Find the parametrization of the intersection of two surfaces.

Question

I'm having trouble figuring this problem out since the $3x^2 + z^2$ is throwing me off. Especially because of the $z$.

Find the parametrization for the curve of intersection between the cylinder $3x^2 + z^2 = 4$ and the plane $x-y-z= 0$

Answer 1

There's a couple different ways to go about this. To make life perhaps somewhat simpler, I would note that you can already parametrize the $x$ and $z$ from the cylinder in elliptical coordinates, $x=\sqrt{4/3}\cos(t)$, $z=2\sin(t)$. From there, you could then use the equation of your plane to find $y$ in terms of $t$.

Answer 2

First of all, you can parametrize the cylinder in this way

$$\left(\sqrt{\frac{4}{3}}\sin t, y, 2 \cos t \right)$$

Then by substituting the plane into the cylinder equation and solve for y ,

$3x^2 +(x-y)^2 = 4$

You will get $y= -2 \cos t + \frac{1}{\sqrt{3}}\sin t $ or $y= 2 \cos t + \frac{1}{\sqrt{3}}\sin t $