Show that the intersection of a plane...

Question

Show that the intersection of the plane $z = 2y$ with the elliptic cylinder $\frac{x^2}{5} + y^2 = 1$ is a circle. Find the radius and center of this circle.

Hint: How can one describe a circle in 3D?

At this point, all I have put together is the equation of a cylinder is $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ but I don't see a $z$ in the equation, unless I'm supposed to incorporate it by using $z = 2y$ somehow. If I did that, would I just go $z = 2y$, so $\frac{z}{2} = y$, so $\frac{x^2}{5} + y + \frac{z}{2} = 1$? Now, I realize that it's the equation of an ellipsoid, but I'm trying to describe it in 3D like the hint suggests.

Answer 1

Hint: You may want to consider parametrizing the plane and the elliptic cylinder, and looking at the intersection that way.

Also, the equation you gave, $$\frac{x^2}{5} + y = 1$$ is a parabolic cylinder. $$\frac{x^2}{5} + y^2 = 1$$ is an elliptic cylinder.

Answer 2

The second equation suggests $x=\sqrt5\cos t, y=\sin t$ and the first one suggests $z=2\sin t$, then the parametric equation is $$r(t)=(\sqrt5\cos t, \sin t, 2\sin t) \\ r(t)=(\sqrt5\cos t, \sin t, 2\sin t)=0+\sqrt5\cos t\ \ \vec{i}+\sqrt5\sin t(0,\frac{1}{\sqrt5},\frac{2}{\sqrt5})$$ which is the general equation in the space. The general equation is $$r(t)=c_0+\rho \cos t\vec{u}+\rho \sin t\vec{v}$$ see circle