146 



CONIC SECTIONS. 



Hvi>er- 

 bola. 



Fi.T- 32c 



that the angle at G is common to the triangles EGD, 

 DGF ; these triangles are similar (6. 6. E.) and ED : 

 DF : : EG : GD, therefore ED J : DF 2 : i EG 2 : GD l ; 

 but EG 1 : GD 1 : : EG : GF (2. Cor. 20. 6. E.) that 

 is, as PC to CF (2. 6". E.) and PC : CF : : AC 2 : CF 2 

 (Def. 1. and Cor. 2. 20. 6. E.) therefore ED 2 : DF 2 : : 

 AC*:CF 2 and ED : DF : : AC: CF; hence D is a 

 point in the hyperbola (Def. 1.) and in like manner it 

 may be shewn, that d is a point in the hyperbola. 



And because the point E must, from the position of 

 the directrix, always fall between F and G, therefore 

 DG, the mean proportional between GE and GF, will 

 always be greater than GE, and consequently much 

 greater than GH, the perpendicular from G on D d ; 

 hence whatever be tr.e position of the given line EH, 

 provided that it be parallel to the axis, the circle will 

 always cut it in two points, so that in this case there, is 

 not any limitation to the possibility of constructing 

 the problem. 



Cor. 1. Every straight line parallel to the trans- 

 verse axis meets the opposite branches of the hyperbo- 

 la each in one point and no more. 



Cor. 2. Every straight line pai'allel to the trans- 

 verse axis, and terminating on opposite branches of the 

 hyperbola, is bisected by the conjugate axis, or that 

 axis produced. 



Cor. 3. The two branches of the hyperbola are 

 perfectly alike, so that if applied one on the other they 

 would coincide. 



Prop. VI. 



If from any point Q in the directrix a straight line 

 be drawn to meet the hyperbola in D, and LF / a pa- 

 rallel to the directrix passing through the focus in H, 

 and if another straight line be drawn to D through F, 

 and a perpendicular Q K be drawn to FD ; the lines 

 FH, QKhave to each other the determining ratio. 



Draw DE perpendicular to the directrix, and let DF 

 meet it in I. The triangles DHF, DO I are evidently 

 equiangular (29. 1. E.) and the triangles QIK DIE 

 are also equiangular, for the angles at K and E are 

 right angles, and the angle at I is common to both, 

 therefore (4. 6. E.) FH : FD (: : Q! : ID) : : QK : DE, 

 and by alternation FH : QK : : FD : DE, that is in the 

 determining ratio. 



Cor. 1. If a straight line QH meet the directrix 

 in Q, and the parallel L/in H; and if KF, a line 

 which is not parallel to QH, have such a position that 

 FH has to QK (the distance of KF from Q) the de- 

 termining ratio, the lines QH, KF shall meet at a point 

 D in the hyperbola : For then FD : DE : : FH : QK, 

 that is in the determining ratio ; therefore D is in the 

 hyperbola (Def. 1.) 



Cor. 2. If QF be drawn to the focus, and in FH, 

 the parallel to the directrix, FL and Yl be taken in 

 contrary directions, each to FQ in the determining ra- 

 tio, then every straight line which meets the hyperbo- 

 la will either pass between, or through the points L, I; 

 For since, by hypothesis, FQ : FL : : KQ : FH, and 

 since KQ, cannot exceed FQ; therefore FH cannot 

 exceed FL. 



Cor. 3. Any straight line drawn from Q to inter- 

 sect the parallel L / beyond the limits L, I falls entire- 

 ly without both branches of the hyperbola. 



Prop. VII. Problem. 



Having given a focus, its directrix, and the deter- 

 mining ratio, to find the points in which a straight line 



given by position, and not parallel to the directrix, 

 meets the hyperbola. 



Hyptr- 

 bola. 



