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: 43


Center: $(3,-1)$ Foci: $(3,-1-\sqrt 5),(3,-1+\sqrt 5)$ Vertices: $(3,-4),(3,2)$ See graph

Work Step by Step

We are given the ellipse: $\dfrac{(x-3)^2}{4}+\dfrac{(y+1)^2}{9}=1$ The equation is in the form: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y+1)^2}{a^2}=1$ Identify $h,k,a,b$: $h=3$ $k=-1$ $a^2=9\Rightarrow a=\sqrt 9=3$ $b^2=4\Rightarrow b-=\sqrt 4=2$ Determine $c$: $a^2=b^2+c^2$ $c^2=a^2-b^2$ $c^2=9-4$ $c^2=5$ $c=\sqrt 5$ Determine the center: $(h,k)=(3,-1)$ Determine the foci: $(h,k-c)=(3,-1-\sqrt 5)$ $(h,k+c)=(3,-1+\sqrt 5)$ Determine the vertices: $(h,k-a)=(3,-1-3)=(3,-4)$ $(h,k+a)=(3,-1+3)=(3,2)$ Determine the $x$-intercepts: $(h-b,k)=(3-2,-1)=(1,-1)$ $(h+b,k)=(3+2,-1)=(5,-1)$ Graph the ellipse:
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