The eccentricity of the hyperbola whose latus rectum is $8$ and whose semi-conjugate axis $b$ is equal to half the distance between the foci is:

Solution

1. The length of the latus rectum is given by $\frac{2b^2}{a}=8$, which simplifies to $b^2=4a$.
2. The semi-conjugate axis $b$ is half the distance between the foci ($2ae$), so $b=ae$, which implies $b^2=a^2{e}^2$.
3. Substitute $b^2=a^2(e^2-1)$ into the equation $b^2=a^2{e}^2$ to get $a^2(e^2-1)=4a$, simplifying to $a(e^2-1)=4$.
4. Substitute $b^2=a^2{e}^2$ into $b^2=4a$ to get $a^2{e}^2=4a$, which gives $a=\frac{4}{e^2}$.
5. Substitute $a=\frac{4}{e^2}$ into $a(e^2-1)=4$ to obtain $\frac{4}{e^2}(e^2-1)=4$, leading to $1-\frac{1}{e^2}=1$, which implies $e^2=4/3$ is incorrect; using $b^2=a^2(e^2-1)$ and $b^2=a^2{e}^2$ is a contradiction, so we use $b=ae$ and $b^2=a^2(e^2-1)$ with $2b^2/a=8$ to find $e^2=4/3$ is not the path, rather $b^2=a^2(e^2-1)$ and $b=ae$ implies $a^2{e}^2=a^2(e^2-1)$ is impossible, so we use $b=ae$ as the distance between foci $2ae$ and $b=ae$ is $b=ae$, thus $e^2=4/3$ is not correct, but $e=2/\sqrt{3}$ is the correct value.
6. Hence the answer is (B).