Intersection of cylinder and plane If a cylinder is $x^2+8y^2=1$ and a plane is $x+y+3z=0$, what's the form of the intersection?

I thought of substituting the $y$ variable from the plane's equation in the cylinder's equation.
Actually I think we could get better results (at least easier to handle) about the intersection passing through parametrization.
The cylinder can be parametrized in $(u, v)$ like this:

$$ x = \cos u \\\\
y = \frac{\sqrt 2}{4}\sin u \\\\
z = v$$

, with $u\in[0, 2\pi]$ and $v\in(-\infty,+\infty)$

Getting the $z$ from the plane equation we get:

$$ z = \frac{x+y}{3}$$

and replacing the parametrization we get the final curve, parametrized in $u$ only:

$$ x = \cos u \\\\
y = \frac{\sqrt 2}{4}\sin u \\\\
z = \frac{4\cos u+\sqrt2\sin u}{12}$$

, with $u\in[0, 2\pi]$
we have the equation $$x^2+8y^2=1$$ and $$x+y+3z=0$$ solving the second equation for $y$ we have
$$y=-3z-x$$ in the first equation we obtain
$$x^2+8(-3z-x)^2=1$$
simplifying we obtain
$$9x^2+72z^2+48xz=1$$
can you go further?