Chapter 9 Quadratic Relations and Conic Sections - 9.5 Graph and Write Equations of Hyperbolas - 9.5 Exercises - Skill Practice - Page 646: 30

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The equation in standard form $$\frac{x^2}{144}+\frac{y^2}{256}=1$$ The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical. Identify the vertices, foci, and asymptotes. Note that $a=16$ and $b=12$. The $x^2-term$ is negative, so the transverse axis is vertical and the vertices are at $(0,\pm 16)$ and $(\pm 12,0)$. Draw the hyperbola.