Conservation Laws that apply to a hyperbolic trajectory 
I don't understand why the answer choice is D, I thought for unbound orbits, none of these principles applied?
Could someone clear this up for me?
 A: The question  

Which conservation laws relate to the motion of the asteroid?  

implies that it is the asteroid which is the system under consideration.  
The gravitational force is an attractive central force which acts along the line joining the centre of masses of the star and the asteroid.  
The asteroid has an external force acting on it, the gravitational attraction due to the star, so linear momentum of the asteroid is not conserved.   
The torque, about the star, acting on the asteroid due to the star is zero because the direction of the gravitational attractive force due the star is along the line joining the star and the asteroid, so the angular momentum, about the star, of the asteroid is constant.  
In classical mechanics energy is always conserved and in this example, with no frictional force etc acting, so is mechanical (gravitational potential and kinetic) energy of the asteroid and star system.
The type of orbit depends on the initial conditions and it could have been circular, elliptical, parabolic or hyperbolic. 
A: The conservation laws have nothing to do with actual trajectory, but with the nature of forces acting on the object.
Take for example conservation of energy of some object in gravitational field. The only force acting on the object is (for simplicity, let us assume radial motion):
$$F=-G\frac{mM}{r^2}$$
In very short time, this will produce the change in velocity:
$$\frac{\Delta v}{\Delta t}=-G\frac{M}{r^2},$$ so the kinetic energy changed by:
$$\Delta E_k=\frac{m(v+\Delta v)^2}{2}-\frac{mv^2}{2}\approx mv\Delta v,$$
where we neglected $\Delta v^2$, because we are assuming small time interval and thus this change is also small. Plugging in from the force law:
$$\Delta E_k=-G\frac{mvM}{r^2}\Delta t,$$
Now, can we see the right hand side as negative change in potential energy? That is can we write the right hand side as simple a function of position (r)? In that case the gained/lost kinetic energy would be fully transformed into potential energy.
If you know calculus, you can calculate what kind of potential function you will get. If you don't just use the known result $E_p=-G\frac{mM}{r}$ and see  wheter it works:
$$\Delta E_p=-G\frac{mM}{r+\Delta r}+G\frac{mM}{r}= G\frac{mM\Delta r}{r(r+\Delta r)}\approx G\frac{mM\Delta r}{r^2},$$
But $\Delta r$ is just $v\Delta t$ and plugging it in, you will see it works $\Delta E_k=-\Delta E_p$!
Notice, the only thing we used is the form of the force acting on the object. We did not assume any initial velocity or position, so our formulas work for any scenario in which gravitational (and only gravitational) force acts on the object.
A: every thing is already wrote by @Umaxo, but i prefer to write it like this:
according to NEWTON second law  and your case the force is a function of r:
$$m\,\ddot{r}=F(r)\tag 1$$
multiply (1) with $\dot{r}$
$$m\,\dot{r}\,\ddot{r}=\dot{r}\,F(r)\tag 2$$
with 
$$\dot{r}\,\ddot{r}=\frac{1}{2}\frac{d}{dt}\,\dot{r}^2$$
and
$$\dot{r}=\frac{d\,r}{dt}=v$$
you get:
$$\frac{m}{2}\frac{d}{dt}\,(v^2)=\frac{dr}{dt}\,F(r)$$
$\Rightarrow$
$$\int \frac{m}{2}\,d(v^2)=\int F(r)\,dr+E=-U(r)$$
or 
$$\boxed{\frac{m}{2}\,v^2+U(r)=\text{E}}$$
this is the conservation of energy
the angular momentum is always conserved if the force $F$ is only function of r, for example gravitation force 
