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: 13

Answer

The graph is shown below

Work Step by Step

$4{{x}^{2}}+9{{y}^{2}}=36$ …… (1) Multiply $\frac{1}{36}$on both the sides of equation $4{{x}^{2}}+9{{y}^{2}}=36$. Identifying $a\text{ and }b$ in the equation, $\begin{align} & \frac{4}{36}{{x}^{2}}+\frac{9}{36}{{y}^{2}}=\frac{36}{36} \\ & \frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1 \\ & \frac{{{x}^{2}}}{{{\left( 3 \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( 2 \right)}^{2}}}=1 \end{align}$ Since ${{a}^{2}}>{{b}^{2}}$, the ellipse will be horizontal. Since $a=3\text{ and }b=2$, the x-intercepts are $\left( -3,0 \right)\text{ and }\left( 3,0 \right)$ and the y-intercepts are $\left( 0,-2 \right)\text{ and }\left( 0,2 \right)$. Plot this point and connect them with the oval shaped curve.
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