Page:Encyclopædia Britannica, Ninth Edition, v. 10.djvu/410

Rh 396 G l1} O M theorenls, and often state one tlieorelll only, the reader beillg re- colllniended to go througll tlle reciprocal proof by himself, aild to supply tlle reciprocal thcorellls wllell not given. § 46. We state the theorems ill tlle pencil reciprocal to the last, ETRY strllction in order to deduce fllrther properties fronl it. 'e also solve the rigllt—llalld problem. Here we select two, viz., u,, -u._, of the five given lilies, -u,, ‘(L2, (1, b, c, as bases of two rows, alld tlle points .—,, B,, C, wllcrc (1, 1:, c ellt 11., as correspoliding to tlle points [PROJ ECTIVE. without proving them :— TIleorem.—It two projective ﬁat pencils are concentric, hilt are neither perspective nor en-plalior, then tlle envelope of the planes joining cor- responding rays is a colle of the second class; that is, no lille through the common centre contains lilore than two of the enveloping planes. ’I‘Ileorelll.—If two projective axial pencils lie ill tlle sanle pencil (their axes lllcet ill a poillt), bllt are lleitller perspective nor co-axial, then the loclls of lilies joining correspoildillg planes is a cone of the second order; that is, no plane ill the pencil con- tains lllore than two of these lilies. § 47. Of theorems about colles of secolld order and cones of second class we shall state only very few. 'e point. ollt, however, the following connexion betwecli the curves and collcs illlder colisidera- tion :— The lines which join ally poillt ill sp.lce to the points on a cllrve of the secolld order form a colle of the second order. The planes which join ally poiilt ill space to the lilies enveloping a curve of the second class envelope them- selves a cone of the second class. Every plane section of a cone of the second order is a cilrve of the second order. Every plane section of a cone of the second class is a curve of the second chlss. By its aid, or by the principle of dllality, it will be easy to ob- t.lill theorems about them frolll the theorems about the curves. We prove the first. two projective pencils. A curve of the second order is generated by These pellcils, when joined to the poillt in space, give rise to two projective axial pencils, which gellerate the colic ill question as loclls of the lilies where corrcspolldillg planes meet. § 48. Them-em.— The curve of second order which is gcllerated by two pro- jective tl-at pencils passes through the centres of the two pellcils. Prnu_f.—If S and S’ are the two pencils, then to the ray S5’ or p’ ill the pencil S’ corresponds hi the pen- cil S a ray p, which is diIl'cl-ent from p’, for the pencils are not perspec- tive. Blit p alld 1)’ meet at .‘, go that S is a poillt on the clirvc, alld simi- larly S’. TIlcm'em.—’l‘lle envelope of second class which is gclleratcd by two pro- jcctive rows contains the bases of these rows as enveloping lilies or tangents. Prnn_f.~If s alid s’ are the two rows, then to the point ss’ or 1" as :l poillt ill 3’ corrcspollds ill 8 a poillt P, which is not coincident with 1”, for the rows are not perspective. But I’ alld 1" are joined by s, so that s is one of the enveloping lilies, and similarly 8’. A2, B2, ('2 where (1, 11, c cllt u._,. ,1‘ ‘--—-_____ F- U1 Fig. 15. 'e get then the following solutions of the two 1)l'0lllt'lllS :— Solutl'rm_—Tlll'ollgh the point A draw any twolilles, ll, alld 'u._,(tig. 15), the ﬁrst. ll, to cllt the pencil S, ill a. row _l:,(‘,, the other 11., to clit the pencil .'._. ill a rolv AB._.C._.. Tllesc two rows will be perspective, as the poillt A corresponds to itself, alld the cclltrc of projection will be the poillt S, where the lilies B,B._. alld (',(‘,_, lncet. 'l‘o find now for ally l'.'1_v 4!, ill S, its correspolldiilg ray (I._. ill S._., we deter- mine the poillt D, where (I, cllts 'll,, project this poilit from S to D2 on 142 alld joill S2 to l)._.. This will be tllc required ray (I._. which cuts (I, at sollle Sollltl'on.—Ill the lille a take any two poilits S, alld S._. as centres of pencils (fig. 16), the til-st .', (.-,l:,l',) to project the row '1l,_ tlle Hillel- .‘~‘._» (A._>B._.('._.) to project the row ll... These two pencils will be pl-l-_<pectiv(, the lille S,.-, being the sallle as the Cul‘l‘€‘S]')(llltllll',Z lille .‘._.A._., and the axis ofprojcctioli will be tllc lille ll, lIll'll joins the illtcrscctioll 1-: of S,l:, and S. ._. to theillterscctioll (' of .',(', and S-_ ,3. 'l'o lilld now for any point ll, ill ll, tllc col-rcspondillg point ll._. ill 'u._., we draw .',lI, alld pr--je--t the poiilt D where this lillc cuts ll fl'lllIl .'