I have to calculate a line integral with respect to the arc length in the form $\int_C \varphi\ ds$ where there curve $C$ is the intersection between $x^2 + y^2 = z^2 $ and $y^2 = x$ and goes from $(0,0,0)$ to $(1,1,\sqrt{2})$. I have found that I can write that as $x^2 + x = z^2$ and if I "complete the square" I have $x^2 + x + \frac{1}{4} - \frac{1}{4} = z^2$, which is $(x+\frac{1}{2})^2 - z^2 = \frac{1}{4}$ or $$4(x+\frac{1}{2})^2 - 4z^2 = 1$$, which is the equation of a hyperbola centered at $(\frac{1}{2}, 0)$.
I read on Wikipedia that one can parametrize the hyperbola using $x = h + a \sinh t$ and $z = h + b \cosh t$ but if I do this type of parametrization it's not consistent with the points that are given originally, that is, $(0,0,0)$ to $(1,1,\sqrt{2})$.
Is there another parametrization I could look into? Is the above derivation correct so it can be parametrized?
PS: I know this question has been answered before in this site, but unfortunately the answers given were not very helpful, even though I tried them all.
Let,
$$x=-\frac{1}{2}+\frac{1}{2}\cosh (t)=\sinh^2(\frac{t}{2})$$
$$z=\frac{1}{2}\sinh (t)$$
With $t \in [0,\sinh^{-1} (2\sqrt{2})]$. Can you see why this works?
As for $y$ it follows from $y^2=x$ and the fact that we are on the part of $C$ where $y \geq 0$.