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I'm trying to put this conic on an identifiable form.

$$4x^2-4xy+y^2+20x+40y=0$$

I know that the term $xy$ implies that I need to rotate the conic so that $xy$ vanishes. But I've read on some books but I couldn't figure out how to do it. It seems that there is a system that needs to be solved, and this system involves some trigonometric funcions.

I thought about the following: As the term with $xy$ is going to be eliminated, I guess I should write:

$$4x^2-4xy+y^2+20x+40y=0\\ 4\left(x+\cfrac{5}{2}\right)^2+(y+20)^2-4xy-90=0$$

Perhaps the center of this conic is $(-5/2,-20)$. I guess that knowing the center must be important to something.

Red Banana
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  • The matrix for the quadratic form $4x^2-4xy+y^2$ is $A = \left(\begin{matrix}4 & -2\ -2 & 1\end{matrix}\right)$ right? So to bring it to a form free of $xy$ term you need to diagonalize $A$ (in general). – r9m Jul 03 '15 at 05:37
  • @r9m There are two matrixes. This one you mentioned and:

    $$\begin{pmatrix} {x}&{y}&{1} \end{pmatrix}\begin{pmatrix} {a}&{b/2}&{d/2}\ {b/2}&{c}&{e/2}\ {d/2}&{e/2}&{f} \end{pmatrix}\begin{pmatrix} {x}\ {y}\ {z} \end{pmatrix}$$

     – Red Banana Jul 03 '15 at 05:46
  • @r9m What's the difference of them? I know that this matrix form is associated (can be expanded) to a quadratic form. But I'm not sure what your matrix does. – Red Banana Jul 03 '15 at 05:47

2 Answers2

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Hint The first three terms can together be rewritten as $$(2 x - y)^2,$$ suggesting part of a change of variable that will eliminate the cross-term.

Travis Willse
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  • I need to do a change of variable such that $xy=0$, right? – Red Banana Jul 03 '15 at 05:54
  • You need to make some change of variable from $(x, y)$ to $(u, v)$ so that, once written in $u, v$ the l.h.s. contains no $uv$ term. – Travis Willse Jul 03 '15 at 06:07
  • One silly doubt: In the book I'm reading, it says about eliminating the linear terms, in another place I've read, it seems that the objective is to eliminate the term with $xy$. Are they equivalent? There is also another book I have that says that the objective seems to be to write the conic in the form $ax^2+bxy+cy^2$. I'm a little confused. – Red Banana Jul 03 '15 at 06:27
  • No, the $xy$ term is a particular quadratic monomial, eliminating the linear terms is a separate normalization; once the cross-term has been eliminated, it's enough to apply an appropriate change of coordinates $(x, y) \mapsto (x - a, y - b)$ to eliminate the linear terms (in general this introduces a constant term). – Travis Willse Jul 03 '15 at 06:31
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Putting $x=x'-\frac{g+hy}a$ will not only remove xy term, but will also make the process easier.All the best.

Siong Thye Goh
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Hrimon
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