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: 34


$\dfrac{x^2}{48}+\dfrac{y^2}{64}=1$ See graph

Work Step by Step

We are given the ellipse: Focus: $(0,-4)$ Vertices: $(0,-8),(0,8)$ Because the $x$-coordinates of the foci are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k,a$ using the vertices: $(h,k-a)=(0,-8)\Rightarrow h=0, k-a=-8$ $(h,k+a)=(0,8)\Rightarrow k+a=8$ $\begin{cases} k-a=-8\\ k+a=8 \end{cases}$ $k-a+k+a=-8+8$ $2k=0$ $k=0$ $k+a=8$ $0+a=8$ $a=8$ Determine $c$ using the focus: $(h,k-c)=(0,-4)$ $(0,0-c)=(0,-4)$ $(0,-c)=(0,-4)$ $c=4$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=8^2-4^2$ $b^2=48$ $b=\sqrt{48}=4\sqrt 3$ The equation of the ellipse is: $\dfrac{x^2}{(4\sqrt 3)^2}+\dfrac{y^2}{8^2}=1$ $\dfrac{x^2}{48}+\dfrac{y^2}{64}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse:
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