Answer
$\frac{y^2}{1}-\frac{4y^2}{3}=1$
Work Step by Step
Foci: $(0,±c)=(0,±1)$
$c=1$
Transverse axis is vertical.
Length $=2a=1$
$a=\frac{1}{2}$
$c^2=a^2+b^2$
$b^2=c^2-a^2=1^2-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$
$b=\frac{\sqrt 3}{2}$
Hyperbola with vertoical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{y^2}{1^2}-\frac{y^2}{(\frac{\sqrt 3}{2})^2}=1$
$\frac{y^2}{1}-\frac{4y^2}{3}=1$
