Page:Encyclopædia Britannica, Ninth Edition, v. 6.djvu/307

CONIC SECTIONS 279 PROP. I. To find where an hyperbola of given focus, directrix, and eccen tricity is cut by a straight line parallel to the directrix. Let S (fig. 29) be the focus, XK the directrix, and e the eccen- fricity. Draw SX perpendicular to XR, and divide it internally and ex ternally in the ratio e to I in the points A, A , so that SA : AX = SA : A X=e :1. Fig. 29. It is clear that A will lie between S and X, and A without SX on the side remote from S. Draw AZ at right angles to SX and equal to AS, and join XZ. Let QX be any straight line parallel to the directrix, cutting XZ in Q and the axis in N. With centre S and radius equal to QN, describe a circle cutting QN in P and P ; these will be points on the hyperbola. It is clear that if the point P exists, the point P on the opposite side of the axis also exists, and therefore the hyperbola is sym metrical with respect to the axis. Again, the point P will exist, or, in other words, the circle will cut Q]ST as long as SP or QN is greater than SN, which is always the case as long as the angle QSN is greater than half a right angle. Now, if SL, A Z be drawn at right angles to SX, cutting XZ in L, Z , then (Eucl. vi. 4) Z A : A X = ZA : AX, but SA : A X = SA : AX, and SA = ZA, .-. SA =Z A , from which it is easily seen that the angle Z SA is half a right angle. The whole curve therefore lies without the two lines AZ, A Z . PROP. II. To find where an hyperbola of given focus, directrix, and eccen tricity is cut by a straight line perpendicular to the directrix. Let S (fig. 30) be the focus, XK the directrix, and e the eccen tricity. Draw SX perpendicular to XK, and divide SX in A, A , so that SA : AX = SA : A X = e :1; and draw AR, A R at right angles to SX. Let KQ be any line parallel to the axis, cutting the direc* trix in K. Join SK, cutting AR, A R in R, R , and upon RR as diameter describe a circle cutting KQ in P, P ; these will be points on the curve. Now SR : RK = SA : AX = e : 1 = SA : A X = SR :R K. Therefore by the Lemma in the introduction Fig. 30. SP : PK = SF : P K = SR : RK = e : 1 . Therefore P and P are points on the hyperbola. Now, if the point P exists, the point P also exists, and it is easily seen that the middle point L lies on a straight line bisecting AA at right angles. The curve, therefore, is symmetrical, not only with respect to the axis AA , but also with respect to the line bisecting AA at right angles. The middle point C of AA is called the centre of the curve, from the fact that every straight line through C is bisected at the point. It is evident from what has preceded that if we measure CS = CS and CX =CX, in the opposite direction to CS and CX, and draw X K parallel to XK, the hyperbola might be described with S for focus, X K for directrix, and eccentricity e. The hyperbola therefore has two foci and two directrices. Now, since SP :PK = SF :P K = e : 1, therefore SP - SP : P K - PK = e : 1, but P K-PK = 2LK = 2CX .-. SP -SP = 2c. CX. Now it is easily seen that SF = S P, therefore S P - SP = 2e . CX, or the difference of the focal distances of any point on the curve ia constant. Again SA : AX = SA : A X = e : 1 . . SA -SA : A X-AX = e : 1, but SA -SA = AA and A X-AX = 2CX. .-. SP -SP = SA -SA = AA , or the difference of the focal distances of any point on the hyper- bola is equal to th transverse axis. The point P will always exist, or in other words, the circle on R, R as diameter will always intersect KQ, because R, R are on opposite sides of KQ. Any straight line therefore parallel to the axis cuts the curve, and the curve must take the form given in fig. 31 which shows the centre C, the foci S, S , and the directrices XK, X K . It can easily be shown that the difference of the focal distances of any point on the concave side of either of the two branches (which is called within the curve) is greater than AA , and the dif ference of the focal distances of any point outside the curve is less than AA ; and also that the ratio of the focal distance of any point within the hyperbola to its distance from the corresponding direc trix is less, and the ratio for any point without is greater, than the eccentricity. If a line BCB be drawn perpendicular to ACA , and points B, B taken in it such that CB 2 = CB 2 = CS 2 - CA 2 , then AA is called the transverse axis, and BB the conjugate axis of the hyperbola. If an hyperbola be described with BB for transverse axis, and A A for conjugate axis, then this hyperbola is said to be conjugate to the first one. It is clear that the foci of the conjugate hyperbola will be in BCB at the same distance from C as S and S . An hyperbola can be described mechanically in the following manner : Suppose a bar SQ (fig. 32) to revolve round its extremity S which is fixed ; then if a string of given length, attached to the bar at Q, and also to a fixed point S , be always kept tight by means of a ring P sliding on SQ, a pencil at P would trace a hyperbola whose foci are S, S , and whose transverse axis is equal to the length of the rod minus the length of the string. PROP. III. If a chord PQ intersect in Z the directrix corresponding to the focus S, then SZ will be the external bisector of PSQ if P, Q both lie on the same branch of the hyperbola, and SZ will be the internal bisector of the angle PSQ, if P, Q lie on different branches. It can be shewn exactly as in Prop. iii. on the ellipse that PZ:QZ=SP:PQ, which proves the proposition. COROLLARY. If the point Q moves up to and coincides with P, or in other words, the chord PQ becomes the tangent to the hyper

bola at P, then the angle PSZ will become i right angle.