Answer
(a) Vertices: $V(0,±3)$
Foci: $F(0,±5)$,
Asymptotes: $y=±\frac{3}{4}x$
(b)
Length of the transverse axis:
$2a=6$
(c)
Work Step by Step
$\frac{y^2}{9}-\frac{x^2}{16}=1$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a^2=9$
$a=3$
$b^2=16$
$b=4$
$c^2=a^2+b^2=9+16=25$
$c=5$
(a) Vertices: $V(0,±a)=V(0,±3)$
Foci: $F(0,±c)=F(0,±5)$,
Asymptotes: $y=±\frac{a}{b}x=±\frac{3}{4}x$
(b)
Length of the transverse axis:
$2a=6$
