College Algebra, Chapter 8, 8.3, Section 8.3, Problem 14

Determine the vertices, foci and asymptotes of the hyperbola $\displaystyle 9x^2 - 4y^2 = 36$. Then sketch its graph

If we divide the equation by $36$, we get

$\displaystyle \frac{x^2}{4} - \frac{y^2}{9} = 1$

Notice that the equation has 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 $a^2 = 4$ and $b^2 = 9$, we get $a = 2$ and $b = 3$ and $c = \sqrt{a^2 + b^2} = \sqrt{13}$. Thus, we obtain

vertices $(\pm a, 0) \to (\pm 2, 0)$

foci $(\pm c, 0) \to (\pm \sqrt{13}, 0)$

asymptote $\displaystyle y = \pm \frac{b}{a} x \to y = \pm \frac{3}{2} x$