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(B) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2\sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} =

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$4e^4 - 24e^2 + 35 = 0$

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(B) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since it passes through $(4, -2\sqrt{3})$,we have $\frac{16}{a^2} - \frac{12}{b^2} = 1 \dots (i)$.Given the directrix $x = \frac{a}{e}$,we have $\frac{a}{e} = \frac{4\sqrt{5}}{5} = \frac{4}{\sqrt{5}}$,so $a^2 = \frac{16e^2}{5} \dots (ii)$.Using $b^2 = a^2(e^2 - 1)$,substitute $a^2$ and $b^2$ into $(i)$:$\frac{16}{a^2} - \frac{12}{a^2(e^2 - 1)} = 1$.Substitute $a^2 = \frac{16e^2}{5}$:$\frac{16}{16e^2/5} - \frac{12}{(16e^2/5)(e^2 - 1)} = 1 \Rightarrow \frac{5}{e^2} - \frac{60}{16e^2(e^2 - 1)} = 1$.Multiply by $4e^2(e^2 - 1)$:$20(e^2 - 1) - 15 = 4e^2(e^2 - 1) \Rightarrow 20e^2 - 35 = 4e^4 - 4e^2$.Rearranging gives $4e^4 - 24e^2 + 35 = 0$.

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

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