How can I prove a curve lies on an ellipsoid?

Question

I am trying to prove a curve parameterized by

$$\mathbf{r} (t) = \cos(t) \, \mathbf{i} + \sqrt{2} \sin(t) \, \mathbf{j}-\sin(t) \, \mathbf{k}$$

lies on an ellipsoid. How do I do this?

Answer 1

The general equation of an ellipsoid in $\Bbb R^3$ is

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1, \tag 1$

where

$abc \ne 0; \tag 2$

in order for the curve

$r(t) = \cos t \mathbf i + \sqrt 2 \sin t \mathbf j - \sin t \mathbf k \tag 3$

to lie on a surface (1), it's components

$x(t) = \cos t, \; y(t) = \sqrt 2 \sin t, \; z(t) = -\sin t, \tag 4$

must satisfy (1) for suitable $a, b, c$; thus we have

$\dfrac{\cos^2 t}{a^2} + \dfrac{2\sin^2 t}{b^2} + \dfrac{\sin^2 t}{c^2} = 1; \tag 5$

if we take

$a = 1, \; b = c = \sqrt 3, \tag 6$

then (5) becomes

$\dfrac{\cos^2 t}{1} + \dfrac{2\sin^2 t}{3} + \dfrac{\sin^2 t}{3} = \cos^2 t + \dfrac{3\sin^2 t}{3} = \cos^2 t + \sin^2 t = 1, \tag 7$

which shows that $r(t)$ lies on the ellipsoid

$x^2 + \dfrac{y^2}{3} + \dfrac{z^2}{3} = 1. \tag 8$