The horizon is the largest positive root of $f(r_H)=0$. With $x=r_H^2$,
$$
0=1+\frac{x}{L^2}-\frac\mu x
\quad\Longleftrightarrow\quad
x^2+L^2x-\mu L^2=0.
$$
Choosing the positive root gives
$$
\boxed{r_H^2=\frac{L^2}{2}
\left(\sqrt{1+\frac{4\mu}{L^2}}-1\right).}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 354 / 2 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-354/2/a/i/solution