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: 8

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Given: $4x^2-16y^2=64$ $$\frac{x^2}{16}-\frac{y^2}{4}=1$$ The denominator of $x^2$ is smaller than $y^2$, so the transverse axis is horizontal. Identify the vertices, foci, and asymptotes. Note that $a=4$ and $b=2$. The $x^2-term$ is positive, so the transverse axis is horizontal and the vertices are at $(\pm 4,0)$. Find the foci: $c^2=a^2+b^2=4^2+2^2=20\\ \rightarrow c=2\sqrt 5$ The foci are at $(\pm 2\sqrt 5,0)$ The asymptotes are $y=\pm \frac{1}{2}x$ Draw the hyperbola.