Find the canonical form of conic section for:


So we have \begin{pmatrix} 0 & \frac{1}{2}\\ \frac{1}{2} & 0 \end{pmatrix}

So \begin{vmatrix} \lambda & -\frac{1}{2}\\ -\frac{1}{2} & \lambda \end{vmatrix} $=\lambda^2-\frac{1}{4}=(\lambda-\frac{1}{2})(\lambda+\frac{1}{2})$

So the eigenvalues are $\lambda=\pm\frac{1}{2}$

So the matrix that diagonalize is \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}

So we have $\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 0 & \frac{1}{2}\\ \frac{1}{2} & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}+\begin{pmatrix} -2 & 6 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}+11=0$

Which is $$\frac{x^2}{2}+\frac{y^2}{2}-\frac{8x}{\sqrt{2}}-\frac{4y}{\sqrt{2}}+11=0$$



So how in the answer the got to $$\frac{(x')^2}{2}-\frac{(y')^2}{2}=1$$?

  • $\begingroup$ That matrix is not diagonal $\endgroup$ – Leonardo Vannini Jul 16 '17 at 22:36
  • 1
    $\begingroup$ @LeonardoVannini the matrix that will diagonalized the first, Sorry do not know the term $\endgroup$ – gbox Jul 16 '17 at 22:37

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