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: 27

Answer

$\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ See graph

Work Step by Step

We are given the ellipse: Center: $(0,0)$ Focus: $(3,0)$ Vertex: $(5,0)$ Because the $y$-coordinates of the vertex 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: $(h,k)=(0,0$ $h=0$ $k=0$ Determine $a$ using the vertex: $(h+a,k)=(5,0)$ $(0+a,0)=(5,0)$ $(a,0)=(5,0)$ $a=5$ Determine $c$ using the focus: $(h+c,k)=(3,0)$ $(0+c,0)=(3,0)$ $(c,0)=(3,0)$ $c=3$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=5^2-3^2$ $b^2=16$ $b=4$ The equation of the ellipse is: $\dfrac{x^2}{5^2}+\dfrac{y^2}{4^2}=1$ $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse:
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.