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: 18


Vertices: $(-3,0),(3,0)$ Foci: $(-\sqrt{5},0),(\sqrt{5},0)$ See graph

Work Step by Step

We are given the ellipse: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=9\Rightarrow a=\sqrt{9}=3$ $b^2=4\Rightarrow b=\sqrt 4=2$ $c^2=a^2-b^2=9-4=5$ $c=\sqrt{5}$ The center is: $(h,k)=(0,0)$ Find the vertices: $(h-a,k)=(0-3,0)=(-3,0)$ $(h+a,k)=(0+3,0)=(3,0)$ Find the foci: $(h-c,k)=(0-\sqrt{5},0)=(-\sqrt{5},0)$ $(h+c,k)=(0+\sqrt{5},0)=(\sqrt{5},0)$ Use the value of $b=2$ to find the two points above and below the center: $(h,k-b)=(0,0-2)=(0,-2)$ $(h,k+b)=(0,0+2)=(0,2)$ Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
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