## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 29

#### Answer

- the vertices are $(\pm 4,0)$ and $(0, \pm 2)$ - the foci are $(\pm \sqrt{12},0)$ - the center is $(0,0)$

#### Work Step by Step

The equation $x^2+4y^2=16$ can be written in the form $$\left(\frac{x}{4}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1$$ which is an ellipse with $a=4, b=2$ so $a\gt b$ and hence $c=\sqrt{a^2-b^2}=\sqrt{16-4}=\sqrt{12}$. So, we have: - the vertices are $(\pm 4,0)$ and $(0, \pm 2)$ - the foci are $(\pm \sqrt{12},0)$ - the center is $(0,0)$

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