64. A Conic Section from Five Elements.

Transcription

1 . onic Sction from Fiv Elmnts. To raw a conic sction of which fiv lmnts - points an tangnts - ar known. W consir th thr cass:. Fiv points ar known.. Four points an a tangnt lin ar known.. Thr points an two tangnts ar known. Not. örri consirs othr cass, which ar ssntially "uals" of ths cass. For xampl th cas of two known points an thr known tangnt lins is th ual of abov. Th proofs of, an rly on Pascal s Thorm (No. ), an th proof of th uals on th ual of Pascal s Thorm, namly rianchon s Thorm (No. ). Not. Most gomtry softwar programs inclu a tool to construct a conic through 5 points. This is cas abov.. Lt th fiv points b,,,, 5 in any orr. Lt b a lin through point 5. W will trmin point on an on. First fin th Pascal lin p as th lin through th points of intrsction of opposit sis an 5, an an 5. Lins an also intrsct on p, say at X; thn 9 X. X p 5 l iffrnt lins through 5 will prouc iffrnt points. It is also possibl to construct th tangnt lin to at any of th fiv givn points, say 5. This uss orollary to Pascal s Thorm. Lt p b th lin through points 9 5 an 9 5. Lt X p 9. Thn lin 5X is tangnt to at 5.

2 p X 5 tangnt lin. raw a conic sction of which four points,,, an on tangnt t ar givn. Thr ar two subcass to consir. a. Th tangnt t passs through on of th givn points, say. W consir th tangnt t as th lin conncting two coincint points an 5, so that t 5. Lt b th point of intrsction of conic with an arbitrary lin x through. Thus x. Now raw th Pascal lin p of hxagon 5 as th lin through th points of intrsction of an 5 t, an an x. p x.x t =5.t Th point of intrsction of an 5 lis on p, an it follows that is th point of intrsction of x an th lin through an 9 p.,,, 5 an ar thn fiv known points on conic, an th problm is now ruc to cas. b. Th tangnt t os not pass through any of th givn points. To solv this problm, w us sargus involution thorm (No. ) with th complt quarangl an lin t. Lt, U,, U b th points of intrsction of t with opposit sis,,, rspctivly. Th rciprocal pairs Ÿ, U an Ÿ, U trmin an involution on t (No.

3 ) thir rciprocal pair Ÿ, U can b foun by intrscting opposit sis an with t, an thn th oubl points H an K of th involution can b construct by Stinr s construction (No. 0). Ths ar points of tangncy of t on th conic. Fiv points ar thn known on conic, an th problm is ruc to cas. Thr ar gnrally two conics. If th involution has no oubl points, thr ar no solutions. t ' K ' H '. raw a conic sction of which thr points,, an two tangnts an ar givn. Thr ar thr subcass to consir. a. passs through an passs through. onstruct th Pascal lin p of th "hxagon" 5 in which,, 5 an is on an arbitrary lin through. Not that sis an ar th tangnt lins an rspctivly. p is th lin through th points of intrsction of an 5, an an. Th point of intrsction X of an 5 must b on, an it follows that 5X 9. l p X == == 5= Similarly, with a iffrnt lin m through, construct point 7 on th conic.

4 m == == 7 5= Now that fiv points on th conic ar know, th conic can b "construct" as in cas. == == 7 5= b. passs through, but os not pass through any of th givn points. Th solution uss th scon corollary to sargus involution thorm. (S No..) Lt 9 an E 9. onstruct th oubl points H an K of th involution trmin by th rciprocal pairs Ÿ, an Ÿ, E. Lins H an K intrsct lin at points of tangncy X an Y on th conic, an th problm rucs to subcas a abov.

5 Y K E H X Thr ar no solutions if th involution has no oubl points. c. Nithr of th two tangnts passs through any of th givn points. Lt 9 an E 9. trmin a oubl point P of th involution trmin by th rciprocal pairs Ÿ, an Ÿ,E. P lis on th chor joining points of tangncy (on ) of an (by th scon corollary to sargus Involution Thorm). Lt U 9 an E U 9. trmin a oubl point P of th U involution trmin by th rciprocal pairs Ÿ, an Ÿ, E. P lis U U U on th chor joining points of tangncy of an. Th lin PP is thus th tangncy chor of th last two paragraphs, U an it mts an at thir tangncy points. This givs fiv points on conic, an th conic can b "construct" as in cas. oth involutions must hav a oubl point in orr for to xist. In this cas, thr ar in fact four solutions, sinc ach involution has two oubl points ' P' P E E' P P' 5

6 ' P' P E E' P P' ' P P' E E' P P' ' P P' E E' P