Laborious old conics question
Narasimham
Evaluate a,h,b,f,g in terms of five given points (x1,y1),.....,(x5,y5)
determining a conic section.
Michael Jørgensen
"Narasimham" <math...@hotmail.com> wrote in message
news:1158738042....@e3g2000cwe.googlegroups.com...
> a x^2 + 2 h x y + b y^2 + 2 f x + 2 g y = 1
>
> Evaluate a,h,b,f,g in terms of five given points (x1,y1),.....,(x5,y5)
> determining a conic section.
>
This is simply just a linear system of five equations with five unknowns.
Write it out as a matrix equation, and use standard techniques to solve it.
-Michael.
JEMebius
Michael Jørgensen wrote:
Is the abovementioned problem just this specific problem? or is it an
attempt to formulate the general problem of determining the conic
section passing thru five given points in the plane?
Please be aware that on the one side five points in a plane in general
determine a unique conic section (which may be degenerate), while on the
other side the equation in Narasimham's post is not the most general
quadratic equation in x and y.
To be specific, conic sections consisting of two lines thru (0, 0) (Yes
- they are conic sections!) cannot be represented by this equation.
Johan E. Mebius
JEMebius
Michael Jørgensen wrote:
Is the abovementioned problem just this specific problem? or is it an
attempt to formulate the general problem of determining the conic
section passing thru five given points in the plane?
Please be aware that on the one side five points in a plane in general
determine a unique conic section (which may be degenerate), while on the
other side the equation in Narasimham's post is not the most general
quadratic equation in x and y.
To be specific, conic sections consisting of two lines thru (0, 0) (Yes
- they are conic sections!) cannot be represented by this equation.
Johan E. Mebius
Supplementary remark: On every conic section one can choose five points
which determine uniquely that conic section.
This observation plugs a hole in my argumentation.
Johan E. Mebius
[Mr.] Lynn Kurtz
wrote:
>a x^2 + 2 h x y + b y^2 + 2 f x + 2 g y = 1
>
>Evaluate a,h,b,f,g in terms of five given points (x1,y1),.....,(x5,y5)
>determining a conic section.
What is the point of your post?
--Lynn
Robert Israel
To be less specific, any conic that passes through (0,0) cannot be
represented by this equation.
On the other hand, if four of the points are collinear, you don't
determine a unique conic section.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
David Ziskind
problem of representing a conic in terms of five points that are members
of the conic.
(This post will look best if displayed in a monospace font.)
In the general case, the equation of the conic might have the constant
term equal to 0. That is a special and irregular situation so from this
point on, I will assume we have the normal situation where the constant
term is non-zero. Now, divide through by the constant and obtain a
logically implied formula:
(x,y) is a member of the conic IFF
(there exist a,b,c,d,e such that)
a*x^2 + b*x*y + c*y^2 + d*x + e*y + 1 = 0
Our objective is to express a, b, c, d, e in terms of five points that
satisfy the conic.
We have the following theorem:
(For all x and y)
{
IF [ a*x^2 + b*x*y + c*y^2 + d*x + e*y + 1 = 0 (1)
and a*x_i^2 + b*x_i*y_i + c*y_i^2 + d*x_i + e*y_i + 1 = 0 (2)
for i=1,2,3,4,5
and det[x_1^2 x^1*y_1 y_1^2 x_1 y_1;
x_2^2 x^2*y_2 y_2^2 x_2 y_2;
x_3^2 x^3*y_3 y_3^2 x_3 y_3;
x_4^2 x^4*y_4 y_4^2 x_4 y_4;
x_5^2 x^5*y_5 y_5^2 x_5 y_5] ~= 0] (3)
THEN det[x^2 x*y y^2 x y 1;
x_1^2 x^1*y_1 y_1^2 x_1 y_1 1;
x_2^2 x^2*y_2 y_2^2 x_2 y_2 1;
x_3^2 x^3*y_3 y_3^2 x_3 y_3 1;
x_4^2 x^4*y_4 y_4^2 x_4 y_4 1;
x_5^2 x^5*y_5 y_5^2 x_5 y_5 1] = 0
}
Comment: To avoid struggling with ASCII art, I have written each matrix
above as a sequence of row vectors (commas omitted), with semi-colon
indicating where one should move down and create a new row. In this
approach, the elements in one column will not necessarily stack
vertically, although, as it happens, they will if viewed in a monospace
font.
