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: 53

Answer

$\frac{x^2}{32}+\frac{y^2}{36}=1$

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 $a \gt b$ 2a = length of major axis 2b = length of minor axis c = distance of each focus from the center $a^2=b^2+c^2$ The foci are (0, -2) and (0, 2), which are equidistant from the origin (0, 0). This means that the ellipse has its center at (0, 0). Each focus is 2 units away from the center therefore $c=2$. RECALL: The Eccentricity (e) of an ellipse is given by the formula: $\\e=\frac{c}{a}$ Substitute the known values of $e$ and $c$ to have: $\\e= \frac{c}{a} \\\frac{1}{3} = \frac{2}{a} \\a(1) = 3(2) \\a=6$ Solve for $b$ using the formula $a^2=b^2+c^2$ to have: $a^2=b^2+c^2 \\6^2=b^2+2^2 \\36=b^2+4 \\36-4=b^2 \\32=b^2 \\\sqrt{32}=b \\\sqrt{16(2)}=b^2 \\4\sqrt2=b$ When connected together using a line, the foci form a vertical line. Thus, the major axis is also vertical. Therefore, the equation of the ellipse is: $\\\frac{x^2}{b^2} + \frac{y^2}{a^2}=1 \\\frac{x^2}{(4\sqrt2)^2}+\frac{y^2}{6^2}=1 \\\frac{x^2}{32}+\frac{y^2}{36}=1$
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