Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.3 The Ellipse - 10.3 Assess Your Understanding - Page 657: 38


$\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$ See graph

Work Step by Step

We are given the ellipse: Vertices: $(-5,0),(5,0)$ $c=2$ Because the $y$-coordinates of the center and vertex are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,a$ using the vertices: $(h-a,k)=(-5,0)\Rightarrow h-a=-5,k=0$ $(h+a,k)=(5,0)\Rightarrow h+a=5$ $\begin{cases} h-a=-5\\ h+a=5 \end{cases}$ $h-a+h+a=-5+5$ $2h=0$ $h=0$ $h+a=5$ $0+a=5$ $a=5$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=5^2-2^2$ $b^2=21$ $b=\sqrt{21}$ The equation of the ellipse is: $\dfrac{x^2}{5^2}+\dfrac{y^2}{(\sqrt{21})^2}=1$ $\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse:
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