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# Re: Conic Sections and Quadric Surfaces
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
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