*To*: mathgroup@smc.vnet.net*Subject*: [mg11130] Re: [mg11082] Conic Sections and Quadric Surfaces*From*: BobHanlon@aol.com*Date*: Mon, 23 Feb 1998 21:40:49 -0500

One approach using Mathematica: pts = {{-1,0},{0,1},{2,2},{1,-1},{0,-3}}; eqn = a x^2 + b x y + c y^2 + d x + e y + f == 0; The substitution rules for the given points are subs = {x -> #[[1]], y -> #[[2]]}& /@ pts {{x -> -1, y -> 0}, {x -> 0, y -> 1}, {x -> 2, y -> 2}, {x -> 1, y -> -1}, {x -> 0, y -> -3}} Solving for the first five variables in terms of the sixth variable soln = Solve[eqn /. subs, {a,b,c,d,e}] {{a -> (-29*f)/42, b -> (20*f)/21, c -> -f/3, d -> (13*f)/42, e -> (-2*f)/3}} Substituting this solution into the original equation eqn2 = eqn /. soln {f + (13*f*x)/42 - (29*f*x^2)/42 - (2*f*y)/3 + (20*f*x*y)/21 - (f*y^2)/3 == 0} Consequently, either f=0 or theEqn = eqn2 /. f -> 42 {42 + 13*x - 29*x^2 - 28*y + 40*x*y - 14*y^2 == 0} Note that any value of f other than zero gives an equivalent equation. To plot the equation, the standard add-on package ImplicitPlot is used: Needs["Graphics`ImplicitPlot`"] ImplicitPlot[theEqn, {x, -4, 3}, {y, -4, 3}]; Bob Hanlon FORWARDED MESSAGE: Dear Sir: Ref: Introduction to Linear Algebra- 4th ed. Johnson,Riess and Arnold (page 35) Find the equation of the conic section passing through the five points (-1,0),(0,1),(2,2),(1,-1), (0,-3). Display the graph of the conic. When I try to set up the augmented matrix I get two (2) indeterminate rows which prevents my proceeding with the method outlined in the textbook Linear Algebra with Mathematica by E. Johnson page 58 to 59. How can I get around this? The equation to satisfy (an ellipse) is ax^2 + bxy + cy^2 + dx + ey + f = 0. Is there some other technique I can use to find the equation given several points? I am presently playing around with Fit[ data,{etc}] but I only know how to fit quadratics. Thank you in advance. Manuel Avalos