Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.1 - The Ellipse - Exercise Set - Page 967: 58

Answer

$\frac{(x+1)^2}{16}+\frac{(y-2)^2}{49}=1$, foci: $(-1,2\pm\sqrt {33})$.
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Work Step by Step

Step 1. Rewrite the given equation as $49(x^2+2x+1)+16(y^2-4y+4)=671+49+64=784$ or $\frac{(x+1)^2}{16}+\frac{(y-2)^2}{49}=1$, Step 2. We have $a^2=49, b^2=16$ and $c=\sqrt {a^2-b^2}=\sqrt {33}$. The ellipse is centered at $(-1,2)$ with a vertical major axis. Step 3. We can graph the equation as shown in the figure with foci at $(-1,2\pm\sqrt {33})$.
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