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The conic is $\Gamma(x,y):8x^2+8xy+2y^2-5y=0$.

Since \begin{equation} det(A)=\begin{pmatrix} 8&4&0\\4&2&-\frac{5}{2}\\0&-\frac{5}{2}&0\end{pmatrix}=-50\neq 0 \end{equation} \begin{equation} det(Q)=\begin{pmatrix} 8&4\\4&2\end{pmatrix}=0 \end{equation} $\Gamma$ is an parabola. By searching the eigenvalues of $Q$ I get: \begin{equation} det(Q-\lambda I)=\begin{vmatrix} 8-\lambda&4\\4&2-\lambda\end{vmatrix}=\lambda(\lambda-10)=0 \end{equation} i.e. $\lambda_{1}=0$ and$\lambda_{2}=10$. The eigenspaces are $E(\lambda_{1})=\mathcal{L}(-\frac{1}{2},1)$ while $E(\lambda_{2})=\mathcal{L}(2,1)$. The basis of such spaces are evidently orthogonal basis of their respective eigenspace. The first one, $v=(-\frac{1}{2},1)$ can be normalized by dividing it by $\frac{\sqrt{5}}{2}$ while the second, $w=(2,1)$, by $\sqrt{5}$. Hence, the rotation matrix $R$ is: \begin{equation} \tilde{R}=\begin{pmatrix} -\frac{1}{2}&2\\1&1\end{pmatrix} \begin{pmatrix}\frac{2}{\sqrt{5}}\\\frac{1}{\sqrt{5}} \end{pmatrix} =\begin{pmatrix} -\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix} \end{equation}
This matrix has determinant $-1$ while it has to be $det(R)=1$, thus I exchange sign: \begin{equation} R=\begin{pmatrix} \frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\-\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix} \end{equation}
It follows that: \begin{equation} \begin{cases} x=\frac{1}{\sqrt{5}}(\tilde{x}+2\tilde{y})\\y=\frac{1}{\sqrt{5}}(-2\tilde{x}+\tilde{y})\end{cases} \end{equation} and, by substituting in the equation of the conic and simplifying, I get: \begin{equation} \Gamma(\tilde{x},\tilde{y}):10\tilde{y}^2+2\sqrt{5}\tilde{x}-\sqrt{5}\tilde{y}=0 \end{equation} Then I should translate to obtain the canonic form (by completing the squares). Even if I'm pretty convinced of the procedure and that I checked the calculations with the PC, I do not understand why I have a term in $x$ instead that in $x^2$. Have I made some mistakes? I'd really appreciate a check of the procedure and/or any advise, thank you!

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  • $\begingroup$ Using your notation: if rank $\;\,A=3\;$ , and rank $\,Q=1\;$ , the quadratic is a parabola, not a hyperbola. Check this. $\endgroup$ – DonAntonio Jan 29 '17 at 16:32
  • $\begingroup$ First of all: $\Gamma$ ia a parabola and not a hyperbola, secondly after a change of variables you will find a term with $x^2$ instead of $y^2$. $\endgroup$ – InsideOut Jan 29 '17 at 16:34
  • $\begingroup$ You are right, I do not know why I wrote hyperbola! @Legoman I don't understand, why are you saying that I should get $x^2$ instead of $y^2$? $\endgroup$ – M.Giacchello Jan 29 '17 at 16:37
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there are two steps, a rotation and a translation. If we begin with rotation, as you did, we have rotated coordinate system $$ p = \frac{2x+y}{\sqrt 5}, \; \; q = \frac{x-2y}{\sqrt 5}. $$ OR $$ x = \frac{2p+q}{\sqrt 5}, \; \; y = \frac{p-2q}{\sqrt 5}. $$ Also $$ 2x+y = p \sqrt 5. $$ So $$ 2 (2x+y)^2 -5y $$ is $$ 2 (p \sqrt 5)^2 - 5 \left(\frac{p-2q}{\sqrt 5} \right) $$ $$ 10p^2 - p \sqrt 5 + 2 q \sqrt 5 $$ Now we do a translation by completing the square in $p$ only. We have $$ p^2 - \frac{p}{2 \sqrt 5}, $$ so we will use $$ \left( p - \frac{1}{4 \sqrt 5} \right)^2 = p^2 - \frac{p}{2 \sqrt 5} + \frac{1}{80},$$ $$ 10 \left( p - \frac{1}{4 \sqrt 5} \right)^2 = 10 p^2 - p \sqrt 5 + \frac{1}{8},$$ $$ 10p^2 - p \sqrt 5 = 10 \left( p - \frac{1}{4 \sqrt 5} \right)^2 -\frac{1}{8}. $$ The translation is $$ u = p - \frac{1}{4 \sqrt 5}, \; \; v = q, $$ $$ p = u + \frac{1}{4 \sqrt 5}, \; \; q = v. $$ Then $$ 10p^2 - p \sqrt 5 + 2 q \sqrt 5 $$ becomes $$ 10 u^2 -\frac{1}{8} + 2 v \sqrt 5. $$ This is being set to zero, so $$ \sqrt 5 u^2 -\frac{1}{16 \sqrt 5} + v = 0, $$ $$ v = \frac{1}{16 \sqrt 5} -\sqrt 5 \; u^2 $$

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Let us try Lagrange's Method:

$$0=8x^2+8xy+2y^2-5y=8\left(x+\frac y2\right)^2-2y^2+2y^2-5y=8\left(x+\frac y2\right)^2-5y\implies$$

Substitute

$$\begin{cases}x'=x+\cfrac y2\\{}\\ y'=y\end{cases}\;\;\implies\;\text{our quadratic is the parabola}\;\;y'=\frac85x'^2$$

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Quite simply, you chose the wrong rotation. With the rotation that you ended up with, $Q$ is diagonalized as $\pmatrix{0&0\\0&10}$, but for there to be an $\overline{x}^2$ term instead of a $\overline{y}^2$ term in the resulting equation, the non-zero diagonal entry has to be in the upper-left corner instead. So, when constructing the rotation matrix, the eigenvector of $10$ has to come first, i.e., swap the columns of $\overline R$. This will also correct the determinant.

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