VOL. VIII. ] PHILOSOPHICAL TRANSACTIONS. lOQ 



kkqqa , a* — 2qqma ., r t ma — qq ... , 



aa — TTJ—^, h , , . : the square of e H 7—^; which may be re- 



bkk ->r bmm ' kk + 7/im ^ ' k ' •' 



ky 



diiced to a simpler equation, by putting kh + ^^ = pp) and -^ = «, for it then 



becomes the square of e--^ + ^=y^ — -^^ — ~i^- + il' "^^'^^^ 

 equation you will find to answer to the above construction, by applying the cal- 

 culation ; and you will at the same time observe, that to which ever of the 

 lines EA, AB, BE, the analysis is referred, the same sections may be always 

 obtained, though by a longer process, and by very different equations. 



From this construction, by analogy, may be deduced the solution of the 

 other problem, viz. to determine a point, from which a ray may be reflected 

 parallel to any given line. Thus, if B be the given luminous point, A the 

 centre of the circle, (fig. 5,) and the reflected ray be parallel to AE. For this 

 is the same thing as if, in the other problem, the distance of the points A and E 

 be supposed infinite ; in which case the third proportional to EA, FA, becomes 

 nothing, and the points A and V coincide: therefore VX will be equal to AX, 

 and AE parallel to PE. Apply therefore the former construction, and you solve 

 the problem: viz. to the vertex X, transverse diameter VX or AX, and latus 

 rectum equal to it, describe the hyperbola XP, whose ordinates to the diame- 

 ter AX are parallel to AE. 



j4gain, M. Huygens, July 1, 1672, writes as follows. 

 It is true, and indeed wonderful, that the construction I formerly sent you, 

 may also be found by M. Slusius's calculation, after changing qqfor aa -f- ee. 

 But this seems only to happen by chance, nor does the simplicity of the con- 

 struction here appear till after close application to it. 



To repeat Alhazeris Problem, viz. Having given a circle, whose centre is A, 

 its radius AD, and two points B, C ; to find a point H in the circumference of 

 the given circle, whence drawing the lines HB, HC, they shall make equal 

 angles at the circumference (fig. Q). 



Suppose it found; and drawing AM, which bisects the angle BAG, and per- 

 pendicular to it draw HT, BM, CL. Then join AH, and make HE perpendi- 

 cular to it, and let AM meet BH and HC in the points K, G. . 

 Now put AM = a Then because the angles KHE and CHZ or EGH 

 MB = Z) are equal, and as EH A is a right angle, it will be, as 

 AL = c KE to EG, so is KA to AG. And because BM is to 

 LC = n MD, as HF to FK, it will be as BM + HF to HF, 

 Radius AD = c/ so MY toYY., ov h + y.y.-.a ^ x -. ~f=^l ; add FA 



FH =2/ or JT, and it makes KA = -'!'—; — -, 



