Ellipse in Space - 2

First we'll find the vertices of the ellipse, they occur when the curvature attains an extreme point.

The curvature is given by $\kappa = \dfrac{\lVert p' \times p'' \rVert}{\lVert p' \rVert^3}$ , in this case we get $\kappa = \dfrac{\lVert p_1 \times p_2 \rVert}{\lVert -p_1\sin(t) + p_2\cos(t) \rVert^3}$ . Since the numerator is constant, we only need to find the extreme points of the denominator, and for simplicity, instead find the extreme points of $f(t)=\lVert -p_1\sin(t) + p_2\cos(t) \rVert^2 = \lVert p_1 \rVert^2 \sin^2(t) - p_1 \cdot p_2 \sin(2t) + \lVert p_2 \rVert^2 \cos^2(t)$ .

Take the derivative: $f'(t) = \lVert p_1 \rVert^2 \sin(2t) - 2p_1 \cdot p_2 \cos(2t) - \lVert p_2 \rVert^2 \sin(2t)$ . So, the extreme points are given by $\tan(2t) = \dfrac{2 p_1 \cdot p_2}{\lVert p_1 \rVert^2 - \lVert p_2 \rVert^2}$ , i.e., $t_k=\theta + \dfrac{\pi}{2}k$ , for $0 \leq k \leq 3$ , where $\theta = \dfrac{1}{2}\arctan \left( \dfrac{2 p_1 \cdot p_2}{\lVert p_1 \rVert^2 - \lVert p_2 \rVert^2} \right)$ .

So, the four vertices are given by $V_k = p(t_k)$ . Then one pair of opposite vertices is $V_0 = p_0 + p_1 \cos \theta + p_2 \sin \theta$ and $V_2 = p_0 - p_1 \cos \theta - p_2 \sin \theta$ , the other pair is $V_1 = p_0 - p_1 \sin \theta + p_2 \cos \theta$ and $V_3 = p_0 + p_1 \sin \theta - p_2 \cos \theta$ . Then, the lenghts of the semi-minor and semi-major axes (not necessarily in that order) are given by $a=\dfrac{1}{2} \lVert V_0 - V_2 \rVert$ and $b=\dfrac{1}{2} \lVert V_1 - V_3 \rVert$ .

Finally, with the given values of $p_0, p_1$ and $p_2$ , we get that the length of the semi-major axis is $\sqrt{61}+4 \approx 11.810$ and the length of the semi-minor axis is $\sqrt{61}-4 \approx 3.810$ .