Please correct the oversight in 6) ...."The other axis is perpendicular in V" with .... "The other axis is perpendicular in S, described in 7)."
Alberto > ---------- > From: Nicelli Alberto > Sent: lunedì 4 maggio 1998 10.35 > To: 'Arthur Carlson' > Cc: 'sundial@rrz.uni-koeln.de' > Subject: hyperbola , foci and other more > > > Hello Arthur . > > In the last cloudy weekend I' ve studied also the inverse problem : > given a hyperbola or at least 5 points find the foci and asymptotes. > I've got a simple solution in the particular case that among these points > there is a vertex . > The solution is based on this nice theorem valid for all conics: > " the middle points of all parallel diameters of a conic lie on a straight > line ." > A diameter is defined to be that segment generated by two intersections of > a straight line with the conic, therefore the parallel diameters are those > generated by a set of parallel straight lines. > " if the conic is a hyperbola or an ellipse the straight line joining the > middle points of the diameters crosses its simmetry centre ", which is , > for the hyperbola , the intersection of the two asymptotes half way > between the two vertexes. That's all about the theory . > Hereafter I'll apply all that to find foci and asymptotes given at least 5 > points got by shadow observation along several hours . The procedure is > very easy to be applied and self-explaining. > > 1) Draw a however directed set of parallel straight lines on the wall ( > two would be sufficient, but we don't know the exact position of the final > hyperbola . ). > 2) In the morning , when the point of the gnomon's shadow crosses the > straight lines mark these points A, B, C on the wall . > 3) Mark the point V of the shadow in the instant of your local noon : this > is a vertex of the hyperbola. > 4 ) Repeat instructions 2) in the afternoon , so that we mark the > corresponding points ., C', B', A' > 5) Find the middle points on the segments AA' , BB' , CC' . and draw the > straight line joining them. > 6) Draw the straight line joining the base of the gnomon and the vertex of > the hyperbola : this is a simmetry axis of the hyperbola , containing also > the foci. The other axis is perpendicular in V. > 7) The intersection S of the two straight lines drawn in 5) and 6) is the > simmetry centre of the hyperbola > 8) Measure the distance SV : this is the value of the parameter a of the > hyperbola > 9) Draw the perpendicular segment from a whatever marked point , say A , > to the simmetry axis and mark the intersection point P. > 10) Measure the length Y of the segment AP. > 11) Measure the length X of the segment SP. > 12 Calculate the parameter b from the hyperbola equation (X^2/ a^2)-(Y^2/ > b^2) = 1 > 13) Calculate c = (a^2+b^2)^1/2, which is the distance of the two foci > from S on the simmetry axis. > 14)Draw the segment HK of length 2b perpendicular to the simmetry axis in > V, in such way that V results its middle point. > 15)The two straight lines containing the segments SH and SK are the > asymptotes of the hyperbola. > > That's all. > I believe that this method could be useful. > Does it fit your needs ? > > Best Regards > > Alberto Nicelli > Italy (45,5 N ; 7,8 E) > > >
