Problems from Limited and canonical equations of the quadrics

Development:

First, we calculate the main matrix associated with the equation of the quadric and the associated characteristical polynomial.

In fact, see that $$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]\Rightarrow det(A-xI)=-x^3+3x^2-2x=-x(x^2-3x+2)=$$ $$=-x(x-1)(x-2)$$ Therefore, in view of the results, we see that there are two non-zero eigenvalues and one whose value is zero. The equation of the quadric is converted into $$q(x,y,z)=x^2+2y^2+4y+2z+3=0$$ Since we only have linear term for y, we complete squares for this coordinate. Finally, the equation of the quadric becomes $$q(x,y,z)=x^2+2(y+1{)}^2+2z+1\approx x^2+2y^2+2z+1=0$$ Therefore, the limited form is of the parabolic type. Finally, we are going to find the canonical equation.

Let $a=1$ and $b={\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{\sqrt{2}}{2}}$ be two real values, then the quadric is of the form $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}+2z+1=0$$ and this is an elliptical paraboloid.

Solution:

The canonical equation is ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}+2z+1=0$ and this is an elliptical paraboloid.

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