Answer
(a)
Vertices: $V(0,±\frac{1}{2})$
Foci: $F(0,±\frac{\sqrt 5}{2})$
Asymptotes: $y=±\frac{1}{2}x$
(b)
Length of the transverse axis:
$2a=1$
Work Step by Step
$4y^2-x^2=1$
$\frac{y^2}{\frac{1}{4}}-\frac{x^2}{1}=1$
$\frac{y^2}{(\frac{1}{2})^2}-\frac{x^2}{1^2}=1$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a=\frac{1}{2}$
$b=1$
$c^2=a^2+b^2=(\frac{1}{2})^2+1^2=\frac{1}{4}+1=\frac{5}{4}$
$c=\frac{\sqrt 5}{2}$
(a)
Vertices: $V(0,±a)=V(0,±\frac{1}{2})$
Foci: $F(0,±c)=F(0,±\frac{\sqrt 5}{2})$
Asymptotes: $y=±\frac{a}{b}x=±\frac{\frac{1}{2}}{1}x=±\frac{1}{2}x$
(b)
Length of the transverse axis:
$2a=1$
