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: 56

Answer

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

Work Step by Step

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