. Memoirs and proceedings of the Manchester Literary & Philosophical Society. monstrated by combining the equations of thecurves. Let A (Fig. 6) be taken as origin with axis ofx and y parallel to BC and BG ; then if AG —a andBG = b, the equations of the hyperbola and the circle are : xy — ahand (x-ay + (y-l>f = 4(## Therefore tan ^#C=tan $(DBF). Thus the three positions for the intersection point F are such that the ansrle DBF- ABC ABC , 2- ABC Air , + — or + -L- 3 3 3 3 3 These three points are evidently the corners of an equi-lateral triangle. The following method is due to Chasles,8 and
. Memoirs and proceedings of the Manchester Literary & Philosophical Society. monstrated by combining the equations of thecurves. Let A (Fig. 6) be taken as origin with axis ofx and y parallel to BC and BG ; then if AG —a andBG = b, the equations of the hyperbola and the circle are : xy — ahand (x-ay + (y-l>f = 4(## Therefore tan ^#C=tan $(DBF). Thus the three positions for the intersection point F are such that the ansrle DBF- ABC ABC , 2- ABC Air , + — or + -L- 3 3 3 3 3 These three points are evidently the corners of an equi-lateral triangle. The following method is due to Chasles,8 and is alsogiven as an example in Taylors Geometry of Curves,. Fig. 7. Method of Chasles. Ex. 528. If OA and OB {Fig. 7) are the bounding radii of a circular arc AB, then a rectangular hyperbola having s Traite des Sections Coniques, p. 36. 1S65. io Gee AND ADAMSON, Trisecting an Angle. OA for a diameter and passing through the point C ofintersection of OB with the tangent AC to the circle atA, will pass through one of the two points of trisection ofthe arc. To draw this hyperbola : Bisect AC at D andOA at L. Join LD and produce to F, making DF andDE each equal to AD. Draw the rectangle AFCE. Theasymptotes XX} YY\ of the hyperbola will be parallelto AF and AE and intersect at L. Construct the hyper-bola intersecting the circle at P as well as at A and attwo other points, Pv P2, not shown in Fig. 7. Join OP,then the angle POB will be ^ of the angle ^40/?. Proof.—Construct the rectangle PGOH on OP asdiagonal with sides parallel to the asymptotes, and drawthe diagonal GKH. Then, as in the construction ofPappus {Fig. 6): _ 0LX= L 0LK+ LXLK=L0XL+_P0
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectscience, bookyear1888
