Constants in the trajectory equation of a particle with uniform acceleration in relativity In the book "Introducing Einstein's Relativity" by Ray d'Inverno he describes the trajectory equation of a uniformly accelerated point mass particle as $$\frac{\left ( x-x_{0}+\frac{c^{2}}{a} \right )^{2}}{\left ( \frac{c^{2}}{a} \right )^{2}}-\frac{\left ( ct-ct_{0} \right )^{2}}{\left ( \frac{c^{2}}{a} \right )^{2}}=1$$ where $x$ and $ct$ are the space and time coordinate in an inertial frame, $a$ is the uniform acceleration of the particle and $x_{0}$ is the position of the particle at $t_{0}$ which act like the integration constants which is performed to get the equation. My question is, can we take $x_{0}=0$ at $t_{0}=0$?
In the text, it ha been specified that $x_{0}-\frac{c^{2}}{a}=t_{0}=0$ has to be taken. Also, when I drew the hyperbola w.r.t $x_{0}=0$ at $t_{0}=0$, the hyperbola was shifted leftwards and the right wing of it stated from the origin. If so, how will the hyperbola asymptotically reach the $x=ct$ line?
 A: There is a standard derivation of the motion under constant proper acceleration, for example in Chapter 6 of Gravitation in Misner, Thorne and Wheeler. I won't reproduce the details here, but it naturally leads to an equation describing the motion of a particle that starts at the position $x(0) = c^2/a$. So on the usual spacetime diagram the motion looks like:

The red line shows our accelerating object and the blue line shows the trajectory of a photon that starts at $x=0$ at the same moment as our object. The graph is generally drawn this way so that the it's clear how the worldline of the object asymptotically approaches the world line of the photon.
But of course we can make our object start at the origin simply by a shift of the origin, and this gives us:

Drawn this way the worldline of the object asymptotically approaches the worldline of a light ray that started at $x = -c^2/a$ at time zero. As you say in your question this diagram is less intuitively obvious, but the gradient, i.e. the speed of the object, does still approach $c$ in the same way.
Incidentally, if the light ray started farther from the origin than $x = -c^2/a$ it can never catch up with the object. You can actually outrun a light ray provided you have a big enough head start and provided you can maintain your acceleration indefinitely!
