Solution
= Solution
The horizon equation gives
$$
2M=\frac{r_+^3}{L^2}+\frac{Q^2}{r_+}.
$$
With $Q=r_+^2\mu/L$, define
$$
g(z)=1-(1+\mu^2)z^3+\mu^2z^4.
$$
The coordinate changes in the question give
$$
\boxed{ds^2=\frac{L^2}{z^2}
\left[-g(z)d\tau^2+\frac{dz^2}{g(z)}+dX^2+dY^2\right]},
$$
$$
\boxed{A=\mu L(1-z)d\tau}.
$$
The horizon is at $z=1$, and all nontrivial dimensionless dependence is through $\mu$; $L$ remains only as the overall curvature scale.
Solved by gpt-5.6-sol high.