Parametrization of the intersection of a cone and plane. EDITED with new progress updates.
As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:
$z^2 = 2x^2+2y^2$ and
$2x+y+3z=4\implies z=\frac{1}{3}(4-2x-y)$
If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.
I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:
$0 = \frac{1}{9}(4-2x-y)^2-2x^2-2y^2$
My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?
Any help would be greatly appreciated.
Thank you,
Eric
 A: Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.
EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like
$a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$.  Then you can parametrize it as
$$\eqalign{y &= \dfrac{\sin(t)}{\sqrt{c}}\cr x &= -b y + \dfrac{\cos(t)}{\sqrt{a}}
= - \dfrac{b \sin(t)}{\sqrt{c}} + \dfrac{\cos(t)}{\sqrt{a}}\cr}$$
A: As @RobertIsrael suggests, take $z$ as a parameter and solve
$$2x^2+2y^2=z^2,\\2x+y=4-3z.$$
The second equation gives
$$y=4-2x-3z,$$ and plugging in the first
$$10x^2+24xz-32x+18z^2-48z+32=z^2,$$ giving the solutions in $x$ and $y$
$$x=\frac85\pm\frac{\sqrt{-26z^2+96z-64}}{10}-\frac{6z}5,\\
y=\frac45\mp\frac{\sqrt{-26z^2+96z-64}}{5}-\frac{3z}5.$$
Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square
$$-26z^2+96z-64=26\left(\frac{944}{13}-\left(z-\frac{48}{13}\right)^2\right)$$ and set
$$z=\sqrt{\frac{944}{13}}\cos t+\frac{48}{13}.$$
This will result in $x,y,z$ being all three affine combinations of $\cos t,\sin t$.
