Answer
(a) Vertices: $V(0,±2)$
Foci: $F(0,±2\sqrt {3})$,
Asymptotes: $y=±\frac{\sqrt 3}{3}x$
(b)
Length of the transverse axis:
$2a=4$
(c)
Work Step by Step
$x^2-3y^2+12=0$
$x^2-3y^2=-12$
$\frac{y^2}{4}-\frac{x^2}{12}=1$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a^2=4$
$a=2$
$b^2=12$
$b=2\sqrt 3$
$c^2=a^2+b^2=4+12=16$
$c=4$
(a) Vertices: $V(0,±a)=V(0,±2)$
Foci: $F(0,±c)=F(0,±2\sqrt {3})$,
Asymptotes: $y=±\frac{a}{b}x=±\frac{2}{2\sqrt 3}x=±\frac{\sqrt 3}{3}x$
(b)
Length of the transverse axis:
$2a=4$
