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: 57


$\dfrac{(x-4)^2}{5}+\dfrac{(y-6)^2}{9}=1$ See graph

Work Step by Step

We are given the ellipse: Focus: $(4,8)$ Vertices: $(4,3),(4,9)$ Because the $x$-coordinates of the vertex and foci are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k,a$ using the vertices: $(h,k-a)=(4,3)\Rightarrow h=4,k-a=3$ $(h,k+a)=(4,9)\Rightarrow k+a=9$ $h=4$ $\begin{cases} k-a=3\\ k+a=9 \end{cases}$ $k-a+k+a=3+9$ $2k=12$ $k=6$ $k+a=9$ $6+a=9$ $a=3$ Determine $c$ using the focus: $(h,k+c)=(4,8)$ $(4,6+c)=(4,8)$ $6+c=8$ $c=2$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=3^2-2^2$ $b^2=5$ $b=\sqrt{5}$ The equation of the ellipse is: $\dfrac{(x-4)^2}{(\sqrt 5)^2}+\dfrac{(y-6)^2}{(3^2}=1$ $\dfrac{(x-4)^2}{5}+\dfrac{(y-6)^2}{9}=1$ The center is: $(h,k)=(4,6)$ Determine the co-vertices: $h-b,k)=(4-\sqrt{5},6)$ $h+b,k)=(4+\sqrt{5},6)$ Graph the ellipse:
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