Answer
Asymptotes: $y=\displaystyle \pm\frac{1}{2}x$
Foci: $(0,\displaystyle \frac{\sqrt{5}}{2})$ and $(0,-\displaystyle \frac{\sqrt{5}}{2})$.
Work Step by Step
Rewriting 4 as $\displaystyle \frac{1}{\frac{1}{4}},$ the equation becomes
$\displaystyle \frac{y^{2}}{\frac{1}{4}}-\frac{x^{2}}{1}=1$
so the hyperbola has a vertical transverse axis and two asymptotes:
$y=\displaystyle \pm\frac{a}{b}x$
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$a=\displaystyle \frac{1}{2}, \quad b=1$
For the foci,
$c^{2}=a^{2}+b^{2}=\displaystyle \frac{1}{4}+1=\frac{5}{4}$
$c=\displaystyle \frac{\sqrt{5}}{2} \approx $1.118
Vertices:$(0,\displaystyle \frac{1}{2})$ and $(0, -\displaystyle \frac{1}{2})$
Asymptotes:
$y=\displaystyle \pm\frac{a}{b}x=\pm\frac{1}{2}x$
Foci: $(0,\displaystyle \frac{\sqrt{5}}{2})$ and $(0,-\displaystyle \frac{\sqrt{5}}{2})$.
