Krdytkyu

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

$2x+y+3z=4implies z=frac{1}{3}(4-2x-y)$

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

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
$$eqalign{y &= dfrac{sin(t)}{sqrt{c}}cr x &= -b y + dfrac{cos(t)}{sqrt{a}}
= - dfrac{b sin(t)}{sqrt{c}} + dfrac{cos(t)}{sqrt{a}}cr}$$

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 $y$

$$x=frac85pmfrac{sqrt{-26z^2+96z-64}}{10}-frac{6z}5,\
y=frac45mpfrac{sqrt{-26z^2+96z-64}}{5}-frac{3z}5.$$

Seeing the polynomial under the radical, this is an ellipse. If you don't like the double signs, you can complete the square

$$-26z^2+96z-64=26left(frac{944}{13}-left(z-frac{48}{13}right)^2right)$$ 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 $cos t,sin t$.