Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.2 - Ellipses - 11.2 Exercises - Page 797: 49

Answer

$\frac{x^2}{100} + \frac{y^2}{91} = 1$

Work Step by Step

RECALL: The standard equation of an ellipse with whose center is at (0, 0) is: (i) $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ (horizontal major axis) (ii) $\frac{x^2}{b^2} + \frac{y^2}{a^2}=1$ (vertical major axis) where $a \gt b$ a = distance from the center to an endpoint of the major axis 2b = distance from the center to an endpoint of the minor axis 2c = distance between the foci $a^2=b^2+c^2$ The endpoints of the major axis are (-10, 0) and (10, 0), which are equidistant from the origin (0, 0). This means that the ellipse has its center at (0, 0). Note that: 1. Each endpoint of the major axis is 10 units away from the center so $a=10$. 2. The foci are 6 units apart $\longrightarrow 2c=6 \longrightarrow c=3$ 3. The major axis is horizontal. Solve for $b$ using the formula $a^2=b^2=c^2$ to have: $\\a^2 = b^2+c^2 \\10^2=b^2+3^2 \\100=b^2+9 \\91=b^2 \\\sqrt{91}=b$ Therefore, the equation of the ellipse is: $\\\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \\\frac{x^2}{10^2} + \frac{y^2}{(\sqrt{91})^2}=1 \\\frac{x^2}{100} + \frac{y^2}{91} = 1$
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