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\displaystyle P=\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}, A=\begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}.\\[br]PAP = \begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}\begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} = \begin{pmatrix} 15&30 \\ -10&5 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} \\[br]= 5\begin{pmatrix} 3&6 \\ -2&1 \end{pmatrix}\begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix} = 5\begin{pmatrix} 15&0 \\ 0&-5 \end{pmatrix}\\ \\[br][br]\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} , \mathbf{x}_1 = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} .\\[br]\mathbf{x} = P \mathbf{x}_1 + \mathbf{a} \\[br]\mathbf{x} = \begin{pmatrix} 1&2 \\ 2&-1 \end{pmatrix}\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \\[br]\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_1+2y_1+a_1 \\ 2x_1-y_1+a_2 \end{pmatrix} .\\ \\[br][br]\mathbf{x}^T A\mathbf{x} + \mathbf{b}^T \mathbf{x} - 11 = 0 \\[br]\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} -1&8 \\ 8&11 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 18&6 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} - 11 = 0 \\[br]\begin{pmatrix} x&y \end{pmatrix} \begin{pmatrix} -x+8y \\ 8x+11y \end{pmatrix} + 18x+6y-11 = 0 \\[br]-x^2 + 8xy + 8xy + 11y^2 + 18x + 6y - 11 = 0 \\[br]-x^2 + 11y^2 + 16xy + 18x + 6y - 11 = 0. \\[br]\therefore -(x_1+2y_1+a_1)^2 + 11(2x_1-y_1+a_2)^2[br]+ 16(x_1+2y_1+a_1)(2x_1-y_1+a_2) + 18(x_1+2y_1+a_1) + 6(2x_1-y_1+a_2) - 11 = 0. \\[br]\therefore x_1^2 (-1+44+32) + y_1^2 (-4+11-32) + x_1y_1 (-4-44+48) + x_1 (-2a_1 + 44a_2 + 16a_2 + 32a_1 + 18 + 12) + y_1 (-4a_1 - 22a_2 + 32a_2 - 16a_1 + 36 - 6) + (-a_1^2 + 11a_2^2 + 16a_1a_2 + 18a_1 + 6a_2 - 11) = 0 \\[br]\therefore 75x_1^2 - 25y_1^2 + (30a_1+60a_2+30)x_1 + (-20a_1+10a_2+30)y_1 + (-a_1^2 + 11a_2^2 + 16a_1a_2 + 18a_1 + 6a_2 - 11) = 0. \\[br]\text{Solving simultaneously: } 30a_1+60a_2+30=0, \; -20a_1+10a_2+30 = 0 \\[br]\therefore a_2 = -1, a_1 = 1. \\[br]75x_1^2 - 25y_1^2 - 5 = 0 \\[br]3x_1^2 - y_1^2 = \tfrac{1}{5}.
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