I want to find the canonical form of the following conic: $$C: 9x^2+4xy+6y^2-10=0.$$ I've found $C$ is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial $$p(t)=\det \begin{pmatrix} 9-t & 2\\ 2 & 6-t\\ \end{pmatrix} $$ The eigenvalue are $t_1=5,t_2=10$, with associated eigenvectors $(-1,2)$, $(2,1)$. Thus I construct the rotation matrix $R$ by putting in columns the normalized eigenvectors (taking care that $\det(R)=1$):

$$ R=\frac{1}{\sqrt{5}} \begin{pmatrix} 1 & 2\\ -2 & 1\\ \end{pmatrix}$$

Then $(x, y)^t=R(x',y')^t$, and after some computations I find the canonical form $$\frac{1}{2}x'^2+\frac{4}{5}y'^2=1.$$

Question: The solution presented in the book says the canonical form is $$2x'^2+y'^2=2,$$ because the rotation matrix they use is different:

$$ R=\frac{1}{\sqrt{5}} \begin{pmatrix} 2 & -1\\ 1 & 2\\ \end{pmatrix}$$

Why my rotation matrix is wrong? How can I detect which rotation I should construct? From what I understand it suffices to put on columns the eigenvectors and pay attention the determinant is one.

Thanks in advance!


1 Answer 1


If$$q(x,y)=9x^2+4xy+6y^2-10$$and$$x'=\frac1{\sqrt5}(x+2y)\ \text{and}\ y'=\frac1{\sqrt5}(-2x+y),$$then $q(x',y')=5x^2+10y^2-10$. So, $q(x',y')=0\iff\frac12x^2+y^2=1$.

  • $\begingroup$ I see, I miscomputed the coefficient of $y'^2$ (10 and not 8): from that it follows the two canonical form coincide. Thanks! $\endgroup$
    – student
    Dec 18, 2021 at 10:50
  • $\begingroup$ I'm glad I could help. $\endgroup$ Dec 18, 2021 at 10:56

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