Prove the intersection between $z^2=x^2+y^2$ and $x+y+2z=2$ is an ellipse. I want to prove that the intersection between the cone $z^2=x^2+y^2$ and the plane  $x+y+2z=2$ is an elispe in this plane. 
My work:
I suppose to prove it I have to see that the equation $x+y+2\sqrt{x^2+y^2}=2$ can be rewritten as $\frac{x^2}{a}+\frac{y^2}{b}=c$. To do so I have tried to completing the square without any results.

You cannot put it in this form, because the ellipse is not always centered at the origin, or having its major axis on either the $y$ or $x$ axes.
$$\frac{x+y-2}{2}=z$$
$$\left(\frac{x+y-2}{2}\right)^2=x^2+y^2$$
$$x^2+xy-2x+xy+y^2-2y-2x-2y+4=4x^2+4y^2$$
$$3x^2+3y^2-2xy+4x+4y=4$$
This conic is an ellipse.  You can use orthogonal diagonalisation to transform it into standard form.
$$3x^2-2xy+3y^2+4x+4y-4=0$$
$$a=3,\ b=-1,\ c=3, \ d=4$$
$$X=
        \begin{pmatrix}
        x\\
       y\
        \end{pmatrix}
$$
$$ A=   \begin{pmatrix}
        3 & -1\\
        -1 & 3\
\
        \end{pmatrix}
$$
$$v=
\begin{pmatrix}
        4\\
        4\
\\
        \end{pmatrix}
$$
$$X^TAX+v^TX=4$$
Find the eigenvalues and eigenvectors of $A$, use the normalized eigenvectors as the columns of a rotation matrix $P$ with determinant $1$. It becomes relatively straightforward after that.
