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If $(4,0)$ and $(-4,0)$ be the vertices and $(6,0)$ and $(-6,0)$ be the foci of a hyperbola, then its eccentricity is
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$3 / 2$
$\begin{aligned} & \text { Vertices }( \pm 4,0) \equiv( \pm a, 0) \Rightarrow a=4 \\ & \text { Foci }( \pm 6,0) \equiv( \pm a e, 0) \Rightarrow e=\frac{6}{4}=\frac{3}{2}\end{aligned}$
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