To apply the above theorem, we will assume that antecedents (2) and (3)
hold as premises. Also, in (1) set y equal to Y(x), and let function Y be
chosen so that (1) is satisfied identically. We now have as conclusions:
(A) a*x^2 + b*x*Y(x) + c*Y(x)^2 + d*x + e*Y(x) + 1 = 0,
for all x; and
(B) det[x^2 x*Y(x) Y(x)^2 x Y(x) 1;
...; etc ...;
x_5^2 x_5*y_5 y_5^2 x_5 y_5 1] = 0, for all x
However, by Laplace development of the determinant (B), we have:
(C) cofactor(1,1)*x^2 + cofactor(1,2)*x*Y(x)
+ cofactor(1,3)*Y(x)^2 + cofactor(1,4)*x
+ cofactor(1,5)*Y(x) + cofactor(1,6)*1 = 0, for all x
(cofactor(1,n) refers to the cofactors of the matrix which is
written in the conclusion of the above theorem.)
Subtract (C) from (A) and obtain:
(a - cofactor(1,1))*x^2 + (b - cofactor(1,2))*x*Y(x)
+ (c - cofactor(1,3))*Y(x)^2 + (d - cofactor(1,4))*x
+ (e - cofactor(1,5))*Y(x) + (1 - cofactor(1,6)) = 0,
for all x
Several rounds of differentiating w/r to x and backward substitution
suffice to show that:
a = cofactor(1,1) b = cofactor(1,2)
c = cofactor(1,3) d = cofactor(1,4)
e = cofactor(1,5) 1 = cofactor(1,6)
Thus, in evaluation of the final determinant in the above theorem, it is
unnecessary to specifically calculate cofactor(1,6). We already know that
the value is 1. In any event, the first five relations (and (A)) give the
values of a, b, c, d, e (and the equation of the conic) in terms of the
data -- which was the original problem objective.
Proof (of above theorem):
1. Assume as Premises P1, P2, P3, respectively the antecedents (1), (2),
(3) of the above theorem.
2. By virtue of P3, we know that there are multipliers m_1, m_2, ..., m_5
satisfying the vector equation:
(x^2, x*y, y^2, x, y) =
Sum{1=1 to 5: m_i*(x_i^2, x_i*y_i, y_i^2, x_i, y_i)}
3. Expanding this vector equation, we obtain five scalar equations:
x^2 =m_1*x_1^2 + m_2*x_2^2 + m_3*x_3^2 + m_4*x_4^2 + m_5*x_5^2
x*y =m_1*x_1*y_1+ m_2*x_2*y_2+ m_3*x_3*y_3+ m_4*x_4*y_4+ m_5*x_5*y_5
... and three more parallel equations
4. Substitute the equations 3.1, 3.2, ..., 3.5 into P1. Rearrange the
result and obtain:
Sum{i=1 to 5: m_i*(a*x_i^2 + b*x_i*y_i + c*y_i^2 + d*x_i + e*y_i)}
+1 = 0
5. By P2, m_i*(a*x_i^2 + b*x_i*y_i + c*y_i^2 + d*x_i + e*y_i) = -1,
for i=1,2,3,4,5
6. Substitute (5) into (4), obtain:
Sum{i=1 to 5: m_i*(-1)}
+1 = 0
7. Thus: 1 = m_1*1 + m_2*1 + m_3*1 + m_4*1 + m_5*1
8. When we append (7) to the five equations of (3), we see we have
six equations which collectively express the vector
(x^2, x*y, y^2, x, y, 1) as a linear combination of the five
vectors (x_i^2, x_i*y_i, y_i^2, x_i, y_i, 1); i=1,2,3,4,5.
