How to parametrise an ellipse

Question

I have a curve shaped like an ellipse with centre (2,-1,1). The furthest points from the centre on the curve are (1,0,0) and (3,-2,2) and the closest points from the centre of the ellipse to the curve are (3,-1,0) and(1,-1,2).

The curve lies on the plane x+2y+z=1

I need to be able to parametrise this, but I'm having great difficulty starting. Any help would be greatly appreciated, thanks :)

Answer 1

The direction from the centre to one of the furthest points is $(1,-1,1)$ and to one of the closest points, $(1,0,-1)$. It is a good idea to check that these are perpendicular. A parametric form of the ellipse is $$(x,y,z)=(2,-1,1)+(1,-1,1)\cos\theta+(1,0,-1)\sin\theta\ ,\quad 0\le\theta\le2\pi\ .$$

Answer 2

$\mathbf x = \mathbf u \cos\theta + \mathbf v \sin\theta+(2,-1,1)$

Is an elipse

now find $\mathbf u$ and $\mathbf v$:

$\mathbf u = (3,-2,2) - (2,-1,1) = (1,-1,1)\\ \mathbf v = (3,-1,0) - (2,-1,1) = (1,0,-1)$

Answer 3

In geometrical terms, your ellipse can naturally be considered as the intersection of

• the (vertical) circular cylinder with equation $(x-2)^2+(y+1)^2=1$ and

• the plane with equation $x+2y+z=1.$

It is thus not surprising that the equations given by @David in particular amounts to equations in cylindrical coordinates.