Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX C - Graphs of Second-Degree Equations - C Exercises - Page A 23: 36

Answer

$\frac{x^2}{9}+\frac{y^2}{25} = 1$

Work Step by Step

We can write the general form of an ellipse: $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$ Since the center is the origin, we can rewrite the equation of the ellipse as follows: $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ We can use the two data pairs to make two equations: equation 1: $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ $\frac{(1)^2}{a^2}+\frac{(-10\sqrt{2}/3)^2}{b^2} = 1$ $\frac{1}{a^2}+\frac{200}{9b^2} = 1$ $\frac{-4}{a^2}+\frac{-800}{9b^2} = -4$ equation 2: $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ $\frac{(-2)^2}{a^2}+\frac{(5\sqrt{5}/3)^2}{b^2} = 1$ $\frac{4}{a^2}+\frac{125}{9b^2} = 1$ We can add both sides of equation 1 and equation 2: $\frac{-800}{9b^2}+\frac{125}{9b^2} = -4+1$ $\frac{-675}{9b^2} = -3$ $b^2 = \frac{675}{27}$ $b^2 = \frac{225}{9}$ $b^2 = \frac{75}{3}$ $b^2 = 25$ $b = 5$ We can use equation 1 to find $a$: $\frac{1}{a^2}+\frac{200}{9b^2} = 1$ $\frac{1}{a^2}+\frac{200}{9(5)^2} = 1$ $\frac{1}{a^2}+\frac{8}{9} = 1$ $\frac{1}{a^2}=\frac{1}{9}$ $a^2 = 9$ $a = 3$ We can write the equation of the ellipse: $\frac{x^2}{9}+\frac{y^2}{25} = 1$
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