Answer
(a)
Vertices: $V(±a,0)=V(±\frac{1}{3},0)$
Foci: $F(±c,0)=F(±\frac{5}{12},0)$
Asymptotes: $y=±\frac{b}{a}x=±\frac{\frac{1}{4}}{\frac{1}{3}}x=±\frac{3}{4}x$
(b)
Length of the transverse axis:
$2a=\frac{2}{3}$
(c)
Work Step by Step
$9x^2-16y^2=1$
$\frac{x^2}{\frac{1}{9}}-\frac{y^2}{\frac{1}{16}}=1$
$\frac{x^2}{(\frac{1}{3})^2}-\frac{y^2}{(\frac{1}{4})^2}=1$
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$a=\frac{1}{3}$
$b=\frac{1}{4}$
$c^2=a^2+b^2=\frac{1}{9}+\frac{1}{16}=\frac{25}{144}$
$c=\frac{5}{12}$
(a)
Vertices: $V(±a,0)=V(±\frac{1}{3},0)$
Foci: $F(±c,0)=F(±\frac{5}{12},0)$
Asymptotes: $y=±\frac{b}{a}x=±\frac{\frac{1}{4}}{\frac{1}{3}}x=±\frac{3}{4}x$
(b)
Length of the transverse axis:
$2a=\frac{2}{3}$
