The problem is as follows.
Let $S_1$=$x^2+y^2-z^2=1, z>0$ and $S_2=z^2-2x+y=0$. And define $c$ is curve which is line of intersection of $S_1$ and $S_2$. Find the curvature of $C$ at $P=(1,1,1) $
I tried to parametrization of C by calculating $S_1$ and $S_2$ BUT it has the root on z-axis, i think it may have problem about regularity ... In fact, i calculate $x= 1+$$ 3\over 2$$cost$ and $y=$$-1 \over 2$ $+$ $3 \over 2$ $sint$... please the full explanation the solution of this problem.
So far we found
$x(t) = 1 + \frac{3}{2} \cos t$
$y(t) = -\frac{1}{2} + \frac{3}{2} \sin t$
Substituting in any of $S1$ or $S2$, we get
$z(t)^2 = \frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t$
So, $z(t) = \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}$
$r(t) = (1 + \frac{3}{2} \cos t, -\frac{1}{2} + \frac{3}{2} \sin t, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t} \,)$
$r'(t) = (-\frac{3}{2} \sin t, \frac{3}{2} \cos t, - \frac{3(\sin t + \frac{1}{2} \cos t)} {2 \, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}} \, \,)$
$r''(t) = (-\frac{3}{2} \cos t, -\frac{3}{2} \sin t, -\frac{3(\cos t - \frac{1}{2} \sin t)} {2 \, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}} - \frac{9(\sin t + \frac{1}{2} \cos t)^2} {4 \, (\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t)^{3/2}} \, \,)$
Using $r(t)$, we know that at point $(1, 1, 1), t = \frac{\pi}{2}$.
So, $r'(t) = (-\frac{3}{2}, 0, -\frac{3}{2})$
$|r'(t)| = \frac{3}{\sqrt2}$
$r''(t) = (0, -\frac{3}{2}, -\frac{3}{2})$
Curvature $k = \displaystyle \frac{|r'(t) \times {r''(t)}|}{|r'(t)|^3}$
Cross product of $r'(t)$ and ${r''(t)} = (-\frac{9}{4}, -\frac{9}{4}, \frac{9}{4})$
$|r'(t) \times {r''(t)}| = \frac{9 \sqrt3}{4}$
So, curvature $k = \frac{1}{\sqrt 6}$