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(D) Given focus $S = (-5, 0)$ and directrix $x = -9/5$. Since the focus is on the $x$-axis,the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.$a

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$189$

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(D) Given focus $S = (-5, 0)$ and directrix $x = -9/5$. Since the focus is on the $x$-axis,the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.$ae = 5$ and $\frac{a}{e} = \frac{9}{5}$.Multiplying these: $a^2 = 9 \Rightarrow a = 3$.Then $e = 5/3$. Since $b^2 = a^2(e^2 - 1)$,$b^2 = 9(\frac{25}{9} - 1) = 16$,so $b = 4$.The hyperbola equation is $\frac{x^2}{9} - \frac{y^2}{16} = 1$.For point $(\alpha, 2\sqrt{5})$ on the hyperbola: $\frac{\alpha^2}{9} - \frac{20}{16} = 1$ $\Rightarrow \frac{\alpha^2}{9} = 1 + \frac{5}{4} = \frac{9}{4}$ $\Rightarrow \alpha^2 = \frac{81}{4}$.The focal distances of a point $P(x, y)$ are $r_1 = |ex - a|$ and $r_2 = |ex + a|$.The product $p = |e^2x^2 - a^2| = |\frac{25}{9} \cdot \frac{81}{4} - 9| = |\frac{225}{4} - 9| = |\frac{225 - 36}{4}| = \frac{189}{4}$.Thus,$4p = 4 \cdot \frac{189}{4} = 189$.

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