Parametric equation of path on ellipsoid

Question

I am having trouble finding a parametric equation for the path $K$, with starting-point $(1,-1,0)$ and end-point $(-1,1,0)$, which is determined by the equations $$\begin{cases} 2x^2 + 2y^2 +z^2 = 4 \\ x + y + z =0 \\ z \geq 0 \end{cases}$$

I know the standard parametric equation for an ellipsoid but I seem to be stuck. Any help is greatly appreciated.

Answer 1

Obviously we need only 1 parameter to describe a line whether curved or straight. Then adding up any extra parameter causes dependence between them. First take parameters $\phi,\theta$ such that: $$x=\sqrt 2 sin(\theta)cos(\phi)$$ $$y=\sqrt 2 sin(\theta)sin(\phi)$$ $$z=2cos(\theta)$$ This satisfies first equation with one more parameter. To discover the dependence, substitute the above representations in the 2nd equation: $$x+y+z=0\to sin(\theta)cos(\phi)+sin(\theta)sin(\phi)+\sqrt 2 cos(\theta)=0$$ which yields to $$-cot\theta=sin(\phi+{\pi\over 4})$$ Now note the following:

$$x+y=2sin\theta sin(\phi+{\pi\over 4})$$

$$x+y=2sin\theta cos(\phi+{\pi\over 4})$$

Throughout the path from $startpoint$ to $endpoint$ we have $-{\pi\over 4}\le\phi\le{3\pi\over 4}$ therefore $0\le\phi+{\pi\over 4}\le \pi$ and then $cos(\phi+{\pi\over 4})\ge 0$ which leads to:

$$cos(\phi+{\pi\over 4})=\sqrt {1-cot^2(\theta)}$$ This enables us to get to the parametric description:

$$x+y=-2cos\theta$$

$$x-y=2\sqrt {1-2cos^2\theta}$$

$$z=2cos\theta$$

which finally gives us:

$$x=-cos\theta+\sqrt {1-2cos^2\theta}$$

$$y=-cos\theta-\sqrt {1-2cos^2\theta}$$

$$z=2cos\theta$$

$${\pi\over 4} \le \theta \le {\pi\over 2}$$

There is an alternative way to represent this description:

$$x=-t+\sqrt {1-2t^2}$$

$$y=-t-\sqrt {1-2t^2}$$

$$z=2t$$

$$0 \le t \le \sqrt 2$$

Answer 2

$$ \eqalign{ & \left\{ \matrix{ 2x^{\,2} + 2y^{\,2} + z^{\,2} = 4 \hfill \cr x + y + z = 0 \hfill \cr 0 \le z \hfill \cr} \right. \cr & \left\{ \matrix{ 0 \le z = - x - y \hfill \cr 3x^{\,2} + 3y^{\,2} + 2xy = 2\left( {x^{\,2} + y^{\,2} + 2xy} \right) + \left( {x^{\,2} + y^{\,2} - 2xy} \right) = 4 \hfill \cr} \right. \cr & \left\{ \matrix{ 0 \le z = - \left( {x + y} \right) \hfill \cr {{\left( {x + y} \right)^{\,2} } \over {\left( {\sqrt 2 } \right)^{\,2} }} + {{\left( {x - y} \right)^{\,2} } \over {2^{\,2} }} = 1 \hfill \cr} \right. \cr & \left\{ \matrix{ x + y = - z = - \sqrt 2 \cos \theta \hfill \cr x - y = 2\sin \theta \hfill \cr - \pi /2 \le \theta \le \pi /2 \hfill \cr} \right. \cr} $$