Page:Encyclopædia Britannica, Ninth Edition, v. 6.djvu/304

Rh 276 CONIC SECTIONS A chord, tangent, and normal are defined exactly in the same words as in the case of the parabola. To find where an ellipse of given focus, directrix, and eccentricity is cut by a straight line parallel to the directrix. Let S (fig. 13) be the focus, XK the directrix, and e the eccen tricity. -; Draw SX perpendicular to XK, and divide it internally and externally in the ratio e : 1 in the points A, A, so that SA :AX = SA : A X-:1. It is clear that A and A will both be on the same side of X as S is, be cause the eccentricity e is less than 1. Draw AZ at right angles to SX and equal to AS ; and join XZ. Let QN 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 ellipse ; for we have SP: XN = QN:XN = ZA: AX = SA:AX = e:l. It is clear that if this point P exists, the point P on the opposite side of the axis also exists, and therefore the ellipse is symmetrical with respect to the axis. Again, the point P will exist, or, in other words, the circle will cut QN 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 anle. The whole curve therefore lies between the two lines AZ, A Z. PROP. II. To find where an ellipse of given focus, directrix, and eccentricity is cut by a straight line perpendicular to the directrix. Let S (fig. 14) be the focus, XK the directrix, and e the eccen tricity. Draw SX perpendicular to XK, and K divide SX in A, A, so that and draw AR, A R at right angles to SX. Let KQ be any line parallel to the axis cutting the directrix 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. For SR:RK = SA:AX = e :1 = SA : A X = SR :R K. Therefore by the Lemma in the introduction SP:PK = SP :P K = SR:RK = e :1. Therefore P and P are points on the ellipse. Now, if the point P exists, the point P also exists, and it is easily seen that the middle point L of PP lies on a straight line bise ct- ing AA at right angles. The curve therefore is symmetrical, not only with respect to the axis AA, but also with respect to this 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 ellipse might be described with S for focus, X K for directrix, and eccentricity e. The ellipse, therefore, lias two foci and two directrices. Now, since SP : PK-SF : FK- : 1. Therefore SP + SP : PK + FK = e : 1, but PK + P K = 2KL = 2CX. .-. SP + SP = 2e.CX. Now, it is easily seen that SP = S P; therefore SP + S P = 2e. CX ; or the sum of the focal distances of any point on the curve is con stant. Again SA : AX = SA : A X = c : 1 therefore SA + SA : AX + A X = : 1. But AX + A X = 2CX, . . SA + SA And SA = S A, therefore SP + S P = SA -J- -AA j or the sum of the focal distances of any point on the ellipse is equal to the major axis. The point P will exist, or, in other words, the circle on RR as diameter will intersect KQ, if OL is less than OR when is the middle point of RR. Now SX:SK = CA:OR, . . OR.SX = SK.CA; and OL-OC+CL-KX. Therefore the point exists if, , KX. SC + KX. SX&amp;lt;SK. CA ; i.e., if KX.CX&amp;lt;SK.CA; if KX.CX S &amp;lt;SK.CA Z ; &amp;lt;(SX 8 +KX 2 ).CA; if KX 2. (CX 2 - CA 2 )&amp;lt;SX 2. CA 2 ; when CX 2 - CA 2 : CA 2 = SX 2 : CB 2. When KX lias this value the points P, P coincide in CL, say at B or B. It appears therefore that the ellipse lies wholly within a certain rectangle, and that its general shape is of the form given in fig. 15 which shews the centre C, the foci S, S, and directrices XK and X K. It can easily be shown that the sum of the focal distances of any point within the ellipse is less X than, and the sum of any point without greater than, AA, and also that the ratio of the focal 6 B Fig. 15. distance of any point within the ellipse to its distance from the corresponding directrix is less than, and the ratio for any point with out greater than, the eccentricity. An ellipse can be described mechanically in the following manner. If an endless string be placed over two small fixed pegs S, S and be kept tight so as to form a triangle PSS, then a pencil at P would trace out an ellipse whose foci are S, S, and whose major axis is equal to the length of the string minus the distance between the pegs. PROP. III. If a chord PQ (fig. 16) intersect in Z the directrix corresponding to the focus S, then SZ will be the external bisector of the angle PSQ. Join SP, SQ ; and draw PM, QN perpendicular to the directrix. Then because the triangles PMZ, QNZ are similar, PZ:QZ = PM:QN = SP:SQ &quot;I- . . (Eucl. vi. A) SZ bisects the ex ternal angle of the triangle PSQ. COROLLARY. If the point Q moves up to and coincides with P, or, in other words, the chord PQ becomes the tangent. .LX ~-^-^ to the ellipse at P, then the angle PSZ Fi&amp;lt;* 16 will become a right angle. PROP. IV. The foot of the perpendicular from the focus on the tan gen t always lies on the circle described on the major axis as diameter. Let PZ (fig. 17) be the tangent at P, meeting the directrix in Z. Let S be the corresponding focus. Join SP, SZ, and draw PM, SY, perpendicular to the directrix and the tangent respectively. Join YX, SM. Because the angles PMZ, PSZ are right angles, a circle will cir cumscribe PSZM ; and because the angles SYZ, SXZ are right angles, a circle will circumscribe SYZX. Therefore angle SYX = angle SZX = supplement of angle SZM = angle, SPM ; and angle SXY = angle SZY- angle SMP. Therefore the triangles SYX, SPM are similar, and SY:YX-SP:PM-SA:AX. Therefore the locus of Y is a circle on AA as diameter. (Lemma, Introduction.)