becomes CN PN: AC: BC, or CN: PN :: AC: BC (B); compare the two proportions (A) and (B), and we have Aa: AC: BC: AC, or Aa=BC. Draw therefore Aa at right angles to AM, and make it equal to BC, join Ca, and this gives the position of the asymptote XCZ. In the same manner, by making Ab= BC, and joining Cb, we determine the position of the asymptote UCY; indeed, from what has been proved, it appears, that if a parallelogram acdb be described about the major and minor axes, the asymptotes will be merely the prolongation of the diagonals of such parallelogram. (99.) These lines, XCZ, UCY, will also be asymptotes to the conjugate hyperbolas; for by a similar process of reasoning it might be shewn, that the position of their asymptotes would be determined by drawing perpendiculars Ba, Ob at B and O, and making Ba and Ob each equal to AC. Thus these four hyperbolas are inclosed as it were between their asymptotes; and by producing the ordinates to meet these asymptotes, new properties of the curves will arise, which we shall now proceed to investigate. PROPERTY 16. (100.) Let the ordinate Pp be produced to meet the asymptotes in the points L, 1; then the rectangle of the parts U parts of it intercepted between the curve and the asymptotes g & dividendo, AC': CN-CA' :: LN-PN': PN': or AC: LN-PN':: CN-CA': PN', R .. LN-PN'=BC'. But LN'-PN2=LN—PN × LN+PN=PL× Pl; .. PLX Pl=BC'=Aa'. For the same reason, pl x pL= Ab3=Aa3. Hence PLX Pl=plxpL. (101.) COR. Draw any other ordinate Qq, and produce it to meet the asymptotes in X and Y, then will QX×QY=Aa'; hence we have QXxQY=PL×Pl. PROPERTY 17. (102.) Draw any diameter PCG, and produce it to g; draw the ordinate QT to that diameter, and produce it to meet the asymptotes in R, r; then QR× Qr=Tr× TR. Through the points P, Q, draw LI, XY perpendicular to the axis of the hyperbola, and draw the tangent ef at P. By sim. triangles QXR, PLe; QrY, Pfl; we have Qx: QR :: PL : and QY: Qr : Pl: Pe, Pf; •. QX× QY: QRxQr :: PLxPl: Pex Pf. But by Art. 101. QX × QY=PL x Pl; hence QR × Qr =Pex Pf. In the same manner, by drawing an ordinate through T perpendicular to the axis, it might be shewn that Trx TR =Pex Pf: hence QRxQr=Tr×TR. (103.) (*) See Fig. in page 62; in which figure, as well as in the figure of page 64, the line GCvg should have passed through the point P. (103.) COR. 1. Since Qr=QT+Tr, and TR=QT +QR, we have or QRXQT+QRxTr=TrxQT+Trx QR. Subtract QRx Tr from each side of this latter equation, and there results QRXQT=Trx QT, from which it appears that QR=Tr; in the same manner it may be (104.) COR. 2. Since the triangles eCf, RCr are similar, and since the diameter G Cg bisects ef in P, it will bisect Rr in v; hence v Rur; and as QR=Tr, we have vQ=vT, i. e. the diameter GCg bisects all its ordinates. (105.) COR. 3. Hence vR2 —vQ'=Pe. For vR-vQ' =vR−vQ × v R +vQ=vR¬vQxvR+vT=QR × RT =QRX Qr (for RT=QR+QT = Tr+QT = Qr). But QRxQr=Pex Pf= (103) Pe'; ..vR' —vQ'=Pe3. PROPERTY 18. (106.) Join AB, and let it cut the asymptote XCZ in S; draw PD parallel to the asymptote UCY, cutting the asymptote XCZ in R; then CR× RP=AS'. (Fig. in page 66.) Since the diagonals BA, a C of the parallelogram a BCA are equal and bisect each other in the point S, the lines SA, SC, SB, Sa are equal; hence the 4 SAC= L SCA; but SCALACY, .. L SAC= L ACY, and consequently AB is parallel to UCY. If therefore Pr, Am are drawn parallel to the asymptote XCZ, then PRCr, ASCm will be parallelograms, and Pr will be equal to CR, and Am to SC. and By sim. triangles Prl, Amb; PRL, A Sa; we have Pr (CR): Pl Am (SC): Ab, RP : PL :: SA : Aa; .. CRX RP: PLxPI :: SCX SA: Aax Ab. But PLX Pl Aax Ab; hence CRX RPSCX SA=SAo. (107.) COR. 1. Since XCZ is likewise an asymptote to the conjugate hyperbola, by a similar process of reasoning it might be shewn that CRX RD SB'=SA'; hence ན ་་ CR |