Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 947: 82


The simplified form is, $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{25}=1$.

Work Step by Step

The provided equation is: $25{{x}^{2}}+16{{y}^{2}}=400$ Bring it to the standard form of an ellipse, as below: $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ So, divide the equation by $400$ on both sides to get: $\begin{align} & \frac{25{{x}^{2}}}{400}+\frac{16{{y}^{2}}}{400}=\frac{400}{400} \\ & \frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{25}=1 \end{align}$ Hence, the required simplified form of the provided equation is $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{25}=1$.
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