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: 63


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

Work Step by Step

We are given the ellipse: Center: $(1,2)$ Vertex: $(4,2)$ Point on the graph: $(1,5)$ 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 $a$ using the vertex: $(h+a,k)=(4,2)$ $(1+a,2)=(4,2)$ $1+a=4$ $a=3$ Determine $b$ using the point $(1,5)$: $\dfrac{(1-1)^2}{a^2}+\dfrac{(5-2)^2}{b^2}=1$ $\dfrac{9}{b^2}=1$ $b^2=9$ $b=3$ The equation of the ellipse is: $\dfrac{(x-1)^2}{3^2}+\dfrac{(y-2)^2}{3^2}=1$ $\dfrac{(x-1)^2}{9}+\dfrac{(y-2)^2}{9}=1$ As $a=b$ the equation represents a circle. The center is: $(h,k)=(1,2)$ Determine the vertices and co-vertices: $(h-a,k)=(1-3,2)=(-2,2)$ $(h+a,k)=(4,2)$ $(h,k-b)=(1,2-3)=(1,-1)$ $(h,k+b)=(1,2+3)=(1,5)$ Graph the circle:
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