Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.2 - Ellipses - 11.2 Exercises - Page 797: 9

Answer

Vertices: $(\pm 5,0)$; Major Horizontal axis length =10 Minor vertical axis length =6 $r=\sqrt {5^2-3^2}=4$; Foci:$(\pm 4,0)$ The eccentricity $e$ is given as: $e=\dfrac{4}{5}$

Work Step by Step

The general form of the equation of an ellipse can be defined as: $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ Here, Vertices: $(\pm p,0)$; Major Horizontal axis length =2p Minor vertical axis length =2q Foci:$(\pm r,0)$ and $r=\sqrt {p^2-q^2}$ The eccentricity $e$ is given as: $e=\dfrac{r}{p}$ Re-arrange the given equation in the genreal form of an ellipse as follows: $\dfrac{x^2}{(5)^2}+\dfrac{y^2}{(3)^2}=1$ Here, we have $p=5,q=3$ Now, Vertices: $(\pm 5,0)$; Major Horizontal axis length =10 Minor vertical axis length =6 $r=\sqrt {5^2-3^2}=4$; Foci:$(\pm 4,0)$ The eccentricity $e$ is given as: $e=\dfrac{4}{5}$
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