Simple enough really. But this exact exercise is giving me trouble, you will see shorty.
Suppose $\Gamma$ is the curve, that comes from an intersection between: $$ z^2=x^2-y^2 \tag1 \label1$$ and $$ x^2+y^2+2z^2=3 \tag2 \label2.$$
Calculate $\vec{T},\vec{N},$and$\vec{B}$, as well as the torsion and flexion curvature in the point $(1,0,-1)$
The actual exercise is quite straightforward. All I need is $\vec{r}(t)$ then plug the derivatives intoeuqations for those vectors as well as the torsion and flexion.
And so I went forth: $x=r\cos(t)$,$y=r\sin(t)$, $x^2+y^2=r^2$;
From $\ref2$ I get $\tag3 \label3 r^2+2z^2=3 \\ 2z^2=3-r^2 $ from $\ref1$ $\tag4 \label4 z^2=r^2(\cos^2(t)-sin^2(t))$
Now if I combine what I know from $\ref3$ and $\ref4$ I get my r(t) (note, not the vector) giving me: $$\tag5 \label5 r(t)=\left(\frac{3}{1+2\cos^2(t)-2\sin^2(t)}\right)^{1/2}$$ and with $\ref5$ finally (note: for simplicity will write $r(t)$ as $r$): $$\tag6 \label6 \vec{r}(t)=(r\cos(t),r\sin(t),r(\cos^2(t)-\sin^2(t))^{1/2})$$
All seems fine but when I try to check for the point $(1,0,-1)$ I simply can't my fit my $\vec{r}(t)$ to it. So I expect I did something wrong here. At closest I get $\vec{r}(0)=(1,0,1)$
Rewrite the first equation as $x^2=y^2+z^2$. This surface is a cone with the $x$-axis as its axis of symmetry. Its cross-sections perpendicular to the $x$-axis are circles, which suggests trying the parameterization $y=x\sin t$, $z=x\cos t$. Plugging this into the equation of the ellipsoid gives $$x^2(\sin^2 t+2\cos^2 t+1)=3,$$ from which $$x^2 = {6\over5+\cos2t}.$$ There are actually two intersection curves, but we want the one on which the point $(1,0,-1)$ lies, so take the positive square root, yielding the parameterization $$\mathbf r(t) = \sqrt{6\over5+\cos2t}(1,\sin t,\cos t).$$ I’ll leave it to you to check that this in fact satisfies both equations. Solving $\mathbf r(t)=(1,0,-1)$ for $t$ produces $t=\pi$ and the rest, as you say, is a matter of plugging the function into formulas.