Constructing the matrix whose rows are the six vectors mentioned,
we have by a fundamental result for determinants, that the
determinant of the matrix is zero. This is the result we wished
to establish as the conclusion of the above theorem.
--------------------
In order to prove this result as easily as possible, I have utilized the
requirement (3) above (in order to guarantee the existence of the m_i's).
Another form is given in the statement of this theorem in Korn & Korn,
"Mathematical Handbook for Scientists and Engineers", Dover, 2000, ISBN
0-486-41147-8. In place of my requirement (3), they use: no four of the
five given data points are collinear. See K&K Section 2.4-11.
David Ziskind
zis...@ntplx.net
David Ziskind
This is a correction to my reply article posted Sept. 21, 2006 with
identifiers <zis...@ntplx.net> and
<news:op.tf8vj...@tnt-33-178.ct.dialin.ntplx.com>. Please apply the
correction following.
In step 5 of the Proof section,
CHANGE:
By P2, m_i*(a*x_i^2 + b*x_i*y_i + c*y_i^2 + d*x_i + e*y_i) = -1,
for i=1,2,3,4,5
TO:
By P2, a*x_i^2 + b*x_i*y_i + c*y_i^2 + d*x_i + e*y_i = -1,
for i=1,2,3,4,5
David Ziskind
zis...@ntplx.net
Narasimham
> In article <45115E1...@xs4all.nl>, JEMebius <jeme...@xs4all.nl> wrote:
> >
> >
> >Michael Jørgensen wrote:
> >
> > >"Narasimham" <math...@hotmail.com> wrote in message
> > >news:1158738042....@e3g2000cwe.googlegroups.com...
> > >
> > >> a x^2 + 2 h x y + b y^2 + 2 f x + 2 g y = 1
>
> represented by this equation.
> Robert Israel
Actually it tickles me that this equation cannot represent a conic
passing through (0,0) .To me this elementary situation does not appear
that way always.
In general, positive, zero and negative values for say T > Tmin are
all admissible for real conics in a x^2 + 2 h x y + b y^2 + 2 f x + 2
g y = T. ( T = Tmin obtained when quadriatic equation discriminant is
zero). For The case T = 0 the conic passes through the origin, no
matter that a contradiction 0 = 1 is produced on substitution for
values at origin. As an example conics 2 x ^2 + y^2 + 3 x + y = T,
T = (1, 0, -1,-1.3) all represent ellipses as long as T > Tmin (
here approx -1.37)
The contradiction is merely a matter of appearances.The discriminant or
invariant a b - h^2 does not change sign when a,b,h change sign along
with linear terms we have implicitly/automatically/ unwittingly
accepted a zero value for T to pass through corrosponding to the conic
passing through the origin. It is glaringly seen in the presence of
linear terms.
Mathematica implicit plots have come out reasonably well for this case
that is an apparent bottleneck.. conics through origin, and pair of
straight lines.
Contrary to JEMebius assertion for the degenerate case when it is
factorizable as a product of two linear equations representing two
straight lines is also well known. ( IIRC, det[ ( a, h,g),(h, b, f),
(g, f, c) ] = 0 ).
It appears this simple incongruence existed from Descartes times..
there should be a better way to write it out. Let me have your views.
Regards
Narasimham
Narasimham
Thanks for the solution. It was possible to numerically plot any conic
using the determinant equation, correctly passing through the given
points.Narasimham
Narasimham
Actually re-posted the point now brought out under new title: "power of
conic".
http://groups.google.com/group/sci.math/browse_frm/thread/db4d0b7fdf5ab5e4?hl=en