Finding The Canonical Form Of Conic Section

Question

let $x^2-2xy+3y^2-4x+5y-6=0$, find the canonical form of conic section

I have came to $$(2+\sqrt{2})(x')^2+(2-\sqrt{2})(y')^2+\frac{1-4\sqrt{2}}{\sqrt{4+\sqrt{8}}}x'+\frac{1+4\sqrt{2}}{\sqrt{4-\sqrt{8}}}y'=6$$

How do I finish the process?

Answer 1

From the answer I see that the canonical form does not have any $x y$ terms.

A conic section $ A x^2 + B y^2 + C x y + D x + E y + F =0$ can be expressed in rotated local coordinates $x'$ and $y'$ with $$\begin{align} x & = x' \cos \theta - y' \sin\theta \\ y & = x' \sin \theta + y' \cos\theta \end{align}$$

Use the above in the equation for the conic and set the coefficent of $x' y'$ to zero

$$ 2 C \cos^2 \theta + 2 (B-A) \sin \theta \cos\theta -C =0 $$ $$ C \cos(2\theta) + (B-A) \sin (2 \theta) = 0 $$ $$ \tan (2 \theta) = \frac{C}{A-B} $$ $$ \theta = \frac{1}{2} \tan^{-1} \left( \frac{C}{A-B} \right) $$

Now use this angle in the expression to get your final equation.