Answer
(a) Vertices: $V(0,±6)$
Foci: $F(0,±2\sqrt {13})$,
Asymptotes: $y=±\frac{3}{2}x$
(b)
Length of the transverse axis:
$2a=12$
(c)
Work Step by Step
$4y^2-9x^2=144$
$\frac{y^2}{36}-\frac{x^2}{16}=1$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a^2=36$
$a=6$
$b^2=16$
$b=4$
$c^2=a^2+b^2=36+16=52$
$c=2\sqrt {13}$
(a) Vertices: $V(0,±a)=V(0,±6)$
Foci: $F(0,±c)=F(0,±2\sqrt {13})$,
Asymptotes: $y=±\frac{a}{b}x=±\frac{6}{4}x=±\frac{3}{2}x$
(b)
Length of the transverse axis:
$2a=12$
