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: 20


Vertices: $(0,-4),(0,4)$ Foci: $(0,-\sqrt{15}),(0,\sqrt{15})$ See graph

Work Step by Step

We are given the ellipse: $x^2+\dfrac{y^2}{16}=1$ Put the equation in standard form: $\dfrac{x^2}{1}+\dfrac{y^2}{16}=1$ The equation is in the form: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k,a,b,c$: $h=0$ $k=0$ $a^2=16\Rightarrow a=\sqrt{16}=4$ $b^2=1\Rightarrow b=\sqrt 1=1$ $c^2=a^2-b^2=16-1=15$ $c=\sqrt{15}$ The center is: $(h,k)=(0,0)$ Find the vertices: $(h,k-a)=(0,0-4)=(0,-4)$ $(h,k+a)=(0,0+4)=(0,4)$ Find the foci: $(h,k-c)=(0,0-\sqrt{15})=(0,-\sqrt{15})$ $(h,k+c)=(0,0+\sqrt{15})=(0,\sqrt{15})$ Use the value of $b=1$ to find the two points to the left and right of the center: $(h-b,k)=(0-1,0)=(-1,0)$ $(h+b,k)=(0+1,0)=(1,0)$ Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
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