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: 59


$\dfrac{(x-2)^2}{16}+\dfrac{(y-1)^2}{7}=1$ See graph

Work Step by Step

We are given the ellipse: Foci: $(5,1),(-1,1)$ Length of major axis: $8$ Because the $y$-coordinates of the foci are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,c$ using the foci: $(h-c,k)=(-1,1)\Rightarrow h-c=-1,k=1$ $(h+c,k)=(5,1)\Rightarrow h+c=5$ $k=1$ $\begin{cases} h-c=-1\\ h+c=5 \end{cases}$ $h-c+h+c=-1+5$ $2h=4$ $h=2$ $h+c=5$ $2+c=5$ $c=3$ Determine $a$ using the length of the major axis: $2a=8$ $a=4$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=4^2-3^2$ $b^2=7$ $b=\sqrt{7}$ The equation of the ellipse is: $\dfrac{(x-2)^2}{4^2}+\dfrac{(y-1)^2}{(\sqrt 7)^2}=1$ $\dfrac{(x-2)^2}{16}+\dfrac{(y-1)^2}{7}=1$ The center is: $(h,k)=(2,1)$ Determine the vertices and co-vertices: $(h-a,k)=(2-4,1)=(-2,1)$ $(h+a,k)=(2+4,1)=(6,1)$ $(h,k-b)=(2,1-\sqrt 7)$ $(h,k+b)=(2,1+\sqrt 7)$ Graph the ellipse:
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