Does anyone know of an integral transform which transforms the normal form $Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fzx = 0$ to the form $ax^2 + by^2 + cz^2 = 0$ ?
Thanks in advance.
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If $D$ is a multiple of $2\gcd(A,B)$, try substituting $x\leftarrow x+uy$, $y\leftarrow y+vx$ where $2(Au+Bv)=D$. – Hagen von Eitzen Jul 20 '13 at 12:21
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It is a two step iterative algorithm:
- Take the first variable $x$. If $x^2$ appears there with non-zero coefficient, complete the square with $x^2$ and the double product $xy$ with the next variable. Continue with the next variables.
- If $x^2$ is not there, only the double product with the next variable $xy$, then make the substitution $x=u+v$, $y=u-v$. This makes the term $u^2$ appear. Go to step $1$ with the variable $u$.
This will give you a change of variable with rational coefficients. You will have to multiply the original form by a convenient factor to clear denominators.
Example:
$$xy+y^2+z^2.$$
The order to input the variables to the algorithm could be other, but let us do it with $x$ as the first variable.
We need step $2$ because there is no $x^2$. We get $$(u+v)(u-v)+(u-v)^2+z^2=2u^2-2uv+z^2.$$
Now $u$ is our first variable. We go to step $1$.
$$2(u^2-uv+v^2/4)-v^2/4+z^2=2(u-v/2)^2-v^2/4+z^2.$$
In the new variables $z_1=u-v/2$, $z_2=v$, $z_3=z$ (where $u=(x+y)/2$ and $v=(x-y)/2$) we get $$2z_1^2-z_2^2/4+z_3^2.$$
Notice we can, in this case, multiply the whole form by $16$ and get rid of the denominators in the change of variable.
OR.
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Thanks @RGB for your answer. One more thing, If I want to recover the values of the original variables after solving $ax^2 + by^2 + cz^2 = 0$. How can I keep track of the transformations I performed? – thilinarmtb Jul 20 '13 at 12:55
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Just keep the changes of variables you did. The transformation in the second step is there, written. The one in the first step is just the expression you get inside the square when you complete it. – OR. Jul 20 '13 at 12:57
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I also found a two step transformation that will do the required thing. First we make the transformation $x \rightarrow x - (By + Cz)/(2A)$ (if $A = 0$ then we switch the variables). This will result in an equation of the form $A'x^2 + D'y^2 + E'yz + F'z^2 = 0$ after clearing the denominators. Then the transform $y \rightarrow y - E'z/(2D')$ would give us the form $ax^2 + by^2 + cz^2 = 0$.
thilinarmtb
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