Chapter 9 Quadratic Relations and Conic Sections - 9.5 Graph and Write Equations of Hyperbolas - 9.5 Exercises - Quiz for Lessons 9.4-9.5 - Page 648: 3

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Given $$36x^2+9y^2=324$$ Rewrite $$\frac{x^2}{9}+\frac{y^2}{36}=1$$ The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is vertical. Identify the vertices, foci, and asymptotes. Note that $a=6$ and $b=3$. The $x^2-term$ is negative, so the transverse axis is vertical and the vertices are at $(0,\pm 6)$. Find the foci: $c^2=a^2-b^2=6^2-3^2=27\\ \rightarrow c=3\sqrt 3$ The foci are at $(0,\pm 3\sqrt 3)$