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Consider the conic bundle: $\mathcal{C_{\lambda,\mu}} : \lambda(4x_0^2+3x_1^2-4x_ox_2-2x_1x_2)+\mu(4x_0^2+5x_1^2+4x_0x_1-8x_0x_2-4x_1x_2)=0$

Find the locus of the centers of those conics and write the equation.

How to find it? I did this, but I think I was following the wrong way:

The matrix which represents the bundle is:

$ \begin{bmatrix} 4\lambda+4\mu & 2\mu & -2\lambda-4\mu \\ 2\mu & 3\lambda+5\mu & -\lambda-2\mu \\ -2\lambda-4\mu & -\lambda-2\mu & 0 \\ \end{bmatrix} $

from the theory I know that I can find the center of a conic solving this system:

$ \begin{bmatrix} c_{11} & c_{12} \\ c_{12} & c_{22} \\ \end{bmatrix} $ $ \begin{bmatrix} x \\ y \\ \end{bmatrix} $ = $ \begin{bmatrix} -c_{01} \\ -c_{02} \\ \end{bmatrix} $ replacing: $ \begin{bmatrix} 3\lambda+5\mu & -\lambda-2\mu \\ -\lambda-2\mu & 0 \\ \end{bmatrix} $ $ \begin{bmatrix} x \\ y \\ \end{bmatrix} $ = $ \begin{bmatrix} -2\mu \\ 2\lambda+4\mu \\ \end{bmatrix} $

then $ \left[ \begin{array}{cc|c} 3\lambda+5\mu&-\lambda-2\mu&-2\mu\\ -\lambda-2\mu&0&2\lambda+4\mu \end{array} \right] $ , after passages: $ \left[ \begin{array}{cc|c} 1&0&-2\\ 0&1&-2(\lambda+3\mu) \end{array} \right] $.

So the coordinates of the center should be $[1, -2, -2(\lambda+3\mu)]$. Is this right? There are other ways? Thank you

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  • $\begingroup$ In an affine where $(x_0:x_1:x_2)=(x:y:1)$ i.e. $\lambda(4x^2+5y^2+4yx-8x-4y)+\mu (4x^2+3y^2-4x-2y)$ you can differentiate wrt $x$ and $y$ and solve the resulting system to get: $(x,y) = (\frac{3\mu+4\lambda}{6\mu+4\lambda}, \frac{\mu}{3\mu+2\lambda})$. $\endgroup$ – Jan-Magnus Økland May 15 '18 at 19:43
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    $\begingroup$ The center is the pole (as in pole-polar relation) to the line at infinity. How is your line at infinity represented? Suppose it is $[1:0:0]$, then just multiply that with the adjugate matrix, i.e. pick the first column of the cofactor matrix. $\endgroup$ – ccorn May 15 '18 at 19:57
  • $\begingroup$ Related keyword: Nine-point conic. $\endgroup$ – ccorn May 15 '18 at 20:12
  • $\begingroup$ You can do a simple sanity check of your solution: for $\lambda=1$, $\mu=0$, does it produce the center of the first conic? Does $\lambda=0$, $\mu=1$ produce the center of the second? $\endgroup$ – amd May 16 '18 at 0:44
  • $\begingroup$ @amd the easiest the better, thank you! And thank to everyone. $\endgroup$ – F.inc May 17 '18 at 17:15

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