Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 9 - Analytic Geometry - Section 9.3 The Ellipse - 9.3 Assess Your Understanding - Page 677: 51

Answer

center $(1,-2)$, foci $(1,-2-\sqrt 5),(1,-2+\sqrt 5)$, vertices $(1,-5),(1,1)$ See graph.

Work Step by Step

1. $9x^2+4y^2-18x+16y-11=0\Longrightarrow 9(x-1)^2+4(y+2)^2=36$ or $\frac{(x-1)^2}{4}+\frac{(y+2)^2}{9}=1$, 2. we can find center $(1,-2)$ with a vertical major axis, $a=3, b=2$, $c=\sqrt {a^2-b^2}=\sqrt {9-4}=\sqrt 5$, 3. thus foci $(1,-2\pm\sqrt 5)$ or $(1,-2-\sqrt 5),(1,-2+\sqrt 5)$, vertices $(1,-2\pm3)$ or $(1,-5),(1,1)$ 4. See graph.
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