Prove the intersection between $z^2=x^2+y^2$ and $x+y+2z=2$ is an ellipse.

Question

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.

Answer 1

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.

Answer 2

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.