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: 30


$\dfrac{x^2}{3}+\dfrac{y^2}{4}=1$ See graph

Work Step by Step

We are given the ellipse: Center: $(0,0)$ Focus: $(0,1)$ Vertex: $(0,-2)$ Because the $x$-coordinates of the vertex 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: $(h,k)=(0,0$ $h=0$ $k=0$ Determine $a$ using the vertex: $(h,k-a)=(0,-2)$ $(0,0-a)=(0,-2)$ $(0,-a)=(0,-2)$ $a=2$ Determine $c$ using the focus: $(h,k+c)=(0,1)$ $(0,0+c)=(0,1)$ $(0,c)=(0,1)$ $c=1$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=2^2-1^2$ $b^2=3$ $b=\sqrt 3$ The equation of the ellipse is: $\dfrac{x^2}{(\sqrt 3)^2}+\dfrac{y^2}{2^2}=1$ $\dfrac{x^2}{3}+\dfrac{y^2}{4}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse:
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