Hello I have a question about reduction quadratic form to a canonical form, I have this quadratic form:
$ C=x^2+9y^2-4x+18y+4=0$
I proceeded in this way to solve it
$A=\begin{bmatrix}1&0&-2\\0&9&9\\-2&9&4\end{bmatrix}$
$ |A|33= \begin{bmatrix}1&0\\0&9\end{bmatrix}$
=> 9>0 I have an ellipse
$Q(x,y)=x^2+9y^2 $
I find the diagonal form:
$\begin{bmatrix}l-1&0\\0&l-9\end{bmatrix}$ => l=1,l=9
my problem is that when I go to compute orthogonal bases of V1,V9. I
$x = 0$
and $y = 0$ in both cases and do not know how to proceed able to help me? I hope I explained
Best Answer
What about completing squares?:
$$0=x^2+9y^2-4x+18y+4=(x-2)^2-4+9(y+1)^2-9+4\implies$$
$$(x-2)^2+9(y-1)^2=9\iff\frac{(x-2)^2}{3^2}+(y-1)^2=1$$
Indeed, an ellipse