Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.2 Conic Sections: Ellipses - 13.2 Exercise Set - Page 861: 16

Answer

The graph is shown below
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Work Step by Step

$9{{x}^{2}}+16{{y}^{2}}=144$ …… (1) Multiply $\frac{1}{144}$on both the sides of equation $9{{x}^{2}}+16{{y}^{2}}=144$. Identifying $a\text{ and }b$ in the equation, $\begin{align} & \frac{9}{144}{{x}^{2}}+\frac{16}{144}{{y}^{2}}=\frac{144}{144} \\ & \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1 \\ & \frac{{{x}^{2}}}{{{\left( 4 \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( 3 \right)}^{2}}}=1 \end{align}$ Since ${{a}^{2}}>{{b}^{2}}$, the ellipse will be horizontal. Since $a=4\text{ and }b=3$, the x-intercepts are $\left( -4,0 \right)\text{ and }\left( 4,0 \right)$ and the y-intercepts are $\left( 0,-3 \right)\text{ and }\left( 0,3 \right)$. Plot these points and connect them with the oval shaped curve.
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