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Given is the conic section $x^2 +xy + y^2 +2x +3y -3 = 0$.

The following tasks:

1.) What is the coordinate matrix $A_1 = M_{\beta} (\sigma) $ of the bilinearform?

2.) do the transformation and determine the normal form

3.) whats the centre and what are the axis?

To number 1.) a conic section is given by $ax^2 + 2bxy + cy^2 + 2dx + 2ey + f = 0$. So I have $(a,b,c,d,e,f) = (1,\frac{1}{2},1,1,\frac{3}{2},-3)$ and thus the following matrix

$A_1 := M_{\beta} (\sigma) = \begin{pmatrix} a&b&d \\ b&c&e \\ d&e&f \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} & 1 \\ \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & \frac{3}{2} & -3 \end{pmatrix}$

Furthermore, I define: $A = \begin{pmatrix} a&b \\ b&c \end{pmatrix} $ $p = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$

To number 2:)

$det A_1 = -1$

$det A = \frac{3}{4}$

=> With these information I know that the conic section is an ellipsoid!

Now substitution:

$x = y +q$ whereas $q = - \frac{1}{2} A^{-1} p $

We get $q = - \frac{1}{2} \cdot \frac{4}{3} \cdot \begin{pmatrix} 1 & - \frac{1}{2} \\ - \frac{1}{2} & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 3 \end{pmatrix} = (- \frac{1}{3}, - \frac{4}{3} ) =: M$

Here I had $A^{-1} = \frac{1}{det A} \cdot adj(A) = \frac{4}{3} \cdot \begin{pmatrix} 1 & -\frac{1}{2} \\ - \frac{1}{2} & 1 \end{pmatrix}$

And now we have

$f* = f - \frac{1}{4} ^tp A^{-1} p = ... = - \frac{16}{3} $

For A I get the following eigenvalues $l_1 = 0.5, l_2 = 1,5$

By normalizing the eigenvectors I get: $T = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ with diagonal matrix $D = \begin{pmatrix} \frac{3}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}$

So the normal form is:

$ - \frac{l_1}{f*}z_1^2 - \frac{l_2}{f*}z_2^2 = 1 $ ->

$\frac{9}{32}z_1^2 + \frac{3}{32}z_2^2 = 1 $

To number 3:)

The centre is given by q. The axis are:

$ a = \sqrt{\frac{32}{9}} , b = \sqrt{\frac{32}{3}}$

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This looks fine up until the last sentence about the axis. I'd be inclined to say that the axis is a VECTOR, but you've written a number. The numbers you wrote are the lengths of what I'd call the axes, so I'd be inclined to call them the "semidiameters".

But if we assume that your book/prof/whatever uses the word "axis" to mean "min/max distance to center", then all your work looks OK to me.

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  • $\begingroup$ Ah, your're absolutely right - I'm sure the books means LENGTH. Thank you very much for this very fast comment =) $\endgroup$ – Vazrael Nov 6 '14 at 12:56
  • $\begingroup$ Let me give one more bit of feedback: you posted a long and clear writeup, but didn't actually ask a question. If you start with "I'm working on this problem and would like someone to verify that I'm on the right track: ...", it'll really help us know what to look for as we read. :) $\endgroup$ – John Hughes Nov 6 '14 at 12:58
  • $\begingroup$ Yes, you're again absolutely right. In fact, I was a bit fuzzy in my head (since doing maths for 6 hours now - last exercises for tomorrow) and I totally forgot the standard things. I apologize and say once again: Thank you very much =) $\endgroup$ – Vazrael Nov 6 '14 at 13:57

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