# Rotation matrix to construct canonical form of a conic

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.

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, $$q(x',y')=0\iff\frac12x^2+y^2=1$$.
• I see, I miscomputed the coefficient of $y'^2$ (10 and not 8): from that it follows the two canonical form coincide. Thanks! Dec 18, 2021 at 10:50