Case 1. Let XY (Fig. 33. No. 1. and No. 2.) the line pi- 

 given by position, pass through F the focus : draw FQ Nog, 

 perpendicular to XY, meeting the directrix in Q, and 

 in LF /, the line passing through the focus parallel to 

 the directrix, take FL and f I each to FQ in the de- 

 termining ratio, and join QL and Q/; then, if neither 

 of the angles FQL, FQ/ be a right angle, as in Fig. 

 33. No. 1, the lines QL, QZ will meet XY in two 

 points D, d, which' will be points in the curve (as is 

 evident from Cor. 1. Prop. VI.) But if one of the 

 angles FQ / be a right angle, as in Fig. 33. No. 2. there 

 will be only one intersection D. To determine the po- 

 sition which the line XY must have, in order that this 

 may happen, produce it to meet the directrix in V, and 

 draw FP perpendicular to the directrix : The triangle 

 FPV being similar to VFQ or FQZ (8. 6. E.) we have 

 FV : FP : : F l~. FQ, that is in the determining ratio. 

 Let M be an intersection of L I and the curve, and 

 FM will have to FP the same ratio (Cor. to Def. 1 — 4) 

 therefore FV=FM, and V is the point in which a circle 

 described on F as a centre, with FM as a radius, meets, 

 the directrix. 



Case 2. Next let the line given by position meet the Ffjj. 

 parallel to the directrix in H, (Fig. 34. No. 1. and 2.) Nos. 

 so that QF being diawn to the focus, and FL and FZ 

 taken to FQ in the determining ratio, it may fall be- 

 tween F and either of the points L,|Z. In QF take QN, 

 so that HF may have to QN the determining ratio, 

 that is, the ratio of FL to QF, then it is manifest that 

 QN is less than QF. On Q as a centre, with QN as a 

 radius, describe a circle, and draw FK, F k tangents 

 to it, then if neither of these is parallel to the line 

 QH, as in Fig. 34. No. 1. they will meet it in two 

 points D, d, which will be points in the hyperbola, as 

 is evident from Prop. 6. Cor. 1. But if, as in Fig 34. 

 No. 2. one of the tangents F k be parallel to the line 

 QH, there will be only one intersection D. To deter- 

 mine the position which the line QH ought to have> 

 that this may happen, let F k meet the directrix in V, 

 and let FP be perpendicular to the directrix. The tri- 

 angles FPV, QV k are evidently equiangular, therefore 

 FV : FP : : VQ or FH : Qk, that is as FM to FP 

 (Cor. to Def. 1—4.) hence FM = FV ; and V is that 

 point in which the directrix meets the circumference of 

 a circle, of which FM is the radius and KQ, the line 

 given by position, is parallel to FV. 



Case 3. Lastly, let the line given by position, and 

 which meets the directrix in Q, pass through one of 

 the points I ( Fig. 34. No. 1 . ) which are so taken that 

 F I or FL has to FQ the determining ratio : Draw 

 FD' perpendicular to FQ, and if it is not parallel to 

 Q I it will meet Q / in some point D' which will be 

 a point in the hyperbola (1. Cor. Prop. 6.) But if, 

 (as in Fig. 3. No. 2,), FQ I is a right angle. FD will 

 be parallel to Q /, and so the line Q / will not meet 

 the curve at all. To determine the position of Q in 

 this case, draw FV perpendicular to FQ meeting the 

 directrix in V. Then the triangle FPV is similar to 

 VFQ or FQ / (8. G. E. ) therefore FV : FP : : F I: FQ 

 that is in the determining ratio ; therefore FV = FM, 

 and so V is the intei'section of the directrix and a circle 

 whose radius is FM, and Ql> is parallel to FV. 



Cor. 1. The points L, / being determined as in the 

 second and third cases, (Fig. 3k No. 1. and No. 2.), 

 any straight line drawn from Q to pass between them, 

 will either cut the hyperbola in one point only, or in 



33. 

 1,2. 



34. 



1,*. 



