Answer
(a) Vertices: $V(±4,0)$
Foci: $F(±2\sqrt {7},0)$,
Asymptotes: $y=±\frac{\sqrt 3}{2}x$
(b)
Length of the transverse axis:
$2a=8$
(c)
Work Step by Step
$\frac{x^2}{16}-\frac{y^2}{12}=1$
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$a^2=16$
$a=4$
$b^2=12$
$b=2\sqrt 3$
$c^2=a^2+b^2=16+12=28$
$c=2\sqrt {7}$
(a) Vertices: $V(±a,0)=V(±4,0)$
Foci: $F(±c,0)=F(±2\sqrt {7},0)$,
Asymptotes: $y=±\frac{b}{a}x=±\frac{2\sqrt 3}{4}x=±\frac{\sqrt 3}{2}x$
(b)
Length of the transverse axis:
$2a=8$
