2.conica-non-centro.tex

\subsection{Conica non a centro}
$\mathfrak{C}:\ x^2+4xy+4y^2-10\sqrt{5}x=0$

%$
%A = \left[
%	\arraycolsep=2.0pt\def\arraystretch{1.0}
%	\begin{array}{@{}cc@{}}
%		1 & 2 \\
%		2 & 4
%	\end{array}
%\right]
%\
%B = \left[
%	\arraycolsep=2.0pt\def\arraystretch{1.0}
%	\begin{array}{@{}ccc@{}}
%		1          & 2 & -5\sqrt{5} \\
%		2          & 4 & 0 \\
%		-5\sqrt{5} & 0 & 0 \\
%	\end{array}
%\right]
%$
\begin{tabular}{l}
	$I_1 = \tr A = 5$ \\
	$I_2 = \det A = 0$ \\ 
	$I_3 = \det B = -500$
\end{tabular}
\begin{tabular}{l}
	$I_2 = 0 \Rightarrow$ conica non a centro. \\
	$I_3 \neq 0 \Rightarrow$ parabola	
\end{tabular}

%Calcolo autospazi di $A$ \\
$p(\lambda) = \det(A-\lambda I) = -\lambda(5-\lambda)$
$\ \Rightarrow \lambda_1 = 0, \lambda_2 = 5$
$V_0 %= \Lin 
%\begin{bmatrix}
%	2 \\[-0.3em]
%	-1 \\
%\end{bmatrix}
= \Lin
\begin{bmatrix}
	2/\sqrt{5} \\[-0.3em]
	-1/\sqrt{5} \\
\end{bmatrix}
\quad
V_5 %= \Lin 
%\begin{bmatrix}
%	1 \\[-0.3em]
%	2 \\
%\end{bmatrix}
= \Lin
\begin{bmatrix}
	1/\sqrt{5} \\[-0.3em]
	2/\sqrt{5} \\
\end{bmatrix}
$
\quad
%Trovo la matrice U ortogonale che diagonalizza A e eseguo rotazione:
%(nota che $U \in SO(2) \Rightarrow \det U = 1$ e $U$ ortonormale)
$
%\begin{bmatrix}
%	x \\[-0.3em]
%	y \\
%\end{bmatrix}
%= U
%\begin{bmatrix}
%	\tilde{x} \\[-0.3em]
%	\tilde{y} \\
%\end{bmatrix}
U =
\left[
	\arraycolsep=2.0pt\def\arraystretch{1.0}
	\begin{array}{@{}cc@{}}
		2/\sqrt{5} & 1/\sqrt{5} \\
		-1/\sqrt{5} & 2/\sqrt{5}
	\end{array}
\right]
%\begin{bmatrix}
%	\tilde{x} \\[-0.3em]
%	\tilde{y} \\
%\end{bmatrix}
$ \\
$
\Rightarrow \tilde{A} = U^TAU = \left[
	\arraycolsep=2.0pt\def\arraystretch{1.0}
	\begin{array}{@{}cc@{}}
		0 & 0 \\
		0 & 5
	\end{array}
\right]
,\quad
\tilde{\vec{b}} = U^T\vec{b} = \begin{bmatrix}
	-10 \\[-0.3em]
	-5 \\
\end{bmatrix}
,\quad
\tilde{c} = c
$ \\
$\Rightarrow \tilde{\mathfrak{C}}:\ 5\tilde{y}^2 - 20\tilde{x} - 10\tilde{y} = 0$ %e completamento dei quadrati
$\Rightarrow \tilde{\mathfrak{C}}:\ 5(\tilde{y}-1)^2 - 20(\tilde{x}+1/4) = 0$ \\
$\Rightarrow 
\begin{bmatrix}
	x \\[-0.3em]
	y \\
\end{bmatrix}
= U
\begin{bmatrix}
	\tilde{x} \\[-0.3em]
	\tilde{y} \\
\end{bmatrix}
= U
\begin{bmatrix}
	x'-1/4 \\[-0.3em]
	y'+1 \\
\end{bmatrix}
= U 
\begin{bmatrix}
	x' \\[-0.3em]
	y' \\
\end{bmatrix}
+
\begin{bmatrix}
	1/(2\sqrt{5}) \\[-0.3em]
	9/(4\sqrt{5}) \\
\end{bmatrix}
$

$\Rightarrow \mathfrak{C}':\ 5y'^2 - 20x' = 0$
$\Rightarrow \mathfrak{C}':\ y'^2 - 4x' = 0$