## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 14

#### Answer

Equation of the ellipse: ${\left( {\frac{{x - 16}}{{12}}} \right)^2} + {\left( {\frac{y}{{4\sqrt 5 }}} \right)^2} = 1$ #### Work Step by Step

From the vertices $\left(4,0\right)$ and $\left(28,0\right)$ we obtain the center of the ellipse at $C = \left( {\frac{{28 + 4}}{2},0} \right) = \left( {16,0} \right)$. Since the vertices are $\left(4,0\right)$ and $\left(28,0\right)$, the semimajor axis is 12. The ellipse centered at the origin would have equation ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$. If the center is translated to $C=\left(16,0\right)$, the equation has the form: (1) ${\ \ \ }$ ${\left( {\frac{{x - 16}}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$. Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the ellipse as if it is in standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the ellipse when it is centered at $C=\left(16,0\right)$. Since the semimajor axis is $12$, in standard position the focal vertices are at $\left( { \pm a,0} \right) = \left( { \pm 12,0} \right)$. By Theorem 4, the eccentricity for an ellipse is $e = \frac{c}{a}$. So, $e = \frac{2}{3} = \frac{c}{{12}}$, ${\ \ }$ $c=8$. $b = \sqrt {{a^2} - {c^2}} = \sqrt {{{12}^2} - {8^2}} = \sqrt {80} = 4\sqrt 5$ Substituting $a$ and $b$ in equation (1) gives ${\left( {\frac{{x - 16}}{{12}}} \right)^2} + {\left( {\frac{y}{{4\sqrt 5 }}} \right)^2} = 1$

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