College Algebra, Chapter 8, 8.3, Section 8.3, Problem 20

Determine the vertices, foci and asymptotes of the hyperbola $\displaystyle 9x^2 - 16y^2 = 1$. Then sketch its graph

We can rewrite the equation as

$\displaystyle \frac{x^2}{\displaystyle \frac{1}{9}} - \frac{y^2}{\displaystyle \frac{1}{16}} = 1$

to get the form $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Since the $x^2$-term is positive, then the hyperbola has a horizontal transverse axis; its vertices and foci are located on the $x$-axis. Since $\displaystyle a^2 = \frac{1}{9}$ and $\displaystyle b^2 = \frac{1}{16}$, we get $\displaystyle a = \frac{1}{3}$ and $\displaystyle b = \frac{1}{4}$ and $\displaystyle c = \sqrt{a^2 + b^2} = \frac{5}{12}$. Thus, we obtain

vertices $\displaystyle (\pm a, 0) \to \left( \pm \frac{1}{3}, 0 \right)$

foci $\displaystyle (\pm c, 0) \to \left( \pm \frac{5}{12}, 0 \right) $

asymptote $\displaystyle y = \pm \frac{b}{a} x \to y = \pm \frac{\displaystyle \frac{1}{4}}{\displaystyle \frac{1}{3}} x = \pm \frac{3}{4} x
$