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 862: 25

Answer

The graph is shown below
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Work Step by Step

$16{{x}^{2}}+25{{y}^{2}}=1$ Now write $16=\frac{1}{\frac{1}{16}}$ and $25=\frac{1}{\frac{1}{25}}$, so rewrite the equation. $\frac{{{x}^{2}}}{\frac{1}{16}}+\frac{{{y}^{2}}}{\frac{1}{25}}=1$ Standard equation of an Ellipse is: $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ Identify $a\text{ and }b$ by comparing the equation with the standard equation, $\begin{align} & \frac{{{x}^{2}}}{\frac{1}{16}}+\frac{{{y}^{2}}}{\frac{1}{25}}=1 \\ & \frac{{{x}^{2}}}{{{\left( \frac{1}{4} \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( \frac{1}{5} \right)}^{2}}}=1 \end{align}$ Since $a=\frac{1}{4}\text{ and }b=\frac{1}{5}$, the x-intercepts are $\left( -\frac{1}{4},0 \right)\text{ and }\left( \frac{1}{4},0 \right)$ and the y-intercepts are $\left( 0,-\frac{1}{5} \right)\text{ and }\left( 0,\frac{1}{5} \right)$. Since ${{a}^{2}}>{{b}^{2}}$, the ellipse will be horizontal. Plot this point and connect them with the oval shaped curve.
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