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: 45


Center: $(-5,4)$ Foci: $(-5-2\sqrt 3,4),(-5+2\sqrt 3,4)$ Vertices: $(-9,4),(-1,4)$ See graph

Work Step by Step

We are given the ellipse: $(x+5)^2+4(y-4)^2=16$ Put the equation in standard form: $\dfrac{(x+5)^2}{16}+\dfrac{4(y-4)^2}{16}=1$ $\dfrac{(x+5)^2}{16}+\dfrac{(y-4)^2}{4}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Identify $h,k,a,b$: $h=-5$ $k=4$ $a^2=16\Rightarrow a=\sqrt{16}=4$ $b^2=4\Rightarrow b=\sqrt 4=2$ Determine $c$: $a^2=b^2+c^2$ $c^2=a^2-b^2$ $c^2=16-4$ $c^2=12$ $c=\sqrt {12}$ $c=2\sqrt 3$ Determine the center: $(h,k)=(-5,4)$ Determine the foci: $(h-c,k)=(-5-2\sqrt 3,4)$ $(h+c,k)=(-5+2\sqrt 3,4)$ Determine the vertices: $(h-a,k)=(-5-4,4)=(-9,4)$ $(h+a,k)=(-5+4,4)=(-1,4)$ Determine the $y$-intercepts: $(h,k-b)=(-5,4-2)=(-5,2)$ $(h,k+b)=(-5,4+2)=(-5,6)$ Graph the ellipse:
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