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: 62


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

Work Step by Step

We are given the ellipse: Center: $(1,2)$ Focus: $(1,4)$ Point on the graph: $(2,2)$ Because the $x$-coordinates of the center and focus are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k$ using the center: $(h,k)=(1,2)$ $h=1$ $k=2$ Determine $c$ using the focus: $(h,k+c)=(1,4)$ $(1,2+c)=(1,4)$ $2+c=4$ $c=2$ We get the relation between $a$ and $b$: $a^2=b^2+c^2$ $a^2-b^2=c^2$ $a^2-b^2=2^2$ $a^2-b^2=4$ Use the point $(2,2)$ to determine another relation between $a$ and $b$: $\dfrac{(2-1)^2}{b^2}+\dfrac{(2-2)^2}{a^2}=1$ $\dfrac{1}{b^2}=1$ $b^2=1$ $b=1$ Determine $a$: $a^2-b^2=4$ $a^2-1^2=4$ $a^2=5$ $a=\sqrt{5}$ The equation of the ellipse is: $\dfrac{(x-1)^2}{1^2}+\dfrac{(y-2)^2}{(\sqrt 5)^2}=1$ $\dfrac{(x-1)^2}{1}+\dfrac{(y-2)^2}{5}=1$ The center is: $(h,k)=(1,2)$ Determine the vertices and co-vertices: $(h,k-a)=(1,2-\sqrt{5})$ $(h,k+a)=(1,2+\sqrt{5})$ $(h-b,k)=(1-1,2)=(0,2)$ $(h+b,k)=(1+1,2)=(2,2)$ Graph the ellipse:
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