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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2 \sqrt{3})$,we have $\frac{16}{a^2} - \fr

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$\frac{7}{2}$

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(B) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2 \sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} = 1$. Using $b^2 = a^2(e^2 - 1)$,we get $\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$,which simplifies to $16(e^2 - 1) - 12 = a^2(e^2 - 1) \Rightarrow 16e^2 - 28 = a^2(e^2 - 1) \quad (i)$. The directrix is $x = \frac{a}{e}$,so $\frac{a}{e} = \frac{4}{\sqrt{5}} \Rightarrow a^2 = \frac{16e^2}{5} \quad (ii)$. Substituting $(ii)$ into $(i)$: $16e^2 - 28 = \frac{16e^2}{5}(e^2 - 1)$. Multiplying by $5$: $80e^2 - 140 = 16e^4 - 16e^2 \Rightarrow 16e^4 - 96e^2 + 140 = 0$. Dividing by $4$: $4e^4 - 24e^2 + 35 = 0$. Factoring the quadratic in $e^2$: $(2e^2 - 7)(2e^2 - 5) = 0$. Thus,$e^2 = \frac{7}{2}$ or $e^2 = \frac{5}{2}$. Given the options,$e^2 = \frac{7}{2}$ is the correct value.

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2 \sqrt{3})$ is $\sqrt{5}x = 4$ and $e$ is its eccentricity,then $e^2 =$

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