of CD, so is the square of CF taken from the first, to the square of CA taken from the second: therefore the remaining rectangle EHF is to the remaining rectangle ADa, as the square of CF to the square of CA (cor. 19. 5. Elem ;) and, by division, the excess of the rectangle EHF above the rectangle ADa, is to A Da, as the (6. 2. Elem.) rectangle AFa to the square of CA: but (by the preceding prop. and 4. 3.) the square of GD is to the rectangle A Da, as the square of CB is to that of CA; and inversely, the square of CA is to the square of CB, as the rectangle ADa is to the square of GD. Cor. The squares of straight lines drawn perpendicular to the transverse axis from points in an hyperbola, or in opposite hyperbolas, are to one another, as the rectangles contained by the segments intercepted between those straight lines and the vertices of the transverse axis; as was shown in the ellipsis (1. cor. 6. 2.) PROP. VIII. THEOR. If from a point in an hyperbola a straight line be drawn perpendicular to the second axis; the square of the second axis is to the square of the transverse, as the sum of the squares of half the second axis, and of its segment between the perpendicular and the centre, is to the square of the perpendicular. From a point G (fig. 3. 4.) of an hyperbola draw GN perpendicular to the second axis Bb; the square of Bb is to the square of Aa as the sum of the squares CB, CN to the square of GN. 0 Because, by the preceding, the square of CA is to the square of CB, as the rectangle ADa is to the square of GD: therefore, inversely, and by proposition 12. b. 5. Elem. the square of CB is to the square of CA, as the sum of the squares of CB, GD is to the square of CA, together with the rectangle ADa, that is, as the sum of the squares of CB, CN is to the square of CD or GN. Cor. Hence, if from two points of an hyperbola, or of opposite hyperbolas, perpendiculars be drawn to the second axis, the square of the one perpendicular is to the square of the other, as the sum of the squares of half the second axis, and of the distance between the former perpendicular and the centre, is to the sum of the squares of half the second axis, and of the distance between the latter and the centre. PROP. IX. THEOR. A straight line terminated both ways by an hyperbola, or opposite hyperbolas, and parallel to either axis, is bisected by the other axis; or, what is the same thing, the axes are conjugate diameters. First, let the straight line DE (fig. 5.) be parallel to the second axis Bb, and meet the transverse in F; and thus the square of DF is to the square of EF as the rectangle AFa is to the (cor. 7. 3.) rectangle AFa: therefore DF, FE are equal. Next, let DG be parallel to the transverse axis Aa, and meet the second axis Bb in K; and thus the square of DK is to the square of KG, as the sum of the squares of CB, CK is to the sum of the same squares (cor. preced.) of CB, CK: therefore DK, GK are equal. PROP. X. THEOR. A straight line terminated both ways by an hyperbola, or opposite hyperbolas, and bisected by either axis, is parallel to the other axis. First, let DE (fig. 5.) be bisected by the transverse axis in F; and draw DK, EL parallel to the same axis, and meeting the second axis in the points K, L; then, because DF, FE are equal, KC, CL are also equal: but the square of DK is to the square of EL, as the squares of CB, CK together, to the squares of CB, CL together; therefore DK, EL are equal, and they are parallel: consequently DE, KL are also parallel (33. 1. Elem.) Next, let DG be bisected by the second axis in the point K, and draw DF, GM parallel to the same axis, and meeting the transverse axis in F, M; then, because DK, GK are equal, FC, CM are likewise equal; and, of consequence, FA, aM are equal: now the square of DF is to the square of GM, as the rectangle AFa to the rectangle AMa; but the rectangles AFa, AMa are equal; and therefore the straight lines DF, GM are equal, and they are parallel; consequently DG, FM are likewise parallel (33. 1. Elem.) Cor. It is manifest from the demonstration, that the straight lines DF, GM, which are parallel to either axis Bb, and cut off, between the centre and the points where they meet the other axis, equal segments FC, MC, are also equal. In the same manner, DK, EL are equal, which are parallel to the axis Aa, and cut off the equal segments CK, CL. And the contrary: if DF, GM are equal to each other, and parallel to Bb, they cut off equal segments FC, MC. In like manner, if DK, EL be equal to each other, and parallel to Aa, they cut off equal segments CK, CL. PROP. XI. THEOR. Any straight line perpendicular to the transverse axis, and meeting it below the vertex, will meet the hyperbola in two points. 1 Let DC (fig. 6. 7.) be perpendicular to the transverse axis Aa, and meet it in C, below the vertex A; then DC meets the hyperbola in two points. Let E, F be the foci; and from C, place CG equal to CF, the distance between C and the nearest focus; and from the other focus place EK (fig. 6.) equal to the transverse axis Aa. If then, the point C be below the focus F, it is evident, that EK is less than EG: but in the other case where the point C is above F; since Aa, EK (fig. 7.) are equal, AK is equal to aE, that is, to AF; and, by hypothesis, FC is less than FA; twice FC is, therefore, less than twice FA, that is, FG is less than FK; and thus EK is less than EG: make then as EK to EF, so is EG (fig. 6. 7.) to a fourth proportional EH; and since EK is less than each of the two EF, EG, and, of consequence, much less than EH; therefore EK, EH are (25. 5. Elem.) together greater than EF, EG together. From these unequals take away twice EK, and KH will be greater than KF and KG together, that is, than twice KC; for CF is equal to CG: hence, if KH is bisected in L, KL will be greater than KC; and therefore the point L falls below the straight line CD; and a circle described from the centre E, with the distance EL, will necessarily meet CD in two points D, d. Describe, from the centre D, distance DF, another circle, which (3. 3. Elem.) will pass through the point G; join DE, and let this circle meet it in the points M, N; and because EK is to EF, as EG to EH, the rectangle HEK is equal to the rectangle FEG, that is, to the rectangle MEN (cor. 36. 3. Elem ;) and ED, EL are equal; and thus their squares are equal; from which take away the equal rectangles MEN, HEK, and the remaining square of DM, or DF is equal to the remaining square of KL (6. 2. Elem.) consequently DM and KL are equal, and which being taken from the equals ED, EL, the remainder EM is equal to the remainder EK, or the transverse axis Aa; and EM is the excess of DE above DF: therefore the point D is in the hyperbola. In like manner it may be demonstrated, that the point d is in the hyperbola. |