Hence (Prop. XII., B. II.) CA: CE2::CT: CE; and, by division (Prop. VII., B. II.), CA: CE-CA':: CT: ET. (1) Again, by Prop. X., CT: CB:: CB: CE' or DE. Hence (Prop. XII., B. II.), CB2 : DE2 : : CT' : DE. But, by similar triangles, therefore CT: DE:: CT: ET; CB2 : DE2 : : CT : ET. Comparing proportions (1) and (2), we have 'CA': CE'-CA' :: CB2 : DE2. (2) But CE-CA' is equal to AEX EA' (Prop. X., B. IV.); hence CA: CB:: AEXEA: DE. In the same manner it may proved that CB2 : CA2 : : BE'× E'B': D/E/2. CA: CB2: CE-CA: DE'. Therefore, the square, &c. Cor. 1. Cor. 2. The squares of the ordinates to either axis, are to each other as the rectangles of their abscissas. Cor. 3. If a circle be described on the major axis, then any tangent to the circle, is to the corresponding ordinate in the hyperbola, as the major axis is to the minor axis. For, by the Proposition, CA: CB::AEXEA': DE2. But AEX EA' is equal to GE (Prop. XXVIII., B. IV.). Therefore or CA': CB' :: GE' : DE', CA: CB::GE: DE. G BI PROPOSITION XIII. THEOREM. The latus rectum is a third proportional to the major and Hence or AC BC BC: LF, Therefore, the latus rectum, &c. PROPOSITION XIV. THEOREM. If from the vertices of two conjugate diameters, ordinates are drawn to either axis, the difference of their squares will be equal to the square of half the other axis. BE Let DD', EE' be any two conjugate diameters, DG and EH ordinates to the major axis drawn from their vertices, in which case, CG and CH will be equal to the ordinates to the T minor axis drawn from the same points; then we shall have CA-CG-CH', and CB'-EH'-DG'. Let DT be a tangent to the curve at D, and ET' a tangent at E. Then, by Prop. X., CGXCT is equal to CA, or CHXCT'; whence Hence HTA G E' CG: CH:: CT: CT; or, by similar triangles, :: CH: GT. CH' GTXCG =(CG-CT) XCG =CG'-CG ×CT that is =CG-CA' (Prop. X.) ; CA'-CG-CH. In the same manner it may be proved that CB'=EH'-DG'. Therefore, if from the vertices, &c. Cor. 1. CH' is equal to CG-CA2; that is, CG ×GT; hence (Prop. XII., Cor. 1), CĂ2 : CB2 : :CG×GT:DG3. Cor. 2. By Prop. XII., By composition, Hence CB2: CA:: EH-CB': CH'. CB: CA:: EH: CA+CH' or CG'. PROPOSITION XV. THEOREM. The difference of the squares of any two conjugate diameters, is equal to the difference of the squares of the axes. Let DD', EE' be any two conjugate diameters; then we shall have DD"-EE-AA"-BB'. Draw DG, EH ordinates to the ma- A jor axis. Then, by the preceding Proposition, CG-CH2-CA', EH'-DG'=CB'. BE HA E' B and Hence or that is, CG'+DG-CH-EH-CA'-CB',· CD2-CE'-CA'-CB'; DD-EE—AA”—BB'. Therefore, the difference of the squares, &c. PROPOSITION XVI. THEOREM. The parallelogram formed by drawing tangents through the vertices of two conjugate diameters, is equal to the rectangle of the axes. Let DED'E' be a parallelogram, formed by drawing tangents to the conjugate hyperbolas through the vertices of two conjugate diameters DD', EE'; its area is equal to AA' XBB'. Let the tangent at D meet the major axis in T; join ET, and draw the ordinates DG, EH. Then, by Prop. XIV., Cor. 2, we have or But hence or CA: CB :: CG: EH. BE A D HTA E' B CT CA: CA: CG (Prop. X.); : CA X CB is equal to CTX EH, which is equal to twice the triangle CTE, or the parallelogram DE; since the triangle and parallelogram have the same base CE, and are between the same parallels. Hence 4CA× CB or AA'× BB' is equal to 4DE, or the parallelogram DED'E'. Therefore, the parallelogram, &c. PROPOSITION XVII. THEOREM. If from the vertex of any diameter, straight lines are drawn to the foci, their product is equal to the square of half the conjugate diameter. (1) But, by Prop. XVI., AC × BC=EC > DK; whence and (2) AC or DL: DK :: EC: BC, Comparing proportions (1) and (2), we have FDXFD: FG × F'H:: EC2 : BC2. But FGF'H is equal to BC (Prop. 1X.); hence FDxF'D s equal to EC. Therefore, if from the vertex, &c. If a tangent and ordinate be drawn from the same point of an hyperbola to any diameter, half of that diameter will be a mean proportional between the distances of the two intersections from the center. Let a tangent EG and an ordinate EH be drawn from the same point E of an hyperbola, meeting the diameter CD produced; then we shall have CG: CD:: CD : CH. Produce GE and HE to meet the major axis in K and L; drav DT a tangent to the curve at the point D, and draw DMrallel to GK. Also, draw the ordinates EN, DO. By Prop. XIV., Cor. 1, CA': CB:: CO×OT: DO', K ::CNxNK; EN' G D Hence TL KANM COXOT: CNxNK:: DO': EN' :: OT: NL, by similar triangles. (1) Also, by similar triangles, OT: NL:: DO: EN :: OM: NK. (2) Multiplying together proportions (1) and (2) (Prop. XI., B. II.), and omitting the factor OT' in the antecedents, and NKXNL in the consequents, we have and, by division, CO: CN :: OM : NL; CO: CN :: CM: CL. CO: CN:: CK: CT. Also, by Prop. X., CK x CN=CA'-CTX CO; hence Comparing proportions (3) and (4), we have (3) (4) CK: CM:: CT: CL. The square of any diameter, is to the square of its conjugate, us the rectangle of its abscissas, is to the square of their ordinate. Let DD', EE' be two conjugate diameters, and GH an ordinate to DD'; then DD: EE":: DH×HD': GH'. Draw GTT' a tangent to the curve at the point G, and draw GK an ordinate to EE'. Then, by Prop. XVIII., CT: CD :: CD: CH, and CD2: CH:: CT: CH (Prop. XII., B. II.). K D G Τ E' |