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: 44


Center: $(-4,-2)$ Foci: $(-4-\sqrt 5,-2),(-4+\sqrt 5,-2)$ Vertices: $(-4,0),(-4,4)$ See graph

Work Step by Step

We are given the ellipse: $\dfrac{(x+4)^2}{9}+\dfrac{(y+2)^2}{4}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Identify $h,k,a,b$: $h=-4$ $k=-2$ $a^2=9\Rightarrow a=\sqrt{9}=3$ $b^2=4\Rightarrow b=\sqrt 4=2$ 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: $(h,k)=(-4,-2)$ Determine the foci: $(h-c,k)=(-4-\sqrt 5,-2)$ $(h+c,k)=(-4+\sqrt 5,-2)$ Determine the vertices: $(h-a,k)=(-4-3,-2)=(-7,-2)$ $(h+a,k)=(-4+3,-2)=(-1,-2)$ Determine the $y$-intercepts: $(h,k-b)=(-4,-2-2)=(-4,-4)$ $(h,k+b)=(-4,-2+2)=(-4,0)$ Graph the ellipse:
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