ca'. But by the proposition ABP : 4Ca' : : AF. FB : FG”. Wherefore, AB :p :: AF. FB : FGʻ. The quantity p which remains constant for different points in the curve is ihe parameter of the transverse diameter AB (Def. 13). Wherefore the transverse diameter of the hyperbola is to its parameter as the rectangle under the abscissas of any point is to the square of the corresponding ordinate. PROPOSITION III. In the hyperbolu, the square of the distance of either focus from the centre is equal to the sum of the squares of the semi-diameters. Let AB be the transverse diameter of the hyperbola, ab its conjugate, F and f the foci, C the centre, and GF an ordinate at the focus F. Then by Prop. 2 and Euc. 11. 6, CA”: Ca :: AF. FB: GF :: CF CAP: GF But F being the focus, GF = } p (Def. 14), and therefore by Def. 13 AC° : C'a' :: Cu? : & pe or GF*. Hence CF CA Ca, or CF = CA' + Car. |