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: 58


$\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{5}=1$ See graph

Work Step by Step

We are given the ellipse: Foci: $(1,2),(-3,2)$ Vertex: $(-4,2)$ Because the $y$-coordinates of the vertex and foci are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ Determine $h,k,c$ using the foci: $(h-c,k)=(-3,2)\Rightarrow h-c=-3,k=2$ $(h+c,k)=(1,2)\Rightarrow h+c=1$ $k=2$ $\begin{cases} h-c=-3\\ h+c=1 \end{cases}$ $h-c+h+c=-3+1$ $2h=-2$ $h=-1$ $h+c=1$ $-1+c=1$ $c=2$ Determine $k, a$ using the vertex: $(h-a,k)=(-4,2)$ $(-1-a,k)=(-4,2)$ $-1-a=-4\Rightarrow a=3$ $k=2$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=3^2-2^2$ $b^2=5$ $b=\sqrt{5}$ The equation of the ellipse is: $\dfrac{(x+1)^2}{3^2}+\dfrac{(y-2)^2}{(\sqrt 5)^2}=1$ $\dfrac{(x+1)^2}{9}+\dfrac{(y-2)^2}{5}=1$ The center is: $(h,k)=(-1,2)$ Determine the vertices and co-vertices: $(h-a,k)=(-4,2)$ $(h+a,k)=(-1+3,2)=(2,2)$ $h,k-b)=(-1,2-\sqrt{5})$ $h,k+b)=(-1,2+\sqrt{5})$ Graph the ellipse:
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