Answer
See graph, $\frac{x^2}{9}+\frac{(y-2)^2}{16}=1$
Work Step by Step
1. See graph for $(x,y)=(3sin(t), 4cos(t)+2)$ over $0\le t\le2\pi$.
2. With $sin(t)=\frac{x}{3}, cos(t)=\frac{y-2}{4}$, we have $\frac{x^2}{9}+\frac{(y-2)^2}{16}=1$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 9 - Analytic Geometry - Cumulative Review - Review Exercises - Page 719: 33
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