Location of event horizon in moving black hole Say a black hole is travelling at $c/2$, does the shape of the event horizon change? What about the location of the event horizon? If it is travelling at a hypothetical $c$, does the event horizon simply resemble a light cone?
 A: The spacetime of an ultrarelativistic observer moving (at nearly c) in the vicinity of a black hole can be found by preforming a procedure known as the Aichelburg-Sexl Ultraboost.$^1$ 
The metric is given, in Brinkmann coordinates, by:
$$ds^{2}=-8m\,\delta (u)\,\log r\,du^{2}+2\,du\,dv+dr^{2}+r^{2}\,d\theta ^{2}$$
The event horizon does, in fact, remain spherical in this case, even for the ultrarelativistic limit.
This is, of course, in the simple case of a schwarzschild (spherically symmetric) black hole.
It is much more interesting (and much more complex) to consider the case of a realistic, rotating (Kerr) black hole. In this case the event horizon is not a sphere to begin with, but an oblate spheroid. Also, in this case there are several other surfaces of interest, such as the inner horizon as well as the boundary of the ergosphere (the surface on which the azimuthal coordinate becomes timelike). A broad discussion is given in Soares, 2019. 

$^1$ Equivalently, this is the spacetime produced by an ultrarelativistic black hole hurdling towards an observer
A: It is convenient to go from Schwarzschild coordinates
$$
ds^2 = \left( 1 - \frac{r_s}{r}\right) c^2 dt^2 - \left( 1 - \frac{r_s}{r}\right)^{-1}dr^2 - r^2 d\Omega
$$
to Isotropic coordinates $r\rightarrow r^2(1+ r_s/4r)^4$, then to quasi-Minkowski using the usual spherical-cartesian coordinate transformation to get
$$
ds^2 = -\left( \frac{1 - r_s/4r}{1 + r_s/4r} \right)^2 c^2 dt^2  + \left( 1 + r_s/4r\right)d\vec{x}^2
$$
Now you can finally use the familiar Lorentz boost formula
\begin{align*}
ct & \rightarrow \gamma\left(ct - \vec{\beta}\cdot\vec{x}\right) \\
\vec{x} & \rightarrow \vec{x} - \gamma \vec{\beta}ct + \frac{\gamma-1}{\beta^2}\left(\vec{\beta}\cdot\vec{x}\right)\vec{\beta}\\
\Rightarrow r^2 &\rightarrow \left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2
\end{align*}
Only now you can start asking physical questions using the new coordinate system. For example let's look at the event horizon. In the old coordinates (pre boosted isotropic quasi-minkowski) requiring $g^{00}=0$ gave us the spherical surface surface $r=r_s/4$. Now let's look at our new $g^{00}$
\begin{align*}
g^{00} = \gamma^2\left( -\left( \frac{1-r_s/4r'}{1+r_s/4r'}\right)^2 + \beta^2 \left(1+r_s/4r' \right) \right)
\end{align*}
So using $g^{00}=0$ the horizon in the new coordinate system is found to be 
$$
r' = \frac{r_s}{4}\frac{1+\beta}{1-\beta}
$$
$\Rightarrow$
$$
\left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2 = \left( \frac{r_s}{4}\frac{1+\beta}{1-\beta} \right)^2
$$
Now since our new coordinates are isotropic the metric is unchanged if we had aligned the $z-$axis with $\vec{\beta}$, in which case you get an ellipsoid with moving center along the $z-$axis $(x_0,y_0,z_0) =(0,0,vt)$ and contracted along the $z$-axis.
Note: proof read for arithmetic mistakes.
A: The event horizon is by definition a set of points in spacetime that is independent of the observer. Therefore its location is independent of the state of motion of the black hole and the observer relative to each other.
We could ask whether the moving observer thinks the event horizon is spherical. This is not a question that is meaningful in general relativity, unless you specify what kind of observations are involved. When we define spherical symmetry in GR, we take pains to do it in a way that is independent of coordinates or observers. You might think that you could take the intersection of the event horizon with a surface of simultaneity for a particular observer, but GR doesn't define that kind of surface of simultaneity.
If you want to define how the horizon appears to an observer, one natural way to do it would be to use optical observations. In SR, the optical silhouette of a sphere is always a disk, regardless of the state of motion, so I suspect that the same would be true for the silhouette of a black hole. This does seem to be true, as far as I can tell, in figures 25-28 of Riazuelo, https://arxiv.org/abs/1511.06025 , although it's not especially clear. I've done some similar simulations myself, but I only did radial motion, so I haven't actually seen what happens for transverse motion.
But of course, even if we established that optical measurements showed the silhouette of a Schwarzschild black hole as a disk for an observer in any state of motion, that isn't an answer as to the question of whether the event horizon appears length contracted, since the result is the same as the SR result for the silhouette of a sphere. Perhaps there is some other type of observation that you could define that would give a natural sense in which this becomes a meaningful question, but it's not immediately obvious to me what type of observation that would be.
