1$ or $\mc{H}(u_H) <1$. This, in turn, corresponds to whether $a$ is real or imaginary. In this paper, we shall only work with the prolate solution implying $a \in \mathbb{R}$. The temperature $T$ and the entropy density $s$ can easily be found as, \bea T&=&-\frac{1}{4 \pi} \mc{F}'_{H}\sqrt{\mc{B}_H}\\ \nn &=& \sqrt{\mc{B}_H}\left(\frac{e^{-\frac{\phi_H}{2}}}{16 \pi u_H} (16+a^2 e^{7 \frac{\phi_H}{2}}u_H^2)-\frac{e^{2 \phi_H}Q^2 u_H^5}{24 \pi}\right)\\ s&=&\frac{N^2 e^{-\frac{5}{4}\phi_H}}{2 \pi u_H^3} \eea where $Q$ is a constant related to the $\U(1)$ charge density on the gauge theory side and the prime indicates derivative with respect to the radial coordinate. In the above we have introduced the notations: $\mc{F}(u_H) \equiv \mc{F}_{H}$ and so on. The gauge field has only one non-vanishing component, \be A_t(u)=-\int_{u_H}^{u}Q \sqrt{\mc{B}}e^{\frac{3}{4}\phi}udu \ee supplemented by the constraint that it vanishes on the horizon, i.e., $A_t(u_H)=0$ whence the $\U(1)$ chemical potential is obtained as \be \mu=\int_{0}^{u_H}Q \sqrt{\mc{B}}e^{\frac{3}{4}\phi}udu. \ee Note that setting $a=0$ we recover the temperature of the RN-AdS black brane temperature, \be T_{a=0}=\frac{1}{2 \pi u_H} (2-q^2) \ee \looseness=-1 where we have defined $q \equiv \f{u_H^3 Q}{2\sqrt{3}}$. As $q \rightarrow \sqrt{2}$ we approach the extremal RN-AdS solution which sets an upper bound on $q$ as $q^2_{\text{max}}=2$. It has been argued in~\cite{Cheng-ml-2014qia} that as one switches on a non-trivial dilaton profile with $a>0$ the horizon of the anisotropic RN-AdS black brane is always greater than its isotropic cousin signifying that the extremal limit can not be accessed. The metric functions $\mc{F}, \mc{B}, \mc{H}$ and the dilaton profile $\phi(u)$ can be found out analytically only in the small anisotropy limit by perturbing around the $a=0$ solution. Otherwise, they can be obtained numerically for any value of $a$. In figure~\ref{metricfn} we provide numerical plots of these functions for some sample values of the parameters.\footnote{See appendix~\ref{app} for details of the numerical scheme followed.} \begin{figure} \centering \subfloat[]{\includegraphics[width=7.2cm,height=5.6cm, angle=-0]{metricfn1.eps}} \subfloat[]{\includegraphics[width=7.2cm,height=5.7cm, angle=-0]{metricfn2.eps}} ]]>

0$. However, generically, $P_1 P_3-N$ can become negative in some range $u_1 <\delta u < u_2$ where $0< u_1 < u_2u_c$. The simultaneous vanishing of the numerator and the denominator has the consequence that the zeros of $P_1, P_3$ and the two zeros of $N$ coincide which permits us to solve for the two momenta independently. First of all, setting the second term in $P_1 P_3-N$ to be zero we find $u_c$ \be \label{uc} \mc{B}_c\mc{F}_c-v^2(\mc{H}_c\cos^2 \psi +\sin^2 \psi)=0 \ee where $\mc{B}_c \equiv \mc{B}(u_c)$ and so on. Using eq.~(\ref{uc}) in either $P_1$ or $P_3$ yields a relation between the two momenta \be \Pi_1 \mc{H}_c \cos \psi=\Pi_3 \sin \psi. \ee Finally, setting the first term in $P_1P_3-N$ to zero yields the exact expressions for the two momenta \bea \Pi_1=\f{v \sin \psi}{u_c^2}, \qquad \Pi_3=\f{\mc{H}_c v \cos \psi}{u_c^2}. \eea Hence, the drag force $\vec{F}$ exerted by the medium upon the massive quark moving with constant velocity $\vec{v}=v\{\sin \psi, \cos \psi\}$ is given by\footnote{Strictly speaking, the external force $\vec{F}$ is the negative of the drag force. But we shall be sloppy in our convention and refer to $\vec{F}$ itself to be the drag force.} \be \vec{F}=\{F_1,F_3\}=\{\Pi_1, \Pi_3\}=\f{\sqrt{\la}}{2\pi}\f{v}{u_c^2}\{\sin \psi, \mc{H}_c \cos \psi \} \ee Also note that we have reinstated the factor $1/2 \pi \a'=\sqrt{\la}/2\pi$ in our expression invoking, in the process, the gauge/string dictionary. Further, it is to be observed that the curious feature of the drag force not being aligned with the velocity as found in the anisotropic medium in~\cite{Chernicoff-ml-2012iq}, continues to hold here, as expected. In the simplifying limit $a \ra 0$ and $q \ra 0$ the expression above reduces to the drag force in $\mc{N}=4$ SYM plasma, \be \vec{F}_{a=0,q=0}=\{F_1,F_3\}=F_0\{ \sin \psi, \cos \psi\} \ee where \be F_0=\f{\pi}{2}\sqrt{\la}T^2\f{v}{\sqrt{1-v^2}} \ee and we have used, \be u_c^2=\f{\sqrt{1-v^2}}{\pi^2 T^2}. \ee ]]>