Reduction of a quadratic form to a canonical

Question

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

Answer 1

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

Answer 2

This quadratic form doesn’t have any $xy$ terms, so it’s not going to need a rotation, but it does have linear terms, so a translation will be needed to put it into canonical form. Since you’ve already determined that the quadratic form is elliptical, you can find the center of the ellipse by differentiating and setting the result to zero: $${\partial\over\partial x}(x^2+9y^2-4x+18y+4)=2x-4=0\implies x=2 \\ {\partial\over\partial y}(x^2+9y^2-4x+18y+4)=18y+18=0\implies y=-1.$$ To get the coefficients of the translated equation, you can either multiply out the application of this translation to your matrix $A$, or take a short-cut: translation to the center of the ellipse makes the linear terms vanish but leaves the quadratic part unchanged, so the transformed matrix will have the form $$\begin{bmatrix}1&0&0\\0&9&0\\0&0&F\end{bmatrix}.$$ The new $F$ is simply the value of the quadratic form evaluated at $(2,-1)$, i.e., $2^2+9(-1)^2-4\cdot2+18(-1)+4=-9$. Thus, the equation is transformed into $(x-2)^2+9(y+1)^2=9$ or $${(x-2)^2\over3^2}+(y+1)^2=1.$$