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: 61


$\dfrac{(x-1)^2}{10}+\dfrac{(y-2)^2}{1}=1$ See graph

Work Step by Step

We are given the ellipse: Center: $(1,2)$ Focus: $(4,2)$ Point on the graph: $(1,3)$ Because the $y$-coordinates of the center and focus are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k$ using the center: $(h,k)=(1,2)$ $h=1$ $k=2$ Determine $c$ using the focus: $(h+c,k)=(3,0)$ $(0+c,0)=(3,0)$ $(c,0)=(3,0)$ $c=3$ We get the relation between $a$ and $b$: $a^2=b^2+c^2$ $a^2-b^2=c^2$ $a^2-b^2=3^2$ $a^2-b^2=9$ Use the point $(1,3)$ to determine another relation between $a$ and $b$: $\dfrac{(1-1)^2}{a^2}+\dfrac{(3-2)^2}{b^2}=1$ $\dfrac{1}{b^2}=1$ $b^2=1$ $b=1$ Determine $a$: $a^2-b^2=9$ $a^2-1^2=9$ $a^2=10$ $a=\sqrt{10}$ The equation of the ellipse is: $\dfrac{(x-1)^2}{(\sqrt{10})^2}+\dfrac{(y-2)^2}{1^2}=1$ $\dfrac{(x-1)^2}{10}+\dfrac{(y-2)^2}{1}=1$ The center is: $(h,k)=(1,2)$ Determine the vertices and co-vertices: $(h-a,k)=(1-\sqrt{10},2)$ $(h+a,k)=(1+\sqrt{10},2)$ $(h,k-b)=(1,2-1)=(1,1)$ $(h,k+b)=(1,2+1)=(1,3)$ Graph the ellipse:
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