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: 60


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

Work Step by Step

We are given the ellipse: Vertices: $(2,-1),(2,5)$ $c=2$ Because the $x$-coordinates of the vertices 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: $(h,k-a)=(2,-1)\Rightarrow h=2,k-a=-1$ $(h,k+a)=(2,5)\Rightarrow k+a=5$ $h=2$ $\begin{cases} k-a=-1\\ k+a=5 \end{cases}$ $k-a+k+a=-1+5$ $2k=4$ $k=2$ $k+a=5$ $2+a=5$ $a=3$ 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: $\dfrac{(x-2)^2}{(\sqrt 5)^2}+\dfrac{(y-2)^2}{(3^2}=1$ $\dfrac{(x-2)^2}{5}+\dfrac{(y-2)^2}{9}=1$ The center is: $(h,k)=(2,2)$ Determine the co-vertices: $h-b,k)=(2-\sqrt{5},2)$ $h+b,k)=(2+\sqrt{5},2)$ Graph the ellipse:
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