Answer
$\displaystyle \frac{x^{2}}{3}-y^{2}=1$
Work Step by Step
Foci on the x-axis: $\quad \displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}$
Foci: $\quad (\pm c, 0)$
Vertices: $\quad (\pm a, 0)$
Asymptotes: $\quad \displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0 \quad$ or $\quad y=\displaystyle \pm\frac{b}{a}x$
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Given the foci, $ c^{2}=4\quad \Rightarrow\quad a^{2}+b^{2}=4$
Given the asymptotes, $\quad \displaystyle \Rightarrow\quad(\frac{b}{a})^{2}=\frac{1}{3}\quad \Rightarrow\quad a^{2}=3b^{2}$
Substituting into $a^{2}+b^{2}=2$,
$4b^{2}=4$,
$b^{2}=1,\quad(a^{2}=3)$
The equation is $\displaystyle \frac{x^{2}}{3}-\frac{y^{2}}{1}=1$
or
$\displaystyle \frac{x^{2}}{3}-y^{2}=1$