._. It follows that every lilic ill one of the two pellcils cilts the curve in two points, viz., ollce at tllc celltrc S of tile pencil, alld once where it cuts its corresponding ray ill the other pencil. These two points, however, coincide, if the line is cllt by its correspollding lille at S itself. Tllc lille )2 ill S, which (‘0l'l‘c.'~'p0ll(lS to the line SS’ in S’, is therefore tllc only liilc through S which has but one poillt ill common with the curve, or wlliell cuts the cllrve ill two coinchlelzt points. Such a line is called a tangent to tllc eurve, touching the latter at the poilit S, which is called the poillt of contact. Ill the same nlanner we get ill the reciprocal investigation the result that through.evcry point ill one of the rows, say ill .9, two tangents may be drawn to the curve, the ollc being s, the other the lille joining the point to its corresponding poillt in s’. Tllerc is, however, one point P ill s for which these two lines coincide. Such a poillt in one of the tangents is called the “ poillt of contact” of the tangent. We tllus get- to u._.. This will give the rel,llil-ed point D9, alld the lilie ll joillill-,: I), to D2 will be :1 new tan;_.ellt to the cllrve. poillt D on the curve. Theorem.—To the liile joining the centres of the projective pencils as a. lille ill one pencil corresponds ill the Theoreln.—To the point of inter- section of the bases of two projective rows as a poillt in one row corre- sponds ill the other the point of con.- otlier the tangent at its centre. tact of its base. § 49. Two projective pencils are detcrnlilled if tllrcc pairs of cor- responding lines are given. Hence if a.,, b,, c, are three lilies ill a pencil S,, alld a2, (7,, c, the corresponding lilies ill a projective pcn- cil S.,,_, the correspondence alld therefore the cilrve of the second order generated by the points of intersection of corresponding rays is deterlllillcd. Of this cllrvc we know the two centres S, and S,, and the three points a,a.,,, b,b.,, c,c.,,, hence ﬁve poillts ill all. This and the reciprocal considerations enable us to solve the following two problems :— Problem.—-To construct a curve of the second class, of which five tangents u,, 112, a, b, c are given. Problem.—-To construct a curve of the second order, of which five points S,, S,, A, B, C are given. Ill order to solve the left-hand problelll, we take two of tllc given poillts, say S, alld S,, as centres of pencils. These we make pro- jective by taking tllc rays a,, b,, c,, which joill S, to A, B, C respectively, as corresponding to tile rays a2, 1),, c._,, which joill S, to A, B, C respectively, so that three rays nleet their correspond- ing rays at the given points A, B, C. This deterlnilles the correspolldcnce of the pencils which will generate a curve of the second order passing through A, ll, C alld through the cclltrcs S, and S._,, hellcc through the five given points. To find more points on the curve we have to construct for ally ray ill S, the correspolid- ing ray ill S,. This has been dolle ill § 36. int we repeat the coil- § 50. Tllese collstructions prove, when rightly iliterpreted, very ililportant properties of tile cilrves ill question. lf ill ﬁg. 15 we draw ill the pcllcil S, the ray 1', which passes through the auxiliary cclltrc S, it will be found that the corl'e.spoll:l- illg ray It, cuts it on ita- TIleorcm.—Ill the above construc- tion the bases of the auxiliary rows u, and uz cllt the curve where they cut the rays S23 and S,S respec- Hcnce—- Theorem.—In the above collstl ne- tion (fig. 16) tlle tangents totlle cllrve from the cclltrcs of the auxiliary pell- cils S, and S, are tllc lilies which pass t,i'e|y_ tllrough 11211. alld u,u respectively. As A is ally given poillt on the curve, and 11, ally lille through it, we have solved the problenls :— I’roblem.—-To ﬁnd the second point ill which any line through a known point on the curve cuts the curve. If we determine in S, (ﬁg. 15) the ray correspondillg to the ray S,S, ill S,, we get the tallgent at S,. Silllilarly we can dctcrnlllle tllc point of contact of tile tangents u, or ‘N2 ill fig. 16. §5l. If five points are given, of which not tllrcc are ill a lille, tllen we can, as has just been shown, always draw a curve of the second order througll tllenl, we select two of the points as centres _of projective pencils, and then one such cllrve is determined. It will e presently shown that we get always the sanle curve if two other points are takell as centres of II('ll('llS, that therefore five poillts defer- P-roblem.-—To find the seeolld tall- gent which can he drawll from ally point in a given tangent to the cllr'e.