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