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: 23


Vertices: $(-2\sqrt 2,0),(2\sqrt 2,0)$ Foci: $(-\sqrt 6,0),(\sqrt 6,0)$ See graph

Work Step by Step

We are given the ellipse: $4y^2+x^2=8$ Put the equation in standard form: $\dfrac{4y^2}{8}+\dfrac{x^2}{8}=1$ $\dfrac{x^2}{8}+\dfrac{y^2}{2}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=8\Rightarrow a=\sqrt{8}=2\sqrt 2$ $b^2=2\Rightarrow b=\sqrt 2$ $c^2=a^2-b^2=8-2=6$ $c=\sqrt{6}$ The center is: $(h,k)=(0,0)$ Find the vertices: $(h-a,k)=(0-2\sqrt 2,0)=(-2\sqrt 2,0)$ $(h+a,k)=(0+2\sqrt 2,0)=(2\sqrt 2,0)$ Find the foci: $(h-c,k)=(0-\sqrt 6,0)=(-\sqrt 6,0)$ $(h+c,k)=(0+\sqrt 6,0)=(\sqrt 6,0)$ Use the value of $b=\sqrt 2$ to find the two points above and below the center: $(h,k-b)=(0,0-\sqrt 2)=(0,-\sqrt 2)$ $(h,k+b)=(0,0+\sqrt 2)=(0,\sqrt 2)$ Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
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