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.
 A: 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.
A: Kind of a trick here: the quadratic object is the standard double cone.  Any plane that intersects it must be a conic section; this plane does not cross the origin so it's not degenerate, and its maximum slope of $\sqrt 2 / 2$ is between 0 and 1 so it's an ellipse.
