I have to identify which conic is:
$$x^2 -2xy +y^2 -5\sqrt{2}x+3\sqrt{2}y+10 = 0$$
The method my book uses is by translation and rotation. It basically proves that if we want to translate na equation
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
Then, we can set $D = 0$ and $E=0$. If there is a translation that will do it, then, it has to satisfy:
$$Ah + \frac{B}{2}k + \frac{D}{2} = 0$$ $$\frac{B}{2}h + Ck + \frac{E}{2} = 0$$
If you find a solution, then the new equation of the translated conic is:
$$Au^2 + Buv + Cv^2 + \frac{D}{2}h + \frac{E}{2}k + F = 0$$ *as you can see, $A, B$ and $C$ are not affected by rotation which was translated by $h$ in the x axis, and $k$ in the y axis.
Then, in order to eliminate the $uv$ term, called $B$, the book also proves that if we want to rotate the conic, we can set $B=0$ and then find the necessary angle $\theta$ that will do it. The new equation should be:
$$A't^2 + C'w^2 + \frac{D}{2}h + \frac{E}{2}k + F = 0$$
where
$$A' + C' = A + C$$ $$A'-C' = \frac{B}{\sin(2\theta)}$$
We can find $$\sin(2\theta)$$ by the relation:
$$cotg(2\theta) = \frac{a-c}{b}$$ and then use some trig identities to find $\sin(2\theta)$
As you can see, I rotated the equation where $E = 0$ and $F = 0$. If we couldn't zero out these coefficients, we would have to rotate the original equation, to get:
$$A'x^2 + C'y^2 + (D\cos\theta + E\sin\theta) + (-D\sin\theta + E\sin\theta) + F$$
(as you can see, the independente term is not affected by rotation, and the terms A' and C' can be found the same way as in the last equation)
I think I'm too sleepy, because my conic has no solution to translation, and when I rotate I get:
$$u^2 + v^2 -2u + 8v +10 = 0$$
which is not the answer... Maybe somebody could help me :)
Here's the answer
It looks like the exercise translated to $V$, but this point does not satisfies any of the two equations. So, neither $E$ or $D$ are $0$ :( I can't understand
To start check $\Delta=B^2-4AC$. $\Delta>0$: hyperbobla or the degenerate forms. $\Delta=0$: parabola or the degenerate forms. $\Delta<0$: ellipse, circle or the degenerate forms.
If $\Delta=0$ you cannot cancel both term by translation. You can start by rotation, rotate the axes by $\theta$, we have the new coordinate $(u,v)=(x\cos\theta+y\sin\theta,-x\sin\theta+y\cos\theta)$ or $(x,y)=(u\cos\theta-v\sin\theta,v\sin\theta+u\cos\theta)$
From the criteria of canceling $xy$ term you already know $\theta=\pi/4$, we have $(x,y)=(\frac{u-v}{\sqrt2},\frac{u+v}{\sqrt2})$. The conic becomes $\frac{u^2-2uv+v^2+2v^2-2u^2+u^2+2uv+v^2}{2}-5(u-v)+3(u+v)+10=0$ $$v^2-u+4v+5=0$$
Now translate it to cancel the linear $v$ term. This can be done by solving equation, or alternatively, complete the square. $$u-1=(v+2)^2$$, let $t=u-1$, $w=v+2$ which means move the origin from $(0,0)$ to $(1,-2)$
$$t=w^2$$