Answer
(a)
Vertices: $V(±a,0)=V(±\sqrt 2,0)$
Foci: $F(±c,0)=F(±\sqrt {10},0)$
Asymptotes: $y=±\frac{b}{a}x=±\frac{\sqrt 2}{2\sqrt 2}x=±\frac{1}{2}x$
(b)
Length of the transverse axis:
$2a=4\sqrt 2$
(c)
Work Step by Step
$x^2-4y^2-8=0$
$x^2-4y^2=8$
$\frac{x^2}{8}-\frac{y^2}{2}=1$
$\frac{x^2}{(2\sqrt 2)^2}-\frac{y^2}{(\sqrt 2)^2}=1$
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$a=2\sqrt 2$
$b=\sqrt 2$
$c^2=a^2+b^2=8+2=10$
$c=\sqrt {10}$
(a)
Vertices: $V(±a,0)=V(±\sqrt 2,0)$
Foci: $F(±c,0)=F(±\sqrt {10},0)$
Asymptotes: $y=±\frac{b}{a}x=±\frac{\sqrt 2}{2\sqrt 2}x=±\frac{1}{2}x$
(b)
Length of the transverse axis:
$2a=4\sqrt 2$
