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: 22


Vertices: $(-3\sqrt 2,0),(3\sqrt 2,0)$ Foci: $(-4,0),(4,0)$ See graph

Work Step by Step

We are given the ellipse: $x^2+9y^2=18$ Put the equation in standard form: $\dfrac{x^2}{18}+\dfrac{9y^2}{18}=1$ $\dfrac{x^2}{18}+\dfrac{y^2}{2}=1$ The equation is in the form: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=18\Rightarrow a=\sqrt{18}=3\sqrt 2$ $b^2=2\Rightarrow b=\sqrt 2$ $c^2=a^2-b^2=18-2=16$ $c=\sqrt{16}=4$ The center is: $(h,k)=(0,0)$ Find the vertices: $(h-a,k)=(0-3\sqrt 2,0)=(-3\sqrt 2,0)$ $(h+a,k)=(0+3\sqrt 2,0)=(3\sqrt 2,0)$ Find the foci: $(h-c,k)=(0-4,0)=(-4,0)$ $(h+c,k)=(0+4,0)=(4,0)$ Use the value of $b=2$ 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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