College Algebra, Chapter 8, 8.3, Section 8.3, Problem 18

Determine the vertices, foci and asymptotes of the hyperbola $\displaystyle x^2 - 2y^2 = 3$. Then sketch its graph

If we divide the equation by 3, we get

$\displaystyle \frac{x^2}{3} - \frac{2y^2}{3} = 1$

Now, we can rewrite the equation as

$\displaystyle \frac{x^2}{3} - \frac{y^2}{\displaystyle \frac{3}{2}} = 1$

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 $a^2 = 3$ and $\displaystyle b^2 = \frac{3}{2}$, we get $a = \sqrt{3}$ and $\displaystyle b = \sqrt{\frac{3}{2}}$ and $\displaystyle c = \sqrt{a^2 + b^2} = \frac{3 \sqrt{2}}{2}$. Thus, we obtain

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

foci $\displaystyle (\pm c, 0) \to \left( \pm \frac{3 \sqrt{2}}{2}, 0 \right) $

asymptote $\displaystyle y = \pm \frac{b}{a} x \to y = \pm \frac{\sqrt{\frac{3}{2}}}{\sqrt{3}} x = \pm \frac{1}{\sqrt{2}} x
$